(Python) Find indices of values ​​in two arrays that are equal to values ​​in two other arrays

Question

I have the following 4 arrays, and I want to get indexes of values ​​that are equal in arrays A and X, corresponding to values ​​at the same position in B and Y. So, for the next example

    import numpy as np
    A = np.asarray([400.5, 100,  700,   200,  15, 900])
    B = np.asarray([500.5, 200,  500, 600.5,   8, 999])
    X = np.asarray([400.5, 700,  100,   300,  15, 555, 900])
    Y = np.asarray([500.5, 500,600.5,   100,   8, 555, 999])

I want to get two arrays with indexes:

indAB = [0 2 4 5]

• 0, because 400.5 and 500.5 in & B are also in X & Y at position 0
• 2, because 700 and 500 in B & B are also in X & Y in position 2
• 4, because 15 and 8 in A & B are also in X & Y at position 4
• 5, because 900 and 999 in A&B are also in X & Y at position 5

indXY = [0 1 4 6]

• 0, 1, 4, and 6 are found similarly to indAB, but instead of X & Y instead.

indAB - B, X Y, indXY - X Y, B.

:

    def indices(a,b):
        setb = set(b)
        ind = [i for i, x in enumerate(a) if x in setb]
        return ind

    iA = np.asarray(indices(A,X))
    iB = np.asarray(indices(X,A))
    iX = np.asarray(indices(B,Y))
    iY = np.asarray(indices(Y,B))

    def CommonIndices(a,b):
        return np.asarray(list(set(a) & set(b)))

    indAB = CommonIndices(iA,iX)
    indXY = CommonIndices(iB,iY)

    print(indAB) # returns = [0 2 4 5]
    print(indXY) # returns = [0 1 2 4 6]

[0 1 2 4 6] indXY, . 2 , 600.5 Y B, 200 100 B () .

, - . !

Answer 1

numpy_indexed ( : - ) . , - NlogN. , , :

import numpy as np
import numpy_indexed as npi

A = np.asarray([400.5, 100,  700,   200,  15, 900])
B = np.asarray([500.5, 200,  500, 600.5,   8, 999])
X = np.asarray([400.5, 700,  100,   300,  15, 555, 900])
Y = np.asarray([500.5, 500,600.5,   100,   8, 555, 999])

AB = np.stack([A, B], axis=-1)
XY = np.stack([X, Y], axis=-1)

# casting the AB and XY arrays to npi.index first is not required, but a performance optimization; without this each call to npi.indices would have to re-index the arrays, which is the expensive part
AB = npi.as_index(AB)
XY = npi.as_index(XY)
# npi.indices(list, items) is a vectorized nd-equivalent of list.index(item)
indAB = npi.indices(AB, XY, missing='mask').compressed()
indXY = npi.indices(XY, AB, missing='mask').compressed()

, . , npi.intersection(XY, AB); , .

Answer 2

:

import numpy as np

A = np.asarray([400.5, 100,  700,   200,  15, 900])
B = np.asarray([500.5, 200,  500, 600.5,   8, 999])
X = np.asarray([400.5, 700,  100,   300,  15, 555, 900])
Y = np.asarray([500.5, 500,600.5,   100,   8, 555, 999])

AB = np.stack([A, B], axis=-1)
XY = np.stack([X, Y], axis=-1)

eq = AB[:, np.newaxis, :] == XY[np.newaxis, :, :]
eq = np.logical_and.reduce(eq, axis=-1)

indAB, = np.where(np.logical_or.reduce(eq, axis=1))
indXY, = np.where(np.logical_or.reduce(eq, axis=0))

print("indAB", indAB)
print("indXY", indXY)

:

indAB [0 2 4 5]
indXY [0 1 4 6]

AB XY - A B X Y, , "" . eq AB XY; np.newaxis AB XY ( , AB 1 XY 0). == . np.logical_and.reduce , "" ( A X B Y), np.logical_or.reduce , - AB XY XY AB. , np.where .

, len(A) x len(X) x 2, , , .

, . " ", ( - ). " ", , :

import numpy as np

MAX_SIZE = 2  # Biggest array will be MAX_SIZE x MAX_SIZE x 2

A = np.asarray([400.5, 100,  700,   200,  15, 900])
B = np.asarray([500.5, 200,  500, 600.5,   8, 999])
X = np.asarray([400.5, 700,  100,   300,  15, 555, 900])
Y = np.asarray([500.5, 500,600.5,   100,   8, 555, 999])

AB = np.stack([A, B], axis=-1)
XY = np.stack([X, Y], axis=-1)

maskAB = np.full(len(AB), False, dtype=bool)
maskXY = np.full(len(XY), False, dtype=bool)

for iAB in range(0, len(AB), MAX_SIZE):
    pAB = np.expand_dims(AB[iAB:iAB + MAX_SIZE], axis=1)
    for iXY in range(0, len(XY), MAX_SIZE):
        pXY = np.expand_dims(XY[iXY:iXY + MAX_SIZE], axis=0)
        eq = pAB == pXY
        eq = np.logical_and.reduce(eq, axis=-1)
        maskAB[iAB:iAB + MAX_SIZE] |= np.logical_or.reduce(eq, axis=1)
        maskXY[iXY:iXY + MAX_SIZE] |= np.logical_or.reduce(eq, axis=0)

indAB, = np.where(maskAB)
indXY, = np.where(maskXY)

print("indAB", indAB)
print("indXY", indXY)

:

indAB [0 2 4 5]
indXY [0 1 4 6]

MAX_SIZE 2, , , , (, MAX_SIZE = 10000 ). MAX_SIZE , .

Answer 3

Here is an alternative method. I dare say that this is relatively clear, it should be effective through the use of sets, and this requires only memory O( len(A) + len(X) ).

numpy not even required, but can be used for arrays.

from collections import defaultdict

A = [400.5, 100, 700, 200, 15, 900]
B = [500.5, 200, 500, 600.5, 8, 999]
X = [400.5, 700, 100, 300, 15, 555, 900]
Y = [500.5, 500, 600.5, 100, 8, 555, 999]

def get_indices(values):
    d = defaultdict(set)
    for i, value in enumerate(values):
        d[value].add(i)
    return d

iA, iB, iX, iY = [get_indices(values) for values in [A, B, X, Y]]
print(iA)
# {400.5: {0}, 100: {1}, 200: {3}, 900: {5}, 700: {2}, 15: {4}}
print(iX)
# {400.5: {0}, 100: {2}, 300: {3}, 900: {6}, 555: {5}, 700: {1}, 15: {4}}

for i, (a, b) in enumerate(zip(A, B)):
    common_indices = iX[a] & iY[b]
    if common_indices:
        print("A B : %d" % i)
        print("X Y : %d" % common_indices.pop())
        print()

#   A B : 0
#   X Y : 0

#   A B : 2
#   X Y : 1

#   A B : 4
#   X Y : 4

#   A B : 5
#   X Y : 6