Element-wise minimum of multiple vectors in numpy

Question

I know that in numpy I can compute the element-wise minimum of two vectors with

numpy.minimum(v1, v2)

What if I have a list of vectors of equal dimension, V = [v1, v2, v3, v4] (but a list, not an array)? Taking numpy.minimum(*V) doesn't work. What's the preferred thing to do instead?

Answer 1

Convert to NumPy array and perform ndarray.min along the first axis -

np.asarray(V).min(0)

Or simply use np.amin as under the hoods, it will convert the input to an array before finding the minimum along that axis -

np.amin(V,axis=0)

Sample run -

In [52]: v1 = [2,5]

In [53]: v2 = [4,5]

In [54]: v3 = [4,4]

In [55]: v4 = [1,4]

In [56]: V = [v1, v2, v3, v4]

In [57]: np.asarray(V).min(0)
Out[57]: array([1, 4])

In [58]: np.amin(V,axis=0)
Out[58]: array([1, 4])

If you need to final output as a list, append the output with .tolist().

Answer 2

*V works if V has only 2 arrays. np.minimum is a ufunc and takes 2 arguments.

As a ufunc it has a .reduce method, so it can apply repeated to a list inputs.

In [321]: np.minimum.reduce([np.arange(3), np.arange(2,-1,-1), np.ones((3,))])
Out[321]: array([ 0.,  1.,  0.])

I suspect the np.min approach is faster, but that could depend on the array and list size.

In [323]: np.array([np.arange(3), np.arange(2,-1,-1), np.ones((3,))]).min(axis=0)
Out[323]: array([ 0.,  1.,  0.])

The ufunc also has an accumulate which can show us the results of each stage of the reduction. Here's it's not to interesting, but I could tweak the inputs to change that.

In [325]: np.minimum.accumulate([np.arange(3), np.arange(2,-1,-1), np.ones((3,))])
     ...: 
Out[325]: 
array([[ 0.,  1.,  2.],
       [ 0.,  1.,  0.],
       [ 0.,  1.,  0.]])