Question

Is it possible to perform min/max in-place assignment with NumPy multi-dimensional arrays without an extra copy?

Say, a and b are two 2D numpy arrays and I would like to have a[i,j] = min(a[i,j], b[i,j]) for all i and j.

One way to do this is:

a = numpy.minimum(a, b)

But according to the documentation, numpy.minimum creates and returns a new array:

numpy.minimum(x1, x2[, out])
Element-wise minimum of array elements.
Compare two arrays and returns a new array containing the element-wise minima.

So in the code above, it will create a new temporary array (min of a and b), then assign it to a and dispose it, right?

Is there any way to do something like a.min_with(b) so that the min-result is assigned back to a in-place?

Answer 1

numpy.minimum() takes an optional third argument, which is the output array. You can specify a there to have it modified in place:

In [9]: a = np.array([[1, 2, 3], [2, 2, 2], [3, 2, 1]])

In [10]: b = np.array([[3, 2, 1], [1, 2, 1], [1, 2, 1]])

In [11]: np.minimum(a, b, a)
Out[11]: 
array([[1, 2, 1],
       [1, 2, 1],
       [1, 2, 1]])

In [12]: a
Out[12]: 
array([[1, 2, 1],
       [1, 2, 1],
       [1, 2, 1]])