Python convert decimal to hex

Question

I have a function here that converts decimal to hex but it prints it in reverse order. How would I fix it?

def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print(n)
    else:
        x =(n%16)
        if (x < 10):
            print(x), 
        if (x == 10):
            print("A"),
        if (x == 11):
            print("B"),
        if (x == 12):
            print("C"),
        if (x == 13):
            print("D"),
        if (x == 14):
            print("E"),
        if (x == 15):
            print ("F"),
        ChangeHex( n / 16 )

See also: [Python: hex conversion always two digits](http://stackoverflow.com/q/11676864/562769) — Martin Thoma, Jan 12 '17 at 20:28
Answers have to do with reinventing the wheel. For an answer to converting an integer to hexadecimal representation, see the [builtin "hex"](https://docs.python.org/3/library/functions.html#hex) — user, Jan 11 '21 at 00:49

score 103 · Answer 1 · edited Nov 22 '22 at 08:22

103

What about this:

hex(dec).split('x')[-1]

Example:

>>> d = 30
>>> hex(d).split('x')[-1]
'1e'

By using -1 in the result of split(), this would work even if split returned a list of 1 element.

edited Nov 22 '22 at 08:22

Halo

1,730
1
8
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answered Nov 27 '12 at 23:37

Rich

1,031
2
7
2

3

https://stackoverflow.com/a/35750576/8437974 [Martijn Pieter's](https://stackoverflow.com/users/100297/martijn-pieters) `format(15, 'x')` – Germa Vinsmoke Apr 21 '19 at 10:23

Joseph Lisee · Answer 2 · 2013-12-04T21:40:43.243

48

This isn't exactly what you asked for but you can use the "hex" function in python:

>>> hex(15)
'0xf'

edited Dec 04 '13 at 21:40

answered Apr 26 '11 at 20:47

Joseph Lisee

3,439
26
21

print hex(15); TypeError: 'str' object is not callable – Moirisa Dikaiosýni Jun 28 '16 at 03:07
3

@Babbit Then you're doing something wrong. You probably named a variable hex or something. – Zizouz212 Oct 07 '16 at 20:42

score 26 · Accepted Answer · answered Apr 26 '11 at 20:47

26

If you want to code this yourself instead of using the built-in function hex(), you can simply do the recursive call before you print the current digit:

def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print n,
    else:
        ChangeHex( n / 16 )
        x =(n%16)
        if (x < 10):
            print(x), 
        if (x == 10):
            print("A"),
        if (x == 11):
            print("B"),
        if (x == 12):
            print("C"),
        if (x == 13):
            print("D"),
        if (x == 14):
            print("E"),
        if (x == 15):
            print ("F"),

answered Apr 26 '11 at 20:47

Sven Marnach

574,206
118
941
841

7

@Apostolos No, I don't think it is. What makes you think so? However, I think it's useful for the original poster of the question to learn how to fix their code, even though it's not the solution i'd recommend to go with in "real" code. – Sven Marnach Jan 06 '18 at 07:22
ChangeHex( n / 16 ) or ChangeHex( n // 16 ) ? – sortas Dec 20 '18 at 14:53
2

@sortas `n // 16` in Python 3. This answer was written for Python 2, and in that version the two are equivalent. (I tried to make the minimal change to the code in the question to make it work.) – Sven Marnach Dec 20 '18 at 15:43

score 26 · Answer 4 · edited Nov 21 '20 at 21:08

I think this solution is elegant:

def toHex(dec):
    digits = "0123456789ABCDEF"
    x = (dec % 16)
    rest = dec // 16
    if (rest == 0):
        return digits[x]
    return toHex(rest) + digits[x]

numbers = [0, 11, 16, 32, 33, 41, 45, 678, 574893]
print [toHex(x) for x in numbers]
print [hex(x) for x in numbers]

This output:

['0', 'B', '10', '20', '21', '29', '2D', '2A6', '8C5AD']
['0x0', '0xb', '0x10', '0x20', '0x21', '0x29', '0x2d', '0x2a6', '0x8c5ad']

score 20 · Answer 5 · edited Jun 17 '14 at 15:49

20

I use

"0x%X" % n

where n is the decimal number to convert.

edited Jun 17 '14 at 15:49

OGHaza

4,795
7
23
29

answered Dec 27 '13 at 18:40

thanos

723
7
9

score 15 · Answer 6 · answered Nov 26 '14 at 08:39

15

If without '0x' prefix:

'{0:x}'.format(int(dec))

else use built-in hex() funtion.

answered Nov 26 '14 at 08:39

nikioa

771
5
6

See also: [Python: hex conversion always two digits](http://stackoverflow.com/q/11676864/562769) – Martin Thoma Jan 12 '17 at 20:28

score 15 · Answer 7 · answered Oct 19 '16 at 20:29

Python's string format method can take a format spec.

