使用 numpy 对数组进行排序以适合另一个数组

Question

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.

Answer 1

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)

Answer 2

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])

Answer 3

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()]