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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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.! https://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 <= integer:
        raise ValueError("Error: Number of bits is not sufficient to \
                          represent the integer. Increase bits parameter.")

    #Initialise binary list
    binary_list = []
    remainder = integer
    for i in range(bits-1, -1, -1):
        #If current bit value is less than or equal to the remainder of 
        #the integer then bit value is 1.
        if 2**i <= remainder:
            binary_list.append(1)
            #Subtract the current bit value from the integer.
            remainder = remainder - 2**i
        else:
            binary_list.append(0)

    return binary_list

Example of how to use it:

get_binary_list_from_decimal(1, 3)
#Return will be [0, 0, 1]

answered Aug 31, 2016 at 22:29

theQuestionMantheQuestionMan

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def nToKBit(n, K=64):
   output = [0]*K

   def loop(n, i):
       if n == 0: 
           return output
       output[-i] = n & 1
       return loop(n >> 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 <= 0: break
    print "Binary #: ", dec2bin(f)
    print "bin(int(f)): ", bin(int(f)) # for comparison

Chris

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answered Apr 2, 2014 at 8:44

2

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