使用 numpy 构建两个数组的所有组合的数组 [英] Using numpy to build an array of all combinations of two arrays

问题描述

在尝试用它做任何复杂的事情之前，我试图遍历 6 参数函数的参数空间来研究它的数值行为，所以我正在寻找一种有效的方法来做到这一点.

I'm trying to run over the parameters space of a 6 parameter function to study its numerical behavior before trying to do anything complex with it, so I'm searching for an efficient way to do this.

我的函数将 6 维 numpy 数组中给出的浮点值作为输入.我最初尝试做的是:

My function takes float values given in a 6-dim numpy array as input. What I tried to do initially was this:

首先，我创建了一个函数，该函数接受 2 个数组并生成一个包含来自两个数组的所有值组合的数组:

First, I created a function that takes 2 arrays and generate an array with all combinations of values from the two arrays:

from numpy import *
def comb(a,b):
    c = []
    for i in a:
        for j in b:
            c.append(r_[i,j])
    return c

然后，我使用 reduce() 将其应用于同一数组的 m 个副本:

Then, I used reduce() to apply that to m copies of the same array:

def combs(a,m):
    return reduce(comb,[a]*m)

最后，我这样评估我的函数:

Finally, I evaluate my function like this:

values = combs(np.arange(0,1,0.1),6)
for val in values:
    print F(val)

这行得通，但方式太慢了.我知道参数的空间很大，但这不应该那么慢.在这个例子中，我只采样了 10⁶(一百万)个点，仅仅创建数组 values 就花了超过 15 秒.

This works but it's way too slow. I know the space of parameters is huge, but this shouldn't be so slow. I have only sampled 10⁶ (a million) points in this example and it took more than 15 seconds just to create the array values.

您知道使用 numpy 执行此操作的更有效方法吗?

Do you know any more efficient way of doing this with numpy?

如果需要，我可以修改函数 F 接受它的参数的方式.

I can modify the way the function F takes it's arguments if it's necessary.

推荐答案

在较新版本的 numpy (>1.8.x) 中，numpy.meshgrid() 提供了更快的实现:

In newer version of numpy (>1.8.x), numpy.meshgrid() provides a much faster implementation:

@pv 的解决方案

In [113]:

%timeit cartesian(([1, 2, 3], [4, 5], [6, 7]))
10000 loops, best of 3: 135 µs per loop
In [114]:

cartesian(([1, 2, 3], [4, 5], [6, 7]))

Out[114]:
array([[1, 4, 6],
       [1, 4, 7],
       [1, 5, 6],
       [1, 5, 7],
       [2, 4, 6],
       [2, 4, 7],
       [2, 5, 6],
       [2, 5, 7],
       [3, 4, 6],
       [3, 4, 7],
       [3, 5, 6],
       [3, 5, 7]])

numpy.meshgrid() 以前只有 2D，现在可以进行 ND.在这种情况下，3D:

numpy.meshgrid() use to be 2D only, now it is capable of ND. In this case, 3D:

In [115]:

%timeit np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3)
10000 loops, best of 3: 74.1 µs per loop
In [116]:

np.array(np.meshgrid([1, 2, 3], [4, 5], [6, 7])).T.reshape(-1,3)

Out[116]:
array([[1, 4, 6],
       [1, 5, 6],
       [2, 4, 6],
       [2, 5, 6],
       [3, 4, 6],
       [3, 5, 6],
       [1, 4, 7],
       [1, 5, 7],
       [2, 4, 7],
       [2, 5, 7],
       [3, 4, 7],
       [3, 5, 7]])

请注意，最终结果的顺序略有不同.

Note that the order of the final resultant is slightly different.

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