Think Python/Answers
Chapter 1
[edit | edit source]See below for Chapter 1 exercises.
Exercise 1.2
[edit | edit source]1.) How many seconds are there in 42 minutes 42 seconds?
>>> (42 * 60) + 42
2562 seconds
2.) How many miles are there in 10 kilometers? Hint there are 1.61 kilometers in a mile.
>>> 10/1.61
6.21 miles
3.) If you run a 10 kilometer race in 42 minutes 42 seconds, what is your average time per mile? What is your average speed in miles per hour? (Hint: there are about 1.61 kilometers in a mile.)
>>> 10 / 1.61 # convert kilometers to miles
6.211180124223602
>>> (42 * 60) + 42 # convert time to seconds
2562
>>> 2562 / 6.211180124223602 # what is your average time (seconds) per mile
412.482
>>> 412.482 / 60 # what is your average time (minutes) per mile
6.874700000000001
>>> 60 / 6.874700000000001 # miles per hour
8.727653570337614
How about another way
>>> 10 / 42.7 # avg kilometers per minute
0.23419203747072598
>>> 0.23419203747072598 * 60 # kilometers per hour
14.05152224824356
>>> 14.05152224824356 / 1.61 # convert to M.P.H
8.727653570337614
or a one-liner
>>> (10 / 1.61) / (42.7 / 60) # (distance in miles) / (time in hours)
8.727653570337614 # miles/hour
Chapter 2
[edit | edit source]Exercise 2.1
[edit | edit source]If you type an integer with a leading zero, you might get a confusing error:
>>> zipcode = 02492
^
SyntaxError: invalid token
Other number seem to work, but the results are bizarre:
>>> zipcode = 02132
>>> print zipcode
1114
So python is assuming you want to convert an octal number to a decimal number. In the base 8 numbering system where valid numbers are 0, 1, 2, 3, 4, 5, 6 and 7.
Base 8: 00 01 02 03 04 05 06 07 10 11 12 13 14 15 16 17 20 21 22 23 24
Base 10: 00 01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20
Every 8 numbers we increment the left hand columns. This means that the right most column is the number of 'ones'. The one to the left of that is a tally of the number of 'eights', the one next to that is a tally of a full column of 'eight' times the 'eight column' - 64. The one next to that is 64*8 - 512 and so on. For more information read Base Eight math.
That is why zipcode = 02492 is invalid as the digit 9 is not a valid octal number. We can do the conversion manually as follows:
>>> print 02132
1114
>>> (2*512)+(1*64)+(3*8)+(2*1)
1114
>>>
Exercise 2.2
[edit | edit source]The volume of a sphere with radius is 4/3 π . What is the volume of a sphere with radius 5?
>>> pi = 3.1415926535897932
>>> r = 5
>>> 4/3*pi*r**3 # This is the wrong answer
392.69908169872411
>>> r = 5.0 # Radius can be a float here as well, but is not _necessary_.
>>> 4.0/3.0*pi*r**3 # Using floats give the correct answer
523.5987755982989
>>>
Suppose the cover price of a book is $24.95, but bookstores get a 40% discount. Shipping costs $3 for the first copy and 75 cents for each additional copy. What is the total wholesale cost for 60 copies?
$24.95 Cost
$9.98 Discount per book
$14.97 Cost per book after discount
60 Total number of books
$898.20 Total cost not inc delivery
$3.00 First book delivery
59 Remaining books
$0.75 Delivery cost for extra books
$44.25 Total cost for extra books
$47.25 Total Delivery cost
$945.45 Total Bill
This answer is wrong because 40.0/100.0 return wrong value 0.40000000000000002 for more info see IEEE 754 (Standard for Floating-Point Arithmetic)
>>> (24.95-24.95*40.0/100.0)*60+3+0.75*(60-1)
945.44999999999993
>>> 24.95*0.6*60+0.75*(60-1)+3
945.45
You can use the decimal module to maintain precision.
from decimal import Decimal
..
...
def wholesale_cost(args):
d = 1 - args.get('discount')/100
purchase = Decimal(args.get('cost') * d * 60)
return purchase + Decimal(args.get('delivery'))
args = {'cost': 24.95, 'discount': 40, 'delivery': 3.00+0.75*59}
Another solution using functions as well as input prompts:
# Total wholesale book cost calculator
cover_price = 24.95
number_of_books = int(input("How many books do you want to order at wholesale? "))
def ship_cost (number_of_books):
if number_of_books == 1:
return (number_of_books * 3) # Cost of shipping one book is $3
else:
return (3 + (number_of_books - 1) * 0.75) # Each additional copy of the book is $0.75 to ship
def discounted_price(number_of_books):
return(cover_price - (cover_price * .4)) # There is a 40% discount on wholesale book sales
def wholesale_cost(number_of_books):
return ((discounted_price(number_of_books) * number_of_books) + ship_cost(number_of_books))
print("The cost of buying and shipping", number_of_books, "books is $",round(wholesale_cost(number_of_books), 2))
If I leave my house at 6:52 am and run 1 mile at an easy pace (8:15 per mile), then 3 miles at tempo (7:12 per mile) and 1 mile at easy pace again, what time do I get home for breakfast?
