Numpy: set one specific element of each column based on indexing by array

Question

On a smaller scale compared to what I need, here's an example of what I'm looking to do:

>>> a
array([[  21,   22,   23,   24,   25,   26,   27],
       [  56,   57,   58,   59,   60,   61,   62],
       [  14,   15,   16,   17,   18,   19,   20],
       [   7,    8,    9, 1010,   11,   12,   13],
       [  42,   43,   44,   45,   46,   47,   48],
       [  63,   64,   65,   66,   67,   68,   69],
       [   0,    1,    2,    3,    4,    5,    6],
       [  49,   50,   51,   52,   53,   54,   55],
       [  28,   29,   30,   31,   32,   33,   34],
       [  35,   36,   37,   38,   39,   40,   41]])
>>> indices = a.argmax(axis=0)
>>> indices
array([5, 5, 5, 3, 5, 5, 5])
>>> b = np.zeros(a.shape)
>>> b[indices] = 1.0
>>> b # below is the actual output, not what I want
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  1.,  1.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  1.,  1.,  1.,  1.,  1.,  1.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.]])

But what I actually need is:

>>> b
array([[ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1.,  1.,  1.,  0.,  1.,  1.,  1.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.]])

Numpy indexing can get extremely complicated and it's a little difficult to put the above into words, so hopefully someone can understand what I'm looking for. Essentially it's to set a 1 wherever there's a max of a column and zero elsewhere. How would I go about doing this?

Andrey Sobolev · Accepted Answer · 2015-04-24 11:47:21Z

7

From the docs:

If the number of objects in the selection tuple is less than N , then : is assumed for any subsequent dimensions.

In your selection there only one array, so you get every row from indices to be equal to 1. To overcome that, you need column indices. I guess this will do the trick:

b[indices, np.arange(a.shape[1])] = 1.0

edited Apr 24, 2015 at 11:47

answered Apr 23, 2015 at 7:18

Andrey Sobolev

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3 Comments

Julien Spronck Over a year ago

nice :-) i knew there was a one-liner somewhere

mjois Over a year ago

Aha, thanks! I feel like I understand numpy indexing a little better now.

rayryeng Over a year ago

Thanks for this. I come from MATLAB, so doing 0:a.shape[1] I thought would actually produce the column indices but this performs slicing. In MATLAB, the : operator does both slicing and generate a vector of indices so that was where my trouble came from. Now I know that I have to manually produce the column indices. Thank you!

Julien Spronck · Accepted Answer · 2015-04-23 07:18:57Z

0

There might be a way without a loop but you can always do:

indices = a.argmax(axis=0)
b = np.zeros(a.shape)
for j,ind in enumerate(indices):
    b[ind, j] = 1

answered Apr 23, 2015 at 7:18

Julien Spronck

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