在 Python 中使用正则表达式替换匹配前的符号

Question

我有这样的字符串：

text1 = ('SOME STRING,99,1234 FIRST STREET,9998887777,ABC')
text2 = ('SOME OTHER STRING,56789 SECOND STREET,6665554444,DEF')
text3 = ('ANOTHER STRING,#88,4321 THIRD STREET,3332221111,GHI')

期望的输出：

SOME STRING 99,1234 FIRST STREET,9998887777,ABC
SOME OTHER STRING,56789 SECOND STREET,6665554444,DEF
ANOTHER STRING #88,4321 THIRD STREET,3332221111,GHI

我的想法：使用正则表达式查找出现的 1-5 位数字，可能前面有一个符号，位于两个逗号之间且后面没有 space 和字母，然后用这个匹配替换而不是前面的逗号。类似于：

text.replace(r'(,\d{0,5},)','.........')

Answer 1

如果您使用 regex 模块而不是 re，那么可能：

import regex
str = "ANOTHER STRING,#88,4321 THIRD STREET,3332221111,GHI"
print(regex.sub(r'(?<!^.*,.*),(?=#?\d+,\d+)', ' ', str))

如果您确定前瞻中的模式后没有其他子字符串，您可以使用 re。

import re
str = "ANOTHER STRING,#88,4321 THIRD STREET,3332221111,GHI"
print(re.sub(r',(?=#?\d+,\d+)', ' ', str))

Answer 2

如果 SOME STRING、SOME OTHER STRING 和 ANOTHER STRING 从不包含逗号，则更易于阅读：

text1.replace(",", " ", 1)

只是将第一个逗号替换为 space

Answer 3

简单而有效：

my_pattern = r"(,)(\W?\d{0,5},)"

p = re.compile(my_pattern)

p.sub(r" ", text1) # 'SOME STRING 99,1234 FIRST STREET,9998887777,ABC'
p.sub(r" ", text2) # 'SOME OTHER STRING,56789 SECOND STREET,6665554444,DEF'
p.sub(r" ", text3) # 'ANOTHER STRING #88,4321 THIRD STREET,3332221111,GHI'

具有非捕获组和详细编译的辅助模式：

my_pattern = r"""
    (?:,)           # Non-capturing group for single comma.
    (\W?\d{0,5},)   # Capture zero or one non-ascii characters, zero to five numbers, and a comma
"""

# re.X compiles multiline regex patterns
p = re.compile(my_pattern, flags = re.X)

# This time we will use  to implement the first captured group
p.sub(r" ", text1)
p.sub(r" ", text2)
p.sub(r" ", text3)

在 Python 中使用正则表达式替换匹配前的符号

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