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I tried to find in stackoverflow a thread answering this question, but I could not find. Thus, if it is duplicate, please provide the link.
The use case is very common:
I have two arrays: X which contains two dimensional datapoints and y which contains labels either 0 or 1.
X has shape (307, 2)
y has shape (307, 1)
I want to have all rows in X where the corresponding row in y has value of 1.
I tried the following code:
X[y==1]
But it raises the following error: 

IndexError: boolean index did not match indexed array along dimension 1; dimension is 2 but corresponding boolean dimension is 1

How can I do that?

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  • You could try X[y, :] Commented Aug 21, 2019 at 12:17
  • @MadPhysicist This gives a totally different array -> shape = (307, 1, 2). This is not what I am looking for. Just the rows where the corresponding rows in y have a value of 1 -> shape = (9, 2) Commented Aug 21, 2019 at 12:27
  • @MadPhysicist And y is not an array of boolean values as described in the question. Thus to mask, you have to write a condition which then results in the IndexError I also mentioned in the question and have found the reason which is stated in the answer to this question Commented Aug 21, 2019 at 12:44
  • X[y.ravel().astype(np.bool), :] Commented Aug 21, 2019 at 12:57

1 Answer 1

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I have found the following way:

X[np.where(np.any(y==1, axis=1))]

I also found out that the reason for the above error is that y has two dimensions. The following code will work, too, and uses masking which has a better performance:

y = y.reshape(-1)
X[y==1,:]
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1 Comment

np.any(y==1, axis=1) is equivalent to y.ravel(). The call to where makes this fancy indexing, which is much less efficient than just using the mask.

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