Think Python/Answers - Wikibooks, open books for an open world (2024)

Think Python

Contents

  • 1 Chapter 1
  • 2 Chapter 2
    • 2.1 Exercise 2.1
    • 2.2 Exercise 2.2
  • 3 Chapter 3
    • 3.1 Exercise 3.1
    • 3.2 Exercise 3.3
  • 4 Chapter 4
    • 4.1 4.3 Exercise 1
    • 4.2 4.3 Exercise 2
    • 4.3 4.3 Exercise 3
  • 5 Chapter 5
    • 5.1 Exercise 5.2
    • 5.2 Exercise 5.3
    • 5.3 Exercise 5.3
  • 6 Chapter 9
    • 6.1 Exercise 9.1
    • 6.2 Exercise 9.2
    • 6.3 Exercise 9.3
    • 6.4 Exercise 9.4
  • 7 Chapter 10
    • 7.1 Exercise 10.1
    • 7.2 Exercise 10.2
    • 7.3 Exercise 10.3
    • 7.4 Exercise 10.4
    • 7.5 Exercise 10.5
  • 8 Chapter 11
    • 8.1 Exercise 11.1
    • 8.2 Exercise 11.2
    • 8.3 Exercise 11.3
    • 8.4 Exercise 11.4
  • 9 Chapter 12
    • 9.1 Exercise 12.1
    • 9.2 Exercise 12.2
    • 9.3 Exercise 12.3
  • 10 Chapter 13
    • 10.1 Exercise 13.7
  • 11 Chapter 14
    • 11.1 Exercise 14.3
    • 11.2 Exercise 14.5
  • 12 Chapter 15
    • 12.1 Exercise 15.1
    • 12.2 Exercise 15.3
  • 13 Chapter 16
    • 13.1 Exercise 16.1
    • 13.2 Exercise 16.2
    • 13.3 Exercise 16.3
    • 13.4 Exercise 16.4
    • 13.5 Exercise 16.5
    • 13.6 Exercise 16.6
    • 13.7 Exercise 16.7
    • 13.8 Exercise 16.8
  • 14 Chapter 17
    • 14.1 Exercise 17.8
  • 15 Chapter 3.5
    • 15.1 calculator
    • 15.2 palindrome
    • 15.3 sum of all digits
    • 15.4 Exercise 18.5
  • 16 Appendix B
    • 16.1 Exercise B.1
    • 16.2 Exercise B.3

Chapter 1

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See below for Chapter 1 exercises.

Exercise 1.2

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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

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Exercise 2.1

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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 zipcode1114

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 24Base 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 021321114>>> (2*512)+(1*64)+(3*8)+(2*1)1114>>>

Exercise 2.2

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The volume of a sphere with radius Think Python/Answers - Wikibooks, open books for an open world (1) is 4/3 π Think Python/Answers - Wikibooks, open books for an open world (2).What is the volume of a sphere with radius 5?

>>> pi = 3.1415926535897932>>> r = 5>>> 4/3*pi*r**3 # This is the wrong answer392.69908169872411>>> 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 answer523.5987755982989>>>

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 BillThis 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)+3945.45You 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 calculatorcover_price = 24.95number_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 shipdef discounted_price(number_of_books): return(cover_price - (cover_price * .4)) # There is a 40% discount on wholesale book salesdef 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 / 60.0easy_pace_hr = (8 + 15 / 60.0 ) / 60.0tempo_pace_hr = (7 + 12 / 60.0) / 60.0running_time_hr = 2 * easy_pace_hr + 3 * tempo_pace_hrbreakfast_hr = start_time_hr + running_time_hrbreakfast_min = (breakfast_hr-int(breakfast_hr))*60breakfast_sec= (breakfast_min-int(breakfast_min))*60print ('breakfast_hr', int(breakfast_hr) )print ('breakfast_min', int (breakfast_min) )print ('breakfast_sec', int (breakfast_sec) )>>>

Chapter 3

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Exercise 3.1

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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.


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

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You can see my solution at http://thinkpython.com/code/grid.py.

