Think python（第二版）习题代码

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3-2：

def do_twice(f, s1):
	f(s1)
	f(s1)

def print_spam(s2):
	print(s2)

def print_twice(s3):
	print(s3)
	print(s3)

def do_four(f1,f2,s4):
	f1(f2,s4)
	f1(f2,s4)

do_twice(print_twice, 'aaa')

do_four(do_twice, print_spam, 'spam')

3-3.1:

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

def push1():
	print('+ - - - -', end = ' ')  # 默认情况下print会自动换行，可以在结尾打印一个空格改变这一行为

def push2():
	print('|        ', end = ' ')

def do_twice(f):
	f()
	f()

def do_four(f):
	f()
	f()
	f()
	f()

def row1():
	do_twice(push1)
	print('+')

def row2():
	do_twice(push2)
	print('|')

row1()
do_four(row2)
row1()
do_four(row2)
row1()

3-3.2:

#!/usr/bin/env python3
# -*- coding: utf-8 -*-

def push1():
	print('+ - - - -', end = ' ')  # 默认情况下print会自动换行，可以在结尾打印一个空格改变这一行为

def push2():
	print('|        ', end = ' ')

def do_twice(f):
	f()
	f()

def do_four(f):
	f()
	f()
	f()
	f()

def row1():
	do_four(push1)
	print('+')

def row2():
	do_four(push2)
	print('|')

row1()
do_four(row2)
row1()
do_four(row2)
row1()
do_four(row2)
row1()
do_four(row2)
row1()

5-1：

import time

time_now = time.time()
minutes = time_now // 60
hours = minutes // 60
days = hours // 24
hour = hours % 24
minutes = minutes % 60
sec = time_now % 60 #秒 seconds

print('time_now =',time_now)
print('days =',days,'\n' ,'hours =',hours,'\n','minutes =',minutes,'\n','sec =',sec)

5-2：

def check_fermat(a, b, c, n): #5-2-1的函数
	a = a ** n
	b = b ** n
	c = c ** n
	if a + b == c and n > 2:
		print('Oh，my gad！ fermat is wrong!')
	else:
		print('No, it not true')

def input_user(): #5-2-2的函数
	a = input('Please enter the value of a: \n')
	b = input('Please enter the value of b: \n')
	c = input('Please enter the value of c: \n')
	n = input('Please enter the value of n: \n')
	check_fermat(int(a), int(b), int(c), int(n))

input_user()

5-3:

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def is_triangle(a, b, c): #5-3-1
	if a>=b+c or b>=a+c or c>=a+b:
		print('No\n')
	else:
		print('Yes\n')

def input_user(): #5-3-2
	a = input('Please enter the value of a: \n')
	b = input('Please enter the value of b: \n')
	c = input('Please enter the value of c: \n')
	is_triangle(int(a), int(b), int(c))

input_user()

6-2：

def ack(m,n):
	if m == 0:
		return n+1
	elif m>0 and n==0:
		return ack(m-1, 1)
	elif m>0 and n>0:
		return ack(m-1, ack(m, n-1))

print(ack(3, 4))

##对于很大的数字m和n，递归调用的层数过多，系统分配的栈空间溢出，从而出现错误提示

6-3：

def is_palindrome(word):
	lenth = len(word)
	if lenth == 0 or lenth == 1:
		return True
	if word[0] == word[-1]:
		return is_palindrome(word[1:-1])
	return False


ans = is_palindrome('asa')
print(ans)

6-4：

def is_power(a, b):
	if a == 1:
		return True
	if a == 0:
		return False
	if a%b==0:
		return is_power(a/b, b)
	return False

ans = is_power(9, 3)
print(ans)
# 1是任何数的乘方，a**n表示a的n次方，a为底数，注意底数不能为0

6-5:

def gcd(a, b):
	if b == 0:
		return a
	return gcd(b, a%b)

print(gcd(12, 15))

7-1：

import math

def mysqrt(a, x):
	while True:
		y = (x + a/x) / 2
		if abs(y-x) < 0.000000000000000001:
			break
		x = y
	return x

def test_square_root(a, x):
	my_ans = mysqrt(a, x)
	math_ans = math.sqrt(a)
	diff = abs(my_ans - math_ans)
	print(a, my_ans, math_ans, diff)


print('a    mysqrt(a)    math.sqrt(a)    diff')
print('-    ---------    ------------    ----')
test_square_root(1, 1)
test_square_root(2, 2)
test_square_root(3, 2)
test_square_root(4, 2)
test_square_root(5, 2)
test_square_root(6, 2)
test_square_root(7, 2)
test_square_root(8, 2)
test_square_root(9 ,2)

