How to Think Like a Computer Scientist: Learning with Python 2nd Edition/Solutions
The following section contains answers to the exercises in the Book.
Chapter 1
Exercise 1:
print("5 ** 5 is", 5**2) print("9 * 5 is", 9 * 5) print("15 / 12 is", 15 / 12) print("12 / 15 is", 12 / 15) print("15 // 12 is", 15 // 12) print("12 // 15 is", 12 // 15) print("5 % 2 is", 5 % 2) print("9 % 5 is", 9 % 5) print("15 % 12 is", 15 % 12) print("12 % 15 is", 12 % 15) print("6 % 6 is", 6 % 6) print("0 % 7 is", 0 % 7)
Exercise 2:
time_start=14 # Use 24 hour clock. Makes life a lot easier. wait=51 time_hours= time_start + wait days=time_hours//24 clock=time_hours-(days*24) print("The alarm goes of at", str(clock)+":00")
Exercise 3:
# 24 hour clock t1=input("Enter the current time's hour using 24 hour clock: ") t1=int(t1) wait=input("Enter wait in hours for next alarm: ") wait=int(wait) t2_hrs=wait+t1 days=t2_hrs//24 t2_time=t2_hrs-(days*24) print("The alarm goes of at", str(t2_time)+":00")
Chapter 3
Solution 4
def cat_n_times(s, n):
string=s print string*n
Chapter 4 Excercise 5
def dispatch(choice): if choice == 'a': function_a() elif choice == 'b': function_b() elif choice == 'c': function_c() else: print "Invalid choice." def function_a(): print "function_a was called ..." def function_b(): print "function_b was called ..." def function_c(): print "function_c was called ..." choice = raw_input ("Please Enter a Function.") #had to ask about this, choice is raw input. print dispatch(choice)
Chapter 4 Excercise 7
def is_divisible_by_3(x): if x % 3 == 0: print x, "This number is divisible by three." else: print x, "This number is not divisible by three." x = input ("Enter a number") print is_divisible_by_3(x) def is_divisible_by_5(y): if y % 5 == 0: print y, "This number is divisible by five." else: print y, "This number is not divisible by five." y = input ("Enter a number") print is_divisible_by_5(y) def is_divisible_by_n(c,z): if c % z == 0: print "Yes", c, "divisible by", z else: print "No", c, "is not divisible by", z c = input ("Enter number") z = input ("Enter another number") print is_divisible_by_n(c,z)
Chapter 5
CH 5 - Solution 1
def compare(a, b): """ >>> compare(5, 4) 1 >>> compare(7, 7) 0 >>> compare(2, 3) -1 >>> compare(42, 1) 1 """ if a > b: return 1 elif a == b: return 0 elif a < b: return -1 if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 2
def hypotenuse(a, b): """ >>> hypotenuse(3, 4) 5.0 >>> hypotenuse(12, 5) 13.0 >>> hypotenuse(7, 24) 25.0 >>> hypotenuse(9, 12) 15.0 """ import math return math.sqrt(a**2 + b**2) # You could also use # return (a**2 + b**2)**0,5 if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 3 Part 1
def slope(x1, y1, x2, y2): """ >>> slope(5, 3, 4, 2) 1.0 >>> slope(1, 2, 3, 2) 0.0 >>> slope(1, 2, 3, 3) 0.5 >>> slope(2, 4, 1, 2) 2.0 """ return float(y2 - y1)/(x2 - x1) if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 3 Part 2
def slope(x1, y1, x2, y2): return float(y2 - y1)/(x2 - x1) def intercept(x1, y1, x2, y2): """ >>> intercept(1, 6, 3, 12) 3.0 >>> intercept(6, 1, 1, 6) 7.0 >>> intercept(4, 6, 12, 8) 5.0 """ m = slope(x1, y1, x2, y2) return float(y1 - m*x1) if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 4
def is_even(n): """ >>> is_even(2) True >>> is_even(3) False >>> is_even(7) False >>> is_even(10) True """ # You could also type just # return n % 2 == 0 if n % 2 == 0: return True else: return False if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 5 Part 1
def is_odd(n): """ >>> is_odd(2) False >>> is_odd(3) True >>> is_odd(7) True >>> is_odd(10) False """ if n % 2 != 0: return True else: return False if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 5 Part 2
def is_even(n): if n % 2 == 0: return True else: return False def is_odd(n): """ >>> is_odd(2) False >>> is_odd(3) True >>> is_odd(7) True >>> is_odd(10) False """ if is_even(n) is True: return False else: return True if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 6
def is_factor(f, n): """ >>> is_factor(3, 12) True >>> is_factor(5, 12) False >>> is_factor(7, 14) True >>> is_factor(2, 14) True >>> is_factor(7, 15) False """ if n % f == 0: return True else: return False if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 7
def is_multiple(m, n): """ >>> is_multiple(12, 3) True >>> is_multiple(12, 4) True >>> is_multiple(12, 5) False >>> is_multiple(12, 6) True >>> is_multiple(12, 7) False """ if m % n == 0: return True else: return False if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 8
def f2c(t): """ >>> f2c(212) 100 >>> f2c(32) 0 >>> f2c(-40) -40 >>> f2c(36) 2 >>> f2c(37) 3 >>> f2c(38) 3 >>> f2c(39) 4 """ return int(round((t - 32.0) * 5.0/9)) if __name__ == '__main__': import doctest doctest.testmod()
CH 5 - Solution 9
def c2f(t): """ >>> c2f(0) 32 >>> c2f(100) 212 >>> c2f(-40) -40 >>> c2f(12) 54 >>> c2f(18) 64 >>> c2f(-48) -54 """ return int(round((t * 9.0/5) + 32.0)) if __name__ == '__main__': import doctest doctest.testmod()
Chapter 6
CH 6 - Solution 1
$ cat ch0601.py
print 'produces', '\n', 'this', '\n', 'output'
$ python ch0601.py
produces
this
output
CH 6 - Solution 2
$ cat ch0602.py
def sqrt(n):
approx = n/2.0
better = (approx + n/approx)/2.0
print better
while better != approx:
approx = better
better = (approx + n/approx)/2.0
return approx
print sqrt(25)
$ python ch0602.py
7.25
5.0
CH 6 - Solution 3
CH 6 - Solution 4
def print_triangular_numbers(x): i = 1 while i <= x: print i, '\t', i*(i+1)/2 i += 1
CH 6 - Solution 5 v1
def is_prime(n): """ >>> is_prime(2) True >>> is_prime(3) True >>> is_prime(4) False >>> is_prime(41) True >>> is_prime(42) False """ count = 2 #so counting starts at 2 #as all numbers can be divided by 1 and remainder = 0; and because nothing can be divided by 0 while count <= n: if n % count == 0 and count != n: #if when n divided by count remainder is 0 #for example when 4 is divided by 2 #and to make sure count != n because any number divided by itself then remainder = 0; #then return False return False #if above is not applicable then return true for doctest to pass elif count == n: return True #to make sure it doesnt exit after one try, and it counts elif n % count != 0: count = count + 1 if __name__ == '__main__': import doctest doctest.testmod()
CH 6 - Solution 5 v2
def is_prime(n): """ >>> is_prime(2) True >>> is_prime(3) True >>> is_prime(4) False >>> is_prime(41) True >>> is_prime(42) False """ count = 2 while count <= n: if n % count == 0 and count != n: return False else: count = count + 1 return True if __name__ == '__main__': import doctest doctest.testmod()
CH 6 - Solution 6 Part 1
What will num_digits(0) return?
def num_digits(n): """ >>> num_digits(12345) 5 >>> num_digits(0) 1 >>> num_digits(-12345) 5 """ count = 0 while n != 0: count = count + 1 n = n / 10 return count #num_digits(0) returns 0 because the while statement is not run, #and therefore it returns count.