"""Solution to Exercise 3.5 on page 27 of Think PythonAllen 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 solutiondef 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 griddef 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(): printdef 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 thefunctions had the same structure: they would do something, dosomething else four times, and then do something else once.So I wrote one_four_one, which takes three functions as arguments; itcalls the first one once, then uses do_four to call the second onefour 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 programoften gives you insight into the problem, which might lead you torewrite the code to reflect the structure of the solution.--- Allen"""print comment# another solutiondef 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 rowsdef 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()-----------------------# mteodordef 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 columnssize=4def beam(): print(" + - - - -"*size, "+")def post(): print(" | "*size, "|")def repeat(func): func() func() func()# manual adjust size for rows# this is 4beam()repeat(post)beam()repeat(post)beam()repeat(post)beam()repeat(post)beam()

Chapter 4

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4.3 Exercise 1

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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

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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

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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

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Exercise 5.2

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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

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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

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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

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Exercise 9.1

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fin = open('words.txt')for line in fin:word = line.strip()if len(word) > 20:print (word)

Exercise 9.2

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fin = open('words.txt')def has_no_e(word): for char in word: if char in 'Ee': return False return True count = 0for line in fin: word = line.strip() if has_no_e(word): count += 1 print (word)percent = (count / 113809.0) * 100print (str(percent)) + "% of the words don't have an 'e'."

Exercise 9.3

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fin = open('words.txt')def avoids(word,letter): for char in word: if char in letter: return False return Trueletter = raw_input('What letters to exclude? ')count = 0for line in fin: word = line.strip() if avoids(word, letter): count += 1 print wordpercent = (count / 113809.0) * 100print str(percent) + "% of the words don't have " + letter + '.'

Exercise 9.4

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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

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Exercise 10.1

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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

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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

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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

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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 = 1while i <= len(x)-2:res.append(x[i])i += 1return res

This can also be done simply with a slice.

>>> def middle(x): return x[1:-1]

Exercise 10.5

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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

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Exercise 11.1

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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

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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)+1return d

Exercise 11.3

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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

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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

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Exercise 12.1

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numbers = (1,2,3)def sumall(numbers): x = 0 for i in numbers: x = x + i print xsumall(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

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import randomdef 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=0final = []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_listi += 2else:final.append(res[i])i += 1if i == len(res)-1:final.append(res[i])return final

or

from random import shuffledef 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

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import stringdef 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) + 1for 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

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Exercise 13.7

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from string import punctuation, whitespace, digitsfrom random import randintfrom bisect import bisect_leftdef process_file(filename):h = dict()fp = open(filename)for line in fp:process_line(line, h)return hdef 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) + 1hist = 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_listdef 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

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Exercise 14.3

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import shelvedef 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 ddef 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

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# 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.requestdef check(zip_code):if zip_code == 'done':return Falseelse: if len(zip_code) != 5: print('\nThe zip code must have five digits!')return Truedef get_html(zip_code):gibberish = urllib.request.urlopen('http://www.uszip.com/zip/' + zip_code)less_gib = gibberish.read().decode('utf-8')return less_gibdef extract_truth(code, key, delimiter):pos = code.find(key) + len(key)nearly_true = code[pos:pos+40]truth = nearly_true.split(delimiter)[0]return truthwhile True:zip_code = input('Please type a zip code (5 digits) or "done" if want to stop:\n')if not check(zip_code): breakcode = get_html(zip_code)invalid_key = '(0 results)'if invalid_key in code:print('\nNot a valid zip code.')continuename_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

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Exercise 15.1

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import mathclass 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 distancep1 = Point()p2 = Point()p1.x = 3p1.y = 2p2.x = 4p2.y = 3print(distance(p1, p2))

Exercise 15.3

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import copydef 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

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Exercise 16.1

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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#formatspecdef 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

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def is_after(t1, t2): return (t1.hour, t1.minute, t1.second) > (t2.hour, t2.minute, t2.second)

Exercise 16.3

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# Comment not by the author: This will give a wrong result, if (time.second + seconds% 60) > 60def 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 Python2def increment(time, seconds):time.second += secondstime.minute += time.second//60time.hour += time.minute//60time.second %= 60time.minute %= 60time.hour %= 24
# A different way of going about itdef 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

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# Solution for Python3# Replace '//' by '/' for Python2from copy import deepcopydef increment(time, seconds):r = deepcopy(time)r.second += secondsr.minute += r.second//60r.hour += r.minute//60r.second %= 60r.minute %= 60r.hour %= 24return r

Exercise 16.5

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class Time(object): """represents the time of day. attributes: hour, minute, second"""time = Time()time.hour = 11time.minute = 59time.second = 30def time_to_int(time): minutes = time.hour * 60 + time.minute seconds = minutes * 60 + time.second return secondsdef int_to_time(seconds): time = Time() minutes, time.second = divmod(seconds, 60) time.hour, time.minute = divmod(minutes, 60) return timedef 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

