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python - 具体正则表达式搜索python

转载 作者:太空宇宙 更新时间:2023-11-04 05:28:06 24 4
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这是我的第一篇文章。当涉及到代码时,我总是来这个论坛寻找答案。

我一直在努力理解 Python 中的正则表达式,但这有点困难。

我有这样的文字:

Name:   Clash1
Distance: -1.341m
Image Location: Test 1_navis_files\cd000001.jpg
HardStatus: New
Clash Point: 3.884m, -2.474m, 2.659m
Date Created: 2016/6/2422:45:09

Item 1
GUID: 6efaec51-b699-4d5a-b947-505a69c31d52
Path: File ->Colisiones_v2015.dwfx ->Segment ->Pipes (1) ->Pipe Types (1) ->Default (1) ->Pipe Types [2463] ->Shell
Item Name: Pipe Types [2463]
Item Type: Shell

Item 2
GUID: 6efaec51-b699-4d5a-b947-505a69c31dea
Path: File ->Colisiones_v2015.dwfx ->Segment ->Walls (4) ->Basic Wall (4) ->Wall 1 (4) ->Basic Wall [2343] ->Shell
Item Name: Basic Wall [2343]
Item Type: Shell

------------------


Name: Clash2
Distance: -1.341m
Image Location: Test 1_navis_files\cd000002.jpg
HardStatus: New
Clash Point: 3.884m, 3.533m, 2.659m
Date Created: 2016/6/2422:45:09

Item 1
GUID: 6efaec51-b699-4d5a-b947-505a69c31d52
Path: File ->Colisiones_v2015.dwfx ->Segment ->Pipes (1) ->Pipe Types (1) ->Default (1) ->Pipe Types [2463] ->Shell
Item Name: Pipe Types [2463]
Item Type: Shell

Item 2
GUID: 6efaec51-b699-4d5a-b947-505a69c31de8
Path: File ->Colisiones_v2015.dwfx ->Segment ->Walls (4) ->Basic Wall (4) ->Wall 1 (4) ->Basic Wall [2341] ->Shell
Item Name: Basic Wall [2341]
Item Type: Shell

------------------

我需要做的是创建一个列表来提取每个文本 block (由 分隔-------------------- --------) 以下内容作为字符串:冲突名称和冲突点。

例如:冲突 1 3.884、3.533、2.659

本人初学Python,对正则表达式了解不多。

谁能给我一些关于使用正则表达式从文本中提取这些值的线索?

我做了这样的事情:

exp = r'(?<=Clash Point\s)(?<=Point\s)([0-9]*)'
match = re.findall(exp, html)

if match:
OUT.append(match)
else:
OUT = 'fail'

但我知道我离我的目标还很远。

最佳答案

如果您正在寻找正则表达式解决方案,您可以想出:

^Name:\s*         # look for Name:, followed by whitespaces
# at the beginning of a line
(?P<name>.+) # capture the rest of the line
# in a group called "name"
[\s\S]+? # anything afterwards lazily
^Clash\ Point:\s* # same construct as above
(?P<point>.+) # same as the other group

参见 a demo on regex101.com .


翻译成 Python 代码,这将是:

import re
rx = re.compile(r"""
^Name:\s*
(?P<name>.+)
[\s\S]+?
^Clash\ Point:\s*
(?P<point>.+)""", re.VERBOSE|re.MULTILINE)

for match in rx.finditer(your_string_here):
print match.group('name')
print match.group('point')

这将输出:

Clash1
3.884m, -2.474m, 2.659m
Clash2
3.884m, 3.533m, 2.659m

参见 a working demo on ideone.com .

关于python - 具体正则表达式搜索python,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/38042750/

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