> Although length-1 axes already being in the data sounds worrying: you have an array that doesn't depend on an axis, but there's an axis anyway to show where i

> Although length-1 axes already being in the data sounds worrying: you have an array that doesn't depend on an axis, but there's an axis anyway to show where it would go if it did? Example: A / A.sum(axis=3, keepdims=True) Make all of the vectors along axis=3 sum up to one. There are other ways of doing it, but this way seems fairly clear to me. The shape of the denominator is the same as the numerator, except for a 1 in position 3. Unfortunately we have to specify `keepdims`, because the default of `False` removes the dimension being summed over, which doesn't work in general. `keepdims=True` is the behavior in Matlab/Octave, so the example becomes A ./ sum(A, 4) with 4=3+1 because Matlab is 1-based like Fortran.