How to add a 2D subarray in an existing 2D array using numpy?

Question

I am having problem to add a subarray to an existing 2D array. I am actually new to numpy and python and coming from MATLAB this was something trivial to do. Note that usually a is a big matrix in my problems.

import numpy as np

a = np.array(arange(16)).reshape(4,4) # The initial array
b = np.array(arange(4)).reshape(2,2) # The subarray to add to the initial array
ind = [0,3] # The rows and columns to add the 2x2 subarray 

a[ind][:,ind] += b #Doesn't work although does not give an error

I saw looking around that the following can work

a[:4:3,:4:3] += b

but how can I define ind beforehand? In addition how to define ind if it consists of more than two numbers that cannot be expressed with a stride? for example ind = [1, 15, 34, 67]

Just to explain why a[ind][:,ind] += b doesn't work, it's because a[ind] (where ind is a sequence of coordinates) makes a copy. Therefore, it does work, but the += is applied to the copy, which is not saved. a winds up being unchanged. Basically, it's because you have use "fancy" indexing twice. @DSM's answer combines it into one "fancy" indexing expression, so += works as expected. Another way to write it is a[ind[:,None], ind[None,:]] += b, but that requires ind to be a numpy array instead of a list. — Joe Kington
– Joe Kington, Commented Jan 9, 2014 at 17:11

DSM · Accepted Answer · 2014-01-09 16:44:03Z

4

One way to handle the general case would be to use np.ix_:

>>> a = np.zeros((4,4))
>>> b = np.arange(4).reshape(2,2)+1
>>> ind = [0,3]
>>> np.ix_(ind, ind)
(array([[0],
       [3]]), array([[0, 3]]))
>>> a[np.ix_(ind, ind)] += b
>>> a
array([[ 1.,  0.,  0.,  2.],
       [ 0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.],
       [ 3.,  0.,  0.,  4.]])

answered Jan 9, 2014 at 16:44

DSM

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