我们如何根据其项目为元组的行的掩码来过滤列？

Let's assume that we have a numpy array which is built like this:

我们假设我们有一个像这样构建的numpy数组:

import numpy as np

data = [(1, 2), (1, 3), (1, 4), (1, 5), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6)]
data1 = [1 for i in data]

table = np.asarray(list(zip(data, data1, data1, data1, data1))).transpose()

which results in:

这导致:

[[(1, 2) (1, 3) (1, 4) (1, 5) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6)]
 [1 1 1 1 1 1 1 1 1 1]
 [1 1 1 1 1 1 1 1 1 1]
 [1 1 1 1 1 1 1 1 1 1]
 [1 1 1 1 1 1 1 1 1 1]]

Now there is another list test = [(1, 2), (1, 3), (1, 4)].
I want to filter the columns in the table if the tuples in first row do not match the tuples in test list.
I want it to result like this:

现在有另一个列表test = [(1,2),(1,3),(1,4)]。如果第一行中的元组与测试列表中的元组不匹配,我想过滤表中的列。我希望它结果如下:

[[(1, 2) (1, 3) (1, 4)]
 [1 1 1]
 [1 1 1]
 [1 1 1]
 [1 1 1]]

I tried this code:

我试过这段代码:

mask = np.in1d(table[0, :], test)
table = table[:, mask]
print(table)

but it resulted an empty list.
Any suggestions? Thank you

但它导致了一个空列表。有什么建议?谢谢

2 个解决方案

#1

#use a bool array to select columns
table[:,np.array([e in test for e in table[0]])]
Out[306]: 
array([[(1, 2), (1, 3), (1, 4)],
       [1, 1, 1],
       [1, 1, 1],
       [1, 1, 1],
       [1, 1, 1]], dtype=object)

#2

Not quite as elegant, but I find this more intuitive:

不太优雅,但我发现这更直观:

tb = set(table[0,:].reshape(table[0,:].size))
table[:,[i for i, t in enumerate(tb) if tuple(t) in test]]