CH 6 - Solution 6 Part 2
Modify it to return 1 for this case.
def num_digits(n): """ >>> num_digits(12345) 5 >>> num_digits(0) 1 >>> num_digits(-12345) 5 """ if n == 0: return 1 #the program checks if n == 0. if it's 0, it simply returns 1. #if not, count below is initialized, then, after the calculation, #count is returned with the final result count = 0 while n != 0: count = count + 1 n = n / 10 return count
CH 6 - Solution 6 Part 3
Why does a call to num_digits(-24) result in an infinite loop (hint: -1/10 evaluates to -1)?
>>> -24/10 -3 >>> -3/10 -1 >>> -1/10 -1 >>>
The loop will end when n = 0, and per above -1/10 == -1, causing an infinite loop.
CH 6 - Solution 6 Part 4
def num_digits(n): """ >>> num_digits(12345) 5 >>> num_digits(0) 1 >>> num_digits(-12345) 5 """ # >>> num_digits(-100) #The code must be checked with this value for assurance # 3 # >>> num_digits(-9999) #The code must be checked with this value for assurance # 4 count = 0 if n == 0: return 1 elif n < 0: n = -n while n: count = count + 1 n = n/10 return count else: while n: count = count + 1 n = n/10 return count if __name__ == '__main__': import doctest doctest.testmod()
CH 6 - Solution 7
def num_even_digits(n): """ >>> num_even_digits(123456) 3 >>> num_even_digits(2468) 4 >>> num_even_digits(1357) 0 >>> num_even_digits(2) 1 >>> num_even_digits(20) 2 """ count = 0 while n != 0: digit = n % 2 if digit == 0: count = count + 1 n = n / 10 return count if __name__ == '__main__': import doctest doctest.testmod()
CH 6 - Solution 8
def print_digits(n): """ >>> print_digits(13789) 9 8 7 3 1 >>> print_digits(39874613) 3 1 6 4 7 8 9 3 >>> print_digits(213141) 1 4 1 3 1 2 """ while n != 0: print n % 10, n = n/10 if __name__ == '__main__': import doctest doctest.testmod()
CH 6 - Solution 9
def sum_of_squares_of_digits(n): """ >>> sum_of_squares_of_digits(1) 1 >>> sum_of_squares_of_digits(9) 81 >>> sum_of_squares_of_digits(11) 2 >>> sum_of_squares_of_digits(121) 6 >>> sum_of_squares_of_digits(987) 194 """ sumof = 0 while n != 0: sumof += (n % 10)**2 n = n/10 return sumof if __name__ == '__main__': import doctest doctest.testmod()
Chapter 7
CH 7 - Question 1
Modify:
prefixes = "JKLMNOPQ" suffix = "ack" for letter in prefixes: print letter + suffix
so that Ouack and Quack are spelled correctly.
CH 7 - Notes regarding Question 1
Note that for letter in prefixes: is a substitution of traversal, for example:
index = 0 while index < len(fruit): letter = fruit[index] print letter index += 1 #fruit = "banana" #while index is less than 6. #6 is the lenght of fruit #letter = fruit[index] #Since index = 0, "b" is equal to letter in loop 1 #letter is printed #1 is added to whatever the value of index is #the loop continues until index < 6
is substituted by:
for char in fruit: print char
CH 7 - Solution 1 v1
prefixes = "JKLMNOPQ" suffix = "ack" for letter in prefixes: if letter == "O" or letter == "Q": print letter + "u" + suffix else: print letter + suffix
CH 7 - Solution 1 v2
prefixes = "JKLMNOPQ" suffix = "ack" for letter in prefixes: if letter in "OQ": print letter + "u" + suffix else: print letter + suffix
CH 7 - Question 2
Encapsulate:
fruit = "banana" count = 0 for char in fruit: if char == 'a': count += 1 print count
in a function named count_letters, and generalize it so that it accepts the string and the letter as arguments.
CH 7 - Solution 2
def count_letters(strng, letter): count = 0 for char in strng: if char == letter: count += 1 return count
CH 7 - Question 3
Now rewrite the count_letters function so that instead of traversing the string, it repeatedly calls find (the version from Optional parameters), with the optional third parameter to locate new occurences of the letter being counted.
CH 7 - Solution 3
def find(strng, ch, start=0): index = start while index < len(strng): if strng[index] == ch: return index index += 1 return -1 #for example strng = "banana" #letter = "a" #x = find(strng, letter, start) will return 1 #we need to modify count_letters so it returns 3 #the start variable can thus be used to our advantage #the loop will end at the last letter of the string by returning -1 def count_letters(strng, letter, start=0): count = 0 x = find(strng, letter, start) while x != -1: count += 1 x = find(strng, letter, x + 1) return count
Tracing the program:
#in our traceback, x = 1 #x = find(strng, letter, 2) #x above will be assigned new value find(strng, letter, 2) >>> find("banana", "a", 1) 1 >>> find("banana", "a", 2) 3 >>> find("banana", "a", 4) 5 >>> find("banana", "a", 6) -1 >>>
CH 7 - Solution 4
Re: the versions of is_lower:
def is_lower(ch): return string.find(string.lowercase, ch) != -1
def is_lower(ch): return ch in string.lowercase
def is_lower(ch): return 'a' <= ch <= 'z'
CH 7 - Question 5
Create a file named stringtools.py and put the following in it:
def reverse(s): """ >>> reverse('happy') 'yppah' >>> reverse('Python') 'nohtyP' >>> reverse("") '' >>> reverse("P") 'P' """ if __name__ == '__main__': import doctest doctest.testmod()
Add a function body to reverse to make the doctests pass.
CH 7 - Solution 5
def reverse(s): """ >>> reverse('happy') 'yppah' >>> reverse('Python') 'nohtyP' >>> reverse("") '' >>> reverse("P") 'P' """ #we need to use len(s) to make it output letters #length = len(happy) = 6 #print last = happy[length - 1] will return y #we need to make it such that [length - 2] and so on #we need the while loop to stop when length - x = 0 #if len(s) is 6 x = 1 while x <= len(s): length = len(s) lastoutput = s[length - x] print lastoutput, x += 1
#Why is x <= len(s)? #because if x < len(s): >>> reverse("happy") y p p a >>> #with x<= len(s) >>> reverse("happy") y p p a h >>>
Now inorder to make the doctest pass we need to enclose our ouput in a string. Here we use a placeholder string: lastoutput = "".
