How to find indices of a numpy reordered array?

Question

Let's say I have a numpy sorted array:

arr = np.array([0.0, 0.0],
               [0.5, 0.0],
               [1.0, 0.0],
               [0.0, 0.5],
               [0.5, 0.5],
               [1.0, 0.5],
               [0.0, 1.0],
               [0.5, 1.0],
               [1.0, 1.0])

and suppose I do a nontrivial operation on it, so that I have a new array that is the same as the old one, but in a different order:

arr2 = np.array([0.5, 0.0],
                [0.0, 0.0],
                [0.0, 0.5],
                [1.0, 0.0],
                [0.5, 0.5],
                [1.0, 0.5],
                [0.0, 1.0],
                [1.0, 1.0],
                [0.5, 1.0])

The question arises: how to get indexes where each element arr2is placed in arr. In other words, I need a method that takes both arrays and returns an array of the same length as arr2, but with the index of the element arr. For example, the first element of the returned array will be the index of the first element arr2in arr.

where_things_are(arr2, arr) 
return : array([1, 0, 3, 2, 4, 5, 6, 8, 7])

Is there such a function that already exists in numpy?

EDIT:

I tried:

np.array([np.where((arr == x).all(axis=1)) for x in arr2])

, , - : numpy?

EDIT2:

, arr2 (, ). , , , .

Answer 1

- . , . , find_map_sorted, , , .

UPDATE: OP, , .

import numpy as np

def invperm(p):
    q = np.empty_like(p)
    q[p] = np.arange(len(p))
    return q

def find_map(arr1, arr2):
    o1 = np.argsort(arr1)
    o2 = np.argsort(arr2)
    return o2[invperm(o1)]

def find_map_2d(arr1, arr2):
    o1 = np.lexsort(arr1.T)
    o2 = np.lexsort(arr2.T)
    return o2[invperm(o1)]

def find_map_sorted(arr1, arrs=None):
    if arrs is None:
        o1 = np.lexsort(arr1.T)
        return invperm(o1)
    # make unique-able
    rdtype = np.rec.fromrecords(arrs[:1, ::-1]).dtype
    recstack = np.r_[arrs[:,::-1], arr1[:,::-1]].view(rdtype).view(np.recarray)
    uniq, inverse = np.unique(recstack, return_inverse=True)
    return inverse[len(arrs):]

x1 = np.random.permutation(100000)
x2 = np.random.permutation(100000)
print(np.all(x2[find_map(x1, x2)] == x1))

rows = np.random.random((100000, 8))
r1 = rows[x1, :]
r2 = rows[x2, :]
print(np.all(r2[find_map_2d(r1, r2)] == r1))

rs = r1[np.lexsort(r1.T), :]
print(np.all(rs[find_map_sorted(r2), :] == r2))

# lose ten elements
print(np.all(rs[find_map_sorted(r2[:-10], rs), :] == r2[:-10]))

Answer 2

numpy Broadcasting:

In [10]: ind = np.where(arr[:, None] == arr2[None, :])[1]

In [11]: ind[np.where(np.diff(ind)==0)]
Out[11]: array([1, 0, 3, 2, 4, 5, 6, 8, 7])

, , , 2, , . :

In [96]: np.where(arr[:, None] == arr2[None, :])
Out[96]: 
(array([0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3,
        3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7,
        7, 7, 8, 8, 8, 8, 8, 8]),
 array([0, 1, 1, 2, 3, 6, 0, 0, 1, 3, 4, 8, 0, 1, 3, 3, 5, 7, 1, 2, 2, 4, 5,
        6, 0, 2, 4, 4, 5, 8, 2, 3, 4, 5, 5, 7, 1, 2, 6, 6, 7, 8, 0, 4, 6, 7,
        8, 8, 3, 5, 6, 7, 7, 8]),
 array([1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1,
        0, 0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 1, 0, 0, 1, 1,
        0, 1, 0, 0, 1, 0, 1, 1]))

, diff 0.

Answer 3

:

[ np.where(np.logical_and((arr2==x)[:,1], (arr2==x)[:,0])==True)[0][0] for x in arr]

, 2D: .

arr2 = np.array([[0.5, 0.0],
[0.0, 0.0],
[0.0, 0.5],
[1.0, 0.0],
[0.5, 0.5],
[1.0, 0.5],
[0.0, 1.0],
[1.0, 1.0],
[0.5, 1.0]])

Answer 4

numpy_indexed package ( : ) ; npi.indices - ndarray- list.index.

import numpy_indexed as npi
idx = npi.indices(arr, arr2)

This returns a list of indexes for which arr [idx] == arr2. If arr2 contains elements that are not in arr, a ValueError is raised; but you can control this with the “missing” kwarg.

To answer your question whether this function is included in numpy; yes, in the sense that numpy is a complete ecosystem. But in fact, if you consider the number of lines of code necessary for its effective, correct and general nature.