meson

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Author	SHA1	Message	Date
Dylan Baker	a8173630ea	Don't use len() to test emptiness vs not emptiness 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.	8 years ago
Jussi Pakkanen	0a14ad63de	Add simple uninstall target. Closes #753 .	8 years ago

Author

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Dylan Baker

a8173630ea

Don't use len() to test emptiness vs not emptiness

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.

8 years ago

Jussi Pakkanen

0a14ad63de

Add simple uninstall target. Closes #753 .

8 years ago

2 Commits (197594146cd588b4c6027a3bcc218e8c00b70a4d)