convert integer to binary

Question

I have an integer which I want to convert to binary and store the string of bits in an one-dimensional array starting from the right. For example, if the input is 6 then it should return an array like [1,1,0]. How to do it in python?

Answer 1

Solution

Probably the easiest way is not to use bin() and string slicing, but use features of .format():

'{:b}'.format(some_int)

How it behaves:

>>> print '{:b}'.format(6)
110
>>> print '{:b}'.format(123)
1111011

In case of bin() you just get the same string, but prepended with "0b", so you have to remove it.

Getting list of ints from binary representation

EDIT: Ok, so do not want just a string, but rather a list of integers. You can do it like that:

your_list = map(int, your_string)

Combined solution for edited question

So the whole process would look like this:

your_list = map(int, '{:b}'.format(your_int))

A lot cleaner than using bin() in my opinion.

• nneonneo: +1; now that {:b} exists, it's time to get rid of bin(x)[2:] abuse.

Answer 2

>>> map(int, bin(6)[2:])
[1, 1, 0]

If you don't want a list of ints (but instead one of strings) you can omit the map component and instead do:

>>> list(bin(6)[2:])
['1', '1', '0']

---

Relevant documentation:

• bin
• list
• map

• Tadeck: I wonder why use bin() and slicing instead of existing other options (namely binary formatting in string's .format()). Could you explain this choice?
• arshajii: @Tadeck Either method will work of course, bin just seemed more appropriate to me in this case.

Answer 3

You could use this command:

map(int, list(bin(YOUR_NUMBER)[2:]))

What it does is this:

• bin(YOUR_NUMBER) converts YOUR_NUMBER into its binary representation
• bin(YOUR_NUMBER)[2:] takes the effective number, because the string is returned in the form '0b110', so you have to remove the 0b
• list(...) converts the string into a list
• map(int, ...) converts the list of strings into a list of integers

Answer 4

You can use the bin function if you have Python>= 2.6:

list(bin(6))[2:]

Edit: oops, forgot to convert items to int:

map(int, list(bin(6))[2:])

Answer 5

In modern Python you can (>python2.5):

>>> bin(23455)
'0b101101110011111'

Discard the first '0b':

>>> [ bit for bit in bin(23455)[2:] ]
['1', '0', '1', '1', '0', '1', '1', '1', '0', '0', '1', '1', '1', '1', '1']

Everything together:

def get_bits(number):
 return [ int(bit) for bit in bin(number)[2:] ]

In 2.5 you will get an NameError: name 'bin' is not defined.

Answer 6

Others answers use bin() for that. It works, but I find that using string operations to do mathematics is a bit... ehm... lame:

def tobits(x):
 r = []
 while x:
 r.append(x & 1)
 x >>= 1
 return r

The tobits(0) will return an empty list. That may be nice or not, depending on what you'll do with it. So if needed treat it as a special case.

Answer 7

You may use numpy.unpackbits.

Here is the detailed link with examples: https://numpy.org/doc/stable/reference/generated/numpy.unpackbits.html