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Suppose I have an array of integers x and I want to do the following:

  1. get an array unique_x of the unique values of x
  2. build an array y in which y[i] is the index in unique_x of the value x[i].

I managed to do this as follows:

import numpy as np

unique_x = np.unique(x)
y = np.zeros_like(x)
for k, value in unique_x:
   indices, = np.nonzero(x == value)
   y[indices] = k

My question is: is there a way to do this using only numpy builtin functions and some slicing? I have a feeling that this loop won't be as fast as it could be if this was done with numpy builtins.

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If np.unique() does what you want it to (returns only the first occurence of each element; it doesn't return only those elements which occur once), you can do use the return_index argument:

In [1]: x = array([1, 1, 2, 3, 4, 4, 5, 6, 2, 1])
In [2]: unique_x, unique_indexes = np.unique(x, return_index=True)
In [3]: unique_x
Out[3]: array([1, 2, 3, 4, 5, 6])
In [4]: unique_indexes
Out[4]: array([0, 2, 3, 4, 6, 7])

(x does not need to be sorted, but unique_x will be). If you want the indicies necessary to reconstruct x from unique_x, on the other hand, you can use the return_inverse argument, as pointed out by @xtofl:

In [5]: unique_x, unique_inverse = np.unique(x, return_inverse=True)
In [6]: unique_inverse
Out[6]: array([0, 0, 1, 2, 3, 3, 4, 5, 1, 0])
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3 Comments

does x have to be sorted for this to work? (would it work with e.g. [1, 2, 1])
... I believe what OP wants is return_inverse=True.
I've edited my answer with some more information for the OP - thanks.

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