7-2:

import math

def eval_loop():
	ans = ''
	while True:
		print('Please enter your words:\n')
		word = input()
		if word == 'done':
			return ans
			break
		ans = eval(word)
		print(ans)

print(eval_loop())

7-3:

import math

def estimate_pi():
	cnt = 2*math.sqrt(2) / 9801
	k = 0
	tie = 0
	while True:
		item = math.factorial(4*k) * (1103+26390*k) / (math.factorial(k) ** 4) / (396 ** (4*k))
		tie = tie + item
		k = k + 1
		if item < 1e-15:
			break
	pi = 1 / (tie*cnt)
	print('my_ans ==',pi)
	print('math.pi_ans ==',math.pi)

estimate_pi() 

8-2：

'''
count(str,beg,end)返回str在string中出现的次数，如果beg或者end指定则返回指定范围内str出现的次数
'''
word = 'banana'
cnt = word.count('a') 
print(cnt)

8-3:

def is_palindrome(word):
	new_word = word[::-1]
	if new_word == word:
		return True
	else:
		return False

print(is_palindrome('asa'))

9-7：

def is_abecedarian(word):
	if len(word) < 6:
		return False
	one = 0
	tow = 2
	three = 4
	while three < len(word)-1:
		if word[one] == word[one+1] and word[tow] == word[tow+1] and word[three] == word[three+1]:
			return True
		one += 1
		tow += 1
		three += 1
	return False

sum = 0
num = 0

fin = open(r'D:\python\words.txt')## 我使用的是本章第一节给出的单词列表
for line in fin:
	word = line.strip()
	sum = sum + 1
	if is_abecedarian(word):
		num += 1
		print(word)

ratio = num / sum * 100
print('sum =',sum)
print('num =',num)
print('The ratio of words without e to total words is:',ratio,'%' )

9-8:

def palindrome(num, start, lenth):
	s = str(num)[start:start+lenth]
	return s == s[::-1]

def check(num):
	if palindrome(num,2,4) and palindrome(num+1,1,5) and palindrome(num+2,1,4) and palindrome(num+3,0,6):
		return True
	return False

def check_all():
	num = 100000
	while num < 999996:
		if check(num):
			print(num)
		num += 1

check_all()

9-9：

'''
每隔十一年会出现一次，一共出现8次,就是：

20	31	42	53	64	75	86	97
02	13	24	35	46	57	68	79

现在的年纪就是57岁，代码就不写了，假定一下母子之间的年龄差距，暴力遍历就好了
'''

'''
zfill()方法：	Python zfill() 方法返回指定长度的字符串，原字符串右对齐，前面填充0。

zfill()方法语法：str.zfill(width) 	width -- 指定字符串的长度。原字符串右对齐，前面填充0。
'''

##示例：

str = "this is string example from runoob....wow!!!"
print ("str.zfill : ",str.zfill(40))
print ("str.zfill : ",str.zfill(50))

##输出结果：

str.zfill :  this is string example from runoob....wow!!!
str.zfill :  000000this is string example from runoob....wow!!!

10-1:

def nested_sum(t):
	ans = 0
	for x in t:
		ans += sum(x)
	return ans

t = [[1,2], [3], [4,5,6]]
print(nested_sum(t)) 

10-2:

def cumsum(t):
	ans = 0
	ans_t = []
	for x in t:
		ans += x
		ans_t = ans_t + [ans]
	return ans_t

t = [1, 2, 3]
print(cumsum(t))

10-3:

def middle(t): ##新建一个列表
	return t[1:-1]

t = [1, 2, 3, 4]
print(middle(t)) 

10-4:

def chop(t):##修改一个list，切片的操作会新建一个list
	del t[0]
	del t[-1]

t = [1, 2, 3, 4]
chop(t)
print(t)

10-5：

def is_sorted(t):
	return t == sorted(t) 
'''
Python list内置sort()方法用来排序，也可以用python内置的全局sorted()方法来对可迭代的序列排序生成新的序列。
sort不会产生一个新的list，仅仅是用于这个list对象的方法，而sorted是函数，返回一个新的list
'''

t = [1, 2, 2]
print(is_sorted(t))

t= ['b', 'a']
print(is_sorted(t))

