例
in
[1,2,3,4,5,6,7,8]
out
[(1,2),(3,4),(5,6),(7,8)]
求优雅写法...
谢谢
in
[1,2,3,4,5,6,7,8]
out
[(1,2),(3,4),(5,6),(7,8)]
求优雅写法...
谢谢
This topic created in 4458 days ago, the information mentioned may be changed or developed.
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1
zip(list[::2], list[1::2])
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2
a = [1,2,3,4,5,6,7,8]
zip(a[0:-1:2], a[1:-1:2]) |
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exoticknightAug 24, 2014
4
binux果然是python高手……
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wy315700Aug 24, 2014
5
学习了，顺便收藏
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6
这个解决方案很多，像zip、itertools等等
我遇到这种问题常在数据分析中，通常都已经import了numpy、pandas，所以常用其中的方法来变形数列～ |
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magineAug 24, 2014
7
虽然有点不礼貌，但是建议楼主先读一下《python基础教程》之类的书。
当然如果没打算长期使用就当我没说。 |
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8
b=iter(a)
[(x, b.next()) for x in b] 速度比上面的zip快，尤其是当a很大的时候 Python 2.7.3 >>> timeit('zip(a[::2], a[1::2])', setup='a=[1,2,3,4,5,6,7,8]', number=10000) 0.00456690788269043 >>> timeit('[(x, b.next()) for x in b]', setup='a=[1,2,3,4,5,6,7,8];b=iter(a)', number=10000) 0.0014100074768066406 >>> timeit('zip(a[::2], a[1::2])', setup='a=range(100000)', number=10000) 21.61440110206604 >>> timeit('[(x, b.next()) for x in b]', setup='a=range(100000);b=iter(a)', number=10000) 0.022505998611450195 |
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hit9Aug 24, 2014
9
>>> lst = [1,2,3,4,5,6,7,8]
>>> zip(*[iter(lst)]*2) [(1, 2), (3, 4), (5, 6), (7, 8)] |
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hit9Aug 24, 2014
10
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starsoiAug 24, 2014
11
@hit9 zip(*[iter(lst)]*2) 太慢了
>>> timeit('zip(*[iter(a)]*2)', setup='a=range(100000);b=iter(a)', number=10000) 18.37099289894104 受你的启发，这样可读性更高，而且也超快: b=iter(a) zip(b,b) >>> timeit('zip(b,b)', setup='a=range(100000);b=iter(a)', number=10000) 0.022648096084594727 |
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hit9Aug 25, 2014
13
两者的开销差貌似只有" 放到list中再unpack到args中"
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