Numpy: Add rows from one matrix to another by index

Question

In short I want to index into a matrix and add to each row.

In this example the first row (indexed by the 0) should get [1,1,1] added to it. Then the second row (indexed by the 1) should get [2, 2, 2] added to it. Finally the first row (indexed by the third 0) should get [3, 3, 3] added to it.

>>> a = np.array([np.array([1,2,3]), np.array([4,5,6])])
>>> a
array([[1, 2, 3],
       [4, 5, 6]])
>>> a[np.array([0,1,0]), :] += np.array([np.array([1,1,1]), np.array([2,2,2]), np.array([3,3,3])])

Desired:

>>> a
array([[5, 6, 7],
       [6, 7, 8]])

Actual:

>>> a
array([[4, 5, 6],
       [6, 7, 8]])

Edit 2:

As per comments below the solution runs slowly. From a portion of the code where I'm just adding 0 to test the speed:

print y.shape
print dW.shape
np.add.at(dW, (y, slice(None)), 0)

Yields:

(49000,)
(10, 3073)

And takes about 21 seconds. Without the np.add.at line the rest of the code takes about 1 second. y.npy dW.npy

could you verbally explain what your addition should mean? In fact, you're trying to replace row 0 twice in your assignment, so I think it's a bit hard to understand. — Marcus Müller
– Marcus Müller, Commented Jan 19, 2015 at 16:53
Document this well in your code. Otherwise it's a lurking bug/feature. — hpaulj
– hpaulj, Commented Jan 19, 2015 at 18:39

gg349 · Accepted Answer · 2015-01-19 17:19:01Z

7

This is a known problem of numpy, explained well here:

For example, a[[0,0]] += 1 will only increment the first element once because of buffering, whereas add.at(a, [0,0], 1) will increment the first element twice.

numpy solves the problem using add.at(). Example:

a = array([1,2,3])
add.at(a,[0,0],4) # now a = array([9, 2, 3])

In this case we want this to work for a multidimensional array:

a = np.array([np.array([1,2,3]), np.array([4,5,6])])
np.add.at(a,([0,1,0],slice(None)),array([[1,1,1],[2,2,2],[3,3,3]]))

The result is:

array([[5, 6, 7], [6, 7, 8]])

I guess you mistyped a 7 for a 6.

answered Jan 19, 2015 at 17:19

gg349

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6 Comments

ibash Over a year ago

Thanks! You got me unstuck!

ibash Over a year ago

Is there anyway to do this in a vectorized way?

gg349 Over a year ago

this is the vectorized way

ibash Over a year ago

Hmm, I'm using this in an implementation and it's 5x slower than just using loops. It might be the slice(None) -- I'm still getting familiar with numpy, is that necessary?

gg349 Over a year ago

This is interesting. Yes it could be because the operation is on a 2D array. And yes I think you really need the slice. Please post an example so that we can work on it.

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