如何在二维NumPy数组中查找另一数组相同行的位置？

你的代码问题出在逐元素比较时，会把部分元素匹配的行索引也提取出来，同时每个完全匹配的行包含3个元素，导致对应索引重复3次。要找到整行匹配的索引，需要先判断每行所有元素是否都满足近似相等，再提取对应索引。

方法一：利用np.all结合np.isclose #

这是最适合浮点数场景的方案，能处理精度问题：

import numpy as np

df1 = np.array([[0.000000,0.000000,0.000000],
[0.090000,0.000000,0.000000],
[0.190000,0.000000,0.000000],
[0.280000,0.000000,0.000000],
[0.380000,0.000000,0.000000],
[0.470000,0.000000,0.000000],
[0.570000,0.000000,0.000000],
[0.660000,0.000000,0.000000],
[0.760000,0.000000,0.000000],
[0.850000,0.000000,0.000000]])

df2 = np.array([[0.470000,0.000000,0.000000],
[0.570000,0.000000,0.000000],
[0.660000,0.000000,0.000000],
[0.760000,0.000000,0.000000],
[0.850000,0.000000,0.000000]
])

# 检查df1每行与df2每行是否所有元素都近似相等
row_matches = np.all(np.isclose(df1[:, np.newaxis], df2), axis=2)
# 提取匹配的df1行索引
df3 = np.argwhere(row_matches)[:, 0]

print(df3)
# 输出：[5 6 7 8 9]

代码说明： #

1. df1[:, np.newaxis]将df1转换为(10,1,3)的三维数组，和df2(5,3)广播后逐元素比较，得到(10,5,3)的布尔数组。
2. np.all(..., axis=2)沿着列维度取逻辑与，得到(10,5)的数组，其中row_matches[i,j]为True表示df1第i行和df2第j行完全匹配。
3. np.argwhere(row_matches)返回所有匹配的(i,j)坐标对，取第一列就是df1的行索引，正好对应df2每行的位置。

方法二：字典映射（适合无精度问题的场景） #

如果可以确保浮点数没有精度误差，也可以用字典建立行到索引的映射：

import numpy as np

# 构建df1行到索引的映射（注意浮点数精度问题，必要时可先round）
index_map = {tuple(np.round(row, 6)): idx for idx, row in enumerate(df1)}
# 遍历df2获取对应索引
df3 = np.array([index_map[tuple(np.round(row, 6))] for row in df2])

print(df3)
# 输出：[5 6 7 8 9]

注意： #

直接用tuple(row)可能因浮点数精度（如0.47实际存储为0.46999999999999997）导致匹配失败，所以建议先对浮点数做近似处理（如保留6位小数）。

内容的提问来源于stack exchange，提问作者jiaming li