Converting values of Existing Numpy ndarray to tuples

Question

Let's say I have a

numpy.ndarray

with

shape (2,3,2)

as below,

arr = np.array([[[1,3], [2,5], [1,2]],[[3,3], [6,5], [5,2]]])

I want to reshape it in such a way that:

arr.shape == (2,3)

arr == [[(1,3), (2,5), (1,2)],[(3,3), (6,5), (5,2)]]

and
each value of

arr

is a size 2

tuple

The reason I want to do this is that I want to take the minimum along axis 0 of the 3dimensional array, but I want to preserve the value that the min of the rows in paired with.

arr = np.array(

  [[[1, 4],

    [2, 1],

    [5, 2]],



   [[3, 3],

    [6, 5],

    [1, 7]]])



print(np.min(arr, axis=0))

>>> [[1,3], 

     [2,1],

     [1,2]]

>>>Should be

    [[1,4],

     [2,1],

     [1,7]]

If the array contained tuples, it would be 2 dimensional, and the comparison operator for minimize would still function correctly,
so I would get the correct result. But I haven't found any way to do this besides iterating over the arrays, which is inefficient and obvious in implementation.

Is it possible to perform this conversion efficiently in numpy?

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Answer 1

First, let's find out which pairs to take:

first_eq = arr[0,:,0] == arr[1,:,0]
which_compare = np.where(first_eq, 1, 0)[0]
winner = arr[:,:,which_compare].argmin(axis=0)

Here, first_eq is True where the first elements match, so we would need to compare the second elements. It's [False, False, False] in your example. which_compare then is [0, 0, 0] (because the first element of each pair is what we will compare). Finally, winner tells us which of the two pairs to choose along the second axis. It is [0, 0, 1].

The last step is to extract the winners:

arr[winner, np.arange(arr.shape[1])]

That is, take the winner (0 or 1) at each point along the second axis.

Converting values of Existing Numpy ndarray to tuples

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