Given a self-indexing (not sure if this is the correct term) numpy array, for example:
a = np.array([3, 2, 0, 1])
This represents this permutation (=> is an arrow):
0 => 3
1 => 2
2 => 0
3 => 1
I'm trying to make an array representing the inverse transformation without doing it "manually" in python, that is, I want a pure numpy solution. The result I want in the above case is:
array([2, 3, 1, 0])
Which is equivalent to
0 <= 3 0 => 2
1 <= 2 or 1 => 3
2 <= 0 2 => 1
3 <= 1 3 => 0
It seems so simple, but I just can't think of how to do it. I've tried googling, but haven't found anything relevant.
The inverse of a permutation p of np.arange(n) is the array of indices s that sort p, i.e.
p[s] == np.arange(n)
must be all true. Such an s is exactly what np.argsort returns:
>>> p = np.array([3, 2, 0, 1])
>>> np.argsort(p)
array([2, 3, 1, 0])
>>> p[np.argsort(p)]
array([0, 1, 2, 3])