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A simple Burrows-Wheeler transform function in python
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#! /usr/bin/env python | |
""" | |
A simple Burrows-Wheeler transform function in python. | |
Algorithm presented in: | |
Burrows M, Wheeler DJ: A Block Sorting Lossless Data Compression Algorithm. | |
Technical Report 124. Palo Alto, CA: Digital Equipment Corporation; 1994. | |
USAGE: bwt.py [-h] [-i INDEX] STRING | |
""" | |
import argparse | |
def bw_transform(s): | |
n = len(s) | |
m = sorted([s[i:n]+s[0:i] for i in range(n)]) | |
I = m.index(s) | |
L = ''.join([q[-1] for q in m]) | |
return (I, L) | |
from operator import itemgetter | |
def bw_restore(I, L): | |
n = len(L) | |
X = sorted([(i, x) for i, x in enumerate(L)], key=itemgetter(1)) | |
T = [None for i in range(n)] | |
for i, y in enumerate(X): | |
j, _ = y | |
T[j] = i | |
Tx = [I] | |
for i in range(1, n): | |
Tx.append(T[Tx[i-1]]) | |
S = [L[i] for i in Tx] | |
S.reverse() | |
return ''.join(S) | |
if __name__ == "__main__": | |
parser = argparse.ArgumentParser( | |
description='Transforms strings using Burrows-Wheeler transform') | |
parser.add_argument('-i', '--index', metavar="INDEX", help='Performs ' + | |
'inverse BWT. A transform index must be specified as an INDEX value.') | |
parser.add_argument('STRING', type=str, help='Transform this string.') | |
args = parser.parse_args() | |
if args.index: | |
print(bw_restore(args.index, args.STRING)) | |
else: | |
print("%d %s" % bw_transform(args.STRING)) |
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Hello, nice and clean code. But, 'index' call is expensive, It would be better to get rid of it, like this: