Convert hex to binary

For solving the left-side trailing zero problem:

my_hexdata = "1a"

scale = 16 ## equals to hexadecimal

num_of_bits = 8

bin(int(my_hexdata, scale))[2:].zfill(num_of_bits)

It will give 00011010 instead of the trimmed version.

Use Built-in format() function and int() function It's simple and easy to understand. It's little bit simplified version of Aaron answer

int()

int(string, base)

format()

format(integer, # of bits)

Example

# w/o 0b prefix
>> format(int("ABC123EFFF", 16), "040b")
1010101111000001001000111110111111111111

# with 0b prefix
>> format(int("ABC123EFFF", 16), "#042b")
0b1010101111000001001000111110111111111111

# w/o 0b prefix + 64bit
>> format(int("ABC123EFFF", 16), "064b")
0000000000000000000000001010101111000001001000111110111111111111

See also this answer

# Python Program - Convert Hexadecimal to Binary
hexdec = input("Enter Hexadecimal string: ")
print(hexdec," in Binary = ", end="")    # end is by default "\n" which prints a new line
for _hex in hexdec:
    dec = int(_hex, 16)    # 16 means base-16 wich is hexadecimal
    print(bin(dec)[2:].rjust(4,"0"), end="")    # the [2:] skips 0b, and the 

Just use the module coden ^{(note: I am the author of the module)}

You can convert haxedecimal to binary there.

1. Install using pip

pip install coden

2. Convert

a_hexadecimal_number = "f1ff"
binary_output = coden.hex_to_bin(a_hexadecimal_number)

The converting Keywords are:

• hex for hexadeimal
• bin for binary
• int for decimal
• _to_ - the converting keyword for the function

So you can also format: e. hexadecimal_output = bin_to_hex(a_binary_number)

The binary version of ABC123EFFF is actually 1010101111000001001000111110111111111111

For almost all applications you want the binary version to have a length that is a multiple of 4 with leading padding of 0s.

To get this in Python:

def hex_to_binary( hex_code ):
  bin_code = bin( hex_code )[2:]
  padding = (4-len(bin_code)%4)%4
  return '0'*padding + bin_code

Example 1:

>>> hex_to_binary( 0xABC123EFFF )
'1010101111000001001000111110111111111111'

Example 2:

>>> hex_to_binary( 0x7123 )
'0111000100100011'

Note that this also works in Micropython :)

HEX_TO_BINARY_CONVERSION_TABLE = {
                              '0': '0000',

                              '1': '0001',

                              '2': '0010',

                              '3': '0011',

                              '4': '0100',

                              '5': '0101',

                              '6': '0110',

                              '7': '0111',

                              '8': '1000',

                              '9': '1001',

                              'a': '1010',

                              'b': '1011',

                              'c': '1100',

                              'd': '1101',

                              'e': '1110',

                              'f': '1111'}

def hex_to_binary(hex_string):
    binary_string = ""
    for character in hex_string:
        binary_string += HEX_TO_BINARY_CONVERSION_TABLE[character]
    return binary_string

when I time hex_to_binary("123ade")

  %timeit hex_to_binary("123ade")

here is the result:

 316 ns ± 2.52 ns per loop

Alternatively, you could use "join" method:

def hex_to_binary_join(hex_string):
    hex_array=[]
    for character in hex_string:
        hex_array.append(HEX_TO_BINARY_CONVERSION_TABLE[character])
    return "".join(hex_array)

I timed this too:

    %timeit hex_to_binary_join("123ade")
       397 ns ± 4.64 ns per loop 

no=raw_input("Enter your number in hexa decimal :")
def convert(a):
    if a=="0":
        c="0000"
    elif a=="1":
        c="0001"
    elif a=="2":
        c="0010"
    elif a=="3":
        c="0011"
    elif a=="4":
        c="0100"
    elif a=="5":
        c="0101"
    elif a=="6":
        c="0110"
    elif a=="7":
        c="0111"
    elif a=="8":
        c="1000"
    elif a=="9":
        c="1001"
    elif a=="A":
        c="1010"
    elif a=="B":
        c="1011"
    elif a=="C":
        c="1100"
    elif a=="D":
        c="1101"
    elif a=="E":
        c="1110"
    elif a=="F":
        c="1111"
    else:
        c="invalid"
    return c

a=len(no)
b=0
l=""
while b<a:
    l=l+convert(no[b])
    b+=1
print l