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def time_to_int(time):minutes = time.hour * 60 + time.minuteseconds = minutes * 60 + time.secondreturn secondsdef int_to_time(seconds):time = Time()minutes, time.second = divmod(seconds, 60)time.hour, time.minute = divmod(minutes, 60)return timedef mul_time(time, factor):seconds = time_to_int(time)seconds *= factorseconds = int(seconds)return int_to_time(seconds)def average_pace(time, distance):return mul_time(time, 1/distance)

Exercise 16.7

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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 formatprint('{}.{}.{}'.format(date.day, date.month, date.year))def is_leap_year(year): # http://en.wikipedia.org/wiki/Leap_year#Algorithmif year % 4 == 0:if year % 100 == 0:if year % 400 == 0:return Truereturn Falsereturn Truereturn Falsedef 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 366return 365def date_to_int(date):days = 0for 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 - 1return daysdef int_to_date(days):date = Date()date.year = 1next_days = 365while days >= next_days:date.year += 1days -= next_daysnext_days = days_of_year(date.year)date.month = 1next_days = 31month_days = month_list(date.year)while days >= next_days:date.month += 1days -= next_daysnext_days = month_days[date.month - 1]date.day = days + 1return datedef 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, 2012print_date(d1)d2 = increment_date(d1, 7)print_date(d2)

Exercise 16.8

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1. Use the datetime module to write a program that gets the current date and prints the day of the week.

from datetime import datedef 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 datetimedef 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

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Exercise 17.8

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2.

from visual import scene, spherescene.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 COLORSfrom visual import scene, spheredef 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

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calculator

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#recursion or recursiveprint "\nINDEX\n""\nC=1 for addition\n""\nC=2 for substraction\n""\nC=3 for multiplication\n""\nC=4 for division\n""\nC=5 for to find modulus\n""\nC=6 to find factorial\n"C=input("Enter your choice here: ")def add(x,y):c=x+yprint x,"+",y,"=",cdef sub(x,y):c=x-yprint x,"-",y,"=",cdef mul(x,y):c=x*yprint x,"*",y,"=",cdef div(x,y):c=x/yprint x,"/",y,"=",cdef mod(x,y):c=x%yprint x,"%",y,"=",cif C==6:def f(n):if n==1:print nreturn nelse: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

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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

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def sum_of_n_numbers(number):if(number==0):return 0else: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 loopsdef sum_of_Digits(number):sum=0while number>0:digit=number%10sum=sum+digitnumber=number/10return sumnum=raw_input("enter the number")num=int(num)sum_of_digits=sum_of_Digits(num)print sum_of_digits

Exercise 18.5

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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 = suitself.rank = rankdef __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 > 0class Deck(object):def __init__(self, label = 'Deck'):self.label = labelself.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.labelreturn '\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 randomrandom.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 lclass 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) + 1return self.suitsdef rank_hist(self):self.ranks = {}for card in self.cards:self.ranks[card.rank] = self.ranks.get(card.rank, 0) + 1return self.ranksdef P(self):self.rank_hist()for val in self.ranks.values():if val >= 2:return Truereturn Falsedef TP(self):self.rank_hist()count = 0for val in self.ranks.values():if val == 4:return Trueelif val >= 2 and val < 4:count += 1return count >= 2def TOAK(self):self.rank_hist()for val in self.ranks.values():if val >= 3:return Truereturn Falsedef 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 seqdef STR(self):seq = self.STRseq()return seq != []def FL(self):self.suit_hist()for val in self.suits.values():if val >= 5:return Truereturn Falsedef 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 Truereturn Falsedef FOAK(self):self.rank_hist()for val in self.ranks.values():if val >= 4:return Truereturn Falsedef SFL(self):seq = self.STRseq()if seq == []:return Falsefor 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 Truereturn Falsedef 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 sdef 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 thenth dealt hand will be config. A handconsists of seven cards."""successes = 0for 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 += 1return 1.0*successes/trials#Iterate until first desired config.:if __name__ == '__main__':c = 1while True:deck = Deck()deck.shuffle()hand = PokerHand('Poker Hand %d' % c)deck.move_cards(hand, 5)print handprint hand.SFL()if hand.SFL():print hand.STRseq()breakprint ''c += 1Code by Victor Alvarez

Appendix B

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Exercise B.1

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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

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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_leftdef bisection(sorted_list, item): i = bisect_left(sorted_list, item) if i < len(sorted_list) and sorted_list[i] == item: return i else: return Noneif __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
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