def dec_to_bin(x):
return int(bin(x)[2:])
It's that easy.
I agree with @aaronasterling's answer. However, if you want a non-binary string that you can cast into an int, then you can use the canonical algorithm:
def decToBin(n):
if n==0: return ''
else:
return decToBin(n/2) + str(n%2)
You can also use a function from the numpy module
from numpy import binary_repr
which can also handle leading zeros:
Definition: binary_repr(num, width=None)
Docstring:
Return the binary representation of the input number as a string.
This is equivalent to using base_repr with base 2, but about 25x
faster.
For negative numbers, if width is not given, a - sign is added to the
front. If width is given, the two's complement of the number is
returned, with respect to that width.
n=int(input('please enter the no. in decimal format: '))
x=n
k=[]
while (n>0):
a=int(float(n%2))
k.append(a)
n=(n-a)/2
k.append(0)
string=""
for j in k[::-1]:
string=string+str(j)
print('The binary no. for %d is %s'%(x, string))
For the sake of completion: if you want to convert fixed point representation to its binary equivalent you can perform the following operations:
Get the integer and fractional part.
from decimal import *
a = Decimal(3.625)
a_split = (int(a//1),a%1)
Convert the fractional part in its binary representation. To achieve this multiply successively by 2.
fr = a_split[1]
str(int(fr*2)) + str(int(2*(fr*2)%1)) + ...
You can read the explanation here.
"{0:#b}".format(my_int)
all numbers are stored in binary. if you want a textual representation of a given number in binary, use bin(i)
>>> bin(10)
'0b1010'
>>> 0b1010
10
Without the 0b in front:
"{0:b}".format(int)
Starting with Python 3.6 you can also use formatted string literal or f-string, --- PEP:
f"{int:b}"
Source: Stackoverflow.com