There is an inconsistency in re.sub() when substituting at the end of a string using a prior match with a '*' qualifier: the substitution now occurs twice.  For example:

txt = re.sub(r'\s*\Z', "\n", txt)

This should work like txt.rstrip() + "\n", but beginning in 3.7, the re.sub version now matches twice and changes any non-empty whitespace into "\n\n" instead of "\n". (If there is no trailing whitespace it only matches once.)

The bug is the same if '$' is used instead of '\Z', but it does not happen if an actual character is specified (e.g. a substitution of r'\s*x' does not substitute twice if x has preceding whitespace).

I tested 2.7.17, 3.6.9, 3.7.7, 3.8.2, and 3.9.0a4, and it starts to fail in 3.7.7 and beyond.

Attached is a test program.

Duplicate of Issue39687.

See https://docs.python.org/3/library/re.html#re.sub and https://docs.python.org/3/whatsnew/3.7.html#changes-in-the-python-api.

This is not the same thing because the match is anchored, so it is not adjacent to the prior match -- it is the same match. I think that r'\s*\Z' should behave the same way as r'\s*x' due to the anchor point. The current behavior is matching the same \Z twice.

Another argument in favor of this being a bug, this does not exhibit the same doubling:

txt = ' test'
txt = re.sub(r'^\s*', '^', txt)

That always substitutes once.

Can this bug please be reopened and fixed? This is an anchored substitution, and so should never match more than once.

I just ran into this change in behavior myself.

It's worth noting that the new behavior appears to match perl's behavior:

# perl -e 'print(("he" =~ s/e*\Z/ah/rg), "\n")'
hahah

txt = ' test'
txt = re.sub(r'^\s*', '^', txt)

substitutes once because the * is greedy.

txt = ' test'
txt = re.sub(r'^\s*?', '^', txt)

substitutes twice, consistent with the \Z behavior.