How to convert decimal to binary list in python [duplicate]

stackoverflow.com › questions › 13557937 › how-to-convert-decimal-to-binary-list-in-python

You can probably use the builtin bin function:

bin(8) #'0b1000'

to get the list:

[int(x) for x in bin(8)[2:]]

Although it seems like there's probably a better way...

Answer from mgilson on Stack Overflow

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How to convert decimal to binary list in python - Stack Overflow

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1 of 3

You can probably use the builtin bin function:

bin(8) #'0b1000'

to get the list:

[int(x) for x in bin(8)[2:]]

Although it seems like there's probably a better way...

2 of 3

Try this:

>>> list('{0:0b}'.format(8))
['1', '0', '0', '0']

Edit -- Ooops, you wanted integers:

>>> [int(x) for x in list('{0:0b}'.format(8))]
[1, 0, 0, 0]

Another edit --

mgilson's version is a little bit faster:

$ python -m timeit "[int(x) for x in list('{0:0b}'.format(8))]"
100000 loops, best of 3: 5.37 usec per loop
$ python -m timeit "[int(x) for x in bin(8)[2:]]"
100000 loops, best of 3: 4.26 usec per loop

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python - convert integer to binary - Stack Overflow

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? ... In case of bin() you just get the same string, but prepended with "0b", so you have to remove it. EDIT: Ok, so do not want just a string, but rather a list ... More on stackoverflow.com

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Converting Integers to binary and placing them in a list.

bin() returns a string containing a binary representation of the integer you pass to it. I'm assuming that you already have a list of integers, and you want to produce a list of strings, where the strings are the results of calling bin() on each of the integers in the first list. If any of that is incorrect, please clarify. integers = [1, 2, 3, 4, 5, 6, 7] binary_representations = [bin(x) for x in integers] More on reddit.com

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November 13, 2012

python - Convert elements of a list into binary - Stack Overflow

Suppose I have a list: lst = [0, 1, 0, 0] How can I make python interpret this list as a binary number 0100 so that 2*(0100) gives me 01000? The only way that I can think of is to first make a More on stackoverflow.com

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Thank you for visiting.

Close: I solved it myself.

Thank you very much.

import numpy as LAC
a = [50,51,52]
print(LAC.unpackbits(LAC.array(a , dtype = LAC.uint8).reshape(len(a),1), axis = 1))

For example, when there is a list a = [50,51,52].
I'm looking for a function to convert this to binary numbers all at once.
I already know the algorithm and the coding, but I don't know if there is a function or not.
Therefore, I would like to ask everyone to let me know if the above function exists or not.

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1 of 2

you can use a list comprehension along with the bin() function to convert all the integers in your list to binary. this is how you can do it for your list a: a = [50, 51, 52] binary_list = [bin(x)[2:] for x in a] print(binary_list) This will give you a new list with the binary representations of all the integers in a. The [2:] is used to remove the '0b' prefix that bin() adds.

2 of 2

When you say convert to binary, you mean string representations of the numbers in binary, right? An alternative to bin is format # Format in binary >>> format(42, "b") '101010' # Format in binary in 10 characters, right justified >>> format(42, "10b") ' 101010' # Format in binary in 10 characters with leading 0's >>> format(42, "010b") '0000101010' # Format 50 - 52 in binary, in 8 characters, with leading 0's >>> [format(n, "08b") for n in range(50, 53)] ['00110010', '00110011', '00110100']

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July 2, 2026 - Many programming languages provide built-in ways to convert integers to binary strings. Once a binary string is available, it can be converted into a list of digits. In Python, for example, bin() returns a string beginning with 0b, so the prefix is removed before creating the list:

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1 of 7

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 `int`s 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.

2 of 7

>>> 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

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r/learnpython on Reddit: Converting Integers to binary and placing them in a list.

November 13, 2012 -

I am trying to find a way of converting an integer to binary, and then placing that binary into a list. Is there a way to use the bin() command to perform such a task. If not, how should I approach this problem. I am fairly new to python, but have a decent understanding

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2 of 4

converting an integer to binary This does not actually mean anything. Everything "is binary". What exactly do you want to end up in the list? What are you going to do with it? What is the purpose of having these values? and then placing that binary into a list. What do you mean "place it into a list"? Is there an existing list for some other purpose, that you want to append to? Do you have a list of these integers, and you want to produce the corresponding list of "binary" values? Or just what? Is there a way to use the bin() command to perform such a task. Did you try it? Does calling bin() on a test integer value produce the result you want?

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python - Convert elements of a list into binary - Stack Overflow

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1 of 8

You can use bitwise operators like this:

>>> lst = [0, 1, 0, 0]
 >>> bin(int(''.join(map(str, lst)), 2) << 1)
'0b1000'

2 of 8

This is not a fancy one-liner, but simple and fast.

lst = [0,1,1,0]

num = 0
for b in lst:
    num = 2 * num + b
print(num) # 6

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September 14, 2022 - Now that we have an in-depth ... an integer into its equivalent binary representation. We have already seen how to do so in earlier sections. For example, the number 1234 would be 10011010010. We arrived at this answer by doing the following: 1*2**10 + 0*2**9 + 0*2**8 +1*2**7 + 1*2**6 + 0*2**5 + 1*2**4 + 0*2**3 + 0*2**2 + 1*2**1 + 0*2**0 · So, how did we get the binary numbers in the first place? There are a few ways we can do the conversion in Python...

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performance - How to write a list of integers to a binary file in python - Stack Overflow

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1 of 1

Use a bytearray() object, write that straight to the output file:

with open (output1, "wb") as compdata:
    for row in range(height):
        bytes = bytearray(cobs(delta_rows[row].getByte_List()))
        compdata.write(bytes)

A sequence of integers is interpreted by a bytearray() as a sequence of byte values.

In Python 3, you can use a bytes() type as well, with the same input; you are not mutating the values after creation, after all.

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How to convert a decimal number into a binary list with fixed number of bits in Python 3.8? - Stack Overflow

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1 of 3

You can do with list comprehension.

>>> [int(i) for i in bin(8)[2:].zfill(8)]
[0, 0, 0, 0, 1, 0, 0, 0]

bin(8) return the binary representation of an integer 8 and it's always prefixed with 0b. So bin(8)[2:] is to remove these first two characters(ie, 0b). And then you can use .zfill(8) to pad a numeric string with zeros on the left(with the given width 8)

2 of 3

You could build an f-string then iterate over that with map() as follows:

n = 8

lst = list(map(int, f'{n:08b}'))

print(lst)

Output:

[0, 0, 0, 0, 1, 0, 0, 0]

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Converting list of integers into a binary "string" in python - Stack Overflow

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1 of 5

You want to send a single byte for ALPHA if it's < 256, but two bytes if >= 256? This seems weird -- how is the receiver going to know which is the case...???

But, if this IS what you want, then

x = struct.pack(4*'B' + 'HB'[ALPHA<256] + 4*'B', *data)

is one way to achieve this.

2 of 5

If you know the data and ALPHA position beforehand, it would be best to use struct.pack with a big endian short for that position and omit the 0 that might be overwritten:

def output(ALPHA):
    data = [2,25,0,ALPHA,0,23,18,188]
    format = ">BBBHBBBB"
    return struct.pack(format, *data)
output(101) # result: '\x02\x19\x00\x00e\x00\x17\x12\xbc'
output(999) # result: '\x02\x19\x00\x03\xe7\x00\x17\x12\xbc'

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Close: I solved it myself.

Solution

Getting list of ints from binary representation

Combined solution for edited question

Getting list of `int`s from binary representation