Answer: 7:30 am
How I did it:
>>> start = (6*60+52)*60
>>> easy = (8*60+15)*2
>>> fast = (7*60+12)*3
>>> finish_hour = (start + easy + fast)/(60*60.0)
>>> finish_floored = (start + easy + fast)//(60*60) #int() function can also be used to get integer value, but it hasn't taught yet.
>>> finish_minute = (finish_hour - finish_floored)*60
>>> print ('Finish time was %d:%d' % (finish_hour,finish_minute))
Finish time was 7:30
*** ANOTHER WAY ***
start_time_hr = 6 + 52 / 60.0
easy_pace_hr = (8 + 15 / 60.0 ) / 60.0
tempo_pace_hr = (7 + 12 / 60.0) / 60.0
running_time_hr = 2 * easy_pace_hr + 3 * tempo_pace_hr
breakfast_hr = start_time_hr + running_time_hr
breakfast_min = (breakfast_hr-int(breakfast_hr))*60
breakfast_sec= (breakfast_min-int(breakfast_min))*60
print ('breakfast_hr', int(breakfast_hr) )
print ('breakfast_min', int (breakfast_min) )
print ('breakfast_sec', int (breakfast_sec) )
>>>
Chapter 3
[edit | edit source]Exercise 3.1
[edit | edit source]Python provides a built-in function called len that returns the length of a string, so the value of len('allen') is 5. Write a function named right_justify that takes a string named s as a parameter and prints the string with enough leading spaces so that the last letter of the string is in column 70 of the display.
>>> def right_justify(s):
print (' '*(70-len(s))+s)
>>> right_justify('allen')
allen
>>>
Alternate Solution Using concatenation and repetition
def right_justify(s):
total_length = 70
current_length = len(s)
current_string = s
while current_length < total_length:
current_string = " " + current_string
current_length = len(current_string)
print(current_string)
OUTPUT
>>> right_justify("monty")
monty
Exercise 3.3
[edit | edit source]You can see my solution at http://thinkpython.com/code/grid.py.
"""
Solution to Exercise 3.5 on page 27 of Think Python
Allen B. Downey, Version 1.1.24+Kart [Python 3.2]
"""
# here is a mostly-straightforward solution to the
# two-by-two version of the grid.
def do_twice(f):
f()
f()
def do_four(f):
do_twice(f)
do_twice(f)
def print_beam():
print('+ - - - -', end='')
def print_post():
print('| ', end='')
def print_beams():
do_twice(print_beam)
print('+')
def print_posts():
do_twice(print_post)
print('|')
def print_row():
print_beams()
do_twice(print_posts)
def print_grid():
do_twice(print_row)
print_beams()
print_grid()
____________
# another solution
def do_twice(f):
f()
f()
def do_four(f):
do_twice(f)
do_twice(f)
def print_column():
print '+----+----+'
def print_row():
print '| | |'
def print_rows():
do_four(print_row)
def do_block():
print_column()
print_rows()
def print_block():
do_twice(do_block)
print_column()
print_block()
# nathan moses-gonzales
_________
# straight-forward solution to 4x4 grid
def do_twice(f):
f()
f()
def do_four(f): # not needed for 2x2 grid
do_twice(f)
do_twice(f)
def print_beam():
print('+----', end='')
def print_post():
print('| ', end='')
def print_beams():
do_twice(print_beam)
print('+')
def print_posts():
do_twice(print_post)
print('|')
def print_row():
print_beams()
do_twice(print_posts)
def print_grid2x2():
do_twice(print_row)
print_beams()
def print_beam4():
do_four(print_beam)
print('+')
def print_post4():
do_four(print_post)
print('|')
def print_row4():
print_beam4()
do_twice(print_post4)
def print_grid4x4():
do_four(print_row4)
print_beam4()
print_grid4x4()
-----------------------
# here is a less-straightforward solution to the
# four-by-four grid
def one_four_one(f, g, h):
f()
do_four(g)
h()
def print_plus():
print '+',
def print_dash():
print '-',
def print_bar():
print '|',
def print_space():
print ' ',
def print_end():
print
def nothing():
"do nothing"
def print1beam():
one_four_one(nothing, print_dash, print_plus)
def print1post():
one_four_one(nothing, print_space, print_bar)
def print4beams():
one_four_one(print_plus, print1beam, print_end)
def print4posts():
one_four_one(print_bar, print1post, print_end)
def print_row():
one_four_one(nothing, print4posts, print4beams)
def print_grid():
one_four_one(print4beams, print_row, nothing)
print_grid()
comment = """
After writing a draft of the 4x4 grid, I noticed that many of the
functions had the same structure: they would do something, do
something else four times, and then do something else once.
So I wrote one_four_one, which takes three functions as arguments; it
calls the first one once, then uses do_four to call the second one
four times, then calls the third.
Then I rewrote print1beam, print1post, print4beams, print4posts,
print_row and print_grid using one_four_one.
Programming is an exploratory process. Writing a draft of a program
often gives you insight into the problem, which might lead you to
rewrite the code to reflect the structure of the solution.
