确定两个列表/数组的混洗索引 [英] Determine the shuffled indices of two lists/arrays

问题描述

作为挑战，我给自己这个问题：

As a challenge, I've given myself this problem:

给定2个列表，A和B，其中B是A的洗牌版本，这个想法是弄清洗牌指数。

Given 2 lists, A, and B, where B is a shuffled version of A, the idea is to figure out the shuffled indices.

例如：

A = [10, 40, 30, 2]
B = [30, 2, 10, 40]

result = [2,   3,    0,      1] 
        A[2]  A[3]   A[0]  A[1]
        ||     ||     ||    ||
        30      2     10    40

请注意，相同元素的关系可以任意解决。

Note that ties for identical elements can be resolved arbitrarily.

我想出了一个解决方案，涉及到使用字典来存储索引。这个问题有哪些其他可能的解决方案？使用库的解决方案也有效。 Numpy，pandas，一切都很好。

I've come up with a solution that involves the use of a dictionary to store indices. What other possible solutions does this problem have? A solution using a library also works. Numpy, pandas, anything is fine.

推荐答案

作为对当前解决方案的改进，你可以使用 collections.defaultdict 并避免 dict.setdefault ：

As an improvement over your current solution, you could use collections.defaultdict and avoid dict.setdefault:

from collections import defaultdict

A = [10, 40, 30, 2]
B = [30, 2, 10, 40]

idx = defaultdict(list)
for i, l in enumerate(A):
    idx[l].append(i)

res = [idx[l].pop() for l in B]
print(res)

---

以下是使用给定样本输入的两种方法的时间安排：

---

Here are the timings for the two methods using the sample input given:

用于测试的脚本

from timeit import timeit


setup = """
from collections import defaultdict;
idx1 = defaultdict(list); idx2 = {}
A = [10, 40, 30, 2]
B = [30, 2, 10, 40]
"""

me = """
for i, l in enumerate(A):
    idx1[l].append(i)
res = [idx1[l].pop() for l in B]
"""

coldspeed = """
for i, l in enumerate(A):
    idx2.setdefault(l, []).append(i)
res = [idx2[l].pop() for l in B]
"""

print(timeit(setup=setup, stmt=me))
print(timeit(setup=setup, stmt=coldspeed))

结果

original: 2.601998388010543
modified: 2.0607256239745766

所以看来使用 defaultdict 实际上会产生轻微的速度提升。这实际上是因为 defaultdict 是用C而不是Python实现的。更不用说原始解决方案的属性查找 - idx.setdefault1 - 费用很高。

So it appears that using defaultdict actually yields a slight speed increase. This actually makes since though since defaultdict is implemented in C rather than Python. Not to mention that the attribute lookup of the original solution - idx.setdefault1 - is costly.

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