# [Numpy-discussion] nicest way to apply an arbitrary sequence of row deltas to an array

Neil neilcrighton@gmail....
Mon Dec 21 03:35:08 CST 2009

```George Dahl <george.dahl <at> gmail.com> writes:

>
> Hi everyone,
> I was wondering if anyone had insight on the best way to solve the
> following problem.
>
> Suppose I have a numpy array called U.
> U has shape (N,M)
> Suppose further that I have another array called dU and that
> dU has shape (P,M) and that P has no particular relationship to N, it
> could be larger or smaller.
> Additionally, I have a one dimensional array called idx and
> idx has shape (P,).  Furthermore idx holds integers that are all valid
> indices into the rows of U.  idx almost certainly contains some
> duplicates.
>
> I want,
>
> for k in range(P):
>    U[idx[k],:] += dU[k,:]
>
> Is there a nice vectorized and efficient way to do this without making
> the obvious python for loop I included above?  For what I am doing, P
> is usually quite large.  I am most interested in a clever use of
> numpy/scipy commands and Python and not Cython.
>

I'm also interested to see if there are any answers to this; I came across a
similar problem recently. It would have been convenient to do something like
U[idx] += dU, but this didn't work because there were repeated indices in idx.
Here's a short example that shows the problem:

In [1]: U = np.array([1., 2., 3., 4.])
In [2]: dU = np.array([0.1, 0.1, 0.1, 0.1])
In [3]: idx = np.array([0, 1, 2, 0])
In [4]: U
Out[4]: array([ 1.,  2.,  3.,  4.])
In [5]: U[idx]
Out[5]: array([ 1.,  2.,  3.,  1.])
In [6]: U[idx] += dU
In [7]: U
Out[7]: array([ 1.1,  2.1,  3.1,  4. ])

Ideally U would end up as array([ 1.2,  2.1,  3.1,  4. ])

Neil

```