I need to generate random text strings of a particular format. Would like some ideas so that I can code it up in Python. The format is <8 digit number><15 character string>.
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5 Answers
#!/usr/bin/python
import random
import string
digits = "".join( [random.choice(string.digits) for i in xrange(8)] )
chars = "".join( [random.choice(string.letters) for i in xrange(15)] )
print digits + chars
EDIT: liked the idea of using random.choice better than randint() so I've updated the code to reflect that.
Note: this assumes lowercase and uppercase characters are desired. If lowercase only then change the second list comprehension to read:
chars = "".join( [random.choice(string.letters[:26]) for i in xrange(15)] )
Obviously for uppercase only you can just flip that around so the slice is [26:] instead of the other way around.
4 Comments
Claes Mogren
Nice answer! (I had never even seen random.choice before.)
Gaslight Deceive Subvert
Probably more readable to use string.lowercase and string.uppercase than slicing the list. Also the solution only holds if the OP is satisfied with only ASCII characters, if he wants to generate strings from the whole unicode character set the problem becomes much harder.
Benjamin Hodgson
You can pass a generator expression in as the argument, rather than explicitly write a list comprehension:
digits = "".join(random.choice(string.digits) for i in xrange(8))
Jubbles
Great answer. If you're using Python 3, replace
string.letters with string.ascii_letters and range() can be used instead of xrange().See an example - Recipe 59873: Random Password Generation .
Building on the recipe, here is a solution to your question :
from random import choice
import string
def GenPasswd2(length=8, chars=string.letters + string.digits):
return ''.join([choice(chars) for i in range(length)])
>>> GenPasswd2(8,string.digits) + GenPasswd2(15,string.ascii_letters)
'28605495YHlCJfMKpRPGyAw'
>>>
1 Comment
loretoparisi
be aware of using
ascii_letters for Python3random.sample is an alternative choice. The difference, as can be found in the python.org documentation, is that random.sample samples without replacement. Thus, random.sample(string.letters, 53) would result in a ValueError. Then if you wanted to generate your random string of eight digits and fifteen characters, you would write
import random, string
digits = ''.join(random.sample(string.digits, 8))
chars = ''.join(random.sample(string.letters, 15))
1 Comment
Jeff Evans
In Python 3,
string.letters has been replaced by string.ascii_letters. See: docs.python.org/3.1/library/string.htmlShorter version since python 3.6.2, with random.choices over random.choice which doesn't need a for loop, but instead just pass k, the length of the random string required.
import random
import string
x = ''.join(random.choices(string.ascii_letters + string.digits, k=16))
print(x)
You can also add string.punctuation if you need string with punctuation characters.
Comments
Here's a simpler version:
import random
import string
digits = "".join( [random.choice(string.digits+string.letters) for i in xrange(10)] )
print digits
