numpy.array with elements of different shapes

Question

I want to have a numpy array of two another arrays (each of them has different shape). As I know, for this reason one must use: dtype = object in the definition of the main array.

For example, let us define (in Python 2.7) our arrays as

     a0 = np.arange(2*2).reshape(2,2)
     a1 = np.arange(3*3*2).reshape(3,3,2)
     b = np.array([a0,a1], dtype = object)

This works perfect: b[1] is the same as a1. But if I change the dimension in a0 from (2,2) to (3,3) something strange happens:

     a0 = np.arange(3*3).reshape(3,3)
     a1 = np.arange(3*3*2).reshape(3,3,2)
     b = np.array([a0,a1], dtype = object)

This time b[1] and a1 are not equal, they even have different shapes. What is the reason of this strange behavior?

Perhaps there is a completely different solution for me. But I don't want to use lists or tuples because I want to allow addition such as b + b. It is clear that I can write my own class for this purpose but is there any simpler way?

mdurant · Accepted Answer · 2014-10-20 14:18:57Z

9

If you explicitly want an objects array, you can create an empty array with type object first and assign to it:

x = empty(5, dtype=object)
x[0] = zeros((3,3))
x[1] = zeros((3,2)) #does not merge axes.
x[2] = eye(4)
x[3] = ones((2,2))*2
x[4] = arange(10).reshape((5,2))

>>> x+x
array([array([[ 0.,  0.,  0.],
   [ 0.,  0.,  0.],
   [ 0.,  0.,  0.]]),
   array([[ 0.,  0.],
   [ 0.,  0.],
   [ 0.,  0.]]),
   array([[ 2.,  0.,  0.,  0.],
   [ 0.,  2.,  0.,  0.],
   [ 0.,  0.,  2.,  0.],
   [ 0.,  0.,  0.,  2.]]),
   array([[ 4.,  4.],
   [ 4.,  4.]]),
   array([[ 0,  2],
   [ 4,  6],
   [ 8, 10],
   [12, 14],
   [16, 18]])], dtype=object)

You will have to fill all elements before you can perform arithmetic, or grow the element from size zero using np.append.

edited Oct 20, 2014 at 14:18

answered Oct 19, 2014 at 17:33

mdurant

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2 Comments

cheyp Over a year ago

In your example it is not possible to perform arithmetic for the whole array x. For instance, x+x won't work.

mdurant Over a year ago

Yes you can! You just have to fill all the elements 0..4, as I said (but only gave example for the first two).

tktk · Accepted Answer · 2014-10-19 17:29:57Z

3

I don't think it's a strange behavior, it's the way you use numpy that's weird to me.

numpy merges dimension as much as it can. That's the default behavior and is what expected when working with arrays. In the first example, all the dimensions of a0 and a1 are different. numpy is forced to use only the first dimension. So if we look at b.shape in the first example, we'll see (2,).

On the second example, a0 and a1 has the same dimension size all the way to the last dimension. So numpy merges those levels. If you'd look at b.shape here, you'll see (2,3,3), since the second and third dimension are of the same size.

answered Oct 19, 2014 at 17:29

tktk

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