Numpy where functionality for Julia code?

Question

I read the answer to "What is Julia equivalent of numpy's where function?", but do not yet see how the answer (ifelse) gives the user all the functionality of numpy.where. I have posted example code below:

    A = [0.0 0.9 0.0 0.99 0.0]

    a = 1:length(A)

    #-v- produces [0 1.0 0 1.0 0] as expected, but how to get the index values?
    b = ifelse.(A .- 1.0 .> -1.0, 1.0, 0 )
    #-^- how to get the array [0.9 0.99]? How to remove all zeros from an array?

Any workarounds other than using for loops would be appreciated.

There's no point to using `where` for this even in NumPy. NumPy supports logical indexing, just like Julia. — user2357112, Jun 10 '20 at 22:17
Does `numpy.where` really support what you are asking for? I doesn't look like it from the docs. Can you show a numpy example? — DNF, Jun 11 '20 at 07:34

score 9 · Answer 1 · answered Jun 10 '20 at 22:15

9

I guess you're looking for the functionality of np.where(cond)? That's simply findall(A .> 0).

To get the array [0.9, 0.99], I'd use logical indexing: A[A .> 0].

answered Jun 10 '20 at 22:15

mbauman

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Frames Catherine White · Accepted Answer · 2020-06-12T12:26:30.587

5

Potentially avoiding allocating the masking array would be faster, so filter(x -> x>0, A)

edited Jun 12 '20 at 12:26

answered Jun 10 '20 at 22:18

Frames Catherine White

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Numpy where functionality for Julia code?

2 Answers2