numpy unique without sort [duplicate]

Answer 1

You can do this with the return_index parameter:

>>> import numpy as np
>>> a = [4,2,1,3,1,2,3,4]
>>> np.unique(a)
array([1, 2, 3, 4])
>>> indexes = np.unique(a, return_index=True)[1]
>>> [a[index] for index in sorted(indexes)]
[4, 2, 1, 3]

Answer 2

You could do this using numpy by doing something like this, the mergsort is stable so it'll let you pick out the first or last occurrence of each value:

def unique(array, orderby='first'):
    array = np.asarray(array)
    order = array.argsort(kind='mergesort')
    array = array[order]
    diff = array[1:] != array[:-1]
    if orderby == 'first':
        diff = np.concatenate([[True], diff])
    elif orderby == 'last':
        diff = np.concatenate([diff, [True]])
    else:
        raise ValueError
    uniq = array[diff]
    index = order[diff]
    return uniq[index.argsort()]

This answer is very similar to:

def unique(array):
    uniq, index = np.unique(array, return_index=True)
    return uniq[index.argsort()]

But, numpy.unique uses an unstable sort internally so you're not guaranteed to get any specific index, ie first or last.

I think an ordered dict might also work:

def unique(array):
    uniq = OrderedDict()
    for i in array:
         uniq[i] = 1
    return uniq.keys()