如何在python中创建一个六进制随机数？

Question

I want to create a file of hexadecimal number where everyone is composed of 32 elements like this:

ccddeeff8899aabb4455667fffffff33
e0370734313198a2885a308aaaaaaaa8
7354776f204f6e65204e696bbbbbbb6f
64976fbb4f6e6ee0cc681e6ccccccc77

I try by this code to create my numbers but I don't now how to have 32 elements for each number:

   import random
   Plaintext_file = open("C:\\Users\\user\\Plaintexts.txt", "w")
   for i in range(5):
      i = random.randint(0, 16777215)
      print "%x" % i 
      Plaintext_file.write("%x \n" % i)

The result that I have:

c39ea9
a737a0
d2d352
fcebf1
ade761

I would be very grateful if you could help.

Answer 1

Use binascii.b2a_hex to convert binary data to a line of ASCII characters,

>>> import os,binascii
>>> binascii.b2a_hex(os.urandom(16))
'a8922d48fba3bddd0214a338ce090ea6'

os.urandom(n) returns a string of n random bytes from an OS-specific randomness source.

Answer 2

>>> import uuid
>>> uuid.uuid4().hex
'd734fde6d45e47e99d06f129b5c128f8'

Answer 3

You can use random.choice using the list of accepted hex chars (from string.hexdigits ):

>>> import string
>>> string.hexdigits
'0123456789abcdefABCDEF'
>>> import random
>>> "".join([random.choice(string.hexdigits) for x in range(32)])
'37bAA921dd6BE09eEff45c280D62FFAb'

If you want only the lowercase you case use string.hexdigits[:16] :

>>> import string
>>> string.hexdigits[:16]
'0123456789abcdef'
>>> import random
>>> "".join([random.choice(string.hexdigits[:16]) for x in range(32)])
'805cb6c9b38515b588bfec42613eff9d'