--- Allen
"""
print comment
# another solution
def beam():
plus = "+"
minus = "-"*4
print(plus, minus, plus,minus, plus, minus, plus, minus, plus)
def straight():
straight = "|"
space = " "*4
print(straight, space, straight, space, straight, space, straight, space,
straight, space)
def quad_straight():
straight()
straight()
straight()
straight()
def twice():
beam()
quad_straight()
beam()
quad_straight()
def quad():
twice()
twice()
beam()
quad()
-- :)
------------------
# Without functions.
print("+ - - - - " * 2 + "+")
print("|\t\t | \t\t|\n" * 3 + "|\t\t | \t\t|")
print("+ - - - - " * 2 + "+")
print("|\t\t | \t\t|\n " *3 + "|\t\t | \t\t|")
print("+ - - - - " * 2 + "+")
------------------
Why not using the first solution and adapt it to the number of rows
def do_twice(f):
f()
f()
def do_four(f):
do_twice(f)
do_twice(f)
def print_column():
print '+----+----+----+----+'
def print_row():
print '| | | | |'
def print_rows():
do_four(print_row)
def do_block():
print_column()
print_rows()
def print_block():
do_twice(do_block)
# print_column()
do_twice(do_block)
print_column()
print_block()
-----------------------
# mteodor
def draw_line(bar, middle = ' ', repeat = 2, length = 2):
""" Draw a single line like this:
[ (B M*repeat)*length B]
"""
for k in range(length):
print("%s %s " % (bar, middle*repeat), end='')
print(bar)
def draw_grid(length = 2, height = 2, width = 2):
""" Draw a grid like this:
+ -- + -- +
| | |
| | |
+ -- + -- +
| | |
| | |
+ -- + -- +
where:
* length x height are the table size
* width is the size of a cell/column
"""
for i in range(height):
draw_line('+', '-', width, length)
for j in range(length):
draw_line('|', ' ', width, length)
draw_line('+', '-', width, length)
draw_grid(4, 4, 3)
--------------------------
#kipp
# auto adjust size of columns
size=4
def beam():
print(" + - - - -"*size, "+")
def post():
print(" | "*size, "|")
def repeat(func):
func()
func()
func()
# manual adjust size for rows
# this is 4
beam()
repeat(post)
beam()
repeat(post)
beam()
repeat(post)
beam()
repeat(post)
beam()
Chapter 4
[edit | edit source]4.3 Exercise 1
[edit | edit source]from TurtleWorld import *
world = TurtleWorld()
bob = Turtle()
def square(t):
for i in range(4):
fd(t, 100)
lt(t)
square(bob)
wait_for_user()
4.3 Exercise 2
[edit | edit source]from TurtleWorld import *
world = TurtleWorld()
bob = Turtle()
print(bob)
def square(t, length):
t = Turtle()
for i in range(4):
fd(t, length)
lt(t)
square(bob, 200)
wait_for_user()
4.3 Exercise 3
[edit | edit source]from swampy.TurtleWorld import *
world = TurtleWorld()
bob = Turtle()
print(bob)
def polygon(t, length, n):
for i in range(n):
fd(t, length)
lt(t, 360 / n)
polygon(bob, 50, 8)
wait_for_user()
Chapter 5
[edit | edit source]Exercise 5.2
[edit | edit source]def countdown(a): # A typical countdown function
if a < 0:
print("Blastoff")
elif a > 0:
print(a)
countdown(a - 1)
def call_function(n,a): # The countdown function is called "n" number of times. Any other function can be used instead of countdown function.
for i in range(n):
countdown(a)
call_function(3, 10)
Exercise 5.3
[edit | edit source]def is_triangle(a, b, c):
if a <= b+c:
if b <= a+c:
if c <= a+b:
return 'yes'
else:
return 'no'
else:
return 'no'
else:
return 'no'
is_triangle(1, 12, 1)
'no'
Exercise 5.3
[edit | edit source]def is_triangle(a, b, c):
if a<=b+c and b<=a+c and c<=a+b:
return 'yes'
else:
return 'no'
is_triangle(1, 12, 3)
'no'
Chapter 9
[edit | edit source]Exercise 9.1
[edit | edit source]fin = open('words.txt')
for line in fin:
word = line.strip()
if len(word) > 20:
print (word)
Exercise 9.2
[edit | edit source]fin = open('words.txt')
def has_no_e(word):
for char in word:
if char in 'Ee':
return False
return True
count = 0
for line in fin:
word = line.strip()
if has_no_e(word):
count += 1
print (word)
percent = (count / 113809.0) * 100
print (str(percent)) + "% of the words don't have an 'e'."
Exercise 9.3
[edit | edit source]fin = open('words.txt')
def avoids(word,letter):
for char in word:
if char in letter:
return False
return True
letter = raw_input('What letters to exclude? ')
count = 0
for line in fin:
word = line.strip()
if avoids(word, letter):
count += 1
print word
percent = (count / 113809.0) * 100
print str(percent) + "% of the words don't have " + letter + '.'
Exercise 9.4
[edit | edit source]def uses_only(word, letters):
"""returns true if word is made only out of letters else flase"""
for letter in word:
if letter not in letters:
return False
return True
Chapter 10
[edit | edit source]Exercise 10.1
[edit | edit source]Write a function called nested_sum that takes a nested list of integers and add up the elements from all of the nested lists.
def nested_sum(nestedList):
'''
nestedList: list composed of nested lists containing int.
