Efficient way to filter a binary string in python

Question

I have every long binary string that I would like to filter by a pattern. Here is a working example:

x = b"\x00\x01\x02\x03\x04\x00\x01\x02\x03\x04"
x[1]
y = [x[i] for i in range(len(x)) if not ((i%5 == 4) or (i%5 == 3))]
bytes(y)

It works fine, but I am wondering if there is a better method, performance-wise. I use python 3.4, if that matters.

Pierre · Accepted Answer · 2014-12-23 09:56:48Z

1

You should probably create a generator instead of a list:

bytes(x[i] for i in range(len(x)) if (i % 5) not in [3, 4])

and not:

bytes([x[i] for i in range(len(x)) if (i % 5) not in [3, 4]]) # DON'T DO THIS

This will save you memory (and experience show it will also be a bit faster). I cannot think of another (more efficient) method.

answered Dec 23, 2014 at 0:34

Pierre

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