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字符串列出python中的转换?

[英]string to list conversion in python?

I have a string like : 我有一个字符串:

searchString = "u:sads asdas asdsad n:sadasda as:adds sdasd dasd a:sed eee"

what I want is list : 我想要的是列表:

["u:sads asdas asdsad","n:sadasda","as:adds sdasd dasd","a:sed eee"]

What I have done is : 我所做的是:

values = re.split('\s', searchString)
mylist = []
word = ''
for elem in values:
  if ':' in elem:
    if word:
      mylist.append(word)
    word = elem
  else:
    word = word + ' ' + elem
list.append(word)
return mylist

But I want an optimized code in python 2.6 . 但我想在python 2.6中优化代码。

Thanks 谢谢

Use regular expressions: 使用正则表达式:

import re
mylist= re.split('\s+(?=\w+:)', searchString)

This splits the string everywhere there's a space followed by one or more letters and a colon. 这会将字符串分割到一个空格,后跟一个或多个字母和冒号。 The look-ahead ( (?= part) makes it split on the whitespace while keeping the \\w+: parts 前瞻( (?=部分)使其在空格上分割,同时保持\\w+:部分

You can use "look ahead" feature offered by a lot of regular expression engines. 您可以使用许多正则表达式引擎提供的“向前看”功能。 Basically, the regex engines checks for a pattern without consuming it when it comes to look ahead. 基本上,正则表达式引擎在前瞻时检查模式而不消耗它。

import re
s = "u:sads asdas asdsad n:sadasda as:adds sdasd dasd a:sed eee"
re.split(r'\s(?=[a-z]:)', s)

This means, split only when we have a \\s followed by any letter and a colon but don't consume those tokens. 这意味着,只有当我们有一个\\s后跟任何字母和冒号但消耗这些令牌时才会拆分。

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