python字典序问题实例，python字典实例

文章由Byrx.net分享于2019-03-22 05:03:52评论（687）

python字典序问题实例，python字典实例

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。

Python字典怎排序，给例子

在Python2.7.x版本中, collections类增加了OrderedDict, 用法如下：

在Python2.7.x版本中, collections类增加了OrderedDict, 用法如下：

pywugw@pywugw-laptop:~$ /usr/local/bin/python2.7
Python 2.7b1 (r27b1:79927, Apr 26 2010, 11:44:19)
[GCC 4.4.3] on linux2
Type "help", "copyright", "credits" or "license" for more information.
>>> from collections import OrderedDict
>>> d = {'banana': 3, 'apple':4, 'pear': 1, 'orange': 2}

#按key排序
>>> OrderedDict(sorted(d.items(), key=lambda t: t[0]))
OrderedDict([('apple', 4), ('banana', 3), ('orange', 2), ('pear', 1)])

#按value排序
>>> OrderedDict(sorted(d.items(), key=lambda t: t[1]))
OrderedDict([('pear', 1), ('orange', 2), ('banana', 3), ('apple', 4)])
#按key的长度排序
>>> OrderedDict(sorted(d.items(), key=lambda t: len(t[0])))
OrderedDict([('pear', 1), ('apple', 4), ('orange', 2), ('banana', 3)])

python中字典的排序

>>> d
{'a': 1, 'world': 11, 'z': 9, 'hello': 10}
>>> k=d.keys()
>>> k.sort()
>>> k
['a', 'hello', 'world', 'z']
>>> t=map(lambda key:(key,d[key]),k)
>>> t
[('a', 1), ('hello', 10), ('world', 11), ('z', 9)]

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python字典序问题实例，python字典实例