python - What's the difference between numpy.take and numpy.choose? -

it seems numpy.take(array, indices) , numpy.choose(indices, array) return same thing: subset of array indexed indices.

are there subtle differences between two, or missing more important? , there reason prefer 1 on other?

numpy.take(array, indices) , numpy.choose(indices, array) behave on 1-d arrays, coincidence. pointed out jonrsharpe, behave differently on higher-dimensional arrays.

numpy.take

numpy.take(array, indices) picks out elements flattened version of array. (the resulting elements of course not same row.)

for example,

numpy.take([[1, 2], [3, 4]], [0, 3]) 

returns

array([1, 4]) 

numpy.choose

numpy.choose(indices, set_of_arrays) plucks out element 0 array indices[0], element 1 array indices[1], element 2 array indices[2], , on. (here, array set of arrays.)

for example

numpy.choose([0, 1, 0, 0], [[1, 2, 3, 4], [4, 5, 6, 7]]) 

returns

array([1, 5, 3, 4]) 

because element 0 comes array 0, element 1 comes array 1, element 2 comes array 0, , element 3 comes array 0.

more information

these descriptions simplified – full descriptions can found here: numpy.take, numpy.choose. example, numpy.take , numpy.choose behave when indices , array 1-d because numpy.choose first broadcasts array.