Access array elements by array/list of column indexes

Question

I have 2D array of integers (e.g. A with A.shape (10,5)) and 1D array of column indexes (which generally differ among each other) (e.g. idx with idx.shape (10,)). From i-th row, I would like to acquire the element from array A with column index idx[i]. What would be the best (fastest) solution if the desired output is 1D array of acquired elements or list of those elements?

A = np.arange(50).reshape(10,5)
idx=np.array([2,4,0,0,3,1,3,1,2,4])

Desirable output:

output = [2,9,10,15,23,26,33,36,42,49]

or

output = np.array([2,9,10,15,23,26,33,36,42,49])

can you writte some code you have with your structure?

Bestasttung
– Bestasttung

2015-01-21 09:23:04 +00:00
Commented Jan 21, 2015 at 9:23 — Bestasttung
– Bestasttung, Commented Jan 21, 2015 at 9:23

Jon Clements · Accepted Answer · 2015-01-21 10:05:21Z

5

Using numpy you can take the diagonal of a column index, eg:

np.diag(A[:,idx])
# array([ 2,  9, 10, 15, 23, 26, 33, 36, 42, 49])

answered Jan 21, 2015 at 10:05

Jon Clements

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NDevox · Accepted Answer · 2015-01-21 09:50:28Z

2

You can use enumerate to access the row number you are on, and then use that to access the index you need on idx like so:

output = []

for row in enumerate(A):
    output.append(row[1][idx[row[0]]])

From this I got output == [2, 9, 10, 15, 23, 26, 33, 36, 42, 49]

edited Jan 21, 2015 at 9:50

answered Jan 21, 2015 at 9:44

NDevox

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hpaulj · Accepted Answer · 2015-01-21 20:21:48Z

2

A[np.arange(A.shape[0]), idx]

np.diag(A[:,idx]) works, but does more work than necessary. A[:,idx] is a (10,10)array (in this example), which is then whittled down to a(10,)`.

For this small array, mine is 2x faster. For a (100,50) it is 16x faster.

answered Jan 21, 2015 at 20:21

hpaulj

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