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2018-08-27using urlopen() with explicit timeoutAndrei Antonov1-2/+3
2018-03-12Use os.path.dirname() instead of os.path.split(...)[0]Aleksey Filippov1-1/+1
2017-09-20ghwt: Add a comment at the top briefly explaining what it isLuke Shumaker1-0/+5
2017-05-02Don't use len() to test emptiness vs not emptinessDylan Baker1-1/+1
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.
2016-08-30Use context manager in scripts.Elliott Sales de Andrade1-2/+2
2016-05-05Created an emergency wraptool that downloads directly from Github.Jussi Pakkanen1-0/+107