Returns the sum of all the int in the nested list
'''
newList = []
#Helper function to flatten the list
def flatlist(nestedList):
'''
Returns a flat list
'''
for i in range(len(nestedList)):
if type(nestedList[i]) == int:
newList.append(nestedList[i])
else:
flatlist(nestedList[i])
return newList
flatlist(nestedList)
print sum(newList)
nested_sum(nestedList)
Exercise 10.2
[edit | edit source]Write a function named "capitalize_nested" that takes a nested list of strings and returns a new nested list with all strings capitalized.
>>> def capitalize_nested(l):
def capitalize(s):
return s.capitalize()
for n, i in enumerate(l):
if type(i) is list:
l[n] = capitalize_nested(l[n])
elif type(i) is str:
l[n] = capitalize(i)
return l
Exercise 10.3
[edit | edit source]Write a function that takes a list of numbers and returns the cumulative sum.
>>> def cumulative(l):
cumulative_sum = 0
new_list = []
for i in l:
cumulative_sum += i
new_list.append(cumulative_sum)
return new_list
Exercise 10.4
[edit | edit source]Write a function called middle that takes a list and returns a new list that contains all but the first and last elements.
>>> def middle(x):
res = []
i = 1
while i <= len(x)-2:
res.append(x[i])
i += 1
return res
This can also be done simply with a slice.
>>> def middle(x):
return x[1:-1]
Exercise 10.5
[edit | edit source]Write a function called chop that takes a list and modifies it, removing the first and last elements, and returns None.
>>> def chop(x):
del x[:1]
del x[-1:]
Chapter 11
[edit | edit source]Exercise 11.1
[edit | edit source]Write a function that reads the words in words.txt and stores them as keys in a dictionary. It doesn’t matter what the values are. Then you can use the in operator as a fast way to check whether a string is in the dictionary.
fin = open('words.txt')
englishdict = dict()
def create_diction():
counter = 0
dictionairy = dict()
for line in fin:
word = line.strip()
dictionairy[word] = counter
counter += 1
return dictionairy
Exercise 11.2
[edit | edit source]def invert_dict(s):
i={}
for key in s:
v=s[key]
i.setdefault(v, []).append(key)
return i
Edit: This was not the exercise I found in my edition of 'Think Python', so I've added my answer in case anyone else is curious: Use get to write histogram more concisely. You should be able to eliminate the if statement.
def histogram(s):
d = dict()
for c in s:
d[c] = d.get(c,0)+1
return d
Exercise 11.3
[edit | edit source]Dictionaries have a method called keys that returns the keys of the dictionary, in no particular order, as a list. Modify print_hist to print the keys and their values in alphabetical order.
v = {'p' : 1, 'a' : 1, 'r' : 2, 'o' : 1, 't' : 1}
def print_hist(h):
d = []
d += sorted(h.keys())
for c in d:
print(c, h[c])
OR
v = {'p' : 1, 'a' : 1, 'r' : 2, 'o' : 1, 't' : 1}
def print_hist(h):
for c in sorted(h.keys()):
print c, h[c]
Exercise 11.4
[edit | edit source]Modify reverse_lookup so that it builds and returns a list of all keys that map to v, or an empty list if there are none.
def reverse_lookup(d,v):
l = list()
for c in d:
if d[c] == v:
l.append(c)
return l
Chapter 12
[edit | edit source]Exercise 12.1
[edit | edit source]numbers = (1,2,3)
def sumall(numbers):
x = 0
for i in numbers:
x = x + i
print x
sumall(numbers)
or
def sumall(*t):
x = 0
for i in range(len(t)):
x += t[i]
return x
or
def sumall(*args):
t = list(args)
return sum(t)
or
def sumall(*args):
return sum(args)
Exercise 12.2
[edit | edit source]import random
def sort_by_length(words):
t = []
for word in words:
t.append((len(word),word))
t.sort(reverse=True)
res = []
for length, word in t:
res.append(word)
i=0
final = []
while i <= len(res)-2:
if len(res[i]) == len(res[i+1]):
y_list = [res[i], res[i+1]]
random.shuffle(y_list)
final = final + y_list
i += 2
else:
final.append(res[i])
i += 1
if i == len(res)-1:
final.append(res[i])
return final
or
from random import shuffle
def sort_by_length(words):
r = []
d = dict()
for word in words:
d.setdefault(len(word), []).append(word)
for key in sorted(d, reverse=True):
if len(d[key]) > 1:
shuffle(d[key])
r.extend(d[key])
return r
Exercise 12.3
[edit | edit source]import string
def most_frequent(s):
d = dict()
inv = dict()
for char in s:
if char in string.ascii_letters:
letter = char.lower()
d[letter] = d.get(letter, 0) + 1
for letter, freq in d.items():
inv.setdefault(freq, []).append(letter)
for freq in sorted(inv, reverse=True):
print('{:.2%}:'.format(freq/(sum(list(inv)*len(inv[freq])))), ', '.join(inv[freq]))
Chapter 13
[edit | edit source]Exercise 13.7
[edit | edit source]from string import punctuation, whitespace, digits
from random import randint
from bisect import bisect_left
def process_file(filename):
h = dict()
fp = open(filename)
for line in fp:
process_line(line, h)
return h
def process_line(line, h):
