python - 两个列表之间更快的元素滚动置换交换

Question

我有两个列表，X 和 Y。X = [1,2,3]和 Y = ['A','B','C']我想从列表 A 中获取两个的所有组合，并将其放在第二个列表中的任何位置。
从最初的第二个列表中，我想将所有元素一个一个地放在列表 X 中的任何位置。
结果应该是这样的:

# Original lists: (X = [1,2,3], Y = ['A','B','C'])

[
 ([1, 'A'], ['B', 'C', 2, 3]),  # 2,3 went to list (Y) and 'A' went to list (X)
 ([1, 'B'], ['A', 'C', 2, 3]),  # 2,3 went to list (Y) and 'B' went to list (X)
 ([1, 'C'], ['A', 'B', 2, 3]),  # 2,3 went to list (Y) and 'C' went to list (X)
 ([2, 'A'], ['B', 'C', 1, 3]),  # 1,3 went to list (Y) and 'A' went to list (X)
 ([2, 'B'], ['A', 'C', 1, 3]),  # 1,3 went to list (Y) and 'B' went to list (X)
 ([2, 'C'], ['A', 'B', 1, 3]),  # 1,3 went to list (Y) and 'C' went to list (X)
 ([3, 'A'], ['B', 'C', 1, 2]),  # 1,2 went to list (Y) and 'A' went to list (X)
 ([3, 'B'], ['A', 'C', 1, 2]),  # 1,2 went to list (Y) and 'B' went to list (X)
 ([3, 'C'], ['A', 'B', 1, 2])   # 1,2 went to list (Y) and 'C' went to list (X)
]

我的实现是这样的:

def itwo(l1,l2):
    ln2 = l2[:]
    final = []
    z = list(it.combinations(l1, 2))
    for i in range(len(z)):
        ln1 = [ elem for elem in l1 if elem not in list(z[i])] 
        ln2 = l2 + list(z[i])
        for y in range(len(l2)):
            ln1.append(l2[y])
            ln2.remove(l2[y])
            final.append((ln1, ln2))    
            ln2 = l2 + list(z[i])
            ln1 = [ elem for elem in l1 if elem not in list(z[i])]
    return(final)

如 X有 100 个元素和 Y十个，例如 A = [1, 2, ...., 100]和 B = [A, B, C, D, E, F, G, H, I, J]在我的 PC 中的性能是:

   Ordered by: internal time

   ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    49500    0.910    0.000    0.910    0.000 C:\...\test.py:25(<listcomp>)
     4950    0.086    0.000    0.086    0.000 C:\...\test.py:18(<listcomp>)
        1    0.040    0.040    1.049    1.049 C:\...\test.py:13(itwo)
    99000    0.006    0.000    0.006    0.000 {method 'append' of 'list' objects}
    49500    0.006    0.000    0.006    0.000 {method 'remove' of 'list' objects}
     4951    0.000    0.000    0.000    0.000 {built-in method builtins.len}
        1    0.000    0.000    0.000    0.000 {method 'disable' of '_lsprof.Profiler' objects}


         207903 function calls in 1.049 seconds

我想知道使用另一种方法是否可以更快。

Answer 1

似乎您应该可以使用 itertools.product 来做到这一点。并设置函数以删除嵌套循环:

import itertools as it

X = range(100)
Y = ['A','B','C', 'D', 'E', 'F', 'G', 'H', 'I', 'J']

def itwo_s(l1,l2):
    s1 = set(l1)
    s2 = set(l2)
    
    for l in it.product(l1, l2):
        yield [list(l), [*s2.difference(l), *s1.difference(l)]]


list(itwo_s(X, Y))

定时:

# original:
%timeit itwo(X, Y)
# 1.55 s ± 13.7 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

# code above:
%timeit list(itwo_s(X, Y))
# 2.97 ms ± 179 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

因为您依赖于集合，所以您对子列表进行松散排序以换取速度。不确定这是否重要。