Addition of multiple arrays in python

Question

I have a number of arrays that I wish to broadcast into a single array using addition, which I know can be simply done such that:

a = numpy.array([1,2,3])
b = numpy.array9[4,5,6])
sum = a + b
print(sum)

[5,7,9]

However, I can't hardcode it like in this simple example as I will run my script numerous times with a different number of inputs and so will have a different number of arrays each time. Sometimes, I may have a and b, but other times I may have a, c and d but not b etc.

Using a loop I therefore append the arrays I do have into a list, so that I end up with something like:

newlist = [array([1,2,3,...5,4,3]), 
          array([5,7,2,...4,6,7]),
          array([3,6,2,...4,5,9])]

What would be the most pythonic way of getting a single array from the arrays in 'newlist' which is the addition of the arrays within it, such that (from newlist):

sum = [8,15,7,...14,15,19]

The arrays are all the same shape.

Answer 1

Stick with Numpy array and use its sum() method:

>>> arr = np.array([[1,2,3,5,4,3], 
          [5,7,2,4,6,7],
          [3,6,2,4,5,9]])
>>> arr.sum(axis=0)
array([ 9, 15,  7, 13, 15, 19])

Of course you can do it with Python lists as well but it is going to be slow:

>>> lst = [[1,2,3,5,4,3], 
          [5,7,2,4,6,7],
          [3,6,2,4,5,9]]
>>> map(sum, zip(*lst))
[9, 15, 7, 13, 15, 19]

Answer 2

There is no need to create a 2D array from your pre-existing 1D arrays. It will certainly not be faster than adding them together, e.g. using reduce with np.add:

In [14]: a = [np.random.rand(10) for _ in range(10)]

In [15]: %timeit np.array(a).sum(axis=0)
100000 loops, best of 3: 10.7 us per loop

In [16]: %timeit reduce(np.add, a)
100000 loops, best of 3: 5.24 us per loop

For larger arrays, it is even less advantageous:

In [17]: a = [np.random.rand(1000) for _ in range(1000)]

In [18]: %timeit np.array(a).sum(axis=0)
100 loops, best of 3: 6.26 ms per loop

In [19]: %timeit reduce(np.add, a)
100 loops, best of 3: 2.43 ms per loop

And of course:

In [20]: np.allclose(np.array(a).sum(axis=0), reduce(np.add, a))
Out[20]: True