5

Given the following data array:

d=np.array([10,11,12,13,14])

and another indexing array:

i=np.array([0, 2, 3, 6])

What is a good way of indexing d with i (d[i]) so that instead of index out of bounds error for 6, I would get: 

np.array([10, 12, 13])
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3 Answers 3

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4

Maybe use i[i < d.size]] to get the elements that are less than the length of d:

print(d[i[i < d.size]])
[10 12 13]
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5 Comments

Directly using the mask would also work: d[i<d.size].
@Divakarm thanks, that is a bit nicer but that would give 10,11,12 no or am I misunderstanding something?
@Divakar, no worries, d.size is still nicer than len
I like that, nice and clean.
@And, it is the best I have to offer, there may well be better ways but I don't know them!
4

It is easy to cleanup i before using it:

In [150]: d[i[i<d.shape[0]]]
Out[150]: array([10, 12, 13])

np.take has several modes for dealing with out of bounds indices, but 'ignore' is not one of them.

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1 Comment

Thanks for the hints about take. I found it after posting the question, but indeed there is no "ignore". Padraic Cunningham was first with the main idea though :)
0 
valid_indices = i[(i >= 0) & (i < len(d))]
selected_items = d[valid_indices]

NumPy doesn't seem to provide an error-handling mode that skips invalid indices, probably because skipping them doesn't make much sense when the result array is multidimensional. Instead, just select the elements of i in the valid range and index with those elements.

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