Changing binary characters in string

Question

I would like to convert this list:

a = [['0001', '0101'], ['1100', '0011']]

to:

a' = [['1110', '1010'],['0011','1100']]

In the second example, every character is changed to its opposite (i.e. '1' is changed to '0' and '0' is changed to '1'). The code I have tried is:

for i in a:
    for j in i:
        s=list(j)
        for k in s:
            position = s.index(k)
            if k=='0':
                s[position] = '1'
            elif k=='1':
                s[position] = '0'
        ''.join(s)

But it doen't work properly. What can I do? Thanks

Inbar Rose · Accepted Answer · 2013-11-11 13:55:55Z

2

You can use a function that flips the bits like this:

from string import maketrans

flip_table = maketrans('01', '10')
def flip(s):
    return s.translate(flip_table)

Then just call it on each item in the list like this:

>>> flip('1100')
'0011'

answered Nov 11, 2013 at 13:55

Inbar Rose

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Holy Mackerel · Accepted Answer · 2013-11-11 14:38:24Z

0

[["".join([str(int(not int(t))) for t in x]) for x in d] for d in a]

Example:

>>> a = [['0001', '0101'], ['1100', '0011']]
>>> a_ = [["".join([str(int(not int(t))) for t in x]) for x in d] for d in a]
>>> a_
[['1110', '1010'], ['0011', '1100']]

edited Nov 11, 2013 at 14:38

answered Nov 11, 2013 at 14:33

Holy Mackerel

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Comments

aIKid · Accepted Answer · 2013-11-11 14:59:44Z

0

Using a simple list comprehension:

[[k.translate({48:'1', 49:'0'}) for k in i] for i in a]

48 is the code for "0", and 49 is the code for "1".

Demo:

>>> a = [['0001', '0101'], ['1100', '0011']]
>>> [[k.translate({48:'1', 49:'0'}) for k in i] for i in a]
[['1110', '1010'], ['0011', '1100']]

For Python 2.x:

from string import translate, maketrans
[[translate(k, maketrans('10', '01')) for k in i] for i in a]

edited Nov 11, 2013 at 14:59

answered Nov 11, 2013 at 14:00

aIKid

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eyquem · Accepted Answer · 2013-11-11 15:08:44Z

0

from ast import literal_eval
import re

a = [['0001', '0101'], ['1100', '0011']]

print literal_eval(re.sub('[01]',lambda m: '0' if m.group()=='1' else '1',str(a)))

literal_eval() is said to be safer than eval()

answered Nov 11, 2013 at 15:08

eyquem

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