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python字典序問題實(shí)例

 更新時(shí)間：2014年09月26日 10:17:37   投稿：shichen2014  
這篇文章主要介紹了python字典序問題,是字符串操作一個(gè)比較典型的應(yīng)用,需要的朋友可以參考下

本文實(shí)例講述了python字典序問題，分享給大家供大家參考。具體如下：

問題描述:

將字母從左向右的次序與字母表中的次序相同,且每個(gè)字符最大出現(xiàn)一次..例如:a,b,ab,bc,xyz等都是升序的字符串.現(xiàn)對字母表A產(chǎn)生的所有長度不超過6的升序字符串按照字典充排列并編碼如下:

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

對一個(gè)升序字符串,迅速計(jì)算出它在上述字典中的編碼。

實(shí)現(xiàn)代碼如下:

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

結(jié)果:83682　　

即abcdef在字典中的編碼。

希望本文所述對大家的Python程序設(shè)計(jì)有所幫助。

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