Select elements of numpy array via mask and preserving original dimensions

Question

Hello I have following data

ids = np.concatenate([1.0 * np.ones(shape=(4, 9,)), 
                      2.0 * np.ones(shape=(4, 3,))], axis=1)

logits = np.random.normal(size=(4, 9 + 3, 256))

Now I want to get numpy array only of ids that have 1.0 and I want to get array of size (4,9, 256)

I tried logits[ids == 1.0, :] but I get (36, 256) How I can make slicing without connecting first two dimensions ?

Current dimensions are only example ones and I am looking for generic solution.

What you are trying to do is not possible as it is. But you can fill the value where conditon doesn't get satisfied with something else such that it doesn't affect your further processing. Refer stackoverflow.com/questions/29046162/… — ggaurav
– ggaurav, Commented Jan 17, 2021 at 16:14
In your example, you are only masking the 2nd dimension. So you could reduce the mask to a 1d array: logits[:,(ids==1.0).all(axis=0),:]. It's up to you to supply the added information about the distribution of True, whether that means reshape after, or modifying the mask before hand. — hpaulj
– hpaulj, Commented Jan 17, 2021 at 17:16

jakevdp · Accepted Answer · 2021-01-17 16:15:25Z

1

Your question appears to assume that each row has the same number of nonzero entries; in that case you can solve your problem generally like this:

mask = (ids == 1)
num_per_row = mask.sum(1)

# same number of entries per row is required
assert np.all(num_per_row == num_per_row[0])  

result = logits[mask].reshape(logits.shape[0], num_per_row[0], logits.shape[2])

print(result.shape)
# (4, 9, 256)

answered Jan 17, 2021 at 16:15

jakevdp

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