Array indexing between two arrays

Question

I have a numpy array and corresponding lookup values. From every column of data, I should get maximum position of the corresponding lookup an convert that position into result of data.

I have to do as shown in the picture.

Picture speaks better than my language.

import numpy as np

data = np.array([[0, 2, 1, 4, 3, 2, 4, 4, 1],
                 [1, 1, 2, 0, 3, 4, 4, 2, 1],
                 [2, 2, 1, 4, 4, 1, 4, 4, 4]] )

print (data)

print ()

lookup = np.array([[60, 90, 90, 60, 50],
                  [90, 90, 80, 90, 90],
                  [60, 40, 90, 60, 50]])

print (lookup)

print ()

I did as follows:

data = data.transpose(1,0)

lookup = lookup.transpose(1,0)

results = []       
for d in data:   
   matches = lookup[d].diagonal()
   print (matches)
   idx = np.argmax(matches, axis=0)
   res = d[idx]
   results.append(res)

print ()

print (results)

It worked. But, any better way of doing it?

Can you describe what you are trying to do, in words not in pictures. — Dan
– Dan, Commented Jan 13, 2020 at 17:37

Vicrobot · Accepted Answer · 2020-01-13 18:16:46Z

1

You can use flattening fact:

l = lookup.shape[1]
data1 = (data + [[0],[l], [2*l]]).copy()
matches = lookup.flatten()[data1.flatten()].reshape(data1.shape).T
results = data.T[:, np.argmax(matches, axis = 1)].diagonal()
print(matches)
print(results)

Output

[[60 90 90]
 [90 90 90]
 [90 80 40]
 [50 90 50]
 [60 90 50]
 [90 90 40]
 [50 90 50]
 [50 80 50]
 [90 90 50]]  # matches

[1 2 1 0 3 2 4 2 1]  #results

edited Jan 13, 2020 at 18:16

answered Jan 13, 2020 at 17:51

Vicrobot

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3 Comments

mk1 Over a year ago

you got matches. then, how to get results?

mk1 Over a year ago

[1, 2, 1, 0, 3, 2, 4, 2, 1]

mk1 Over a year ago

getting argmax of the matches you produced and convert into data values. expected result is [1, 2, 1, 0, 3, 2, 4, 2, 1]

Dan · Accepted Answer · 2020-01-13 17:47:46Z

0

I'm not sure I understand you, but maybe this is what you're looking for?

data = data.T
lookup = lookup.T
cols = np.zeros_like(data) + np.array([0, 1, 2])
idx = np.argmax(lookup[data, cols], axis=1)
data[np.arange(data.shape[0]), idx]

answered Jan 13, 2020 at 17:47

Dan

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3 Comments

mk1 Over a year ago

why + np.array([0, 1, 2])?

Dan Over a year ago

just a lazy way to make something like [[0, 1, 2], [0, 1, 2], [0, 1, 2],...]] i.e. the colums for each row. Maybe a better route would be to use np.ndgrid.

Dan Over a year ago

lookup[data, cols] is your matches

Maxpxt · Accepted Answer · 2020-01-13 18:17:30Z

Use numpy's fancy indexing. With it, you can provide arrays of indexes. See example:

data = np.arange(20)
data[[0, 1, 5, 10]]    # array([ 0,  1,  5, 10])

The arrays used for the indexing also broadcast. See example:

data = np.arange(20).reshape(5, 4)
i = np.array([0, 4])
j = np.array([0, 3])
data[i, j]                                 # array([ 0, 19])
i, j = i[:, np.newaxis], j[np.newaxis, :]
data[i, j]                                 # array([[ 0,  3], [16, 19]])

In your case, matches is such that matches[i, j] = lookup[i, data[i, j]]. Using fancy indexing, we do:

i = np.arange(data.shape[-2])[:, np.newaxis]
matches = lookup[i, data]

And the final result is supposed to be result[j] = data[i_max[j], j], with i_max[j] = argmax_i(matches[i, j]), which translates to:

i_max = np.argmax(matches, axis=-2)
j = np.arange(data.shape[-1])
result = data[i_max, j]

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