Maison  >  Article  >  développement back-end  >  python字典序问题实例

python字典序问题实例

WBOY
WBOYoriginal
2016-06-06 11:32:451688parcourir

本文实例讲述了python字典序问题,分享给大家供大家参考。具体如下:

问题描述:

将字母从左向右的次序与字母表中的次序相同,且每个字符最大出现一次..例如:a,b,ab,bc,xyz等都是升序的字符串.现对字母表A产生的所有长度不超过6的升序字符串按照字典充排列并编码如下:

1 2 .. 26 27 28 ...
a b .. z ab ac ..

对一个升序字符串,迅速计算出它在上述字典中的编码。

实现代码如下:

import string
all_letter = string.ascii_lowercase
def gen_dict():
  result = {}
  list_num_one = [ a_letter for a_letter in all_letter ]
  list_num_two = [ i + j for i in all_letter for j in all_letter[all_letter.find(i)+1:]]
  list_num_three = [ i + j + k for i in all_letter 
           for j in all_letter[all_letter.find(i)+1:]
           for k in all_letter[all_letter.find(j)+1:]]
  list_num_four = [ i + j + k + l for i in all_letter 
           for j in all_letter[all_letter.find(i)+1:]
           for k in all_letter[all_letter.find(j)+1:]
           for l in all_letter[all_letter.find(k)+1:]]
  list_num_five = [ i + j + k + l + m for i in all_letter 
           for j in all_letter[all_letter.find(i)+1:]
           for k in all_letter[all_letter.find(j)+1:]
           for l in all_letter[all_letter.find(k)+1:]
           for m in all_letter[all_letter.find(l)+1:]]
  list_num_six = [ i + j + k + l + m + n  for i in all_letter
      for j in all_letter[all_letter.find(i)+1:]
      for k in all_letter[all_letter.find(j)+1:]
      for l in all_letter[all_letter.find(k)+1:]
      for m in all_letter[all_letter.find(l)+1:]
      for n in all_letter[all_letter.find(m)+1:]
      ]
  for key,value in enumerate(list_num_one + list_num_two + list_num_three + list_num_four + list_num_five + list_num_six):
    result.setdefault(key+1,value)
  return result
  
my_dict = gen_dict()
value_to_get = 'abcdef'
for key,value in my_dict.iteritems():
  if value == value_to_get:
    print key

结果:83682  

即abcdef在字典中的编码。

希望本文所述对大家的Python程序设计有所帮助。

Déclaration:
Le contenu de cet article est volontairement contribué par les internautes et les droits d'auteur appartiennent à l'auteur original. Ce site n'assume aucune responsabilité légale correspondante. Si vous trouvez un contenu suspecté de plagiat ou de contrefaçon, veuillez contacter admin@php.cn