Fastest way to add values of one numpy array to a 2D numpy array at particular index

Question

I have a 2D numpy array of zeros, a list of numpy arrays (which can be of different lengths) and a list of indices. I want to add the contents of each of the arrays in the list from the start of the corresponding row indice in the 2D array.

Of course, I can just loop over the arrays. However, I need to perform this operations for many different samples. Therefore, I was wondering whether anybody is aware of a more efficient way to do this.

In [1]: A = np.zeros((5, 5))
   ...: arrays = [np.random.randint(1, 10, size=(1,5)) for i in range(3)]
   ...: indices = [1, 3, 4]
   ...: print(arrays)
Out[1]:
[array([3, 1, 3, 6]), array([4, 9]), array([5, 9, 6])]

Expected output:

array([[0., 0., 0., 0., 0.], 
       [3., 1., 3., 6., 0.], 
       [0., 0., 0., 0., 0.], 
       [4., 9., 0., 0., 0.], 
       [5., 9., 6., 0., 0.]]

Any help would be much appreciated!

Typically how many samples do you have? And typically what's the length of such samples? — Divakar
– Divakar, Commented Aug 28, 2019 at 8:33
The shape of A is roughly (10, 200) and I have approximately 700,000 indices+arrays combinations. The shape of A is constant across all samples. In my current application I loop over all samples and put them in one big array of shape (700000, 10, 200). — Hilde
– Hilde, Commented Aug 28, 2019 at 13:05
You are starting with 2 lists. One of those lists contains arrays that vary in size; so it can't be turned into a 2d array. How else do you deal with lists other than to zip and iterate? — hpaulj
– hpaulj, Commented Aug 28, 2019 at 18:44
If you search on the question of padding a list of arrays, you'll find an advanced method by @Divakar that uses an hstack(array) and [len(i) for i in arrays] to map the arrays onto A in one step. Your indices complicates things, though it might be easier to use indices to remap A after padding. — hpaulj
– hpaulj, Commented Aug 28, 2019 at 18:50

Chris · Accepted Answer · 2019-08-28 08:36:17Z

1

Using zip:

for i, a in zip(indices, arrays):
    A[i, :len(a)] = a

Output:

array([[0., 0., 0., 0., 0.],
       [3., 1., 3., 6., 0.],
       [0., 0., 0., 0., 0.],
       [4., 9., 0., 0., 0.],
       [5., 9., 6., 0., 0.]])

answered Aug 28, 2019 at 8:36

Chris

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1 Comment

Hilde Over a year ago

This approach what I meant with "looping over the arrays". Do you think this is the fastest approach possible?

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