From decimal to binary

"{0:b}".format(154)
'10011010'

From decimal to octal

"{0:o}".format(154)
'232'

From decimal to hexadecimal

"{0:x}".format(154)
'9a'

Format spec docs for Python 2

Format spec docs for Python 3

Tomas · Answer 8 · 2021-10-01T11:23:08.770

2

It is good to write your own functions for conversions between numeral systems to learn something. For "real" code I would recommend to use build in conversion function from Python like bin(x), hex(x), int(x).

edited Oct 01 '21 at 11:23

answered Nov 21 '19 at 10:54

Tomas

71
5

score 2 · Answer 9 · answered Jan 08 '20 at 09:34

2

If you need even numbers of chars to be returned, you can use:

def int_to_hex(nr):
  h = format(int(nr), 'x')
  return '0' + h if len(h) % 2 else h

Example

int_to_hex(10) # returns: '0a'

and

int_to_hex(1000) # returns: '03e8'

answered Jan 08 '20 at 09:34

Piotr

1,777
17
24

Perfect solution, plus the examples. Thank you sir. – Meto Dec 22 '20 at 08:04

score 2 · Answer 10 · answered Aug 18 '21 at 04:34

2

you can use this method which uses slicing

output = hex(15)[2:]

which cuts out the first 2 characters (0x) from the output

answered Aug 18 '21 at 04:34

Hyperx837

773
5
13

score 1 · Answer 11 · answered Jul 29 '17 at 01:17

def main():
    result = int(input("Enter a whole, positive, number to be converted to hexadecimal: "))
    hexadecimal = ""
    while result != 0:
        remainder = changeDigit(result % 16)
        hexadecimal = str(remainder) + hexadecimal
        result = int(result / 16)
    print(hexadecimal)

def changeDigit(digit):
    decimal =     [10 , 11 , 12 , 13 , 14 , 15 ]
    hexadecimal = ["A", "B", "C", "D", "E", "F"]
    for counter in range(7):
        if digit == decimal[counter - 1]:
            digit = hexadecimal[counter - 1]
    return digit

main()

This is the densest I could make for converting decimal to hexadecimal. NOTE: This is in Python version 3.5.1

score 1 · Answer 12 · answered Aug 20 '18 at 17:47

Apart from using the hex() inbuilt function, this works well:

letters = [0,1,2,3,4,5,6,7,8,9,'A','B','C','D','E','F']
decimal_number = int(input("Enter a number to convert to hex: "))
print(str(letters[decimal_number//16])+str(letters[decimal_number%16]))

However this only works for converting decimal numbers up to 255 (to give a two diget hex number).

blackhumdinger · Answer 13 · 2020-05-21T14:03:16.693

1

def tohex(dec):
    x = (dec%16)
    igits = "0123456789ABCDEF"
    digits = list(igits)
    rest = int(dec/16)
    if (rest == 0):
        return digits[x]
    return tohex(rest) + digits[x]

numbers = [0,16,32,48,46,2,55,887]
hex_ = ["0x"+tohex(i) for i in numbers]
print(hex_)

edited May 21 '20 at 14:03

answered May 21 '20 at 13:48

blackhumdinger

11
2

score 1 · Answer 14 · answered Feb 26 '21 at 13:11

I recently made this python program to convert Decimal to Hexadecimal, please check this out. This is my first Answer in stack overflow .

decimal = int(input("Enter the Decimal no that you want to convert to Hexadecimal : "))
intact = decimal
hexadecimal = ''
dictionary = {1:'1',2:'2',3:'3',4:'4',5:'5',6:'6',7:'7',8:'8',9:'9',10:'A',11:'B',12:'C',13:'D',14:'E',15:'F'}

while(decimal!=0):
    c = decimal%16 
    hexadecimal =  dictionary[c] + hexadecimal 
    decimal = int(decimal/16)

print(f"{intact} is {hexadecimal} in Hexadecimal")

When you Execute this code this will give output as:

Enter the Decimal no that you want to convert to Hexadecimal : 2766

2766 is ACE in Hexadecimal

score 0 · Answer 15 · answered Sep 25 '17 at 14:49

A version using iteration:

def toHex(decimal):
    hex_str = ''
    digits = "0123456789ABCDEF"
    if decimal == 0:
       return '0'

    while decimal != 0:
        hex_str += digits[decimal % 16]
        decimal = decimal // 16

    return hex_str[::-1] # reverse the string

numbers = [0, 16, 20, 45, 255, 456, 789, 1024]
print([toHex(x) for x in numbers])
print([hex(x) for x in numbers])

score 0 · Answer 16 · answered Jan 30 '18 at 13:42

hex_map = {0:0, 1:1, 2:2, 3:3, 4:4, 5:5, 6:6, 7:7, 8:8, 9:9, 10:'A', 11:'B', 12:'C', 13:'D', 14:'E', 15:'F'}

def to_hex(n):
    result = ""
    if n == 0:
        return '0'
    while n != 0:
        result += str(hex_map[(n % 16)])
        n = n // 16
    return '0x'+result[::-1]

score 0 · Answer 17 · edited May 09 '19 at 00:30

n = eval(input("Enter the number:"))
def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print(n),
    else:
        ChangeHex( n / 16 )
        x =(n%16)
        if (x < 10):
            print(x), 
        if (x == 10):
            print("A"),
        if (x == 11):
            print("B"),
        if (x == 12):
            print("C"),
        if (x == 13):
            print("D"),
        if (x == 14):
            print("E"),
        if (x == 15):
            print ("F"),

score 0 · Answer 18 · edited Jun 09 '19 at 19:13

This is the best way I use it

hex(53350632996854).lstrip("0x").rstrip("L")
# lstrip helps remove "0x" from the left  
# rstrip helps remove "L" from the right 
# L represents a long number

