Let’s say I have the following numpy array, of 1’s and 0’s exclusively:
import numpy as np example = np.array([0,1,1,0,1,0,0,1,1], dtype=np.uint8)
I want to group all elements into chunks of 3, and replace the chunks by a single value, based on a condition. Let’s say I want [0,1,1] to become 5, and [0,1,0] to become 10. Thus the desired output would be:
[5,10,5]
All possible combinations of 1’s and 0’s in a chunk have a corresponding unique value that should replace the chunk. What’s the fastest way to do this?
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Answer
As the other answers indicated, you can start by reshaping your array (actually, you should probably generate it with the correct shape to begin with, but that’s another issue):
example = np.array([0, 1, 1, 0, 1, 0, 0, 1, 1], dtype=np.uint8) data = example.reshape(-1, 3)
Now running a custom python function over the array is going to be slow, but luckily numpy has your back here. You can use np.packbits to transform each row into a number directly:
data = np.packbits(data, axis=1, bitorder='little').ravel() # [6, 2, 6]
If you wanted 101 to map to 5 and 110 to map to 6, your job is done. Otherwise, you will need to come up with a mapping. Since you have three bits, you only need 8 numbers in the mapping array:
mapping = np.array([7, 4, 3, 8, 124, 1, 5, 0])
You can use data as an index directly into mapping. The output will have the type of mapping but the shape of data:
result = mapping[data] # [5, 3, 5]
You can do this in one line:
mapping[np.packbits(example.reshape(-1, 3), axis=1, bitorder='little').ravel()]