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I have a list of boolean strings. Each string is of length 6. I need to get the complement of each string. E.g, if the string is "111111", then "000000" is expected. My idea is 

bin(~int(s,2))[-6:]
  1. convert it to integer and negate it by treating it as a binary number
  2. convert it back to a binary string and use the last 6 characters.

I think it is correct but it is not readable. And it only works for strings of length less than 30. Is there a better and general way to complement a boolean string?

I googled a 3rd party package "bitstring". However, it is too much for my code.

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  • 1
    Python is not C, What problem are you trying to solve using this approach? Commented Feb 13, 2014 at 10:28
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    "correct but it is not readable" What does that mean O_ó ? The code is hard to understand? Commented Feb 13, 2014 at 10:31
  • @FredrikPihl I am using geng (pallini.di.uniroma1.it), a subroutine of nauty to generate graphs. The format of the output is a string of characters and each character encodes 6 binary bits. Commented Feb 13, 2014 at 10:31
  • This link could help: stackoverflow.com/questions/13492826/ones-complement-python Commented Feb 13, 2014 at 10:32
  • @luk32 Without the explanation of context, I find it confusing myself. Anyway, I should just treat this "complementing" as "replacing" as the two answers suggest. Commented Feb 13, 2014 at 10:36

2 Answers 2

Reset to default
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Well, you basically have a string in which you want to change all the 1s to 0s and vice versa. I think I would forget about the Boolean meaning of the strings and just use maketrans to make a translation table:

from string import maketrans complement_tt = maketrans('01', '10') s = '001001' s = s.translate(complement_tt) # It's now '110110'
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1 Comment

Your idea is super clean!
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Replace in three steps:

>>> s = "111111" >>> s.replace("1", "x").replace("0", "1").replace("x", "0") '000000'
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