numpy unique without sort [duplicate]

Question

How can I use numpy unique without sorting the result but just in the order they appear in the sequence? Something like this?

a = [4,2,1,3,1,2,3,4]

np.unique(a) = [4,2,1,3]

rather than

np.unique(a) = [1,2,3,4]

Use naive solution should be fine to write a simple function. But as I need to do this multiple times, are there any fast and neat way to do this?

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()