How do you convert a number to binary without inbuilt function in python?

This function receives as a parameter an integer and should return a list representing the same value expressed in binary as a list of bits, where the first element in the list is the most significant [leftmost] bit.

My function currently outputs '1011' for the number 11, I need [1,0,1,1] instead.

For example,

>>> convert_to_binary[11]
[1,0,1,1]

asked Nov 23, 2012 at 3:36

user1790201user1790201

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4

def trans[x]:
    if x == 0: return [0]
    bit = []
    while x:
        bit.append[x % 2]
        x >>= 1
    return bit[::-1]

answered Nov 23, 2012 at 3:39

Jun HUJun HU

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Just for fun - the solution as a recursive one-liner:

def tobin[x]:
    return tobin[x/2] + [x%2] if x > 1 else [x]

answered Nov 23, 2012 at 4:05

Óscar LópezÓscar López

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may I propose this:

def tobin[x,s]:
    return [[x>>k]&1 for k in range[0,s]]

it is probably the fastest way and it seems pretty clear to me. bin way is too slow when performance matters.

cheers

answered Jun 11, 2013 at 12:28

3

You can first use the format function to get a binary string like your current function. For e.g the following snippet creates a binary string of 8 bits corresponding to integer 58.

>>>u = format[58, "08b"]
'00111010'

Now iterate the string to convert each bit to an int to get your desired list of bits encoded as integers.

>>>[int[d] for d in u]
[0, 0, 1, 1, 1, 0, 1, 0]

answered Apr 8, 2016 at 21:08

theOnetheOne

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You can use numpy package and get very fast solution:

python -m timeit -s "import numpy as np; x=np.array[[8], dtype=np.uint8]" "np.unpackbits[x]"
1000000 loops, best of 3: 0.65 usec per loop

python -m timeit "[int[x] for x in list['{0:0b}'.format[8]]]"
100000 loops, best of 3: 3.68 usec per loop

unpackbits handles inputs of uint8 type only, but you can still use np.view:

python -m timeit -s "import numpy as np; x=np.array[[124567], dtype=np.uint64].view[np.uint8]" "np.unpackbits[x]"
1000000 loops, best of 3: 0.697 usec per loop

answered Jul 21, 2016 at 5:19

bubblebubble

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This will do it. No sense in rolling your own function if there's a builtin.

def binary[x]:
    return [int[i] for i in bin[x][2:]]

The bin[] function converts to a string in binary. Strip of the 0b and you're set.

answered Nov 23, 2012 at 3:41

Dietrich EppDietrich Epp

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Here is the code for one that I made for college. Click Here for a youtube video of the code.! //www.youtube.com/watch?v=SGTZzJ5H-CE

__author__ = 'Derek'
print['Int to binary']
intStr = input['Give me an int: ']
myInt = int[intStr]
binStr = ''
while myInt > 0:
    binStr = str[myInt % 2] + binStr
    myInt //= 2
print['The binary of', intStr, 'is', binStr]
print['\nBinary to int']
binStr = input['Give me a binary string: ']
temp = binStr
newInt = 0
power = 0
while len[temp] > 0:   # While the length of the array if greater than zero keep looping through
    bit = int[temp[-1]]   # bit is were you temporally store the converted binary number before adding it to the total
    newInt = newInt + bit * 2 ** power  # newInt is the total,  Each time it loops it adds bit to newInt.
    temp = temp[:-1]  # this moves you to the next item in the string.
    power += 1  # adds one to the power each time.
print["The binary number " + binStr, 'as an integer is', newInt]

answered Oct 19, 2014 at 11:54

Derek MCDerek MC

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Padded with length

In most cases you want your binary number to be a specific length. For example you want 1 to be 8 binary digits long [0,0,0,0,0,0,0,1]. I use this myself:

def convert_to_binary[num, length=8]:
    binary_string_list = list[format[num, '0{}b'.format[length]]]
    return [int[digit] for digit in binary_string_list]

answered Feb 16, 2015 at 14:05

Not really the most efficient but at least it provides a simple conceptual way of understanding it...

1] Floor divide all the numbers by two repeatedly until you reach 1

2] Going in reverse order, create bits of this array of numbers, if it is even, append a 0 if it is odd append a 1.

Here's the literal implementation of that:

def intToBin[n]:
    nums = [n]
    while n > 1:
        n = n // 2
        nums.append[n]

    bits = []
    for i in nums:
        bits.append[str[0 if i%2 == 0 else 1]]
    bits.reverse[]
    print ''.join[bits]

Here's a version that better utilizes memory:

def intToBin[n]:
    bits = []

    bits.append[str[0 if n%2 == 0 else 1]]
    while n > 1:
        n = n // 2
        bits.append[str[0 if n%2 == 0 else 1]]

    bits.reverse[]
    return ''.join[bits]

answered Apr 17, 2016 at 18:31

ThinkBonoboThinkBonobo

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Not the pythonic way...but still works:

def get_binary_list_from_decimal[integer, bits]:
    '''Return a list of 0's and 1's representing a decimal type integer.

    Keyword arguments:
    integer -- decimal type number.
    bits -- number of bits to represent the integer.

    Usage example:
    #Convert 3 to a binary list
    get_binary_list_from_decimal[3, 4]
    #Return will be [0, 0, 1, 1]
    '''
    #Validate bits parameter.
    if 2**bits  1, i+1]

   return loop[n, 1]

answered Aug 21, 2017 at 3:08

Converting decimal to binary is a matter of how you are going to use the % and //

def getbin[num]:
    if [num==0]:
        k=[0] 
        return k 
    else:
        s = []
        while[num]:
            s.append[num%2]
            num=num//2
        return s

answered Oct 15, 2018 at 9:04

1

Just sharing a function that processes an array of ints:

def to_binary_string[x]:
    length = len[bin[max[x]][2:]]

    for i in x:
        b = bin[i][2:].zfill[length]

        yield [int[n] for n in b]

Test:

x1 = to_binary_string[[1, 2, 3]]
x2 = to_binary_string[[1, 2, 3, 4]]

print[list[x1]] # [[0, 1], [1, 0], [1, 1]]
print[list[x2]] # [[0, 0, 1], [0, 1, 0], [0, 1, 1], [1, 0, 0]]

answered May 3, 2019 at 8:36

RendicahyaRendicahya

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Convert integer to list of bits with a fixed length :

[int[x] for x in list['{0:0{width}b}'.format[8, width=5]]]

answered Aug 22, 2020 at 10:52

florexflorex

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# dec2bin.py
# FB - 201012057
import math

def dec2bin[f]:
    if f >= 1:
        g = int[math.log[f, 2]]
    else:
        g = -1
    h = g + 1
    ig = math.pow[2, g]
    st = ""    
    while f > 0 or ig >= 1: 
        if f < 1:
            if len[st[h:]] >= 10: # 10 fractional digits max
                   break
        if f >= ig:
            st += "1"
            f -= ig
        else:
            st += "0"
        ig /= 2
    st = st[:h] + "." + st[h:]
    return st

# MAIN
while True:
    f = float[raw_input["Enter decimal number >0: "]]
    if f