The Python docs say:
re.MULTILINE: When specified, the pattern character ‘^’ matches at the beginning of the string and at the beginning of each line (immediately following each newline)… By default, ‘^’ matches only at the beginning of the string…
So what’s going on when I get the following unexpected result?
>>> import re
>>> s = """// The quick brown fox.
... // Jumped over the lazy dog."""
>>> re.sub('^//', '', s, re.MULTILINE)
' The quick brown fox.n// Jumped over the lazy dog.'
Answers:
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Method 1
Look at the definition of re.sub:
re.sub(pattern, repl, string[, count, flags])
The 4th argument is the count, you are using re.MULTILINE (which is 8) as the count, not as a flag.
Either use a named argument:
re.sub('^//', '', s, flags=re.MULTILINE)
Or compile the regex first:
re.sub(re.compile('^//', re.MULTILINE), '', s)
Method 2
re.sub('(?m)^//', '', s)
Method 3
The full definition of re.sub is:
re.sub(pattern, repl, string[, count, flags])
Which means that if you tell Python what the parameters are, then you can pass flags without passing count:
re.sub('^//', '', s, flags=re.MULTILINE)
or, more concisely:
re.sub('^//', '', s, flags=re.M)
All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0