获得整数排列的更有效方法是什么？ [英] More efficient way to get integer permutations?

问题描述

我可以得到这样的整数排列：

I can get integer permutations like this:

myInt = 123456789

l = itertools.permutations(str(myInt))
[int(''.join(x)) for x in l]

是否有更有效的方法在Python中获取整数排列，省去了创建字符串的开销，然后加入生成的元组？对它进行定时，元组加入过程比 list（l）长3倍。

Is there a more efficient way to get integer permutations in Python, skipping the overhead of creating a string, then joining the generated tuples? Timing it, the tuple-joining process makes this 3x longer than list(l).

添加了支持信息

myInt =123456789
def v1(i): #timeit gives 258ms
    l = itertools.permutations(str(i))
    return [int(''.join(x)) for x in l]

def v2(i): #timeit gives 48ms
    l = itertools.permutations(str(i))
    return list(l)

def v3(i): #timeit gives 106 ms
    l = itertools.permutations(str(i))
    return [''.join(x) for x in l]

推荐答案

你可以这样做：

>>> digits = [int(x) for x in str(123)]
>>> n_digits = len(digits)
>>> n_power = n_digits - 1
>>> permutations = itertools.permutations(digits)
>>> [sum(v * (10**(n_power - i)) for i, v in enumerate(item)) for item in permutations]
[123, 132, 213, 231, 312, 321]

这避免了转换为元组和从元组转换，因为它将使用元组中的整数位置来计算其值（例如，（1,2,3）表示 100 + 20 + 3 ）。

This avoids conversion to and from a tuple as it'll use the integer's position in the tuple to compute its value (e.g., (1,2,3) means 100 + 20 + 3).

因为 n_digits 的值在整个过程中是已知且相同的，我认为您还可以优化计算：

Because the value of n_digits is known and the same throughout the process, I think you can also optimize the computations to:

>>> values = [v * (10**(n_power - i)) for i, v in enumerate(itertools.repeat(1, n_digits))]
>>> values
[100, 10, 1]
>>> [sum(v * index for v, index in zip(item, values)) for item in permutations]
[123, 132, 213, 231, 312, 321]

我还认为我们不需要一直拨打 zip（），因为我们不需要该列表：

I also think we don't need to call zip() all the time because we don't need that list:

>>> positions = list(xrange(n_digits))
>>> [sum(item[x] * values[x] for x in positions) for item in permutations]
[123, 132, 213, 231, 312, 321]

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