Second, we remove redundant code:
x = 1
while x <= len(s):
length = len(s)
lastoutput = s[length - x]
per:
length = len(fruit) last = fruit[length-1]
by using negative indices, such as:
fruit[-1] and fruit[-2]
to:
x = 1
while x <= len(s):
lastoutput = s[-x]
def reverse(s): """ >>> reverse('happy') 'yppah' >>> reverse('Python') 'nohtyP' >>> reverse("") '' >>> reverse("P") 'P' """ last = "" x = 1 while x <= len(s): lastoutput = s[-x] last += lastoutput x += 1 return last if __name__ == '__main__': import doctest doctest.testmod()
CH 7 - Solution 6
def reverse(s): last = "" x = 1 while x <= len(s): lastoutput = s[-x] last += lastoutput x += 1 return last def mirror(s): """ >>> mirror("good") 'gooddoog' >>> mirror("yes") 'yessey' >>> mirror('Python') 'PythonnohtyP' >>> mirror("") '' >>> mirror("a") 'aa' """ #we will be calling function reverse #first we need it to ouput everything s_back = reverse(s) return s + s_back if __name__ == '__main__': import doctest doctest.testmod()
CH 7 - Solution 7
def reverse(s): last = "" x = 1 while x <= len(s): lastoutput = s[-x] last += lastoutput x += 1 return last def mirror(s): out = reverse(s) out2 = reverse(out) return out2 + out def remove_letter(letter, strng): """ >>> remove_letter('a', 'apple') 'pple' >>> remove_letter('a', 'banana') 'bnn' >>> remove_letter('z', 'banana') 'banana' >>> remove_letter('i', 'Mississippi') 'Msssspp' """ without_letter = "" for c in strng: if c not in letter: without_letter += c return without_letter if __name__ == '__main__': import doctest doctest.testmod()
CH 7 - Solution 8
def reverse(s): """ >>> reverse('happy') 'yppah' >>> reverse('Python') 'nohtyP' >>> reverse("") '' >>> reverse("P") 'P' """ last = "" x = 1 while x <= len(s): lastoutput = s[-x] last += lastoutput x += 1 return last def mirror(s): """ >>> mirror("good") 'gooddoog' >>> mirror("yes") 'yessey' >>> mirror('Python') 'PythonnohtyP' >>> mirror("") '' >>> mirror("a") 'aa' """ out = reverse(s) out2 = reverse(out) return out2 + out def remove_letter(letter, strng): """ >>> remove_letter('a', 'apple') 'pple' >>> remove_letter('a', 'banana') 'bnn' >>> remove_letter('z', 'banana') 'banana' >>> remove_letter('i', 'Mississippi') 'Msssspp' """ without_letter = "" for c in strng: if c not in letter: without_letter += c return without_letter def is_palindrome(s): """ >>> is_palindrome('abba') True >>> is_palindrome('abab') False >>> is_palindrome('tenet') True >>> is_palindrome('banana') False >>> is_palindrome('straw warts') True """ half = len(s)/2 fronthalf = s[:half] backhalf = s[-half:] #N.B no need to compare middle letter return fronthalf == reverse(backhalf) # We can make this even simpler by not bothering to split the word # into halves. Simplest solution is the single line: # return s == reverse(s) # This will return True if word is the same forwards as backwards. # Although might be slower for v. long words? def count(sub, s): """ >>> count('is', 'Mississippi') 2 >>> count('an', 'banana') 2 >>> count('ana', 'banana') 2 >>> count('nana', 'banana') 1 >>> count('nanan', 'banana') 0 """ count = 0 x = s.find(sub) while x != -1: count += 1 x = s.find(sub, x + 1) return count def remove(sub, s): """ >>> remove('an', 'banana') 'bana' >>> remove('cyc', 'bicycle') 'bile' >>> remove('iss', 'Mississippi') 'Missippi' >>> remove('egg', 'bicycle') 'bicycle' """ x = s.find(sub) if x != -1: out1 = s[:x] y = len(sub) out2 = s[x + y:] return out1 + out2 else: return s def remove_all(sub, s): """ >>> remove_all('an', 'banana') 'ba' >>> remove_all('cyc', 'bicycle') 'bile' >>> remove_all('iss', 'Mississippi') 'Mippi' >>> remove_all('eggs', 'bicycle') 'bicycle' """ while s.find(sub) != -1: s = remove(sub, s) return s if __name__ == '__main__': import doctest doctest.testmod()
CH 7 - Solution 9
>>> "%s %d %f" % (5, 5, 5) '5 5 5.000000' #note %f outputs 6 decimal places of 0s
>>> "%-.2f" % 3 '3.00' #note .2 before f causes decimal output at 2 places (.00)
>>> "%-10.2f%-10.2f" % (7, 1.0/2) '7.00 0.50 ' #note "The - after each % in the converstion specifications indicates left justification. #The numerical values specify the minimum length, #so %-13d is a left justified number at least 13 characters wide." #Here if we count from index 0 @ 7 we will reach to index 10, before #the next string formatting operator
>>> print " $%5.2fn $%5.2fn $%5.2f" % (3, 4.5, 11.2) $ 3.00n $ 4.50n $11.20
CH 7 - Solution 10
>>> "%s %s %s %s" % ('this', 'that', 'something') Traceback (most recent call last): File "<pyshell#24>", line 1, in <module> "%s %s %s %s" % ('this', 'that', 'something') TypeError: not enough arguments for format string
>>> "%s %s %s" % ('this', 'that', 'something') 'this that something'
>>> "%s %s %s" % ('yes', 'no', 'up', 'down') Traceback (most recent call last): File "<pyshell#27>", line 1, in <module> "%s %s %s" % ('yes', 'no', 'up', 'down') TypeError: not all arguments converted during string formatting
>>> "%s %s %s %s" % ('yes', 'no', 'up', 'down') 'yes no up down'
>>> "%d %f %f" % (3, 3, "three") Traceback (most recent call last): File "<pyshell#29>", line 1, in <module> "%d %f %f" % (3, 3, "three") TypeError: float argument required, not str
>>> "%d %f %f" % (3, 3, 3) '3 3.000000 3.000000'
Chapter 8
CH 8 - Solution 4
Not fully complete but...
<syntaxhighlight lang="python"> fruit = "banana" def count_letters(array, n): count = 0 for i in range(len(array)): m = array.find(n,i) if( i == m ): count += 1 print(count) count_letters(fruit,"a")
Chapter 9
CH 9 - Solution 1
x = ['spam!', 1, ['Brie', 'Roquefort', 'Pol le Veq'], [1, 2, 3]] for i in x: print len(i)
>>> 5 Traceback (most recent call last): File "C:\Python26\ch9.py", line 5, in <module> print len(i) TypeError: object of type 'int' has no len() >>>
x = ['spam!', 'one', ['Brie', 'Roquefort', 'Pol le Veq'], [1, 2, 3]] for i in x: print len(i)
>>> 5 3 3 3 >>>
CH 9 - Solution 2
CH 9 - Solution 2.1
""" >>> a_list[3] 42 >>> a_list[6] 'Ni!' >>> len(a_list) 8 """ a_list = [1, 2, 3, 42, 5, 6, 'Ni!', 8] if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 2.2
""" >>> b_list[1:] ['Stills', 'Nash'] >>> group = b_list + c_list >>> group[-1] 'Young' """ b_list = ['Crosby', 'Stills', 'Nash'] c_list = ['Young'] if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 2.3
""" >>> 'war' in mystery_list False >>> 'peace' in mystery_list True >>> 'justice' in mystery_list True >>> 'oppression' in mystery_list False >>> 'equality' in mystery_list True """ mystery_list = ['peace', 'justice', 'equality'] if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 2.4
""" >>> range(a, b, c) [5, 9, 13, 17] """ a = 5 b = 18 c = 4 if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 3
>>> range(10, 0, -2) [10, 8, 6, 4, 2]
What happens if start is less than stop and step is less than 0?
