6-2字符串标识符.修改例 6-1 的 idcheck.py 脚本,使之可以检测长度为一的标识符,并且可以识别 Python 关键字,对后一个要求,你可以使用 keyword 模块(特别是 keyword.kelist)来辅助
#!/usr/bin/env python
def check_id(myInput):
import string
import keyword
alphas = string.letters + '_'
nums = string.digits
alphanums = alphas + nums
key = keyword.kwlist
str_length = len(myInput)
if str_length == 0:
print 'WRONG: zero-lengthed string("%s").'%(myInput)
elif myInput[0] not in alphas:
print '''invalid: first symbol must be alphabetic'''
elif myInput in key:
print '''invalid :Input cannot be a keyword'''
else:
for otherChar in myInput[1:]:
if otherChar not in alphanums:
print '''invalid: remaining symbols must be alphanumertic'''
break
else:
print '%s is logal identifier.' % (myInput)
if __name__ == '__main__':
while True:
myInput = raw_input('Identifier to test?')
check_id(myInput) 6-3.排序
(a) 输入一串数字，从大到小排列之。
(b) 跟 a 一样，不过要用字典序从大到小排列之.。
s = raw_input('please enter a serial of numbers,using SPACE to seperate: \n')
import string
alphas = string.digits + " "
for eachs in s:
if eachs not in alphas:
print 'you may have entered non-digit character.'
break
a = s.strip()
s_input = a.split(" ")
num = []
for i in s_input:
num.append(int(i))
num.sort()
num.reverse()
print 'Sorted result(Big-->Small): ',
for t in num:
print t,
print
print 'acording the directory to get result: ',
for i,j in enumerate(num):
print (i,j), 6-6.字符串.创建一个 string.strip()的替代函数：接受一个字符串，去掉它前面和后面的空格(如果使用 stri
ng.*strip()函数那本练习就没有意义了)
#!/usr/bin/env python
s = raw_input('please enter a string : ')
print 'the strip of string is :',
import string
alpha = ' ' + string.letters
st = []
for c in s :
if c not in alpha:
print 'Error:you may have entered non-string'
exit(0)
break
l = len(s)
i = 0
while i < l:
if s[i] == ' ':
i += 1
else:
break
s = s[i:]
l = len(s) - 1
while l > 0:
if s[l] == ' ':
l -= 1
else:
break
s = s[:l+1]
print s6-7.调试。看一下在例 6.5 中给出的代码(buggy.py)，这个程序有一个很大的问题，比如输入 6,12,20,30,等它会死掉,实际上它不能处理任何的偶数，找出原因。
#!/usr/bin/env python
num_str = raw_input('Enter a number: ')
num_num = int(num_str)
fac_list = range(1, num_num+1)
print "BEFORE:", fac_list
set_fac = set(fac_list)
i = 0
while i < len(fac_list):
if num_num % fac_list[i] == 0:
del fac_list[i]
else:
i = i + 1
print "AFTER:"+str(fac_list)
print "Factors of %d are: "%num_num, str(list(set_fac-set(fac_list)))6-8.列表。给出一个×××值，返回代表该值的英文，比如输入89返回“eight-nine”。附加题：能够返回符合英文语法规则形式，比如输入“89”返回“eighty-nine”。本练习中的值限定在0~1000。
#!/usr/bin/env python
#_*_ coding:utf-8 _*_
alpha = ['one','two','three','forth','five','six','seven','eight','nine','ten','eleven','twelve','thirteen','fourteen',\
'fifteen','sixteen','seventeen','eighteen','nineteen','twenty','']
tendigits = ['twenty','thirty','forty','fifty','sixty','eighty','ninety','']
s_num = raw_input("enter a number: ")
s = int(s_num)
if s < 0:
print 'Error:you may have entered a minus.'
