deleting rows of a numpy array based on uniqueness of a value

Question

let's say I have a bi-dimensional array like that

numpy.array([[0,1,1.2,3],[1,5,3.2,4],[3,4,2.8,4], [2,6,2.3,5]])

I want to have an array formed eliminating whole rows based on uniqueness of values of last column, selecting the row to keep based on value of third column. e.g. in this case i would like to keep only one of the rows with 4 as last column, and choose the one which has the minor value of third column, having something like that as a result:

array [0,1,1.2,3],[3,4,2.8,4],[2,6,2.3,5]

thus eliminating row [1,5,3.2,4]

which would be the best way to do it?

Answer 1

My numpy is way out of practice, but this should work:

#keepers is a dictionary of type int: (int, int)
#the key is the row's final value, and the tuple is (row index, row[2])
keepers = {}
deletions = []
for i, row in enumerate(n):
    key = row[3]
    if key not in keepers:
        keepers[key] = (i, row[2])
    else:
        if row[2] > keepers[key][1]:
            deletions.append(i)
        else:
            deletions.append(keepers[key][0])
            keepers[key] = (i, row[2])
o = numpy.delete(n, deletions, axis=0)

I've greatly simplified it from my declarative solution, which was getting quite unwieldy. Hopefully this is easier to follow; all we do is maintain a dictionary of values that we want to keep and a list of indexes we want to delete.

Answer 2

thank you very much indeed! I think this was really far more complex than my current ability could handle, it's a very smart snippet of code. I think i've understood the logic, and it seems to be rigth with the example array i had provided, but it fails with other arrays, for example:

n = numpy.array([[1,1,1.2,3],[1,5,3.2,3],[3,4,2.8,3],[2,6,2.3,5]])

or even

n = numpy.array([[1.,1.2,3.,2.],[5.,3.2,4.,3.],[4.,2.8,7.,6.],[6.,2.3,5.,3.]])

i really can't see why...