line = line.replace('-', ' ')
for word in line.split():
word = word.strip(punctuation + whitespace + digits)
word = word.lower()
if word != '':
h[word] = h.get(word, 0) + 1
hist = process_file('emma.txt')
def cum_sum(list_of_numbers):
cum_list = []
for i, elem in enumerate(list_of_numbers):
if i == 0:
cum_list.append(elem)
else:
cum_list.append(cum_list[i-1] + elem)
return cum_list
def random_word(h):
word_list = list(h.keys())
num_list = []
for word in word_list:
num_list.append(h[word])
cum_list = cum_sum(num_list)
i = randint(1, cum_list[-1])
pos = bisect_left(cum_list, i)
return word_list[pos]
print(random_word(hist))
Chapter 14
[edit | edit source]Exercise 14.3
[edit | edit source]import shelve
def dict_of_signatures_and_words(filename='words.txt'):
d = dict()
for line in open(filename):
word = line.lower().strip()
signature = ''.join(sorted(word))
d.setdefault(signature, []).append(word)
return d
def db_of_anagrams(filename='anagrams', d=dict_of_signatures_and_words()):
db = shelve.open(filename)
for key, values in d.items():
if len(values)>1:
for index, value in enumerate(values):
db[value]=values[:index]+values[index+1:]
db.close()
def print_contents_of_db(filename='anagrams'):
db = shelve.open(filename, flag='r')
for key in sorted(db):
print(key.rjust(12), '\t<==>\t', ', '.join(db[key]))
db.close()
db_of_anagrams()
print_contents_of_db()
Exercise 14.5
[edit | edit source]# Replace urllib.request with urllib if you use Python 2.
# I would love to see a more elegant solution for this exercise, possibly by someone who understands html.
import urllib.request
def check(zip_code):
if zip_code == 'done':
return False
else:
if len(zip_code) != 5:
print('\nThe zip code must have five digits!')
return True
def get_html(zip_code):
gibberish = urllib.request.urlopen('http://www.uszip.com/zip/' + zip_code)
less_gib = gibberish.read().decode('utf-8')
return less_gib
def extract_truth(code, key, delimiter):
pos = code.find(key) + len(key)
nearly_true = code[pos:pos+40]
truth = nearly_true.split(delimiter)[0]
return truth
while True:
zip_code = input('Please type a zip code (5 digits) or "done" if want to stop:\n')
if not check(zip_code):
break
code = get_html(zip_code)
invalid_key = '(0 results)'
if invalid_key in code:
print('\nNot a valid zip code.')
continue
name_key = '<title>'
name_del = ' zip'
name = extract_truth(code, name_key, name_del)
pop_key = 'Total population</dt><dd>'
pop_del = '<'
pop = extract_truth(code, pop_key, pop_del)
if not 1 < len(pop) < 9:
pop = 'not available'
print('\n' + name)
print('Population:', pop, '\n')
Chapter 15
[edit | edit source]Exercise 15.1
[edit | edit source]import math
class Point(object):
"""represents a point in 2-D space"""
def distance(p1, p2):
distance = math.sqrt((p2.x - p1.x)**2 + (p2.y - p1.y)**2)
return distance
p1 = Point()
p2 = Point()
p1.x = 3
p1.y = 2
p2.x = 4
p2.y = 3
print(distance(p1, p2))
Exercise 15.3
[edit | edit source]import copy
def move_rectangle(rec, dx, dy):
""" does a deep copy and return the newRec with changes applied """
newRec = copy.deepcopy(rec)
newRec.corner.x += dx
newRec.corner.y += dy
return newRec
Chapter 16
[edit | edit source]Exercise 16.1
[edit | edit source]def print_time(t):
print '%.2d:%.2d:%.2d' % (t.hour, t.minute, t.second)
or
# Solution for Python3
# More on string formatting: http://docs.python.org/py3k/library/string.html#formatspec
def print_time(t):
# 0 is a fill character, 2 defines the width
print('{}:{:02}:{:02}'.format(t.hour, t.minute, t.second))
Exercise 16.2
[edit | edit source]def is_after(t1, t2):
return (t1.hour, t1.minute, t1.second) > (t2.hour, t2.minute, t2.second)
Exercise 16.3
[edit | edit source]# Comment not by the author: This will give a wrong result, if (time.second + seconds % 60) > 60
def increment(time, seconds):
n = seconds/60
time.second += seconds - 60.0*n
time.minute += n
m = time.minute/60
time.minute -= m*60
time.hour += m
or
# Solution for Python3
# Replace '//' by '/' for Python2
def increment(time, seconds):
time.second += seconds
time.minute += time.second//60
time.hour += time.minute//60
time.second %= 60
time.minute %= 60
time.hour %= 24
# A different way of going about it
def increment(time, seconds):
# Converts total to seconds, then back to a readable format
time.second = time.hour*3600 + time.minute*60 + time.second + seconds
(time.minute, time.second) = divmod(time_in_seconds, 60)
(time.hour, time.minute) = divmod(time.minute, 60)
Exercise 16.4
[edit | edit source]# Solution for Python3
# Replace '//' by '/' for Python2
from copy import deepcopy
def increment(time, seconds):
r = deepcopy(time)
r.second += seconds
r.minute += r.second//60
r.hour += r.minute//60
r.second %= 60
r.minute %= 60
r.hour %= 24
return r
Exercise 16.5
[edit | edit source]class Time(object):
"""represents the time of day.
