How to create a numpy _object_ array of other numpy arrays of same and different length?

Question

This is my very first question. So let's see if I can explain exactly what I need.

I am given a python list of numpy arrays which can or cannot have different lengths (in one dimension only but this is not important here), e.g.

my_list = [
    np.ones((20, 3, 3)),
    np.ones(( 1, 3, 3)),
    np.ones((20, 3, 3))
]

Now when I do

wrapped_list = np.array(my_list)

I get an object of the following structure

np.array(shape=(3, ), dtype=object)

with the initial three arrays as content. This is what I want. Now the problem:

If my_list contained sublists of identical length, then I get, e.g.

my_list2 = [
    np.ones((20, 3, 3)),
    np.ones((20, 3, 3)),
    np.ones((20, 3, 3))
]

np.array(my_list2)

leads to

np.array(shape=(3, 20, 3, 3), dtype=np.float64)

This is not, what I want. I tried specifying the dtype, like

np.array(my_list, dtype=object)

which will cast all (sub-)arrays to dtype=object.

I think I found a way to go without wrapping at all, but I am curious on how to set the dtype on a np.array without affecting nested numpy arrays.

Christian K. · Accepted Answer · 2016-02-11 14:19:55Z

2

Create an empty object arrary first and fill it with my_list, e.g.:

wrapped_list = np.empty((3,),dtype=object)
wrapped_list[:] = my_list2

answered Feb 11, 2016 at 14:19

Christian K.

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2 Comments

Yes, this does what I want although I tried to put this into a one-liner in a lambda function actually. Is there a way to do that, too?

The right way to avoid long code is to wrap it in function definition. Don't try to use lambda to combine lines.

Eric · Accepted Answer · 2016-09-15 15:42:08Z

0

Another (dumb) way, that at least has the virtue of being a single expression:

wrapped_list = np.array(my_list + [None])[:-1]

answered Sep 15, 2016 at 15:42

Eric

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