Help us collect modern Python NumPy code solutions

Re:Help us collect modern Python NumPy code solutions by PartiallyTyped on Monday March 28, 2022 @09:53AM (#30830475)

That does not zip though. In fact, it creates a 1-dimensional array of objects because the arrays do not have the same number of items.

     >>> a = [np.arange(i) for i in range(1, 11)]

     >>> np.array(a).T

        array([array([0]), array([0, 1]), array([0, 1, 2]), array([0, 1, 2, 3]),

        array([0, 1, 2, 3, 4]), array([0, 1, 2, 3, 4, 5]),

        array([0, 1, 2, 3, 4, 5, 6]), array([0, 1, 2, 3, 4, 5, 6, 7]),

        array([0, 1, 2, 3, 4, 5, 6, 7, 8]),

        array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])], dtype=object)

In fact, if we are to zip them, it should yield:

>>> list(zip(*a))

[(0, 0, 0, 0, 0, 0, 0, 0, 0, 0)]

zip_arrays otoh does give the same result.

>>> zip_arrays(a)

array([[0, 0, 0, 0, 0, 0, 0, 0, 0, 0]])

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Re:Help us collect modern Python NumPy code solutions by johndough on Monday March 28, 2022 @10:32AM (#30830883)

You are completely right. Nevertheless, when I zip over arrays of different length, it is usually a bug instead of being intentional.

A code-golfed version for different array lengths:

    np.c_[[a[:min(map(len,arrays))].T for a in arrays]].T

The trick to get a local variable within a list comprehension statement with

    for l in [foo()]

from your zip_array function is very cool by the way, but I sacrificed it in favor of shorter code.

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Re:Help us collect modern Python NumPy code solutions by PartiallyTyped on Monday March 28, 2022 @01:15PM (#30832905)

I like your your solution, I think using `a[0:min(map(len,arrays))]` (ie with 0) is faster because it returns a slice and not a copy of the array.

I am glad you liked the trick!

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