python - NumPy selecting specific column index per row by using a list of indexes

Question

I'm struggling to select the specific columns per row of a NumPy matrix.

Suppose I have the following matrix which I would call X:

[1, 2, 3]
[4, 5, 6]
[7, 8, 9]

I also have a list of column indexes per every row which I would call Y:

[1, 0, 2]

I need to get the values:

[2]
[4]
[9]

Instead of a list with indexes Y, I can also produce a matrix with the same shape as X where every column is a bool / int in the range 0-1 value, indicating whether this is the required column.

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

I know this can be done with iterating over the array and selecting the column values I need. However, this will be executed frequently on big arrays of data and that's why it has to run as fast as it can.

I was thus wondering if there is a better solution?

Answer 1

If you've got a boolean array you can do direct selection based on that like so:

>>> a = np.array([True, True, True, False, False])
>>> b = np.array([1,2,3,4,5])
>>> b[a]
array([1, 2, 3])

To go along with your initial example you could do the following:

>>> a = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> b = np.array([[False,True,False],[True,False,False],[False,False,True]])
>>> a[b]
array([2, 4, 9])

You can also add in an arange and do direct selection on that, though depending on how you're generating your boolean array and what your code looks like YMMV.

>>> a = np.array([[1,2,3], [4,5,6], [7,8,9]])
>>> a[np.arange(len(a)), [1,0,2]]
array([2, 4, 9])

Hope that helps, let me know if you've got any more questions.

Answer 2

You can do something like this:

In [7]: a = np.array([[1, 2, 3],
   ...: [4, 5, 6],
   ...: [7, 8, 9]])

In [8]: lst = [1, 0, 2]

In [9]: a[np.arange(len(a)), lst]
Out[9]: array([2, 4, 9])

More on indexing multi-dimensional arrays: http://docs.scipy.org/doc/numpy/user/basics.indexing.html#indexing-multi-dimensional-arrays

Answer 3

Recent numpy versions have added a take_along_axis (and put_along_axis) that does this indexing cleanly.

In [101]: a = np.arange(1,10).reshape(3,3)                                                             
In [102]: b = np.array([1,0,2])                                                                        
In [103]: np.take_along_axis(a, b[:,None], axis=1)                                                     
Out[103]: 
array([[2],
       [4],
       [9]])

It operates in the same way as:

In [104]: a[np.arange(3), b]                                                                           
Out[104]: array([2, 4, 9])

but with different axis handling. It's especially aimed at applying the results of argsort and argmax.

Answer 4

A simple way might look like:

In [1]: a = np.array([[1, 2, 3],
   ...: [4, 5, 6],
   ...: [7, 8, 9]])

In [2]: y = [1, 0, 2]  #list of indices we want to select from matrix 'a'

range(a.shape[0]) will return array([0, 1, 2])

In [3]: a[range(a.shape[0]), y] #we're selecting y indices from every row
Out[3]: array([2, 4, 9])

Answer 5

You can do it by using iterator. Like this:

np.fromiter((row[index] for row, index in zip(X, Y)), dtype=int)

Time:

N = 1000
X = np.zeros(shape=(N, N))
Y = np.arange(N)

#@Aशwini चhaudhary
%timeit X[np.arange(len(X)), Y]
10000 loops, best of 3: 30.7 us per loop

#mine
%timeit np.fromiter((row[index] for row, index in zip(X, Y)), dtype=int)
1000 loops, best of 3: 1.15 ms per loop

#mine
%timeit np.diag(X.T[Y])
10 loops, best of 3: 20.8 ms per loop

Answer 6

Another clever way is to first transpose the array and index it thereafter. Finally, take the diagonal, its always the right answer.

X = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9], [10, 11, 12]])
Y = np.array([1, 0, 2, 2])

np.diag(X.T[Y])

Step by step:

Original arrays:

>>> X
array([[ 1,  2,  3],
       [ 4,  5,  6],
       [ 7,  8,  9],
       [10, 11, 12]])

>>> Y
array([1, 0, 2, 2])

Transpose to make it possible to index it right.

>>> X.T
array([[ 1,  4,  7, 10],
       [ 2,  5,  8, 11],
       [ 3,  6,  9, 12]])

Get rows in the Y order.

>>> X.T[Y]
array([[ 2,  5,  8, 11],
       [ 1,  4,  7, 10],
       [ 3,  6,  9, 12],
       [ 3,  6,  9, 12]])

The diagonal should now become clear.

>>> np.diag(X.T[Y])
array([ 2,  4,  9, 12]