妄徒之命 2010-02-13 12:23 采纳率: 100%
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如何在 Python 中使用大写字母和数字随机生成字符串?

我想生成一个大小为 n 的字符串。

它应该由数字和大写英文字母组成,例如:

  • 6U1S75
  • 4Z4UKK
  • U911K4

我怎样才能以一种毕达哥拉斯的方式实现这个目标?

转载于:https://stackoverflow.com/questions/2257441/random-string-generation-with-upper-case-letters-and-digits-in-python

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27条回答

  • perhaps? 2010-02-13 12:26
    关注

    Answer in one line:

    ''.join(random.choice(string.ascii_uppercase + string.digits) for _ in range(N))
    

    or even shorter starting with Python 3.6 using random.choices():

    ''.join(random.choices(string.ascii_uppercase + string.digits, k=N))
    

    A cryptographically more secure version; see https://stackoverflow.com/a/23728630/2213647:

    ''.join(random.SystemRandom().choice(string.ascii_uppercase + string.digits) for _ in range(N))
    

    In details, with a clean function for further reuse:

    >>> import string
    >>> import random
    >>> def id_generator(size=6, chars=string.ascii_uppercase + string.digits):
    ...    return ''.join(random.choice(chars) for _ in range(size))
    ...
    >>> id_generator()
    'G5G74W'
    >>> id_generator(3, "6793YUIO")
    'Y3U'
    

    How does it work ?

    We import string, a module that contains sequences of common ASCII characters, and random, a module that deals with random generation.

    string.ascii_uppercase + string.digits just concatenates the list of characters representing uppercase ASCII chars and digits:

    >>> string.ascii_uppercase
    'ABCDEFGHIJKLMNOPQRSTUVWXYZ'
    >>> string.digits
    '0123456789'
    >>> string.ascii_uppercase + string.digits
    'ABCDEFGHIJKLMNOPQRSTUVWXYZ0123456789'
    

    Then we use a list comprehension to create a list of 'n' elements:

    >>> range(4) # range create a list of 'n' numbers
    [0, 1, 2, 3]
    >>> ['elem' for _ in range(4)] # we use range to create 4 times 'elem'
    ['elem', 'elem', 'elem', 'elem']
    

    In the example above, we use [ to create the list, but we don't in the id_generator function so Python doesn't create the list in memory, but generates the elements on the fly, one by one (more about this here).

    Instead of asking to create 'n' times the string elem, we will ask Python to create 'n' times a random character, picked from a sequence of characters:

    >>> random.choice("abcde")
    'a'
    >>> random.choice("abcde")
    'd'
    >>> random.choice("abcde")
    'b'
    

    Therefore random.choice(chars) for _ in range(size) really is creating a sequence of size characters. Characters that are randomly picked from chars:

    >>> [random.choice('abcde') for _ in range(3)]
    ['a', 'b', 'b']
    >>> [random.choice('abcde') for _ in range(3)]
    ['e', 'b', 'e']
    >>> [random.choice('abcde') for _ in range(3)]
    ['d', 'a', 'c']
    

    Then we just join them with an empty string so the sequence becomes a string:

    >>> ''.join(['a', 'b', 'b'])
    'abb'
    >>> [random.choice('abcde') for _ in range(3)]
    ['d', 'c', 'b']
    >>> ''.join(random.choice('abcde') for _ in range(3))
    'dac'
    
    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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