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

Question

I am trying to convert big integer number to hexadecimal, but in result I get extra \"0x\" in the beginning and \"L\" at the and. Is there any way to remove them. Thanks. The nu

Answer 1

Similar to Praveen's answer, you can also directly use built-in format().

>>> a = 44199528911754184119951207843369973680110397
>>> format(a, 'x')
'1fb62bdc9e54b041e61857943271b44aafb3d'

Answer 2

I think it's dangerous idea to use strip.
because lstrip or rstrip strips 0.

ex)

a = '0x0'
a.lstrip('0x')  

''

result is '', not '0'.

In your case, you can simply use replce to prevent above situation.
Here's sample code.

hex(bignum).replace("L","").replace("0x","")

Answer 3

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.)

Answer 4

Be careful when using the accepted answer as lstrip('0x') will also remove any leading zeros, which may not be what you want, see below:

>>> account = '0x000067'
>>> account.lstrip('0x')
'67'
>>>

If you are sure that the '0x' prefix will always be there, it can be removed simply as follows:

>>> hex(42)
'0x2a'
>>> hex(42)[2:]
'2a'
>>>

[2:] will get every character in the string except for the first two.

Answer 5

A more elegant way would be

hex(_number)[2:-1]

but you have to be careful if you're working with gmpy mpz types, then the 'L' doesn't exist at the end and you can just use

hex(mpz(_number))[2:]

Answer 6

The 0x is literal representation of hex numbers. And L at the end means it is a Long integer.

If you just want a hex representation of the number as a string without 0x and L, you can use string formatting with %x.

>>> a = 44199528911754184119951207843369973680110397
>>> hex(a)
'0x1fb62bdc9e54b041e61857943271b44aafb3dL'
>>> b = '%x' % a
>>> b
'1fb62bdc9e54b041e61857943271b44aafb3d'