Block replace coded numpy 2d array

Question

I have a numpy array that is built with coded sections. The sections come in 2x2 blocks. I have a large dictionary of what those 2x2 blocks should be replaced with. How do I replace those 2x2 codes with values in the dictionary.

info_dict = {
    5: np.array([[1, 0], [1, 0]], "int"),
    6: np.array([[1, 0], [0, 1]], "int"),
    7: np.array([[0, 1], [1, 0]], "int"),
    8: np.array([[1, 1], [0, 0]], "int"),
}
print(np.array([[5, 5, 8, 8], [5, 5, 8, 8], [6, 6, 7, 7], [6, 6, 7, 7]]))
print(np.array([[1, 0, 1, 1], [1, 0, 0, 0], [1, 0, 0, 1], [0, 1, 1, 0]]))

# before (coded)
[[5 5 8 8]
 [5 5 8 8]
 [6 6 7 7]
 [6 6 7 7]]
# after (final matrix)
[[1 0 1 1]
 [1 0 0 0]
 [1 0 0 1]
 [0 1 1 0]]

For reference

#5
[[1 0]
 [1 0]]
#6
[[1 0]
 [0 1]]
#7
[[0 1]
 [1 0]]
#8
[[1 1]
 [0 0]]

Where is the data coming from. Are you open to changing the format? — Mad Physicist
– Mad Physicist, Commented Dec 21, 2021 at 23:48

bb1 · Accepted Answer · 2021-12-21 23:34:45Z

1

One way to do it:

import numpy as np

info_dict = {
    5: np.array([[1, 0], [1, 0]], "int"),
    6: np.array([[1, 0], [0, 1]], "int"),
    7: np.array([[0, 1], [1, 0]], "int"),
    8: np.array([[1, 1], [0, 0]], "int"),
}

a = np.array([[5, 5, 8, 8], [5, 5, 8, 8], [6, 6, 7, 7], [6, 6, 7, 7]])
np.block([[info_dict[b] for b in r] for r in a[::2, ::2]])

It gives:

[[1 0 1 1]
 [1 0 0 0]
 [1 0 0 1]
 [0 1 1 0]]

answered Dec 21, 2021 at 23:34

bb1

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Mad Physicist · Accepted Answer · 2021-12-21 23:51:19Z

0

Assuming your dictionary has a small number of integers in it, and your squares are N to a side, you info_dict as follows:

mapping = np.zeros((max(info_dict) + 1, N, N), dtype=int)
for channel, value in info_dict.items():
    mapping[channel] = value

If you can store the dictionary in this format to begin with, that's even better. Now you can use simple index with some rearrangement to get the result:

after = mapping[before[::N, ::N]].transpose(0, 2, 1, 3).reshape(before.shape)

edited Dec 21, 2021 at 23:51

answered Dec 21, 2021 at 23:45

Mad Physicist

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