生成 10 位密码

Question

So I need to generate a 10-digit password (needs to use the random module) that must contain 2 lower ase letters, 2 uppercase letters, 3 special symbols and 3 numbers all in a random order every time. I've got the random password generator part done but I'm not sure how to restrict it to 2 lower case letters, 2 upper case letters, 3 special symbols and 3 numbers.

This is what I have so far:

import random
import string
lc_letter = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
uc_letter = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
symbols = ["!","@","#","$","%","^","&","*","(",")","_","+","=","-","/",">","<",",",".","?","\\"]
numbers = ["0","1","2","3","4","5","6","7","8","9"]
options = [lc_letter,uc_letter,symbols,numbers]
for i in range(10):
    choice = random.choice(options)
    digit = random.choice(choice)
    print(digit, end = '')

Answer 1

An alternative approach that I will suggest is, Take 2 letters from uppercase, lowercase etc. and then shuffle resulting password using random.shuffle method.

Answer 2

Pick every needed characters first, then shuffle them:

from random import choice as rd
from random import shuffle
import string
lc_letter = ["a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"]
uc_letter = ["A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T","U","V","W","X","Y","Z"]
symbols = ["!","@","#","$","%","^","&","*","(",")","_","+","=","-","/",">","<",",",".","?","\\"]
numbers = ["0","1","2","3","4","5","6","7","8","9"]
options = [
    rd(lc_letter),
    rd(lc_letter),
    rd(uc_letter),
    rd(uc_letter),
    rd(symbols),
    rd(symbols),
    rd(symbols),
    rd(numbers),
    rd(numbers),
    rd(numbers),
]
shuffle(options)
print(''.join(options))

Answer 3

You can use constants from string :

import random
import string

s = ""

for i in range(2):
    s = s + random.choice(string.ascii_lowercase)
for i in range(2):
    s = s + random.choice(string.ascii_uppercase)
for i in range(3):
    s = s + random.choice(string.punctuation)
for i in range(3):
    s = s + random.choice(string.digits)

s = ''.join(random.sample(s, 10))

print(s)

Answer 4

What you can actually do is pick make a list of length 10 like this:

dist = [0, 0, 1, 1, 2, 2, 2, 3, 3, 3]

This list represents the distribution of each index out of your options list. Now you can pick an index in the list:

idx = random.randint(0, len(dist))

You can pick your choice from options[dist[idx]] and pop the value at idx from dist .

Answer 5

You can use random.choice , random.sample , and constants from the string module to obtain randomly generated passwords.

import random
import string
lc_letter = string.ascii_lowercase
uc_letter = string.ascii_uppercase
# Could use string.punctuation here, but it would be different
# as your list doesn't contain semicolons or colons,
# while string.punctuation does.
symbols = ["!","@","#","$","%","^","&","*","(",")","_","+","=","-","/",">","<",",",".","?","\\"]
numbers = string.digits

lc_selection = [random.choice(lc_letter) for _ in range(2)]
uc_selection = [random.choice(uc_letter) for _ in range(2)]
symbol_selection = [random.choice(symbols) for _ in range(3)]
number_selection = [random.choice(numbers) for _ in range(3)]
print(''.join(random.sample(lc_selection + uc_selection + symbol_selection + number_selection, 10)))

Answer 6

An in my opinion better version of Yevgeniy Kosmak's solution (the stand-alone config is clearer to see, the loop avoids code duplication, and using choices instead of choice avoids a loop).

import random
import string

config = [
    (2, string.ascii_lowercase),
    (2, string.ascii_uppercase),
    (3, string.punctuation),  # or use your '!@#$%^&*()_+=-/><,.?\\'
    (3, string.digits),
]

picked = []
for k, options in config:
    picked += random.choices(options, k=k)
random.shuffle(picked)
password = ''.join(picked)

print(password)

Try it online!