问题：在Python中，如何在一行代码中创建n个字符的字符串？
我需要在Python中生成一个包含n个字符的字符串。使用现有的Python库是否可以实现这一目标？例如，我需要一个由10个字母组成的字符串：
string_val = 'abcdefghij'
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'
回答 0
要将同一字母重复10次：
string_val = "x" * 10 # gives you "xxxxxxxxxx"
而且，如果您想要更复杂的东西，例如n随机的小写字母，它仍然只是一行代码（不计算import语句并定义n）：
from random import choice
from string import ascii_lowercase
n = 10
string_val = "".join(choice(ascii_lowercase) for i in range(n))
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))
回答 1
前十个小写字母为string.lowercase[:10]（当然，如果您string先前已导入标准库模块，则为-）。
“使10个字符组成字符串”的其他方法：（'x'*10所有10个字符均为小写xs;-），''.join(chr(ord('a')+i) for i in xrange(10))（又是前十个小写字母），等等，等等；-）。
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;-).
回答 2
如果您只需要任何字母：
'a'*10 # gives 'aaaaaaaaaa'
如果要连续字母（最多26个）：
''.join(['%c' % x for x in range(97, 97+10)]) # gives 'abcdefghij'
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'
回答 3
为什么要“一行”？您可以将任何东西放在一行上。
假设您希望它们以“ a”开头，并且每次增加一个字符（环绕> 26），则显示以下一行：
>>> mkstring = lambda(x): "".join(map(chr, (ord('a')+(y%26) for y in range(x))))
>>> mkstring(10)
'abcdefghij'
>>> mkstring(30)
'abcdefghijklmnopqrstuvwxyzabcd'
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'
回答 4
这可能有点问题，但是对于那些对生成的字符串的随机性感兴趣的人，我的答案是：
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))
请参阅这些 答案和random.py的资料以获取更多见解。
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.
回答 5
如果可以使用重复的字母，则可以使用*运算符：
>>> 'a'*5
'aaaaa'
If you can use repeated letters, you can use the * operator:
>>> 'a'*5
'aaaaa'
