在Python中，如何在一行代码中创建n个字符的字符串？

I need to generate a string with n characters in Python. Is there a one line answer to achieve this with the existing Python library? For instance, I need a string of 10 letters:

string_val = 'abcdefghij'

To simply repeat the same letter 10 times:

string_val = "x" * 10  # gives you "xxxxxxxxxx"

And if you want something more complex, like n random lowercase letters, it’s still only one line of code (not counting the import statements and defining n):

from random import choice
from string import ascii_lowercase
n = 10

string_val = "".join(choice(ascii_lowercase) for i in range(n))

The first ten lowercase letters are string.lowercase[:10] (if you have imported the standard library module string previously, of course;-).

Other ways to “make a string of 10 characters”: 'x'*10 (all the ten characters will be lowercase xs;-), ''.join(chr(ord('a')+i) for i in xrange(10)) (the first ten lowercase letters again), etc, etc;-).

if you just want any letters:

 'a'*10  # gives 'aaaaaaaaaa'

if you want consecutive letters (up to 26):

 ''.join(['%c' % x for x in range(97, 97+10)])  # gives 'abcdefghij'

Why “one line”? You can fit anything onto one line.

Assuming you want them to start with ‘a’, and increment by one character each time (with wrapping > 26), here’s a line:

>>> mkstring = lambda(x): "".join(map(chr, (ord('a')+(y%26) for y in range(x))))
>>> mkstring(10)
'abcdefghij'
>>> mkstring(30)
'abcdefghijklmnopqrstuvwxyzabcd'

This might be a little off the question, but for those interested in the randomness of the generated string, my answer would be:

import os
import string

def _pwd_gen(size=16):
    chars = string.letters
    chars_len = len(chars)
    return str().join(chars[int(ord(c) / 256. * chars_len)] for c in os.urandom(size))

See these answers and random.py‘s source for more insight.

If you can use repeated letters, you can use the * operator:

>>> 'a'*5

'aaaaa'