think python学习心得-（6）利用words.txt来进行填字游戏

练习9-1

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

练习9-2

统计字母表中含有e的单词占有的百分比

      fin = open('C://words.txt')
      counter1 = 0
      counter2 = 0
      def has_no_e(wo):
      for i in range(len(wo)):
      if wo[i-1]=='e':
      return True
      for line in fin:
       word = line.strip()
       flag = has_no_e(word)
       counter1=counter1+1
      if flag == True:
       counter2=counter2+1
      counter1=float(counter1)
      counter2=float(counter2)
      print 100-counter2/counter1*100
  
 

利用下标循环实现输入文字中是否存在禁用字母的鉴别

      def avoids(word,forbiden):
      for i in range(len(forbiden)):
      for j in range(len(word)):
      if word[j-1]==forbiden[i-1]:
      return False
      return True
      print 'Plesase input forbiden alphabet:'
      ban=raw_input('>')
      print 'Please input desired word:'
      test=raw_input('>>')
      print avoids(test,ban)
  
 

检测是否是由可用字符组成的

      def use_only(word,avaliable):
      for i in range(len(word)):
       flag=0
      for j in range(len(avaliable)):
      if word[i-1]==avaliable[j-1]:
       flag =1
      if flag==0:
      return False
      return True
      print 'Plesase input avaliable alphabet:'
      ban=raw_input('>')
      print 'Please input desired word:'
      test=raw_input('>>')
      print use_only(test,ban)
  
 

练习9-5

检测alphebat是否被全部使用

      def uses_all(word,avaliable):
       counter=0
      for i in range(len(avaliable)):
       flag=0
      for j in range(len(word)):
      if word[j-1]==avaliable[i-1]:
       flag =1
      if flag==0:
      return False
      else:
       counter=counter+1
      if counter == len(avaliable):
      return True
      else:
      return False
      print 'Plesase input avaliable alphabet:'
      ban=raw_input('>')
      print 'Please input desired word:'
      test=raw_input('>>')
      print uses_all(test,ban)
  
 

      def is_abecedarian():
       fin = open ('C://words.txt')
       counter=0
      for line in fin:
       flag = 0
      for i in range(len(line)-1):
      if line[i]<=line[i+1]:
       flag=flag+1
      if flag==len(line)-2:
       counter=counter+1
      return counter
      print 'The number of abecedarian words in directionary is'
      print is_abecedarian()
  
 

QNMD车迷天下。。。。不做了

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原文链接：zclhit.blog.csdn.net/article/details/51038066