attributes: hour, minute, second"""
time = Time()
time.hour = 11
time.minute = 59
time.second = 30
def time_to_int(time):
minutes = time.hour * 60 + time.minute
seconds = minutes * 60 + time.second
return seconds
def int_to_time(seconds):
time = Time()
minutes, time.second = divmod(seconds, 60)
time.hour, time.minute = divmod(minutes, 60)
return time
def increment(time, addtime):
seconds = time_to_int(time)
return int_to_time(seconds + addtime)
def print_time (x):
print 'The time is %.2d : %.2d : %.2d' % (x.hour, x.minute, x.second)
print_time (time)
newtime = increment (time, 70)
print_time (newtime)
Exercise 16.6
[edit | edit source]def time_to_int(time):
minutes = time.hour * 60 + time.minute
seconds = minutes * 60 + time.second
return seconds
def int_to_time(seconds):
time = Time()
minutes, time.second = divmod(seconds, 60)
time.hour, time.minute = divmod(minutes, 60)
return time
def mul_time(time, factor):
seconds = time_to_int(time)
seconds *= factor
seconds = int(seconds)
return int_to_time(seconds)
def average_pace(time, distance):
return mul_time(time, 1/distance)
Exercise 16.7
[edit | edit source]Write a class definition for a Date object that has attributes day, month and year. Write a function called increment_date that takes a Date object, date, and an integer, n, and returns a new Date object that represents the day n days after date. Hint: “Thirty days hath September...” Challenge: does your function deal with leap years correctly? See wikipedia.org/wiki/Leap_year.
class Date(object):
"""represents a date.
attributes: day, month, year"""
def print_date(date):
# German date format
print('{}.{}.{}'.format(date.day, date.month, date.year))
def is_leap_year(year):
# http://en.wikipedia.org/wiki/Leap_year#Algorithm
if year % 4 == 0:
if year % 100 == 0:
if year % 400 == 0:
return True
return False
return True
return False
def month_list(year):
if is_leap_year(year):
return [31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
return [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
def days_of_year(year):
if is_leap_year(year):
return 366
return 365
def date_to_int(date):
days = 0
for year in range(1, date.year):
days += days_of_year(year)
month_days = month_list(date.year)
for month in range(1, date.month):
days += month_days[month - 1]
days += date.day - 1
return days
def int_to_date(days):
date = Date()
date.year = 1
next_days = 365
while days >= next_days:
date.year += 1
days -= next_days
next_days = days_of_year(date.year)
date.month = 1
next_days = 31
month_days = month_list(date.year)
while days >= next_days:
date.month += 1
days -= next_days
next_days = month_days[date.month - 1]
date.day = days + 1
return date
def increment_date(date, n):
days = date_to_int(date)
return int_to_date(days + n)
d1 = Date()
d1.day, d1.month, d1.year = 8, 3, 2012
print_date(d1)
d2 = increment_date(d1, 7)
print_date(d2)
Exercise 16.8
[edit | edit source]1. Use the datetime module to write a program that gets the current date and prints the day of the week.
from datetime import date
def current_weekday():
i = date.today()
print i.strftime('%A')
current_weekday()
2. Write a program that takes a birthday as input and prints the user’s age and the number of days, hours, minutes and seconds until their next birthday.
# Python3 solution. Replace "input" by "raw_input" for Python2.
from datetime import datetime
def time_until_birthday():
dob_input = input(('Please enter the date of your birth in '
'the format "mm/dd/yyyy": '))
dob = datetime.strptime(dob_input, '%m/%d/%Y')
now = datetime.now()
if now > datetime(now.year, dob.month, dob.day):
age = now.year - dob.year
next_year = True
else:
age = now.year - dob.year - 1
next_year = False
time_to_birthday = datetime(now.year + next_year,
dob.month, dob.day) - now
days = time_to_birthday.days
hours, remainder = divmod(time_to_birthday.seconds, 3600)
minutes, seconds = divmod(remainder, 60)
print("\nYou are {} years old.".format(age))
print(("You have {0} days, {1} hours, {2} minutes and {3} "
"seconds left until your next birthday.").format(
days, hours, minutes, seconds))
time_until_birthday()
Chapter 17
[edit | edit source]Exercise 17.8
[edit | edit source]2.
from visual import scene, sphere
scene.range = (256, 256, 256)
scene.center = (128, 128, 128)
t = range(0, 256, 51)
for x in t:
for y in t:
for z in t:
pos = x, y, z
color = (x/255., y/255., z/255.)
sphere(pos=pos, radius=10, color=color)
3. Download http://thinkpython.com/code/color_list.py and use the function read_colors to generate a list of the available colors on your system, their names and RGB values. For each named color draw a sphere in the position that corresponds to its RGB values.
# As there currently (2013-04-12) is no function read_colors in color_list.py
# I use a workaround and simply import the variable COLORS from color_list.py.
# I then use the function all_colors() on COLORS to get a list of the colors.
from color_list import COLORS
from visual import scene, sphere
def all_colors(colors_string=COLORS):
"""Extract a list of unique RGB-tuples from COLORS.
