RegEx用于捕获字符串的一部分

Question

I am trying to grab top level Markdown headings (ie, headings beginning with a single hash -- # Introduction) in an .md doc with Python's re library and cannot for the life of me figure this out.

Here is the code I'm trying to execute:

import re

pattern = r"(# .+?\\n)"

text = r"# Title\n## Chapter\n### sub-chapter#### What a lovely day.\n"

header = re.search(pattern, text)
print(header.string)

The result from the print(header.string) is:

# Title\\n## Chapter\\n### sub-chapter#### What a lovely day.\\n whereas I only want # Title\\n

This example on regex101 says it should work, but I can't figure out why it isn't. https://regex101.com/r/u4ZIE0/9

Answer 1

You get that result because you use header.string which is calling .string on a Match object which will give you back the string passed to match() or search () .

The string already has newlines in it:

text = r"# Title\n## Chapter\n### sub-chapter#### What a lovely day.\n"

So if you use your pattern (note that it will also match the newline), you could update your code to:

import re

pattern = r"(# .+?\\n)"
text = r"# Title\n## Chapter\n### sub-chapter#### What a lovely day.\n"
header = re.search(pattern, text)
print(header.group())

Python demo

Note that re.search looks for the first location where the regex produces a match.

Another option to match your value could be matching from the start of the string a # followed by a space and then any character except a newline until the end of the string:

^# .*$

For example:

import re

pattern = r"^# .*$"
text = "# Title\n## Chapter\n### sub-chapter#### What a lovely day.\n"
header = re.search(pattern, text, re.M)
print(header.group())

Python demo

If there can not be any more # following after, you might also use a negated character class to match not a # or a newline:

^# [^#\n\r]+$

Answer 2

I'm guessing that we are wishing to extract the # Title\\n , which in that case, your expression seems to be working fine with a slight modification:

(# .+?\\n)(.+)

DEMO

Test

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"(# .+?\\n)(.+)"

test_str = "# Title\\n## Chapter\\n### sub-chapter#### The Bar\\nIt was a fall day.\\n"

subst = "\\1"

# You can manually specify the number of replacements by changing the 4th argument
result = re.sub(regex, subst, test_str, 1)

if result:
    print (result)

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.