10-6:

def is_anagram(t1, t2):
	t1.sort()
	t2.sort()
	return t1 == t2

t1 = ['a', 'c', 'b', 'd']
t2 = ['b', 'd', 'c', 'a']
print(is_anagram(t1, t2))

t1 = ['a', 'b' ,'b']
t2 = ['b', 'b', 'a']
print(is_anagram(t1, t2))

t1 = ['a', 'b', 'c']
t2 = ['a', 'b', 'd']
print(is_anagram(t1, t2))

10-7:

def has_duplicates(t):
	new_t = list(t)
	new_t.sort()
	for i in range(len(t) - 1):
		if new_t[i] == new_t[i+1]:
			return True
	return False

t = ['a', 'b', 'c']
print(has_duplicates(t))

t = ['a', 'b', 'a']
print(has_duplicates(t))

10-8：

import random

def has_duplicates(t):##查看是否有同一天过生日的同学
	new_t = list(t)
	new_t.sort()
	for i in range(len(t) - 1):
		if new_t[i] == new_t[i+1]:
			return True
	return False

def random_birthday(n):##随机生成23个同学的生日，并储存为一个list
	t = []
	for i in range(n):
		birthday = random.randint(1, 365)
		t.append(birthday)
	return t

def count_match(num_students, num_simulations):##计算23个同学在1000次试验中，有多少次至少有一对同学同一天过生日的
	count = 0
	for i in range(num_simulations):
		t = random_birthday(num_students)
		if has_duplicates(t):
			count += 1
	return count

num_students = 23
num_simulations = 1000
count = count_match(num_students, num_simulations)
print('After %d simulations' % num_simulations)
print('with %d students' % num_students)
print('there were %d simulations with at least one match' % count)
print('The probability of matching success is ',count / num_simulations)

10-9：

import time

def fuc_1():
	fin = open(r'D:\python\words.txt')
	t = []
	for line in fin:
		word = line.strip()
		t.append(word)
	return t

def fuc_2():
	fin = open(r'D:\python\words.txt')
	t = []
	for line in fin:
		word = line.strip()
		t = t + [word]
	return t

start_time = time.time()
t = fuc_1()
elapsed_time = time.time() - start_time
print (len(t))
print (t[:10])
print(elapsed_time)

start_time = time.time()
t = fuc_2()
elapsed_time = time.time() - start_time
print(len(t))
print(t[:10])
print(elapsed_time)

10-10：

import bisect

def make_word_list():
	word_list = []
	fin = open(r'D:\python\words.txt')
	for line in fin:
		word = line.strip()
		word_list.append(word)
	return word_list

def in_bisect(word_list, word):
	if len(word_list) == 0:
		return False
	i = len(word_list) // 2
	if word_list[i] == word:
		return True
	if word_list[i] > word:
		return in_bisect(word_list[:i], word)
	else:
		return in_bisect(word_list[i+1:], word)

def in_bisect_cheat(word_list,word):
	i = bisect.bisect_left(word_list, word) ##在list中查找word的位置,并返回下标
	if i == len(word_list):
		return False
	return word_list[i] == word

word_list = make_word_list()

for word in ['as', 'to', 'for', 'be', 'so', 'zoo','aaaaa']:
	print(word,'in list',in_bisect(word_list, word))

print('\n')

for word in ['as', 'to', 'for', 'be', 'so', 'zoo','aaaaa']:
	print(word,'in list',in_bisect_cheat(word_list, word))

10-11：

def make_word_list():
	word_list = []
	fin = open(r'D:\python\words.txt')
	for line in fin:
		word = line.strip()
		word_list.append(word)
	return word_list

def in_bisect(word_list, word):
	if len(word_list) == 0:
		return False
	i = len(word_list) // 2
	if word_list[i] == word:
		return True
	if word_list[i] > word:
		return in_bisect(word_list[:i], word)
	else:
		return in_bisect(word_list[i+1:], word)

def reverse_pair(word_list,word):
	rev_word = word[::-1]
	return in_bisect(word_list, rev_word)


word_list = make_word_list()

for word in word_list:
	if reverse_pair(word_list, word):
		print(word, word[::-1])