There is a number of questions out here concerning sorting arrays but I was not able to find the one solving my problem (which makes me hope that I'm not thinking way to complicated). So here we go:
I have two arrays with equal values but in different order:
a = np.array([4, 2, 5, 6])
# b = np.random.permutation(a) # general case
b = np.array([5, 2, 6, 4])
I need the indices that sort b to a, so here:
ind = np.array([3, 1, 0, 2])
Note that I do not want to change a.
One approach with broadcasting & np.where -
_,C = np.where(a[:,None] == b)
Sample run -
In [210]: a = np.array([4, 2, 5, 6])
In [211]: b = np.array([5, 2, 6, 4])
In [212]: _,C = np.where(a[:,None] == b)
In [213]: C
Out[213]: array([3, 1, 0, 2], dtype=int64)
a and b are large, this builds a boolean array of len(a) x len(b) which might be excessive depending on the use-case.np.argsort is its own inverse. If y = x[x.argsort()], then x = y[x.argsort().argsort()]. Specifically, you convert a into a sorted array with a.argsort(), and convert from a sorted array to b with b.argsort().argsort(). You can therefore write:
b = a[a.argsort()][b.argsort().argsort()]
In other words:
ind = a.argsort()[b.argsort().argsort()]
b = a[a.argsort()[b.argsort().argsort()]] (note the closing bracket) as is clear from the second snippet.
In this case, a is already sorted. So, all you need are the sorting indices for b.
However, in case a was not sorted, you can do
a = array([2, 4, 5, 6])
b = array([5, 2, 6, 4])
b.argsort()[a.argsort()] # array([1, 3, 0, 2])
This uses the argsort method to return the indices needed to sort b and a.
Another example:
a = array([4, 2, 5, 6])
b = array([5, 2, 6, 4])
b.argsort()[a.argsort()] # array([3, 1, 0, 2])
a = np.array([4, 4, 2, 5, 6]) and b = np.array([5, 2, 4, 6, 4]).argsort.b = a[a.argsort()[b.argsort().argsort()]] as given by @MadPhysicist below (but correcting the misplaced bracket).