elif s <= 20:
print alpha[s - 1]
elif s < 100:
i = s / 10
j = s - i * 10
print tendigits[i - 2]+'-'+alpha[j - 1]
elif s <1000:
a = s / 100
b = (s - a * 100) / 10
c = s - a * 100 - b * 10
if b == 0 and c ==0:
print alpha[a - 1]+' hundred'
elif b == 0 and c != 0:
print alpha[a-1]+' hundred'+' and '+alpha[c-1]
elif b != 0 and c == 0:
print alpha[a-1] + tendigits[b-2]
else:
print alpha[a - 1]+' hundred'+' and '+tendigits[b - 2]+'-'+alpha[c - 1]
elif s == 1000:
print 'one thousand'6-10.字符串。写一个函数，返回一个跟输入字符串相似的字符串，要求字符串的大小写翻转。比如，输入“Mr.Ed”，应该返回“mR.eD”作为输出。
#!/usr/bin/env python
#_*_ coding: utf-8 _*_
def reverseletters():
str_input = raw_input('Input a string : ')
#str_input.upper()
#str_input.lower()
print str_input.swapcase()
if __name__ =='__main__':
reverseletters()6-12. 字符串.
a). 创建一个名字为 findchr() 的函数, 函数声明如下:
def findchr(string, char)
findchr() 要在字符串 string 中查找字符 char, 找到就返回该值的索引,否则返回-1. 不能用string.*find()
或者 string.*index() 函数和方法.
#!/usr/bin/env python
#_*_ coding : utf-8 _*_
def findchr(string,char):
i = 0
l = len(string)
if char not in string:
print -1
else:
while i < l:
if string[i] == char:
print i,
i += 1
if __name__ == '__main__':
string = raw_input("please input a string :")
char = raw_input('a char:')
findchr(string,char)b). 创建另一个叫 rfindchr() 的函数,查找字符 char 最后一次出现的位置. 它跟findchr() 工作类似,不过它是
从字符串的最后开始向前查找的.
#! /usr/bin/env python
#_*_ coding : utf-8 _*_
def rfindchar():
str_input = raw_input('please enter a string :')
str_char = raw_input('please enter a char :')
l = len(str_input)
i = -1
for a in range(i,-l-1,-1):
while i >= -l:
if str_input[a] == str_char:
print a+lc). 创建第三个函数, 名字叫做 subchr(), 声明如下:
def subchr(string, origchar, newchar)
subchr() 跟 findchr() 类似, 不同的是, 如果找到匹配的字符就用新的字符替换原先字符.返回修改后的字符.
#! /usr/bin/env python
#_*_ coding:utf-8 _*_
def subchar(string, origchar, newchar):
l = len(string)
i = 0
while i < l:
if string[i] == origchar:
string = string[:i]+newchar+string[i+1:]
i += 1
print string
if __name__ == '__main__':
string = raw_input('please enter a string :')
origchar = raw_input('input a origchar : ')
newchar = raw_input('input a new char : ')
subchar(string,origchar,newchar)
import string
string.atoi()6-14. 随机数. 设计一个"石头,剪子,布"游戏,有时又叫"Rochambeau", 你小时候可能玩过,下面是规则. 你和你的对手, 在同一
时间做出特定的手势,必须是下面一种: 石头,剪子,布. 胜利者从下面规则中产生,这个规则本身是个悖论.
a). 布包石头;
b). 石头砸剪子
c). 剪子剪破布.
在你的计算机版本中, 用户输入她/他的选项, 计算机找一个随机选项,然后由你的程序来决定一个胜利者或者平手.
注意: 最好的算法是尽量少的使用if语句.
#!/usr/bin/env python
#_*_ coding : utf-8 _*_
def Rochambeau():
text = {0 : 'Cloth', 1 : 'Scissor', 2 : 'Stone'}
import random
mac = random.randrange(0,3,1)
humans = int(raw_input('your choice(0-Cloth, 1-Scissor, 2-Stone):'))
nums = range(0,3,1)
print "You: %s Computer: %s " % (text[humans], text[mac])
print 'the result is : '
if humans not in nums:
print 'Error:Wrong input!'
elif humans-mac == 0:
print 'Draw'
elif humans-mac == 1 or humans-mac == -2:
print 'You win !!!'
else:
print 'Computer win !!!'
if __name__ == '__main__':
while True:
Rochambeau()