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How do I reverse a list or loop over it backwards? (37 answers)

Closed 11 months ago.

在沙尔,制定一份新清单的最佳方式何在,其项目与其他一些清单相同,但恰恰相反? (我不想修改现有名单。)

对我来说,这是一个解决办法:

new_list = list(reversed(old_list))

还可复制<代码>old_list。从而扭转现有的重复:

new_list = list(old_list) # or `new_list = old_list[:]`
new_list.reverse()

是否有更好的选择被我忽视? 如果没有,是否有令人信服的理由(例如效率)使用上述一种做法?

Answer 1

newlist = oldlist[::-1]

The [:-1] slicing (由我的妻子Anna喜欢称作“Marian smiley”;-) means: slice the whole series, with a step of -1, i.e., invert. 它符合所有顺序。

说明 (and) 您提到的替代物相当于“软拷贝”,即:如果这些物品是可变的,而且你向它们打电话,原始清单上所列物品的变数也出现在倒置清单中,反之亦然。如果需要避免,请设一个<代码>copy.deepcopy。 (尽管总是可能昂贵的作业),但在这种情况下,只有<代码>.reverse才是唯一的好选择。

Answer 2

现在请rel=“noreferer”>timeit<>><>a>。缩略语

$ p -m timeit "ol = [1, 2, 3]; nl = list(reversed(ol))"
100000 loops, best of 3: 2.34 usec per loop

$ p -m timeit "ol = [1, 2, 3]; nl = list(ol); nl.reverse();"
1000000 loops, best of 3: 0.686 usec per loop

$ p -m timeit "ol = [1, 2, 3]; nl = ol[::-1];"
1000000 loops, best of 3: 0.569 usec per loop

$ p -m timeit "ol = [1, 2, 3]; nl = [i for i in reversed(ol)];"
1000000 loops, best of 3: 1.48 usec per loop


$ p -m timeit "ol = [1, 2, 3]*1000; nl = list(reversed(ol))"
10000 loops, best of 3: 44.7 usec per loop

$ p -m timeit "ol = [1, 2, 3]*1000; nl = list(ol); nl.reverse();"
10000 loops, best of 3: 27.2 usec per loop

$ p -m timeit "ol = [1, 2, 3]*1000; nl = ol[::-1];"
10000 loops, best of 3: 24.3 usec per loop

$ p -m timeit "ol = [1, 2, 3]*1000; nl = [i for i in reversed(ol)];"
10000 loops, best of 3: 155 usec per loop

<>Update: 检查员G4dget建议的补充清单计算方法。我要说的是,结果本身。

Answer 3

Adjustments

值得为单列时间计算提供基线基准/调整,显示在没有经常不必要的<条码>清单()转换的情况下出现逆转的性能。 <<>编码>操作>在操作时间上增加了26个使用装置,只有在发射器不能接受的情况下才需要。

<>Results>

reversed(lst) -- 11.2 usecs

list(reversed(lst)) -- 37.1 usecs

lst[::-1] -- 23.6 usecs

<>计算:

# I ran this set of 100000 and came up with 11.2, twice:
python -m timeit "ol = [1, 2, 3]*1000; nl = reversed(ol)"
100000 loops, best of 3: 11.2 usec per loop

# This shows the overhead of list()
python -m timeit "ol = [1, 2, 3]*1000; nl = list(reversed(ol))"
10000 loops, best of 3: 37.1 usec per loop

# This is the result for reverse via -1 step slices
python -m timeit "ol = [1, 2, 3]*1000;nl = ol[::-1]"
10000 loops, best of 3: 23.6 usec per loop

http://www.hchr.org。

这些测试的完成时间是reversed(>)高于12.4用途代码[:-1]。

Adjustments

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