The tuples look like (r, g, b), where r, g and b are each integers in
[0, 255].
"""
# split the string into lines and remove irrelevant lines
lines = colors_string.split('\n')[2:-2]
# split the individual lines and remove the names
numbers_only = [line.split()[:3] for line in lines]
# turn strings into ints and rgb-lists into tuples
rgb_tuples = [tuple([int(s) for s in lst]) for lst in numbers_only]
# return a list of unique tuples
return list(set(rgb_tuples))
def make_spheres(color_tuples=all_colors()):
scene.range = (256, 256, 256)
scene.center = (128, 128, 128)
for (r, g, b) in color_tuples:
sphere(pos=(r, g, b), radius=7, color=(r/255., g/255., b/255.))
if __name__ == '__main__':
make_spheres()
Chapter 3.5
[edit | edit source]calculator
[edit | edit source]#recursion or recursive
print "\n INDEX\n""\n C=1 for addition\n""\n C=2 for substraction\n""\n
C=3 for multiplication\n""\n C=4 for division\n""\n C=5 for to find modulus\n""\n C=6 to find factorial\n"
C=input("Enter your choice here: ")
def add(x,y):
c=x+y
print x,"+",y,"=",c
def sub(x,y):
c=x-y
print x,"-",y,"=",c
def mul(x,y):
c=x*y
print x,"*",y,"=",c
def div(x,y):
c=x/y
print x,"/",y,"=",c
def mod(x,y):
c=x%y
print x,"%",y,"=",c
if C==6:
def f(n):
if n==1:
print n
return n
else:
print n,"*",
return n*f(n-1)
n=input("enter your no here: ")
print f(n)
if C==1:
a=input("Enter your first no here: ")
b=input("Enter your second no here: ")
add(a,b)
elif C==2:
a=input("Enter your first no here: ")
b=input("Enter your second no here: ")
sub(a,b)
elif C==3:
a=input("Enter your first no here: ")
b=input("Enter your second no here: ")
mul(a,b)
elif C==4:
a=input("Enter your first no here: ")
b=input("Enter your second no here: ")
div(a,b)
elif C==5:
a=input("Enter your first no here: ")
b=input("Enter your second no here: ")
mod(a,b)
palindrome
[edit | edit source]def first(word):
return word[0]
def last(word):return word[-1]
def middle(word):
return word[1:-1]
def palindrome(word):
if first(word)==last(word):
word = middle(word)
n=len(word)
if n<2:
print "palindrome"
else:
return palindrome(word)
else:
print "not palindrome"
word=raw_input("Enter the string:")
palindrome(word)
sum of all digits
[edit | edit source]def sum_of_n_numbers(number):
if(number==0):
return 0
else:
return number + sum_of_n_numbers(number-1)
num = raw_input("Enter a number:")
num=int(num)
sum = sum_of_n_numbers(num)
print sum
###another answer in case of while loops
def sum_of_Digits(number):
sum=0
while number>0:
digit=number%10
sum=sum+digit
number=number/10
return sum
num=raw_input("enter the number")
num=int(num)
sum_of_digits=sum_of_Digits(num)
print sum_of_digits
Exercise 18.5
[edit | edit source]class Card(object):
suit_names = ['Clubs', 'Diamonds', 'Hearts', 'Spades']
rank_names = [None, 'Ace', '2', '3', '4', '5', '6', '7',
'8', '9', '10', 'Jack', 'Queen', 'King']
def __init__(self, suit = 0, rank = 2):
self.suit = suit
self.rank = rank
def __str__(self):
return '%s of %s' % (Card.rank_names[self.rank],
Card.suit_names[self.suit])
def __cmp__(self, other):
c1 = (self.suit, self.rank)
c2 = (other.suit, other.rank)
return cmp(c1, c2)
def is_valid(self):
return self.rank > 0
class Deck(object):
def __init__(self, label = 'Deck'):
self.label = label
self.cards = []
for i in range(4):
for k in range(1, 14):
card = Card(i, k)
self.cards.append(card)
def __str__(self):
res = []
for card in self.cards:
res.append(str(card))
print self.label
return '\n'.join(res)
def deal_card(self):
return self.cards.pop(0)
def add_card(self, card):
self.cards.append(card)
def shuffle(self):
import random
random.shuffle(self.cards)
def sort(self):
self.cards.sort()
def move_cards(self, other, num):
for i in range(num):
other.add_card(self.deal_card())
def deal_hands(self, num_hands, num_cards):
if num_hands*num_cards > 52:
return 'Not enough cards.'