The result will be an empty list.
Write a rule for the relationships among start, stop, and step.
CH 9 - Solution 4
>>> a = [1, 2, 3] >>> b = a[:] >>> id(a) 18484400 >>> id(b) 18594872 >>> b[0] 1 >>> b[0] = 5 >>> id(a) 18484400 >>> id(b) 18594872 >>> b [5, 2, 3] >>> a [1, 2, 3]
CH 9 - Solution 5
>>> this = ['I', 'am', 'not', 'a', 'crook'] >>> that = ['I', 'am', 'not', 'a', 'crook'] >>> print "Test 1: %s" % (id(this) == id(that)) Test 1: False >>> that = this >>> print "Test 2: %s" % (id(this) == id(that)) Test 2: True >>>
CH 9 - Solution 6
CH 9 - Solution 6.1
""" >>> 13 in junk True >>> del junk[4] >>> junk [3, 7, 9, 10, 17, 21, 24, 27] >>> del junk[a:b] >>> junk [3, 7, 27] """ junk = [3, 7, 9, 10, 13, 17, 21, 24, 27] a = 2 b = (len(junk) - 2) if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 6.2
""" >>> nlist[2][1] 0 >>> nlist[0][2] 17 >>> nlist[1][1] 5 """ nlist = [[1, 2, 17], [4, 5, 6], [7, 0, 9]] if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 6.3
""" >>> import string >>> string.split(message, '??') ['this', 'and', 'that'] """ message = "this??and??that" if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 7
def add_lists(a, b): """ >>> add_lists([1, 1], [1, 1]) [2, 2] >>> add_lists([1, 2], [1, 4]) [2, 6] >>> add_lists([1, 2, 1], [1, 4, 3]) [2, 6, 4] """ new_list = [] for x, elmt1 in enumerate(a): for y, elmt2 in enumerate(b): if x == y: new_list += [elmt1 + elmt2] return new_list #there is a simpler solution: #new_list = [] #for index, value in enumerate(a): # new_list += [a[index], b[index]] #return new_list if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 8
def mult_lists(a, b): """ >>> mult_lists([1, 1], [1, 1]) 2 >>> mult_lists([1, 2], [1, 4]) 9 >>> mult_lists([1, 2, 1], [1, 4, 3]) 12 """ sm = 0 new_list = [] for x, elmt1 in enumerate(a): for y, elmt2 in enumerate(b): if x == y: new_list += [elmt1 * elmt2] q = len(new_list) - 1 while q != -1: sm += new_list[q] q += -1 return sm if __name__ == '__main__': import doctest doctest.testmod() #Simpler Code #new_list = [] #for index, value in enumerate(b): #new_list += [a*value] #return new_list
CH 9 - Solution 9
CH 9 - Solution 9.1
def add_row(matrix): """ >>> m = [[0, 0], [0, 0]] >>> add_row(m) [[0, 0], [0, 0], [0, 0]] >>> n = [[3, 2, 5], [1, 4, 7]] >>> add_row(n) [[3, 2, 5], [1, 4, 7], [0, 0, 0]] >>> n [[3, 2, 5], [1, 4, 7]] """ y = len(matrix[0]) matrix2 = matrix[:] for z in range(1): matrix2 += [[0] * y] return matrix2 if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 9.2
def add_column(matrix): """ >>> m = [[0, 0], [0, 0]] >>> add_column(m) [[0, 0, 0], [0, 0, 0]] >>> n = [[3, 2], [5, 1], [4, 7]] >>> add_column(n) [[3, 2, 0], [5, 1, 0], [4, 7, 0]] >>> n [[3, 2], [5, 1], [4, 7]] """ x = len(matrix) matrix2 = [d[:] for d in matrix] for z in range(x): matrix2[z] += [0] return matrix2 if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 10
def add_matrices(m1, m2): """ >>> a = [[1, 2], [3, 4]] >>> b = [[2, 2], [2, 2]] >>> add_matrices(a, b) [[3, 4], [5, 6]] >>> c = [[8, 2], [3, 4], [5, 7]] >>> d = [[3, 2], [9, 2], [10, 12]] >>> add_matrices(c, d) [[11, 4], [12, 6], [15, 19]] >>> c [[8, 2], [3, 4], [5, 7]] >>> d [[3, 2], [9, 2], [10, 12]] """ new_matrix = [] for i, row in enumerate(m1): new_row = [] for j, m1_value in enumerate(row): m2_value = m2[i][j] new_row += [m1_value + m2[i][j]] new_matrix += [new_row] return new_matrix if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 11
def scalar_mult(n, m): """ >>> a = [[1, 2], [3, 4]] >>> scalar_mult(3, a) [[3, 6], [9, 12]] >>> b = [[3, 5, 7], [1, 1, 1], [0, 2, 0], [2, 2, 3]] >>> scalar_mult(10, b) [[30, 50, 70], [10, 10, 10], [0, 20, 0], [20, 20, 30]] >>> b [[3, 5, 7], [1, 1, 1], [0, 2, 0], [2, 2, 3]] """ new_matrix = [] for row in m: new_row = [] for value in row: new_row += [value*n] new_matrix += [new_row] return new_matrix if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 12
# Once we have obtained lists for the relevant rows and columns # in row_times_column function, we will re-use the mult_lists # function from question 8: def mult_lists(a, b): sum = 0 for i, a_value in enumerate(a): sum += a_value * b[i] return sum def row_times_column(m1, row, m2, column): """ >>> row_times_column([[1, 2], [3, 4]], 0, [[5, 6], [7, 8]], 0) 19 >>> row_times_column([[1, 2], [3, 4]], 0, [[5, 6], [7, 8]], 1) 22 >>> row_times_column([[1, 2], [3, 4]], 1, [[5, 6], [7, 8]], 0) 43 >>> row_times_column([[1, 2], [3, 4]], 1, [[5, 6], [7, 8]], 1) 50 """ m1_row = m1[row] m2_col = [] for m2_row in m2: m2_col += [m2_row[column]] return mult_lists(m1_row, m2_col) #This code may also be used to bypass the mult_lists code def row_times_column(m1, row, m2, column): x = 0 y = 0 for a in m1[row]: t = a * m2[x][column] y += t x += 1 return y def matrix_mult(m1, m2): """ >>> matrix_mult([[1, 2], [3, 4]], [[5, 6], [7, 8]]) [[19, 22], [43, 50]] >>> matrix_mult([[1, 2, 3], [4, 5, 6]], [[7, 8], [9, 1], [2, 3]]) [[31, 19], [85, 55]] >>> matrix_mult([[7, 8], [9, 1], [2, 3]], [[1, 2, 3], [4, 5, 6]]) [[39, 54, 69], [13, 23, 33], [14, 19, 24]] """ #To pass these tests, the function needs to multiply each each m1 row #by each m2 column, to produce a new matrix with with the same number #of rows as m1, and the same number of columns as m2. (This is the way #matrices are commonly multiplied in mathematics). We can use #row_times_column for each row/column combination m1_num_rows = len(m1) m2_num_cols = len(m2[0]) product_matrix = [] for row in range(m1_num_rows): new_row = [] for column in range(m2_num_cols): new_row += [row_times_column(m1, row, m2, column)] product_matrix += [new_row] return product_matrix if __name__ == '__main__': import doctest doctest.testmod()
CH 9 - Solution 13
CH 9 - Solution 14
import string def replace(s, old, new): """ >>> replace('Mississippi', 'i', 'I') 'MIssIssIppI' >>> s = 'I love spom! Spom is my favorite food. Spom, spom, spom, yum!' >>> replace(s, 'om', 'am') 'I love spam! Spam is my favorite food. Spam, spam, spam, yum!' >>> replace(s, 'o', 'a') 'I lave spam! Spam is my favarite faad. Spam, spam, spam, yum!' """ s_without_old = string.split(s, old) s_with_new = string.join(s_without_old, new) return s_with_new if __name__ == '__main__': import doctest doctest.testmod()
