Pythonic way to convert a numpy array into another array with column indices

Question

So, what I want to do, given this input:

a=np.array([[5, 1, 10], [2, 3, 4]])

convert into another np array:

[[(5, 0), (1, 1), (10, 2)], [(2, 0), (3, 1), (4, 2)]]

What's the pythonic way to do this?

EDIT: I was using 1 indexing but 0 indexing also works. Added more context to clarify. Fixed a typo in initialization.

The column indices represent items. The values in each cell represent score for that item. What I want to do is to sort each row in decreasing order based on the score and grab the indices for further processing.

So, if I have the output as explained above, after sorting it'd look like:

[[(10, 2), (5, 0), (1, 1)], [(4,2), (3, 1), (2, 0)]]

and then I can grab the indices:

[[2, 0, 1],[2, 1, 0]]

Thanks

First, what did you try? (Perhaps in a non-pythonic way?) Second, what is the algorithm that you are trying to implement? — DYZ
– DYZ, Commented Mar 19, 2017 at 19:13
so, the column indices represent items and the values represent scores. What I want to do is sort each row using the scores and grab the column indices for further processing. — user2103008
– user2103008, Commented Mar 19, 2017 at 19:16
so, the output for the algorithm would be [[3, 1, 2],[1, 2, 3]] — user2103008
– user2103008, Commented Mar 19, 2017 at 19:17

hpaulj · Accepted Answer · 2017-03-19 19:47:02Z

First your a expression is missing []

In [231]: a=np.array([[5, 1, 10], [2, 3, 4]])  # add extra []
In [232]: a
Out[232]: 
array([[ 5,  1, 10],
       [ 2,  3,  4]])

A list comprehension is the easiest way to produce the shown list

In [233]: [[(n,i+1) for i,n in enumerate(row)] for row in a]
Out[233]: [[(5, 1), (1, 2), (10, 3)], [(2, 1), (3, 2), (4, 3)]]

I could do it by concatenating np.arange(1,4) as well, but the inner elements wouldn't be tuples. I'd have to use a structured array to get that kind of display.

This is the kind of 3d array we'd get with a concatenate:

In [234]: np.array(_)
Out[234]: 
array([[[ 5,  1],
        [ 1,  2],
        [10,  3]],

       [[ 2,  1],
        [ 3,  2],
        [ 4,  3]]])

A structured array with the same tolist() output:

In [244]: alist=[[(n,i+1) for i,n in enumerate(row)] for row in a]
In [245]: a3=np.array(alist, dtype='i,i')
In [246]: a3
Out[246]: 
array([[( 5, 1), ( 1, 2), (10, 3)],
       [( 2, 1), ( 3, 2), ( 4, 3)]], 
      dtype=[('f0', '<i4'), ('f1', '<i4')])

Building direct:

In [254]: a2=np.zeros((a.shape[0],a.shape[1],2),a.dtype)
In [255]: a2[:,:,0]=a
In [256]: a2[:,:,1]=np.arange(1,4)
In [257]: a2
Out[257]: 
array([[[ 5,  1],
        [ 1,  2],
        [10,  3]],

       [[ 2,  1],
        [ 3,  2],
        [ 4,  3]]])

or for the structured case:

In [258]: a2=np.zeros((a.shape[0],a.shape[1]),dtype='i,i')
In [259]: a2['f0']=a
In [260]: a2['f1']=np.arange(1,4)

Same construction, but with the sorted values as described in the edits:

In [281]: idx=np.argsort(-a,axis=1)
In [282]: a1 = -np.sort(-a,1)
In [283]: a2=np.zeros((a.shape[0],a.shape[1]),dtype='i,i')
In [285]: a2['f0']=a1
In [286]: a2['f1']=idx
In [287]: a2
Out[287]: 
array([[(10, 2), ( 5, 0), ( 1, 1)],
       [( 4, 2), ( 3, 1), ( 2, 0)]], 
      dtype=[('f0', '<i4'), ('f1', '<i4')])

Divakar · Accepted Answer · 2017-03-19 19:53:13Z

To get to the final output of corresponding column indices that represent the scores from corresponding rows in descending order, simply use np.argsort along each row by listing axis parameter as 1 or -1 in case you are dealing with multi-dimensional arrays and then flip the columns with [:,::-1] or [...,::-1] for multi-dimensional arrays. Alternative way again with argsort would be to use negated version of input array. This way we won't be dealing with forming the stacked tuples of indices and their scores.

Thus, the two solutions with np.argsort would be -

np.argsort(a,axis=1)[:,::-1]
np.argsort(-a,axis=1)

Sample run -

In [51]: a
Out[51]: 
array([[ 5,  1, 10],
       [ 2,  3,  4]])

In [52]: np.argsort(a,axis=1)[:,::-1]
Out[52]: 
array([[2, 0, 1],
       [2, 1, 0]])

In [53]: np.argsort(-a,axis=1)
Out[53]: 
array([[2, 0, 1],
       [2, 1, 0]])

that solves all that i was trying to do. Thanks for the simple solution. However, since the original question was about converting to column indices, i'll mark the other answer as correct to avoid confusion.

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