需要有关随机字符串生成的帮助 python

Question

What I want is to generate a string in this specific format: l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l With each l and d being a different string or number. The issue is when I try to generate, the whole thing is the same value/string. But I want it different. Here is an example: What I am getting: lll9999l9llll9999l9llll9999l9llll9999l9l What I need: bfb7491w3anfr4530x2zzbg9891u2rbep8421m9s

def id_gen():
    l = random.choice(string.ascii_lowercase)
    d = random.choice(string.digits)
    id = l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l+l+l+l+d+d+d+d+l+d+l
    print(id)

The result:

lll9999l9llll9999l9llll9999l9llll9999l9l

I need this to generate something different:)

Answer 1

This seems to work for me:

def gen_id() :
    pattern = 'lllddddldllllddddldllllddddldllllddddldl'
    digits = [random.choice(string.digits) for i in range(len(pattern))]
    letters = [random.choice(string.ascii_lowercase) for i in range(len(pattern))]
    return ''.join( digits[i] if pattern[i] == 'd' else letters[i] for i in range(len(pattern)) )

testing:

>>> gen_id()
'lnx1066k0hnrd5409d1nhgo1254t6rzyw5165f8v'
>>> gen_id()
'sbc7119f4ythd8845i1afay1900f4wjcv0659b4e'
>>> gen_id()
'yan6228r0nebj5097y7jnwh7065s7osra0391j5f'
>>> 

seems different enough... please, don't forget to import string, random =)

Answer 2

To not consume the random generator, IMHO this is the best solution:

def gen_id(pattern) :
    l = len(pattern)
    d = pattern.count('d')
    digits = random.choices(string.digits, d)
    letters = random.choices(string.ascii_lowercase, l-d)
    return ''.join( digits.pop() if pattern[i] == 'd' else letters.pop() for i in range(l) )

Answer 3

You can use this to get a random combination of letters and digits in the desired order:

def letter():
    return random.choice(string.ascii_lowercase)

def digit():
    return random.choice(string.digits)

def id_gen():
    return letter() + digit() + letter() + letter()  # ldll