使用itertools.chain.from_iterable
:
from itertools import chain
mylist = list(chain.from_iterable(myarray))
演示:
>>> from itertools import chain
>>> myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]
>>> list(chain.from_iterable(myarray))
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']
但是,Hadro 的sum()
解决方案对于较短的样品更快:
>>> timeit.timeit('f()', 'from __main__ import withchain as f')
2.858742465992691
>>> timeit.timeit('f()', 'from __main__ import withsum as f')
1.6423718839942012
>>> timeit.timeit('f()', 'from __main__ import withlistcomp as f')
2.0854451240156777
但itertools.chain
如果输入变大则获胜:
>>> myarray *= 100
>>> timeit.timeit('f()', 'from __main__ import withchain as f', number=25000)
1.6583486960153095
>>> timeit.timeit('f()', 'from __main__ import withsum as f', number=25000)
23.100156371016055
>>> timeit.timeit('f()', 'from __main__ import withlistcomp as f', number=25000)
2.093297885992797