numpy 如何找到相同的行从一个2D数组到另一个2D数组的位置？

w6mmgewl 于 2023-10-19 发布在其他

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import numpy as np

# Create two sample dataframes
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]
])

df3 = np.where(np.isclose(df1[:, np.newaxis], df2))[0]

print(df3)

我想在df1中找到df2的位置，正确答案是[5, 6, 7, 8, 9]，但python输出是[0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3 4 4 4 4 4 4 4 4 4 4 5 5 5 5 5 5 5 5 5 5 5 6 6 6 6 6 6 6 6 6 6 6 7 7 7 7 7 7 7 7 7 7 7 8 8 8 8 8 8 8 8 8 8 8 9 9 9 9 9 9 9 9 9 9 9]，这没有意义。我怎样才能改变代码找到df2在df1中的正确位置？

numpy

来源：https://stackoverflow.com/questions/77152147/how-to-find-the-position-of-same-rows-from-one-2d-array-to-another-2d-array

2条答案

按热度按时间

7gyucuyw1#

如果你只想匹配非零值，你可以删除它们：

np.where(np.isclose(np.where(df1!=0, df1, np.nan)[:, None], df2))[0]

输出量：

array([5, 6, 7, 8, 9])

如果您想要完全匹配，包括零（每行），您可以使用：用途：

out = np.where(np.isclose(df1[:, None], df2).all(2))[0]

输出：array([5, 6, 7, 8, 9])

赞(0）回复(0）举报 2023-10-19

yvt65v4c2#

一种完全匹配的方法：

np.argwhere(np.isin(df1, df2).all(axis=1))

array([[0],[5],[6],[7],[8],[9]], dtype=int64)

赞(0）回复(0）举报 2023-10-19

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numpy 如何找到相同的行从一个2D数组到另一个2D数组的位置？

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