CODEWITHSUNDEEP
pythonhex
user1047517
user1047517

Reputation: 1559

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

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 number is:

44199528911754184119951207843369973680110397865530452125410391627149413347233422
34022212251821456884124472887618492329254364432818044014624401131830518339656484
40715571509533543461663355144401169142245599341189968078513301836094272490476436
03241723155291875985122856369808620004482511813588136695132933174030714932470268
09981252011612514384959816764532268676171324293234703159707742021429539550603471
00313840833815860718888322205486842202237569406420900108504810

In hex I get:

0x2ef1c78d2b66b31edec83f695809d2f86e5d135fb08f91b865675684e27e16c2faba5fcea548f3
b1f3a4139942584d90f8b2a64f48e698c1321eee4b431d81ae049e11a5aa85ff85adc2c891db9126
1f7f2c1a4d12403688002266798ddd053c2e2670ef2e3a506e41acd8cd346a79c091183febdda3ca
a852ce9ee2e126ca8ac66d3b196567ebd58d615955ed7c17fec5cca53ce1b1d84a323dc03e4fea63
461089e91b29e3834a60020437db8a76ea85ec75b4c07b3829597cfed185a70eeaL

Upvotes: 22

Views: 77798

Answers (6)

Jiwon
Jiwon

Reputation: 1218

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 replace to prevent above situation.
Here's sample code.

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

Upvotes: 3

Mislav Jurinić
Mislav Jurinić

Reputation: 1

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:]

Upvotes: -2

matthew
matthew

Reputation: 2254

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.

Upvotes: 0

Alex
Alex

Reputation: 5646

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

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

Upvotes: 11

kindall
kindall

Reputation: 184211

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.)

Upvotes: 41

Praveen Gollakota
Praveen Gollakota

Reputation: 38950

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'

Upvotes: 61

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