Example:

>>> decimal = 53350632996854
>>> hexadecimal = hex(decimal).lstrip("0x")
>>> print(hexadecimal)
3085a9873ff6

if you need it Upper Case, Can use "upper function" For Example:

decimal = 53350632996854
hexadecimal = hex(decimal).lstrip("0x").upper()
print(hexadecimal)
3085A9873FF6

score 0 · Answer 19 · answered Apr 26 '11 at 22:01

Instead of printing everything in the function, you could just allow it to return the value in hex, and do whatever you want with it.

def ChangeHex(n):
    x = (n % 16)
    c = ""
    if (x < 10):
        c = x
    if (x == 10):
        c = "A"
    if (x == 11):
        c = "B"
    if (x == 12):
        c = "C"
    if (x == 13):
        c = "D"
    if (x == 14):
        c = "E"
    if (x == 15):
        c = "F"

    if (n - x != 0):
        return ChangeHex(n / 16) + str(c)
    else:
        return str(c)

print(ChangeHex(52))

There are probably more elegant ways of parsing the alphabetic components of the hex, instead of just using conditionals.

score 0 · Answer 20 · edited Oct 17 '19 at 03:48

In order to put the number in the correct order i modified your code to have a variable (s) for the output. This allows you to put the characters in the correct order.

s=""
def ChangeHex(n):
    if (n < 0):
        print(0)
    elif (n<=1):
        print(n)
    else:
        x =(n%16)
        if (x < 10):
            s=str(x)+s, 
        if (x == 10):
            s="A"+s,
        if (x == 11):
            s="B"+s,
        if (x == 12):
            s="C"+s,
        if (x == 13):
            s="D"+s,
        if (x == 14):
            s="E"+s,
        if (x == 15):
            s="F"+s,
        ChangeHex( n / 16 )

NOTE: This was done in python 3.7.4 so it may not work for you.

score 0 · Answer 21 · answered Feb 13 '22 at 20:08

0

def decimal_to_base(decimal_number, destination_base):
digits_array = []
hexadecimals = {10:"A", 11:"B", 12:"C", 13:"D", 14:"E", 15:"F"}
while decimal_number > 0:
    rest = decimal_number % destination_base
    if rest in hexadecimals:
        rest = hexadecimals[rest]
    digits_array.insert(0, rest)
    decimal_number = decimal_number // destination_base
return digits_array

answered Feb 13 '22 at 20:08

Carlos Sosa

1

1

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Feb 14 '22 at 01:22

score -1 · Answer 22 · answered Apr 26 '14 at 12:23

non recursive approach to convert decimal to hex

def to_hex(dec):

    hex_str = ''
    hex_digits = ('0','1','2','3','4','5','6','7','8','9','A','B','C','D','E','F')
    rem = dec % 16

    while dec >= rem:
        remainder = dec % 16
        quotient = dec / 16
        if quotient == 0:
            hex_str += hex_digits[remainder]
        else:
            hex_str += str(remainder)
        dec = quotient

    return hex_str[::-1] # reverse the string

score -1 · Answer 23 · edited May 04 '18 at 11:39

-1

dec = int(input("Enter a number below 256: "))
hex1 = dec // 16

if hex1 >= 10:
    hex1 = hex(dec)

hex2 = dec % 16
if hex2 >= 10:
    hex2 = hex(hex2)

print(hex1.strip("0x"))

Works well.

edited May 04 '18 at 11:39

MrLeeh

5,321
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answered May 04 '18 at 11:13

TXVXT

1

score -2 · Answer 24 · answered Sep 21 '21 at 08:48

This code is Incomplite/-\ Max input is 159

def DicToHex(lsdthx, number, resault):
   bol = True
   premier = 0
   for i in range(number):
      for hx in lsdthx:
        if hx <= number:
            if number < 16:
                if hx > 9:
                    resault += lsdthx[hx]
                else:
                    resault += str(lsdthx[hx])
                number -= hx
            else:
                while bol:
                    if number - 16 >= 0:
                        number -= 16
                        premier += 1
                    else:
                        resault += str(premier)
                        bol = False
      return resault

dthex = {15:'F', 14:'E', 13:'D', 12:'C', 11:'B', 10:'A', 9:9, 8:8, 7:7,6:6, 5:5, 4:4, 3:3, 2:2, 1:1}
reslt = ''
num = int(input('enter dicimal number : '))
print(DicToHex(dthex, num, reslt))

As it’s currently written, your answer is unclear. Please [edit] to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers [in the help center](/help/how-to-ask). — Community, Sep 21 '21 at 09:16

Python convert decimal to hex

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