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我有一个相对简单的问题，如果不使用循环就无法解决。我很难找出这个问题的正确标题。假设我们有两个 numpy 数组：

array_1 = np.array([[0, 1, 2],
                    [3, 3, 3],
                    [3, 3, 4],
                    [3, 6, 2]])

array_2 = np.array([[0, 0, 0], 
                    [1, 1, 1],
                    [2, 2, 2],
                    [3, 3, 3],
                    [4, 4, 4],
                    [5, 5, 5],
                    [6, 6, 6]])

array_1array_2表示我们想要的行的索引sum。因此，例如，数组中的4第 行应包含与array_1中的所有 s 具有相同行索引的所有行的总和。resultarray_23

在代码中更容易理解：

result = np.empty(array_2.shape)

for i in range(array_1.shape[0]):
    for j in range(array_1.shape[1]):
        index = array_1[i, j]
        result[index] = result[index] + array_2[i]

结果应该是：

[[ 0  0  0]
 [ 0  0  0]
 [ 3  3  3]
 [10 10 10]
 [ 2  2  2]
 [ 0  0  0]
 [ 3  3  3]]

我尝试使用np.einsum，但我需要将数组中的两个元素都用作索引，并将其行用作索引，所以我不确定np.einsum这里是否是最佳路径。

这是我在图形中遇到的问题。array_1表示三角形的顶点索引并array_2表示法线，其中行的索引对应于顶点的索引

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1 回答 1

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Any time you're adding something from a repeated index, normal ufuncs like np.add don't work out of the box because they only process a repeated fancy index once. Instead, you have to use the unbuffered version, which is np.add.at.

Here, you have a pair of indices: the row in array_1 is the row index into array_2, and the element of array_1 is the row index into the output.

First, construct the indices explicitly as fancy indices. This will make it much simpler to use them:

output_row = array_1.ravel()
input_row = np.repeat(np.arange(array_1.shape[0]), array_1.shape[1]).ravel()

You can apply input_row directly to array_2, but you need add.at to use output_row:

output = np.zeros_like(array_2)
np.add.at(output, output_row, array_2[input_row])

You really only use the first four rows of array_2, so it could be truncated to

array_2 = array2[:array_1.shape[0]]

In that case, you would want to initialize the output as:

output = np.zeros_like(array_2, shape=(output_row.max() + 1, array2.shape[1]))
于 2022-01-11T21:32:10.490 回答