You can also do:
def myreplace(s, old, new): """ >>> myreplace('Mississippi', 'i', 'I') 'MIssIssIppI' >>> s = 'I love spom! Spom is my favorite food. Spom, spom, spom, yum!' >>> myreplace(s, 'om', 'am') 'I love spam! Spam is my favorite food. Spam, spam, spam, yum!' >>> myreplace(s, 'o', 'a') 'I lave spam! Spam is my favarite faad. Spam, spam, spam, yum!' """ new_string = "" counter = 0 while counter<len(s): # check if old equals a slice of len(old) if old == s[counter:counter+len(old)]: new_string += new #s[counter+len(old):] counter+=len(old) else: new_string += s[counter] counter += 1 return new_string if __name__ == '__main__': import doctest doctest.testmod(verbose=True)
Chapter 10
CH 10 - Solution 1
CH 10 - Solution 1.1
>>> import calendar >>> year = calendar.calendar(2008) >>> print year 2008 January February March Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 1 2 3 1 2 7 8 9 10 11 12 13 4 5 6 7 8 9 10 3 4 5 6 7 8 9 14 15 16 17 18 19 20 11 12 13 14 15 16 17 10 11 12 13 14 15 16 21 22 23 24 25 26 27 18 19 20 21 22 23 24 17 18 19 20 21 22 23 28 29 30 31 25 26 27 28 29 24 25 26 27 28 29 30 31 April May June Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 1 2 3 4 1 7 8 9 10 11 12 13 5 6 7 8 9 10 11 2 3 4 5 6 7 8 14 15 16 17 18 19 20 12 13 14 15 16 17 18 9 10 11 12 13 14 15 21 22 23 24 25 26 27 19 20 21 22 23 24 25 16 17 18 19 20 21 22 28 29 30 26 27 28 29 30 31 23 24 25 26 27 28 29 30 July August September Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su 1 2 3 4 5 6 1 2 3 1 2 3 4 5 6 7 7 8 9 10 11 12 13 4 5 6 7 8 9 10 8 9 10 11 12 13 14 14 15 16 17 18 19 20 11 12 13 14 15 16 17 15 16 17 18 19 20 21 21 22 23 24 25 26 27 18 19 20 21 22 23 24 22 23 24 25 26 27 28 28 29 30 31 25 26 27 28 29 30 31 29 30 October November December Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su Mo Tu We Th Fr Sa Su 1 2 3 4 5 1 2 1 2 3 4 5 6 7 6 7 8 9 10 11 12 3 4 5 6 7 8 9 8 9 10 11 12 13 14 13 14 15 16 17 18 19 10 11 12 13 14 15 16 15 16 17 18 19 20 21 20 21 22 23 24 25 26 17 18 19 20 21 22 23 22 23 24 25 26 27 28 27 28 29 30 31 24 25 26 27 28 29 30 29 30 31 >>>
CH 10 - Solution 1.2
isleap(year) - Return 1 for leap years, 0 for non-leap years.
>>> from calendar import * >>> isleap(2008) True >>> isleap(2009) False >>>
CH 10 - Solution 2
CH 10 - Solution 2.1
python C:\Python26\Lib\pydoc.py -p 7464
CH 10 - Solution 2.2
There are 35 functions in the math module.
CH 10 - Solution 2.3
The floor function finds the greatest integral value less than or equal to x. The ceil function finds the lowest integeral value greater than or equal to x.
CH 10 - Solution 2.4
CH 10 - Solution 2.5
The two data constants in the math module are: 'e' and 'pi'.
CH 10 - Solution 3
"""
Interface summary:
import copy
x = copy.copy(y) # make a shallow copy of y
x = copy.deepcopy(y) # make a deep copy of y
For module specific errors, copy.Error is raised.
The difference between shallow and deep copying is only relevant for
compound objects (objects that contain other objects, like lists or
class instances).
- A shallow copy constructs a new compound object and then (to the
extent possible) inserts *the same objects* into it that the
original contains.
- A deep copy constructs a new compound object and then, recursively,
inserts *copies* into it of the objects found in the original."""
deepcopy would have come handy in exercises you didn't have to solve regarding object reference, thus no answer is excpected here.
CH 10 - Solution 4
CH 10 - Solution 5
CH 10 - Solution 6
Namespaces are one honking great idea -- let's do more of those!
CH 10 - Solution 7
CH 10 - Solution 8
def matrix_mult(m1, m2):
"""
>>> matrix_mult([[1, 2], [3, 4]], [[5, 6], [7, 8]])
[[19, 22], [43, 50]]
>>> matrix_mult([[1, 2, 3], [4, 5, 6]], [[7, 8], [9, 1], [2, 3]])
[[31, 19], [85, 55]]
>>> matrix_mult([[7, 8], [9, 1], [2, 3]], [[1, 2, 3], [4, 5, 6]])
[[39, 54, 69], [13, 23, 33], [14, 19, 24]]
"""
qr = len(m1)
qc = len(m2[0])
NewMtx = []
for row in range(qr):
newRow = []
for column in range(qc):
newRow.append(row_times_column(m1, row, m2, column))
NewMtx.append(newRow)
return NewMtx
return NewMtx
qr = len(m1)
qc = len(m2[0])
NewMtx = []
for row in range(qr):
newRow = []
for column in range(qc):
newRow.append(row_times_column(m1, row, m2, column))
NewMtx.append(newRow)
CH 10 - Solution 9
CH 10 - Solution 10
def myreplace(old, new, s): """ Replace all occurences of old with new in the string s. >>> myreplace(',', ';', 'this, that, and, some, other, thing') 'this; that; and; some; other; thing' >>> myreplace(' ', '**', 'Words will now be separated by stars.') 'Words**will**now**be**separated**by**stars.' """ old_removed = s.split(old) new_added = new.join(old_removed) return new_added # Shorter version # new_added = new.join(s.split(old))
CH 10 - Solution 11
def cleanword(word): """ >>> cleanword('what?') 'what' >>> cleanword('"now!"') 'now' >>> cleanword('?+="word!,@$()"') 'word' """ cleaned_word = '' for i in range(len(word)): char = word[i] if char.isalpha(): cleaned_word += char return cleaned_word def has_dashdash(s): """ >>> has_dashdash('distance--but') True >>> has_dashdash('several') False >>> has_dashdash('critters') False >>> has_dashdash('spoke--fancy') True >>> has_dashdash('yo-yo') False """ return s.find('--') != -1 def extract_words(s): """ >>> extract_words('Now is the time! "Now", is the time? Yes, now.') ['now', 'is', 'the', 'time', 'now', 'is', 'the', 'time', 'yes', 'now'] >>> extract_words('she tried to curtsey as she spoke--fancy') ['she', 'tried', 'to', 'curtsey', 'as', 'she', 'spoke', 'fancy'] """ if has_dashdash(s): s = myreplace('--', ' ', s) #using myreplace function from Q. 10 words_punc = s.split() cleanlist = [] for word in words_punc: cleanedword = cleanword(word).lower() cleanlist.append(cleanedword) return cleanlist def wordcount(word, wordlist): """ >>> wordcount('now', ['now', 'is', 'time', 'is', 'now', 'is', 'is']) ['now', 2] >>> wordcount('is', ['now', 'is', 'time', 'is', 'now', 'is', 'the', 'is']) ['is', 