According to Python documentation[1] dirname and basename
are defined as follows:
os.path.dirname() = os.path.split()[0]
os.path.basename() = os.path.split()[1]
For the purpose of better readability split() is replaced
by appropriate function if only one part of returned tuple
is used.
[1]: https://docs.python.org/3/library/os.path.html#os.path.split
Meson has a common pattern of using 'if len(foo) == 0:' or
'if len(foo) != 0:', however, this is a common anti-pattern in python.
Instead tests for emptiness/non-emptiness should be done with a simple
'if foo:' or 'if not foo:'
Consider the following:
>>> import timeit
>>> timeit.timeit('if len([]) == 0: pass')
0.10730923599840025
>>> timeit.timeit('if not []: pass')
0.030033907998586074
>>> timeit.timeit('if len(['a', 'b', 'c', 'd']) == 0: pass')
0.1154778649979562
>>> timeit.timeit("if not ['a', 'b', 'c', 'd']: pass")
0.08259823200205574
>>> timeit.timeit('if len("") == 0: pass')
0.089759664999292
>>> timeit.timeit('if not "": pass')
0.02340641999762738
>>> timeit.timeit('if len("foo") == 0: pass')
0.08848102600313723
>>> timeit.timeit('if not "foo": pass')
0.04032287199879647
And for the one additional case of 'if len(foo.strip()) == 0', which can
be replaced with 'if not foo.isspace()'
>>> timeit.timeit('if len(" ".strip()) == 0: pass')
0.15294511600222904
>>> timeit.timeit('if " ".isspace(): pass')
0.09413968399894657
>>> timeit.timeit('if len(" abc".strip()) == 0: pass')
0.2023209120015963
>>> timeit.timeit('if " abc".isspace(): pass')
0.09571301700270851
In other words, it's always a win to not use len(), when you don't
actually want to check the length.
Aleksey Filippov
Jussi Pakkanen
Dylan Baker
Mike Sinkovsky
Igor Gnatenko
Elliott Sales de Andrade