按前 N 个字符对字符串进行排序 [英] Sort strings by the first N characters

问题描述

我有一个这样的文本文件:

2010-02-18 11:46:46.1287 bla2010-02-18 11:46:46.1333 foo2010-02-18 11:46:46.1333 吧2010-02-18 11:46:46.1467 bla

一个简单的排序会交换第 2 行和第 3 行(bar 在 foo 之前)，但我想保留行(具有相同日期/时间)的原始顺序.

如何在 Python 中执行此操作?

解决方案

sorted(array, key=lambda x:x[:24])

示例:

<预><代码>>>>a = ["wxyz", "abce", "abcd", "bcde"]>>>排序(一)['abcd', 'abce', 'bcde', 'wxyz']>>>sorted(a, key=lambda x:x[:3])['abce', 'abcd', 'bcde', 'wxyz']

I have a text file with lines like this:

2010-02-18 11:46:46.1287 bla
2010-02-18 11:46:46.1333 foo
2010-02-18 11:46:46.1333 bar
2010-02-18 11:46:46.1467 bla

A simple sort would swap lines 2 and 3 (bar comes before foo), but I would like to keep lines (that have the same date/time) in their original order.

How can I do this in Python?

解决方案

sorted(array, key=lambda x:x[:24])

Example:

>>> a = ["wxyz", "abce", "abcd", "bcde"]
>>> sorted(a)
['abcd', 'abce', 'bcde', 'wxyz']
>>> sorted(a, key=lambda x:x[:3])
['abce', 'abcd', 'bcde', 'wxyz']

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