Feature request: Set replace flag with tuple for nd array

To be able to vectorise random sampling with/without replacement is would be useful to be able to set the replace flag in random.choice to a tuple matching the shape of size. i.e.

# THIS IS NOT A WORKING CODE
from numpy import np

x = [i for i in range(100)]
k = np.random.poission(0.5, 10)
samples = np.random.choice(x, size=(10, 10), replace=(False, True), p=k)

This would allow for a much neater way to produce a 2D array where each row is a sample from the same population taken without replacement.

[eric-wieser]

I’m not a huge fan of the proposed spelling, but agreed that the feature itself makes sense.

[bashtage]

If it really needed to go into choice I would suggest something like:

choice(a, size=None, replace=True, p=None, axis=0, repeat=None)

Parameters
----------
...
repeat : {int, tuple[int], None}
    Independently repeat the random sampling. If size is not None, the output dimension is
    repeat + size where integer repeat and size values have been transformed into tuples
    (e.g., (size,).) [Further clarification needed if a is not 1d.] 

With this API, your call would be

rng.choice(x, size=10, p=k, repeat=10)

and each row would be an independent sample.

You could also do things like

rng.choice(x, size=(2,2), p=k, repeat=(100,100))

which would produce an output that is (100, 100, 2, 2) which is 10,000 independent samples of 2 by 2 choice.