Question

我有一个DataFrame，其中一列是看起来像这样的字符串行：

Received value 126;AOC;H3498XX from 602
Received value 101;KYL;0IMMM0432 from 229

我想在第二个分号之后删除（或不替换任何内容）部分，使其看起来像

Received value 126;AOC; from 602

但是我要删除的这部分将具有变化且不可预测的长度（始终是A-Z和0-9的组合）。分号和froms将始终存在以供参考。

我正在通过研究以下链接来尝试使用正则表达式：https://docs.python.org/3/library/re.html

import re
for row in df[‘column’]:
    row = re.sub(‘;[A-Z0-9] from’ , ‘; from’, row)

我认为[A-Z0-9]无法合并我想要的变长方面。

Answer 1

将str.replace()与str.split()结合使用的示例：

s = ['126;AOC;H3498XX from 602', '101;KYL;0IMMM0432 from 229']

for elem in s:
    print(elem.replace(elem.split(";",2)[-1].split()[0],''))

输出：

126;AOC; from 602
101;KYL; from 229

编辑：

同样适用于以下示例：

s = ['Received value 126;AOC;H3498XX from 602', 'Received value 101;KYL;0IMMM0432 from 229']

for elem in s:
    print(elem.replace(elem.split(";",2)[-1].split()[0],''))

输出：

Received value 126;AOC; from 602
Received value 101;KYL; from 229

Answer 2

使用模式(Received value \d+;[A-Z]+;)\w+(\s.*?)

例如：

import re

s = ["Received value 126;AOC;H3498XX from 602", "Received value 101;KYL;0IMMM0432 from 229"]

for i in s:
    print( re.sub(r"(Received value \d+;[A-Z]+;)\w+(\s.*?)", r"\1", i) )

输出：

Received value 126;AOC;from 602
Received value 101;KYL;from 229

如何删除可变长度的字符串的一部分

2 个答案: