获取numpy中相同值的元素

Question

存在一个numpy矩阵:

bins =array([[2076., -861.84471079],
[2076., -858.23866597],
[2076., -861.84471079],
...,
[9757., 1847.33889178],
[9757., 1830.39082856],
[9757., -932.14347751]])

我想收集沿1轴的所有元素对于沿0轴的每个唯一值:

bins = array([[2076., [-861.84471079, -858.2386, -861.84471]],
[9757., [1847.33889178, 1830.3908, -932.14]]])

等

我做了如下:

bins = points_OYZ[points_OYZ[:, 0].argsort()]
vals, idx_start, count = np.unique(bins[:, 0], return_counts=True, return_index=True)
binarize_matrix = np.split(bins, idx_start[1:])
q = []
for i in binarize_matrix:
min_index = np.argmin(i[:, 1])
q.append(i[min_index])

我想让它更简单，但我不知道如何，请告诉我!

Answer 1

这是你想要做的吗

import numpy as np
bins = np.array([[2076., -861.84471079],
[2076., -858.23866597],
[2076., -861.84471079],
[9757., 1847.33889178],
[9757., 1830.39082856],
[9757., -932.14347751],
[9760., -159.14347751]])
vals, idx_start, count = np.unique(bins[:, 0], return_counts=True, return_index=True)
output = [[vals[i], bins[idx_start[i]:idx_start[i]+count[i]][:, 1].tolist()] for i in range(len(vals))]
print(output)

输出:

[[2076.0, [-861.84471079, -858.23866597, -861.84471079]], 
[9757.0, [1847.33889178, 1830.39082856, -932.14347751]], 
[9760.0, [-159.14347751]]]