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up vote 2 down vote favorite

I have an array that maps the cells in a game of life. So far I have an int for every single cell. To minimize the amount of reads I want to compress the array to whole 8 cells per 1 integer.

The problem arises when I want to create a bitmap, which expects an array of 0 and 1 like in the previous case.

Is there any quick way in numpy to convert this? I could you a loop for this but there might be a better way.

Example:

[01011100b, 0x0] -> [0, 1, 0, 1, 1, 1, 0 ,0, 0,0,0,0,0,0,0,0]
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I'm not sure how you have 01011100b stored. Can you get the string representation? If so, list() converts a string into a list of its characters. Is that helpful? –  leekaiinthesky yesterday
    
It's an int32, I wrote it in binary to clearly show what I had in mind. Your solution might work though along with hsplit. Wonder if there is a better way without using to string –  Bartlomiej Lewandowski yesterday
    
If you don't go the string route, you'll have to do each place manually, something like this (though using 0b10000000 etc. instead of 0x...): stackoverflow.com/questions/2562308/integer-to-byte-conversion –  leekaiinthesky yesterday
1  
numpy.packbits seems relevant. –  user2357112 yesterday
    
@user2357112 I think unpackbits will solve my problem, you can add this as an answer and I'll accept this in a bit –  Bartlomiej Lewandowski yesterday

1 Answer 1

up vote 2 down vote accepted

numpy.packbits and numpy.unpackbits are a pretty close match for what you're looking for:

>>> numpy.packbits([1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1])
array([225, 184], dtype=uint8)
>>> numpy.unpackbits(array([225, 184], dtype=numpy.uint8))
array([1, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0], dtype=uint8)

Note that numpy.unpackbits(numpy.packbits(x)) won't have the same length as x if len(x) wasn't a multiple of 8; the end will be padded with zeros.

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