4] >>> wordcount('time', ['now', 'is', 'time', 'is', 'now', 'is', 'is']) ['time', 1] >>> wordcount('frog', ['now', 'is', 'time', 'is', 'now', 'is', 'is']) ['frog', 0] """ return [word, wordlist.count(word)] def wordset(wordlist): """ >>> wordset(['now', 'is', 'time', 'is', 'now', 'is', 'is']) ['is', 'now', 'time'] >>> wordset(['I', 'a', 'a', 'is', 'a', 'is', 'I', 'am']) ['I', 'a', 'am', 'is'] >>> wordset(['or', 'a', 'am', 'is', 'are', 'be', 'but', 'am']) ['a', 'am', 'are', 'be', 'but', 'is', 'or'] """ for word in wordlist: count = wordcount(word, wordlist)[1] if count > 1: for a in range(count - 1): wordlist.remove(word) wordlist.sort() return wordlist def longestword(wordset): """ >>> longestword(['a', 'apple', 'pear', 'grape']) 5 >>> longestword(['a', 'am', 'I', 'be']) 2 >>> longestword(['this', 'that', 'supercalifragilisticexpialidocious']) 34 """ longest = 0 for word in wordset: length = len(word) if length > longest: longest = length return longest if __name__ == '__main__': import doctest doctest.testmod()
CH 10 - Solution 12
#sort_fruits.py source = open('unsorted_fruits.txt', 'r') fruits = source.readlines() source.close() fruits.sort() newfile = open('sorted_fruits.txt', 'w') newfile.writelines(fruits) newfile.close()
CH 10 - Solution 13
CH 10 - Solution 14
#mean.py from sys import argv nums = argv[1:] for i, value in enumerate(nums): nums[i] = float(value) mean = sum(nums) / len(nums) print mean
CH 10 - Solution 15
#median.py from sys import argv nums = argv[1:] for i, value in enumerate(nums): nums[i] = float(value) nums.sort() size = len(nums) middle = size / 2 if size % 2 == 0: median = (nums[middle - 1] + nums[middle]) / 2 else: median = nums[middle] if median == float(int(median)): median = int(median) print median
CH 10 - Solution 16
# # countletters.py # def display(i): if i == 10: return 'LF' if i == 13: return 'CR' if i == 32: return 'SPACE' return chr(i) from sys import argv filename = argv[1] infile = open(filename, 'r') text = infile.read() infile.close() counts = 128 * [0] for letter in text: counts[ord(letter)] += 1 filenamesplit = filename.split('.') # splits 'name.txt' -> ['name', 'txt'] count_file = filenamesplit[0] + '_counts.dat' # 'name' -> 'name_counts.dat' outfile = open(count_file, 'w') outfile.write("%-12s%s\n" % ("Character", "Count")) outfile.write("=================\n") for i in range(len(counts)): if counts[i]: outfile.write("%-12s%d\n" % (display(i), counts[i])) outfile.close()
Chapter 11
CH 11 - Solution 1
CH 11 - Solution 2
CH 11 - Solution 3
# #seqtools.py # def remove_at(pos, seq): return seq[:pos] + seq[pos+1:] def encapsulate(val, seq): """ >>> encapsulate((1, 'a'), [0 , 'b']) [(1, 'a')] >>> encapsulate(42, 'string') '42' >>> tup = 1, 2, 3 # NB. Testmod seems to report this >>> encapsulate(5, tup) # as a fail, despite returning the (5,) # correct result? """ if type(seq) == type(""): return str(val) if type(seq) == type([]): return [val] return (val,) def insert_in_middle(val, seq): """ >>> insert_in_middle('two', (1,3)) (1, 'two', 3) >>> insert_in_middle(4, 'two six') 'two 4 six' >>> insert_in_middle((2, 4), [(1, 2), (3, 6)]) [(1, 2), (2, 4), (3, 6)] """ middle = len(seq)/2 return seq[:middle] + encapsulate(val, seq) + seq[middle:] def make_empty(seq): """ >>> make_empty([1, 2, 3, 4]) [] >>> make_empty(('a', 'b', 'c')) () >>> make_empty("No, not me!") '' """ if type(seq) == type(""): return '' if type(seq) == type([]): return [] return () def insert_at_end(val, seq): """ >>> insert_at_end(5, [1, 3, 4, 6]) [1, 3, 4, 6, 5] >>> insert_at_end('x', 'abc') 'abcx' >>> insert_at_end(5, (1, 3, 4, 6)) # NB. Testmod seems to report this as a fail (1, 3, 4, 6, 5) # despite returning the correct result """ return seq + encapsulate(val, seq) def insert_in_front(val, seq): """ >>> insert_in_front(5, [1, 3, 4, 6]) [5, 1, 3, 4, 6] >>> insert_in_front(5, (1, 3, 4, 6)) (5, 1, 3, 4, 6) >>> insert_in_front('x', 'abc') 'xabc' """ return encapsulate(val, seq) + seq def index_of(val, seq, start=0): """ >>> index_of(9, [1, 7, 11, 9, 10]) 3 >>> index_of(5, (1, 2, 4, 5, 6, 10, 5, 5)) 3 >>> index_of(5, (1, 2, 4, 5, 6, 10, 5, 5), 4) 6 >>> index_of('y', 'happy birthday') 4 >>> index_of('banana', ['apple', 'banana', 'cherry', 'date']) 1 >>> index_of(5, [2, 3, 4]) -1 >>> index_of('b', ['apple', 'banana', 'cherry', 'date']) -1 """ for i in range(start, len(seq)): if seq[i] == val: return i return -1 def remove_at(index, seq): """ >>> remove_at(3, [1, 7, 11, 9, 10]) [1, 7, 11, 10] >>> remove_at(5, (1, 4, 6, 7, 0, 9, 3, 5)) (1, 4, 6, 7, 0, 3, 5) >>> remove_at(2, "Yomrktown") 'Yorktown' """ return seq[:index] + seq[index + 1:] def remove_val(val, seq): """ >>> remove_val(11, [1, 7, 11, 9, 10]) [1, 7, 9, 10] >>> remove_val(15, (1, 15, 11, 4, 9)) (1, 11, 4, 9) >>> remove_val('what', ('who', 'what', 'when', 'where', 'why', 'how')) ('who', 'when', 'where', 'why', 'how') """ return remove_at(index_of(val, seq), seq) def remove_all(val, seq): """ >>> remove_all(11, [1, 7, 11, 9, 11, 10, 2, 11]) [1, 7, 9, 10, 2] >>> remove_all('i', 'Mississippi') 'Msssspp' """ while index_of(val, seq) != -1: seq = remove_val(val, seq) return seq def count(val, seq): """ >>> count(5, (1, 5, 3, 7, 5, 8, 5)) 3 >>> count('s', 'Mississippi') 4 >>> count((1, 2), [1, 5, (1, 2), 7, (1, 2), 8, 5]) 2 """ count = 0 for item in seq: if item == val: count += 1 return count def reverse(seq): """ >>> reverse([1, 2, 3, 4, 5]) [5, 4, 3, 2, 1] >>> reverse(('shoe', 'my', 'buckle', 2, 1)) (1, 2, 'buckle', 'my', 'shoe') >>> reverse('Python') 'nohtyP' """ output = make_empty(seq) for item in seq: output = insert_in_front(item, output) return output def sort_sequence(seq): """ >>> sort_sequence([3, 4, 6, 7, 8, 2]) [2, 3, 4, 6, 7, 8] >>> sort_sequence((3, 4, 6, 7, 8, 2)) (2, 3, 4, 6, 7, 8) >>> sort_sequence("nothappy") 'ahnoppty' """ listseq = list(seq) listseq.sort() output = make_empty(seq) for item in listseq: output = insert_at_end(item, output) return output
CH 11 - Solution 4
