How to Use Hex() Without 0X in Python

How to use hex() without 0x in Python?

(Recommended)

Python 3 f-strings: Answered by @GringoSuave

>>> i = 3735928559
>>> f'{i:x}'
'deadbeef'

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Alternatives:

format builtin function (good for single values only)

>>> format(3735928559, 'x')
'deadbeef'

And sometimes we still may need to use str.format formatting in certain situations @Eumiro

(Though I would still recommend f-strings in most situations)

>>> '{:x}'.format(3735928559)
'deadbeef'

(Legacy) f-strings should solve all of your needs, but printf-style formatting is what we used to do @msvalkon

>>> '%x' % 3735928559
'deadbeef'

Without string formatting @jsbueno

>>> i = 3735928559
>>> i.to_bytes(4, "big").hex()
'deadbeef'

Hacky Answers (avoid)

hex(i)[2:] @GuillaumeLemaître

>>> i = 3735928559
>>> hex(i)[2:]
'deadbeef'

This relies on string slicing instead of using a function / method made specifically for formatting as hex. This is why it may give unexpected output for negative numbers:

>>> i = -3735928559
>>> hex(i)[2:]
'xdeadbeef'
>>> f'{i:x}'
'-deadbeef'

How to convert an integer to hexadecimal without the extra '0x' leading and 'L' trailing characters in Python?

Sure, go ahead and remove them.

hex(bignum).rstrip("L").lstrip("0x") or "0"

(Went the strip() route so it'll still work if those extra characters happen to not be there.)

How to turn decimals into hex without prefix `0x`

You have an extra zero because the line should be elif decimal < 16 not 17.

Use format strings¹:

def rgb(r,g,b):
    def hex1(d):
        return '{:02X}'.format(0 if d < 0 else 255 if d > 255 else d)
    return hex1(r)+hex1(g)+hex1(b)

print rgb(16,159,-137)

Output:

109F00

¹https://docs.python.org/2.7/library/string.html#format-specification-mini-language

How to print hex number in little endian, padded zeroes, and no '0x' in python?

Let's assume you've got the part that converts to little-endian working. The part you're asking is how to format a number into a particular string.

That's exactly what the format function is for:

>>> h = 20675
>>> format(h, '04X')
50C3

The full details are in the Format Specification Mini-Language, but briefly:

• 'X' is a type field, meaning "Hex format. Outputs the number in base 16, using upper- case letters for the digits above 9."
• '4' is a width field, which means anything shorter than 4 characters will be padded.
• '0' is a… special case that has no name, apparently, but "Preceding the width field by a zero ('0') character enables sign-aware zero-padding for numeric types. This is equivalent to a fill character of '0' with an alignment type of '='."

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But let's revisit that first sentence. You don't have the byte-swapping right, because you want a 32-bit little-endian number, and you're creating a 16-bit one. That's what 'H' means. If you want 32 bits, you have to use 'I'.

And then, of course, you need to change the width from 4 to 8.

Convert hex string to integer in Python

Without the 0x prefix, you need to specify the base explicitly, otherwise there's no way to tell:

x = int("deadbeef", 16)

With the 0x prefix, Python can distinguish hex and decimal automatically:

>>> print(int("0xdeadbeef", 0))
3735928559
>>> print(int("10", 0))
10

(You must specify 0 as the base in order to invoke this prefix-guessing behavior; if you omit the second parameter, int() will assume base-10.)

How can I format an integer to a two digit hex?

You can use string formatting for this purpose:

>>> "0x{:02x}".format(13)
'0x0d'

>>> "0x{:02x}".format(131)
'0x83'

Edit: Your code suggests that you are trying to convert a string to a hexstring representation. There is a much easier way to do this (Python2.x):

>>> "abcd".encode("hex")
'61626364'

An alternative (that also works in Python 3.x) is the function binascii.hexlify().

Python , Printing Hex removes first 0?

This is happening because hex() will not include any leading zeros, for example:

>>> hex(15)[2:]
'f'

To make sure you always get two characters, you can use str.zfill() to add a leading zero when necessary:

>>> hex(15)[2:].zfill(2)
'0f'

Here is what it would look like in your code:

fc = '0x'
for i in b[0x15c:0x15f]:
    fc += hex(ord(i))[2:].zfill(2)

How to convert a hex string to hex number

Try this:

hex_str = "0xAD4"
hex_int = int(hex_str, 16)
new_int = hex_int + 0x200
print hex(new_int)

If you don't like the 0x in the beginning, replace the last line with

print hex(new_int)[2:]

Generate a range of hex numbers in format

Try this :

for dec_ldev in range(0,16777215):
    hex_ldev = hex(dec_ldev)[2:].zfill(6)
    print("{}{}{}:{}{}:{}{}".format('',*hex_ldev))

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