Python核心编程 第八章练习

欢迎拍砖

8_2

再这个题中，我试图用xrange代替range来解决问题，但是结果输出了‘xrange（2， 26， 4）’这样奇怪的答案。然后才理解了书上的“xrange不产生列表”这句话的含义。
”’python
if name == “main“:
numbers = raw_input(“Please input from, to, incremen”)
lst = numbers.strip().split()
f, t, i = (int(num) for num in lst)
print range(f, t, i)
”’

8_4

此处应该给while…else…一个掌声。

#!/usr/bin/env python
# coding: utf-8

def IsPrime(a):
    b = a/2
    while b > 1:
        if a % b == 0:
            return False
        b -= 1
    else:
        return True

if __name__ == "__main__":
    num = int(raw_input("Please input a number: ").strip())
    if (IsPrime(num)):
        print "The number is a prime."
    else:
        print "The number is not a prime."

8_5

#!/usr/bin/env python
# coding: utf-8

def ShowFactor(a):
    count = a/2 + 1
    lst = [i for i in xrange(1, count) if a % i == 0]
    return lst

if __name__ == "__main__":
    num = int(raw_input("input a number: "))
    lst = ShowFactor(num)
    print lst

8_6

#!/usr/bin/env python
# coding: utf-8

def ShowFactors(a, lst = []):
    print "the lst is : ", lst
    count = a / 2
    for i in range(2, count+1):
        if a % i == 0:
            a = a/i
            lst.append(i)
            return ShowFactors(a, lst)
    else:
        lst.append(a)
        return lst

if __name__ == "__main__":
    num = int(raw_input("Please input a num: ").strip())
    lst = ShowFactors(num)
    print lst

8_7

这道题倒是简单，直接用8_5的答案改动一些就好了

#!/usr/bin/env python
# coding: utf-8

def ShowFactor(a):
    count = a/2 + 1
    lst = [i for i in xrange(1, count) if a % i == 0]
    return lst

def IsPerfect(num):
    lst = ShowFactor(num)
    if sum(lst) == num:
        return 1
    else: 
        return 0

if __name__ == "__main__":
    num = int(raw_input("input a number: "))
    if(IsPerfect(num)):
        print "It's a perfect num"
    else:
        print "not a perfect"

8_8

#!/usr/bin/env python
# coding: utf-8

def Factorial(n):
    factor =  1
    for i in xrange(1, n+1):
        factor *= i
    return factor

if __name__ == "__main__":
    num = int(raw_input("Input a number: "))
    print "N! is: ", Factorial(num)

8_9

这道题要为了提高计算的效率，我用了一个小技巧，每次返回的时候不但返回n-1的fibonacci数，还将n-2一并返回了，因此可以极大程度上避免重复计算。（详见数据结构与算法中有这部分的介绍）

#!/usr/bin/env python
# coding: utf-8

def Fibonacci(n):
    if n == 1:
        #之所以返回两个数是避免重复计算n-2的大小
        return (0, 1)
    Fibn2, Fibn1 = Fibonacci(n-1)
    return Fibn1, Fibn2+Fibn1

def Fib(n):
    #fib2得到的是n-1的Fibonacci数，所以无用
    fib2, fib1 = Fibonacci(n)
    return fib1

if __name__ == "__main__":
    num = int(raw_input("Input a number: "))
    print Fib(num)

8_10

总觉得这道题写的不够pythonic，不知道有没有什么其他更好的写法呢？

#/usr/bin/env python
# coding: utf-8

def CountWord(sentence):
    iVowel = 0
    iConsonant = 0
    strVower = 'aeiou'
    for i in sentence:
        if i.isalpha():
            if i.lower() in strVower:
                iVowel += 1
            else:
                iConsonant += 1
    iWord = len(sentence.split())
    return (iVowel, iConsonant, iWord)

if __name__ == "__main__":
    sentence = raw_input("Input a sentence: ")
    iNum = CountWord(sentence)
    print "The number of vowerl, consonant, word is: %s, %s, %s" %iNum

8_12

#!/usr/bin/env python
# coding: utf-8

def ListPrint():
    print "_" * 20
    iStart = int(raw_input("Enter begin value: "))
    iEnd = int(raw_input("Enter end value: "))
    bIsStr = False
    if 32 < iStart < 128 or 34 < iEnd < 129 or (iStart<33 and iEnd>128):
        bIsStr = True
        print "DEC\t\tBIN\t\tOCT\t\tHEX\t\tASCII"
    else:
        bIsStr = False
        print "DEC\t\tBIN\t\tOCT\t\tHEX"
    print "_" * 30
    for i in xrange(iStart, iEnd):
        if 32128:
            print "%s\t\t%s\t%s\t\t%s\t\t%s" %(i, bin(i), oct(i), hex(i), chr(i))
        else:
            print "%s\t\t%s\t%s\t\t%s" %(i, bin(i), oct(i), hex(i))

if __name__ == "__main__":
    ListPrint()