一个用矩阵查询的小trick

import numpy as np
X = np.array([[0, 1, 0, 1],
            [1, 1, 1, 0],
            [0, 1, 1, 0],
            [0, 0, 0, 1],
            [0, 1, 1, 0],
            [0, 1, 0, 1],
            [1, 0, 0, 1]])
y = np.array([0, 1, 1, 0, 1, 0, 0])
counts = {}
for label in np.unique(y):
    counts[label] = X[y == label].sum(axis=0)
print("{}".format(counts))

//output
{0: array([1, 2, 0, 4]), 1: array([1, 3, 3, 0])}

X[y==label]返回X中条件为true的行，sum（axis=0）做每一列的和，保存到字典中