def recursive_min(nested_num_list): """ >>> recursive_min([2, 9, [1, 13], 8, 6]) 1 >>> recursive_min([2, [[100, 1], 90], [10, 13], 8, 6]) 1 >>> recursive_min([2, [[13, -7], 90], [1, 100], 8, 6]) -7 >>> recursive_min([[[-13, 7], 90], 2, [1, 100], 8, 6]) -13 """ smallest = nested_num_list[0] while type(smallest) == type([]): smallest = smallest[0] for element in nested_num_list: if type(element) == type([]): min_of_elem = recursive_min(element) if smallest > min_of_elem: smallest = min_of_elem else: if smallest > element: smallest = element return smallest
CH 11 - Solution 5
def recursive_count(target, nested_num_list): """ >>> recursive_count(2, [2, 9, [2, 1, 13, 2], 8, [2, 6]]) 4 >>> recursive_count(7, [[9, [7, 1, 13, 2], 8], [7, 6]]) 2 >>> recursive_count(15, [[9, [7, 1, 13, 2], 8], [2, 6]]) 0 >>> recursive_count(5, [[5, [5, [1, 5], 5], 5], [5, 6]]) 6 """ count = 0 for element in nested_num_list: if type(element) == type([]): count += recursive_count(target, element) else: if element == target: count += 1 return count
CH 11 - Solution 6
def flatten(nested_num_list): """ >>> flatten([2, 9, [2, 1, 13, 2], 8, [2, 6]]) [2, 9, 2, 1, 13, 2, 8, 2, 6] >>> flatten([[9, [7, 1, 13, 2], 8], [7, 6]]) [9, 7, 1, 13, 2, 8, 7, 6] >>> flatten([[9, [7, 1, 13, 2], 8], [2, 6]]) [9, 7, 1, 13, 2, 8, 2, 6] >>> flatten([[5, [5, [1, 5], 5], 5], [5, 6]]) [5, 5, 1, 5, 5, 5, 5, 6] """ flat_num_list = [] for element in nested_num_list: if type(element) == type([]): flat_num_list += flatten(element) else: flat_num_list += [element] return flat_num_list
CH 11 - Solution 7
def readposint(prompt = 'Please enter a positive integer: '): while True: posint = raw_input(prompt) try: posint = float(posint) if posint != int(posint): raise ValueError, '%s is not an integer' % posint elif posint < 1: raise ValueError, '%s is not positive' % posint break except: print '%s is not a positive integer. Try again.' % posint return int(posint)
CH 11 - Solution 8
CH 11 - Solution 9
CH 11 - Solution 10
def factorial(n): nshriek = 1 # n factorial is sometimes called while n > 1: # 'n shriek' because of the 'n!' notation nshriek *= n n -= 1 return nshriek
CH 11 - Solution 11
# # litter.py # import os import sys def getroot(): if len(sys.argv) == 1: path = '' else: path = sys.argv[1] if os.path.isabs(path): tree_root = path else: tree_root = os.path.join(os.getcwd(), path) return tree_root def getdirlist(path): dirlist = os.listdir(path) dirlist = [name for name in dirlist if name[0] != '.'] dirlist.sort() return dirlist def traverse(path, t=0): dirlist = getdirlist(path) for num, file in enumerate(dirlist): dirsize = len(dirlist) path2file = os.path.join(path, file) if os.path.isdir(path2file): t += 1 newtrash = open(path2file + '\\trash.txt', 'w') newtrash.close() if getdirlist(path2file): t = traverse(path2file, t) return t if __name__ == '__main__': root = getroot() trashes = traverse(root) if trashes == 1: filestring = 'file' else: filestring = 'files' print '%d trash.txt %s created' % (trashes, filestring)
# # cleanup.py # import os import sys def getroot(): if len(sys.argv) == 1: path = '' else: path = sys.argv[1] if os.path.isabs(path): tree_root = path else: tree_root = os.path.join(os.getcwd(), path) return tree_root def getdirlist(path): dirlist = os.listdir(path) dirlist = [name for name in dirlist if name[0] != '.'] dirlist.sort() return dirlist def traverse(path, t = 0): dirlist = getdirlist(path) for num, file in enumerate(dirlist): dirsize = len(dirlist) path2file = os.path.join(path, file) if file == 'trash.txt': t += 1 os.remove(path2file) elif os.path.isdir(path2file): t += traverse(path2file) return t if __name__ == '__main__': root = getroot() trashed = traverse(root) if trashed == 1: filestring = 'file' else: filestring = 'files' print '%d trash.txt %s deleted' % (trashed, filestring)
Chapter 12
CH 12 Solution 1
# # letter_counts.py # import sys def count_letters(s): letter_counts = {} for letter in s: if letter.isalpha(): letter_counts[letter.lower()] = letter_counts.get(letter.lower(), 0) + 1 return letter_counts def print_counts(letter_counts): letter_items = letter_counts.items() letter_items.sort() for item in letter_items: print item[0], item[1] if len(sys.argv) > 1: letter_counts = count_letters(sys.argv[1]) print_counts(letter_counts) else: print 'No argument supplied'
CH 12 Solution 2
def add_fruit(inventory, fruit, quantity=0): """ Adds quantity of fruit to inventory. >>> new_inventory = {} >>> add_fruit(new_inventory, 'strawberries', 10) >>> new_inventory.has_key('strawberries') True >>> new_inventory['strawberries'] 10 >>> add_fruit(new_inventory, 'strawberries', 25) >>> new_inventory['strawberries'] 35 """ if inventory.has_key(fruit): inventory[fruit] += quantity else: inventory[fruit] = quantity
CH 12 Solution 3
# # alice_words.py # import string filename = 'alice_in_wonderland.txt' countfile = 'alice_counts.txt' def add_word(counts, word): if counts.has_key(word): counts[word] += 1 else: counts[word] = 1 def get_word(item): word = '' item = item.strip(string.digits) item = item.lstrip(string.punctuation) item = item.rstrip(string.punctuation) word = item.lower() return word def count_words(text): text = ' '.join(text.split('--')) #replace '--' with a space items = text.split() #leaves in leading and trailing punctuation, #'--' not recognised by split() as a word separator counts = {} for item in items: word = get_word(item) if not word == '': add_word(counts, word) return counts infile = open(filename, 'r') text = infile.read() infile.close() counts = count_words(text) outfile = open(countfile, 'w') outfile.write("%-18s%s\n" %("Word", "Count")) outfile.write("=======================\n") counts_list = counts.items() counts_list.sort() for word in counts_list: outfile.write("%-18s%d\n" %(word[0], word[1])) outfile.close
The word "alice" occurs 386 times (not including 12 occurences of "alice's")
CH 12 Solution 4
The longest 'word' in the list is "bread-and-butter" with 16 characters. This can be found by adding the following code to the alice_words.py program:
longest = ('', 0) for word in counts: if len(word) > longest[1]: longest = (word, len(word)) print longest
Slightly altering this we can find the longest unhyphenated word, "disappointment", which has 14 characters.
