使用2维数组,看起来像这样:
myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]
Run Code Online (Sandbox Code Playgroud)
每个元素都是一个具有固定长度元素的数组.如何成为这个,
mylist = ['jacob','mary','jack','white','fantasy','clothes','heat','abc','edf','fgc']
Run Code Online (Sandbox Code Playgroud)
这是我的解决方案
mylist = []
for x in myarray:
mylist.extend(x)
Run Code Online (Sandbox Code Playgroud)
我猜应该更简单
用途itertools.chain.from_iterable:
from itertools import chain
mylist = list(chain.from_iterable(myarray))
Run Code Online (Sandbox Code Playgroud)
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']
Run Code Online (Sandbox Code Playgroud)
但是,Haidro的sum()解决方案对于您的较短样本来说更快:
>>> 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
Run Code Online (Sandbox Code Playgroud)
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
Run Code Online (Sandbox Code Playgroud)
>>> myarray = [['jacob','mary'],['jack','white'],['fantasy','clothes'],['heat','abc'],['edf','fgc']]
>>> sum(myarray,[])
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']
Run Code Online (Sandbox Code Playgroud)
要么
>>> [i for j in myarray for i in j]
['jacob', 'mary', 'jack', 'white', 'fantasy', 'clothes', 'heat', 'abc', 'edf', 'fgc']
Run Code Online (Sandbox Code Playgroud)
| 归档时间: |
|
| 查看次数: |
3114 次 |
| 最近记录: |