有效地填充一个多维数组，该数组有许多if else语句

Question

我想以一种特定和有效的方式填充一个4模糊的numpy数组。因为我不太清楚，所以我用if better语句来编写代码，但这看起来不太好，可能很慢，而且我也不能确定我是否考虑到了每一个组合。下面是我不再写的代码：

sercnew2 = numpy.zeros((gn, gn, gn, gn))
for x1 in range(gn):
    for x2 in range(gn):
        for x3 in range(gn):
            for x4 in range(gn):
                if x1 == x2 == x3 == x4: 
                    sercnew2[x1, x2, x3, x4] = ewp[x1]
                elif x1 == x2 == x3 != x4:
                    sercnew2[x1, x2, x3, x4] = ewp[x1] * ewp[x4]
                elif x1 == x2 == x4 != x3:
                    sercnew2[x1, x2, x3, x4] = ewp[x1] * ewp[x3]
                elif x1 == x3 == x4 != x2:
                    sercnew2[x1, x2, x3, x4] = ewp[x1] * ewp[x2]
                elif x2 == x3 == x4 != x1:
                    sercnew2[x1, x2, x3, x4] = ewp[x2] * ewp[x1]
                elif x1 == x2 != x3 == x4:
                    sercnew2[x1, x2, x3, x4] = ewp[x1] * ewp[x3]
                elif ... many more combinations which have to be considered

因此，应该发生的基本情况是，如果所有变量(x1、x2、x3、x4)彼此不同，条目将是：

sercnew2[x1, x2, x3, x4] = ewp[x1]* ewp[x2] * ewp[x3] * ewp[x4]

现在如果假设变量x2和x4是相同的，那么：

sercnew2[x1, x2, x3, x4] = ewp[x1]* ewp[x2] * ewp[x3]

其他示例可以在上面的代码中看到。基本上，如果两个或多个变量是相同的，那么我只考虑其中的一个。我希望模式是清楚的。否则，请让我注意，我会尽量更好地表达我的问题。我很肯定，有一个更聪明的方法来做这件事。希望你能更好地了解并提前感谢:)

Answer 1

真希望我能得到它！这是一个矢量化的方法-

from itertools import product

n_dims = 4 # Number of dims

# Create 2D array of all possible combinations of X's as rows
idx = np.sort(np.array(list(product(np.arange(gn), repeat=n_dims))),axis=1)

# Get all X's indexed values from ewp array
vals = ewp[idx]

# Set the duplicates along each row as 1s. With the np.prod coming up next,
#these 1s would not affect the result, which is the expected pattern here.
vals[:,1:][idx[:,1:] == idx[:,:-1]] = 1

# Perform product along each row and reshape into multi-dim array
out = vals.prod(1).reshape([gn]*n_dims)

Answer 2

我真的不明白你说这些变量是相同的是什么意思，但如果它们确实是相同的，那么你只需要使用set()就行了。

from functools import reduce
from operator import mul
sercnew2 = numpy.zeros((gn, gn, gn, gn))
for x1 in range(gn):
    for x2 in range(x1, gn):
        for x3 in range(x2, gn):
            for x4 in range(x3, gn):
                set_ = [ewp[n] for n in set([x1, x2, x3, x4])]
                sercnew2[x1, x2, x3, x4] = reduce(mul, set_, 1)

它的工作方式是创建一个删除重复项的set()，然后使用reduce函数从set_中选择第一个数字，并将其与1 (初始化器值)相乘，其结果将作为第一个参数传递给reduce，第二个将是set_的第二个条目。抱歉，我解释得不好。

Answer 3

您也可以在单个for循环中这样做。在Divakar的索引列表技巧的基础上，我们需要做的第一件事是，如何只提取4d数组sercnew2中给定元素的唯一索引。

最快的方法之一(参见：https://www.peterbe.com/plog/uniqifiers-benchmark)是使用sets。然后，我们只需将sercnew2初始化为一个1数组，而不是0。

from itertools import product
import numpy as np

sercnew2 = np.ones((gn, gn, gn, gn))
n_dims=4
idx = list(product(np.arange(gn), repeat=n_dims))

for i,j,k,l in idx:
    unique_items = set((i,j,k,l))
    for ele in unique_items:
        sercnew2[i,j,k,l] *= ewp[ele]

编辑:正如@unutbu所建议的，我们也可以使用来自https://stackoverflow.com/a/11146645/5714445的https://stackoverflow.com/a/11146645/5714445函数来加速idx的初始化

Edit2:如果您很难理解来自itertools的product做了什么，那么它提供了所有的排列。例如，假设gn=2将重复维设置为4，您将得到

[0, 0, 0, 0]
[0, 0, 0, 1]
[0, 0, 1, 0]
[0, 0, 1, 1]
[0, 1, 0, 0]
[0, 1, 0, 1]
[0, 1, 1, 0]
[0, 1, 1, 1]
[1, 0, 0, 0]
[1, 0, 0, 1]
[1, 0, 1, 0]
[1, 0, 1, 1]
[1, 1, 0, 0]
[1, 1, 0, 1]
[1, 1, 1, 0]
[1, 1, 1, 1]