I have an numpy matrix, for example:
a = np.array([[1, 2], [3, 4]])
Then, I want to use another numpy matrix with indexes to extend the matrix, for example:
idxes = np.array(([0, 1, 0], [1, 0, 1]]) # the indexes matrix
The operation is like:
result = a[:, idxes] # this is an wrong expression
The result I expected is:
>>result
array([[1, 2, 1],
[4, 3, 4]])
I want to know how to do that.
You'll need to specify the a range for the first (0th) axis.
a[np.arange(len(a))[:,None], idxes]
This intuitively follows the indexing operation, the first row of idxes will index into the first row of a, the second row of idxes will index the second row of a, and so on.
Additionally, the dimensions of arange array need to be expanded from 1D to 2D because idxes is also a 2D array.
A fun way diagonal + take, diagonal here make sure you always slice the row index equal to the columns index items , which will return the just like row-wise out put
np.diagonal(np.take(a, idxes,1)).T
array([[1, 2, 1],
[4, 3, 4]])
Or
np.diagonal(a[:,idxes]).T
array([[1, 2, 1],
[4, 3, 4]])
a[:,idxes] contains many rows that you won't need as well.
a[:, idxes] is a very huge matrix, and the run time is very long. So, it is not a perfect way to do this.
a[np.arange(len(a))[:,None], idxes], I am refraining from posting an answer in case this is a dupe.