longest = ('', 0) for word in counts: if len(word) > longest[1] and word.find('-') == -1 longest = (word, len(word)) print longest
CH 12 Solution 5
CH 12 Solution 6
CH 12 Solution 7
CH 12 Solution 8
CH 12 Solution 9
CH 12 Solution 10
CH 12 Solution 11
CH 12 Solution 12
Chapter 13
CH 13 Solution 1
CH 13 Solution 2
def distance(p1, p2): dx = p2.x - p1.x dy = p2.y - p1.y dsquared = dx**2 + dy**2 result = dsquared**0.5 return result
CH 13 Solution 3
def move_rect(rect, dx, dy): rect.corner.x += dx rect.corner.y += dy
CH 13 Solution 4
def move_rect(rect, dx, dy): import copy new_rect = copy.deepcopy(rect) new_rect.corner.x += dx new_rect.corner.y += dy return new_rect
Chapter 14
CH 14 Solution 1
def print_time(t): print "%i:%i:%i" % (t.hours, t.minutes, t.seconds)
CH 14 Solution 2
def after(t1, t2): return convert_to_seconds(t1) > convert_to_seconds(t2)
CH 14 Solution 3
def increment(time, seconds): sum = convert_to_seconds(time) + seconds newtime = make_time(sum) time.hours = newtime.hours time.minutes = newtime.minutes time.seconds = newtime.seconds >>> increment(t1, 180)
CH 14 Solution 4
def increment(time, seconds): sum = convert_to_seconds(time) + seconds newtime = make_time(sum) return newtime >>> increment(t1, 180) <__main__.Time instance at 0x91ceeac> >>> t1 = increment(t1, 180)
Chapter 15
CH 15 Solution 1
class Time(): # other method definitions here def convertToSeconds(self): minutes = self.hours * 60 + self.minutes seconds = minutes * 60 + self.seconds return seconds
CH 15 Solution 2
def find(str, ch, start=0, end = "None"): index = start if end == "None": end = len(str) while index < end: if str[index] == ch: return index index = index + 1 return -1
Chapter 16
CH 16 Solution 1
class Card: ... def __cmp__(self, other): # check the suits if self.suit > other.suit: return 1 if self.suit < other.suit: return -1 # suits are the same... check ranks if self.rank > other.rank: if other.rank == 1: return -1 #other is Ace (and self is not) return 1 if self.rank < other.rank: if self.rank == 1: return 1 #self is Ace (and other is not) return -1 # ranks are the same... it's a tie return 0
Chapter 17
CH 17 Solution 1
class OldMaidGame(CardGame): ... def printHands(self): for hand in self.hands: print hand
Chapter 18
CH 18 Solution 1
def print_list(node): s = '[' while node: s += str(node.cargo) node = node.next if node: s += ', ' s += ']' print s
Chapter 19
CH 19 Solution 1
>>> print eval_postfix("1 2 + 3 *") 9
CH 19 Solution 2
>>> print eval_postfix("1 2 3 * +") 7
Chapter 20
Solution 1
class listQueue: def __init__(self): self.items = [] def is_empty(self): return self.items == [] def insert(self, item): self.items.append(item) def remove(self): if self.is_empty(): return None return self.items.pop()
class llPriorityQueue: def __init__(self): # NB, self.last is no longer needed because self.length = 0 # the linked list will be ordered with self.head = None # the next item to be removed at the head def is_empty(self): return (self.length == 0) def insert(self, cargo): # insert() needs to keep the list ordered to node = Node(cargo) # make remove() constant time. So insert() if self.head == None: # must traverse the list. Now insert() is self.head = node # linear time instead of remove()! else: if cargo > self.head.cargo: node.next = self.head self.head = node else: smaller = self.head while smaller.cargo >= cargo:# traverse queue until we previous = smaller # find first node with cargo smaller = smaller.next # smaller than the new node if smaller == None: #<--(or we get to the end break # of the list) previous.next = node # insert node ahead of first node.next = smaller # smaller item self.length += 1 def remove(self): if self.is_empty(): return None cargo = self.head.cargo self.head = self.head.next self.length = self.length - 1 return cargo
Chapter 21
CH 21 Solution 1
def print_tree_inorder(tree): if tree == None: return if tree.left: print '(', print_tree_inorder(tree.left), print tree.cargo, print_tree_inorder(tree.right), if tree.right: print ')', # The output from this function is correct and unambiguous, but many # brackets are not really necessary and look messy. For example: >>> token_list =['(',3, '+', 7, ')', '*', '(', 5, '+', 3, ')', '+', 4, '*', 5, 'end'] >>> tree = get_sum(token_list) >>> print_tree_inorder(tree) ( ( ( 3 + 7 ) * ( 5 + 3 ) ) + ( 4 * 5 ) ) # The ideal output would be something like: (3 + 7) * (5 + 3) + 4 * 5
CH 21 Solution 2
def make_token_list(s): token_list = [] for char in s: if char == ' ': continue elif char.isdigit(): token_list.append(int(char)) else: token_list.append(char) token_list.append('end') return token_list
CH 21 Solution 3
CH 21 Solution 4
# ... (only modified functions shown) def dump(tree): # returns string representation of tree in postfix-style # order with commas separating nodes. Leaves are # represented with two preceding commas, corresponding to # the empty tree.left & tree.right attributes. if tree == None: return ',' s = '' s += dump(tree.left) s += dump(tree.right) s += str(tree) + ',' return s def restore_tree(token_list): # Recreate tree from token list generated from save file cargo = token_list.pop() if cargo == '': return right = restore_tree(token_list) left = restore_tree(token_list) tree = Tree(cargo, left, right) return tree def animal(): #start with a singleton root = Tree("bird") # or use save file if it exists try: infile = open(savefile, 'r') s = infile.read() infile.close() token_list = s.split(',') token_list.pop() #remove empty item at end root = restore_tree(token_list) except IOError: pass # loop until the user quits while True: print if not yes("Are you thinking of an animal? "): break # walk the tree tree = root while tree.left != None: prompt = tree.cargo + "? " if yes(prompt): tree = tree.right else: tree = tree.left # make a guess guess = tree.cargo prompt = "Is it a " + guess + "? " if yes(prompt): print "Yatta!" continue # get new information prompt = "What is the animal's name? " animal = raw_input(prompt) prompt = "What question would distinguish a %s from a %s? " question = raw_input(prompt % (animal, guess)).strip('?').capitalize() # add new information to the tree tree.cargo = question prompt = "If the animal were %s the answer would be? " if yes(prompt % animal): tree.left = Tree(guess) tree.right = Tree(animal) else: tree.left = Tree(animal) tree.right = Tree(guess) # save the tree s = dump(root) outfile = open(savefile, 'w') outfile.write(s) outfile.close()