Incrementing a binary number in python

Question

I wonder, how can I increment a string representing a binary number, all the way up to another binary number? (for use in a while loop, for example).

For example I start with "0000" and after 15 incs, I should reach "1111" (in other words: "0000", "0001", "0010", ..., "1111"). At first this problem seemed really simple, but the only solutions I could come up with were quite ridiculous (not pythonic, some might say). Does anyone have an advice?

Thanks in advance!

Is there a reason why you want to start out with strings, and not with integers that you convert to strings for display purposes? — Levon
– Levon, Commented May 20, 2012 at 11:34
@Mark Byers I have no plausible explanation for that. Facepalm. — Py42
– Py42, Commented May 20, 2012 at 11:35

Mark Byers · Accepted Answer · 2012-05-20 11:36:15Z

10

To do what you literally asked for you should convert to an integer, add one, then change back to binary.

x = to_binary(int(x, 2) + 1)

You may find the bin built-in method useful in implementing to_binary, though you will need to modify its output slightly to match your desired result.

However it would be better if possible to not convert back and forth: just store an integer and convert it to binary when you need to display it.

On reading your question carefully though, it seems that all you want to do is to generate the strings '0000', '0001', '0010', etc. If this is correct, then I suggest using itertools.product.

Example:

import itertools
for x in map(''.join, itertools.product('01', repeat=4)):
    print x

0000
0001
0010
...

See it working online: ideone

edited May 20, 2012 at 11:36

answered May 20, 2012 at 11:30

Mark Byers

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1 Comment

Py42 Over a year ago

What an ingenious solution! I would've never come up with that. I'm repeatedly amazed at what Python has to offer. Thank you =]

wim · Accepted Answer · 2012-05-20 11:43:12Z

3

>>> def increment_binary_string(s):
...   return '{:04b}'.format(1 + int(s, 2))
... 
>>> x = '0000'
>>> for _ in xrange(15):
...   print x
...   x = increment_binary_string(x)
... 
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110

edited May 20, 2012 at 11:43

answered May 20, 2012 at 11:33

wim

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4 Comments

Py42 Over a year ago

This is similar to a solution I tried, the problem is that it returns "1" instead of "0001" (I combined that with a padding function that just pads the result with leading zeros).

wim Over a year ago

Oh, ok, for leading zeros you can use str.format (see edit)

wim Over a year ago

By the way, I agree with Mark about the real answer, which is just work with integer, and convert for display when necessary

Py42 Over a year ago

Thanks, I wasn't very familiar with str.format up until now =]

NTUI · Accepted Answer · 2012-06-02 17:03:21Z

2

This is simple to do with bin() and zfill().

>>> for x in range(16):
...     print bin(x)[2:].zfill(4)
... 
0000
0001
0010
0011
0100
0101
0110
0111
1000
1001
1010
1011
1100
1101
1110
1111

Or you can put the values in a list with a comprehension:

>>> [bin(x)[2:].zfill(4) for x in range(16)]
['0000', '0001', '0010', '0011', '0100', '0101', '0110', '0111', '1000', '1001', '1010', '1011', '1100', '1101', '1110', '1111']

answered Jun 2, 2012 at 17:03

NTUI

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Comments

Junuxx · Accepted Answer · 2012-05-20 12:17:30Z

1

Short and simple with the default function bin():

>>> for i in range(15):
...     print bin(i)
...
0b1
0b10
0b11
0b100
0b101
0b110
0b111
0b1000
0b1001
0b1010
0b1011
0b1100
0b1101
0b1110
0b1111

If you don't like that they start with the 0b prefix, you can use print str(bin(i))[2:] instead.

edited May 20, 2012 at 12:17

answered May 20, 2012 at 12:11

Junuxx

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Comments

Nurmukhamed Ubaidullayev · Accepted Answer · 2020-03-06 07:51:50Z

0

If you want use only strings:

def increment(l):
    if len(l) == 0:
        return l

    if l[-1] == '0':
        l = l[:-1] + '1'
    else:
        l = increment(l[:-1]) + '0'

    return l

answered Mar 6, 2020 at 7:51

Nurmukhamed Ubaidullayev

1

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