在 Python 中仅替换正则表达式字符串的一部分的方法
[英]Way to substitute only part of a regex string in Python
问题描述
I am working with a text file that has text laid out like below:
我正在使用一个文本文件，其文本布局如下：
SCN DD1251
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
DD1271 C DD1271 R
DD1351 D DD1351 B
E
SCN DD1271
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
DD1301 T DD1301 A
DD1251 R DD1251 C
SCN DD1301
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
DD1271 A DD1271 T
B
C
D
SCN DD1351
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
A DD1251 D
DD1251 B
C
I am currently using the following regex pattern to match the Node followed by the 5 wide space and following letter like so:我目前正在使用以下正则表达式模式来匹配节点，后跟 5 个宽空格和以下字母，如下所示：
DD1251 B
[A-Z]{2}[0-9]{3}[0-9A-Z] [A-Z]
My goal is to replace the 5 wide space with an underscore to look like so:我的目标是用下划线替换 5 宽空间，如下所示：
DD1251_B
I am trying to achieve this using the following code:我正在尝试使用以下代码来实现这一点：
def RemoveLinkSpace(input_file, output_file, pattern):
with open(str(input_file) + ".txt", "r") as file_input:
with open(str(output_file) + ".txt", "w") as output:
for line in file_input:
line = pattern.sub("_", line)
output.write(line)
upstream_pattern = re.compile(r"[A-Z]{2}[0-9]{3}[0-9A-Z] [A-Z]")
RemoveLinkSpace("File1","File2",upstream_pattern)
However, this results in a text file that looks like the below pattern:但是，这会生成一个类似于以下模式的文本文件：
SCN DD1251
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
_ C DD1271 R
_ D DD1351 B
E
SCN DD1271
UPSTREAM DOWNSTREAM FILTER
NODE LINK NODE LINK LINK
_ T DD1301 A
_ R DD1251 C
My question is, is there a way to still search for the entire regex, but then to only replace the spaces contained within in?我的问题是，有没有办法仍然搜索整个正则表达式，但只替换其中包含的空格？
1 个解决方案
解决方案1
2 已采纳 2021-06-10 01:51:19
We can replace by group, you missed this point.我们可以按组替换，您错过了这一点。 \1 means the first group, \2 second group So in search pattern ([AZ]{2}[0-9]{3}[0-9A-Z]) is first pattern and ([AZ]) is second pattern. \1 表示第一组，\2 第二组 所以在搜索模式中 ([AZ]{2}[0-9]{3}[0-9A-Z]) 是第一个模式， ([AZ]) 是第二个模式.
Also, space between group1 and group 2 exists not 5, just 6. so I search over 5 continue space.此外，group1 和 group 2 之间的空间不是 5，只有 6。所以我搜索了 5 个连续空间。
def RemoveLinkSpace(input_file, output_file, pattern):
with open(str(input_file) + ".txt", "r") as file_input:
with open(str(output_file) + ".txt", "w") as output:
for line in file_input:
line = re.sub(pattern,r"\1_\2", line)
output.write(line)
upstream_pattern = re.compile(r"([A-Z]{2}[0-9]{3}[0-9A-Z])[ ]{5,}([A-Z])")
RemoveLinkSpace("in","out", upstream_pattern)
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