l = []
for i in range(1, num_hands + 1):
hand_i = Hand('Hand %d' % i)
self.move_cards(hand_i, num_cards)
l.append(hand_i)
return l
class Hand(Deck):
def __init__(self, label = ''):
self.cards = []
self.label = label
# 18-6, 1-4:
class PokerHand(Hand):
def suit_hist(self):
self.suits = {}
for card in self.cards:
self.suits[card.suit] = self.suits.get(card.suit, 0) + 1
return self.suits
def rank_hist(self):
self.ranks = {}
for card in self.cards:
self.ranks[card.rank] = self.ranks.get(card.rank, 0) + 1
return self.ranks
def P(self):
self.rank_hist()
for val in self.ranks.values():
if val >= 2:
return True
return False
def TP(self):
self.rank_hist()
count = 0
for val in self.ranks.values():
if val == 4:
return True
elif val >= 2 and val < 4:
count += 1
return count >= 2
def TOAK(self):
self.rank_hist()
for val in self.ranks.values():
if val >= 3:
return True
return False
def STRseq(self):
seq = []
l = STRlist()
self.rank_hist()
h = self.ranks.keys()
h.sort()
if len(h) < 5:
return []
# Accounts for high Aces:
if 1 in h:
h.append(1)
for i in range(5, len(h)+1):
if h[i-5:i] in l:
seq.append(h[i-5:i])
return seq
def STR(self):
seq = self.STRseq()
return seq != []
def FL(self):
self.suit_hist()
for val in self.suits.values():
if val >= 5:
return True
return False
def FH(self):
d = self.rank_hist()
keys = d.keys()
for key in keys:
if d[key] >= 3:
keys.remove(key)
for key in keys:
if d[key] >= 2:
return True
return False
def FOAK(self):
self.rank_hist()
for val in self.ranks.values():
if val >= 4:
return True
return False
def SFL(self):
seq = self.STRseq()
if seq == []:
return False
for list in seq:
list_suits = []
for index in list:
for card in self.cards:
if card.rank == index:
list_suits.append(card.suit)
list_hist = histogram(list_suits)
for key in list_hist.keys():
if list_hist[key] >= 5:
return True
return False
def classify(self):
self.scores = []
hands = ['Pair', 'Two-Pair',
'Three of a Kind', 'Straight',
'Flush', 'Full House',
'Four of a Kind', 'Straight Flush']
if self.P():
self.scores.append(1)
if self.TP():
self.scores.append(2)
if self.TOAK():
self.scores.append(3)
if self.STR():
self.scores.append(4)
if self.FL():
self.scores.append(5)
if self.FH():
self.scores.append(6)
if self.FOAK():
self.scores.append(7)
if self.SFL():
self.scores.append(8)
if self.scores != []:
return hands[max(self.scores)-1]
def STRlist():
s = []
for i in range(0,9):
s.append(range(1,14)[i:i+5])
s.append([10,11,12,13,1])
return s
def histogram(l):
d = dict()
for k in range(len(l)):
d[l[k]] = 1 + d.get(l[k],0)
return d
# 18-6, 5:
def p(config = '', trials = 10000, n = 1):
"""Estimates probability that the
nth dealt hand will be config. A hand
consists of seven cards."""
successes = 0
for i in range(1, trials + 1):
deck = Deck('Deck %d' % i)
deck.shuffle()
box = Hand()
deck.move_cards(box, (n-1)*7)
hand = PokerHand('Poker Hand %d' % i)
deck.move_cards(hand, 7)
if hand.classify() == config:
successes += 1
return 1.0*successes/trials
#Iterate until first desired config.:
if __name__ == '__main__':
c = 1
while True:
deck = Deck()
deck.shuffle()
hand = PokerHand('Poker Hand %d' % c)
deck.move_cards(hand, 5)
print hand
print hand.SFL()
if hand.SFL():
print hand.STRseq()
break
print ''
c += 1
Code by Victor Alvarez
Appendix B
[edit | edit source]Exercise B.1
[edit | edit source]Read the Wikipedia page on Big-Oh notation and answer the following questions:
1. What is the order of growth of n³ + n²? What about 1000000n³ + n²? What about n³ + 1000000n²?
- A: All are O(n³)
2. What is the order of growth of (n² + n) · (n + 1)? Before you start multiplying, remember that you only need the leading term.
- A: O(n³)
3. If f is in O(g), for some unspecified function g, what can we say about af + b, where a and b are constants?
- A: a and b are also O(g) since O-notation uses the leading coefficient. Constants are O(1), so they are irrelevant in this case.
4. If f1 and f2 are in O(g), what can we say about f1 + f2?
- A: Assuming the author means that g in this problem is separate from g the last problem, then O(g) remains unchanged.
5. If f1 is in O(g) and f2 is in O(h), what can we say about f1 + f2?
- A: f1 + f2 may or may not match the higher order of growth of either O(g) and O(h). If f1 or f2 are the leading coefficients, then f1 + f2 will have the same order of growth as the the largest order of growth of either O(g) or O(h). If that is not the case, then any information about order of growth of f1 + f2 is unknown to us. i.e. It doesn't matter that they are being added.
6. If f1 is in O(g) and f2 is O(h), what can we say about f1 · f2?
- A: This is different. If f1 and f2 are multiplied (and assuming they each contain an n in them) then O(f1 · f2) has to be O(n) or larger. It could be larger if f1 is n³, for example, but it must at least be O(n).
Exercise B.3
[edit | edit source]Write a function called bisection that takes a sorted list and a target value and returns the index of the value in the list, if it’s there, or None if it’s not.
from bisect import bisect_left
def bisection(sorted_list, item):
i = bisect_left(sorted_list, item)
if i < len(sorted_list) and sorted_list[i] == item:
return i
else:
return None
if __name__ == '__main__':
a = [1, 2, 3]
print(bisection(a, 2)) # expect 1
b = [1, 3]
print(bisection(b, 2)) # expect None
c = [1, 2]
print(bisection(c, 3)) # expect None