How to Think Like a Computer Scientist: Learning with Python 2nd Edition/Solutions/CH 9
Chapter 9 edit
CH 9 - Solution 1 edit
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 edit
CH 9 - Solution 2.1 edit
"""
>>> 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 edit
"""
>>> 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 edit
"""
>>> '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 edit
"""
>>> 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 edit
>>> 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 edit
>>> 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 edit
>>> 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 edit
CH 9 - Solution 6.1 edit
"""
>>> 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 edit
"""
>>> 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 edit
"""
>>> import string
>>> string.split(message, '??')
['this', 'and', 'that']
"""
message = "this??and??that"
if __name__ == '__main__':
import doctest
doctest.testmod()
CH 9 - Solution 7 edit
def add_vectors(u, v):
"""
>>> add_vectors([1, 0], [1, 1])
[2, 1]
>>> add_vectors([1, 2], [1, 4])
[2, 6]
>>> add_vectors([1, 2, 1], [1, 4, 3])
[2, 6, 4]
>>> add_vectors([11, 0, -4, 5], [2, -4, 17, 0])
[13, -4, 13, 5]
"""
new_list = []
for index, value in enumerate(u):
for index2, value2 in enumerate(v):
if index == index2:
new_list += [value + value2]
return new_list
#However, a less complex solution follows:
new_list = []
for index, value in enumerate(u):
new_list += [u[index] + z[index]]
return new_list
if __name__ == '__main__':
import doctest
doctest.testmod()
CH 9 - Solution 8 edit
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 edit
CH 9 - Solution 9.1 edit
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 edit
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 edit
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 edit
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 edit
# 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 edit
CH 9 - Solution 14 edit
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)