如何将列表中的整数仅拆分为个位数? [英] How do a split integers within a list into single digits only?

问题描述

假设我有这样的东西:

    list(range(9:12))

哪个给我一个列表:

    [9,10,11]

但是我希望它像这样:

    [9,1,0,1,1]

哪个整数将每个整数都拆分为一个数字，在不牺牲过多性能的情况下是否有实现这一目标的方法?还是有一种首先生成此类列表的方法?

Which splits every integer into single digits, is there anyway of achieving this without sacrificing too much performance? Or is there a way of generating list like these in the first place?

推荐答案

您可以有效地构建最终结果，而不必使用

You can build the final result efficiently without having to build one large and/or small intermediate strings using itertools.chain.from_iterable.

In [18]: list(map(int, chain.from_iterable(map(str, range(9, 12)))))
Out[18]: [9, 1, 0, 1, 1]

---

In [12]: %%timeit
    ...: list(map(int, chain.from_iterable(map(str, range(9, 20)))))
    ...: 
100000 loops, best of 3: 8.19 µs per loop

In [13]: %%timeit
    ...: [int(i) for i in ''.join(map(str, range(9, 20)))]
    ...: 
100000 loops, best of 3: 9.15 µs per loop

In [14]: %%timeit
    ...: [int(x) for i in range(9, 20) for x in str(i)]
    ...: 
100000 loops, best of 3: 9.92 µs per loop

计时与输入成比例. itertools 版本还有效地使用内存，尽管与list(map(int, ...))一起使用时，它比str.join版本略慢:

Timings scale with input. The itertools version also uses memory efficiently although it is marginally slower than the str.join version if used with list(map(int, ...)):

In [15]: %%timeit
    ...: list(map(int, chain.from_iterable(map(str, range(9, 200)))))
    ...: 
10000 loops, best of 3: 138 µs per loop

In [16]: %%timeit
    ...: [int(i) for i in ''.join(map(str, range(9, 200)))]
    ...: 
10000 loops, best of 3: 159 µs per loop

In [17]: %%timeit
    ...: [int(x) for i in range(9, 200) for x in str(i)]
    ...: 
10000 loops, best of 3: 182 µs per loop

In [18]: %%timeit
    ...: list(map(int, ''.join(map(str, range(9, 200)))))
    ...: 
10000 loops, best of 3: 130 µs per loop

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