How can I adjust the repetition of for loops?-CodePudding

I wanted to think about this problem personally But I know there are experienced people here who have great solutions. I'm trying to create a code number generator and I will improve that to includes all letter cases. But my problem is that for example, for an 8-letter string, I have to copy the for loop eight times, and I can not say how many strings I want by setting a number. Now I want to ask if there is a solution that prevents for for duplication in the code and can only be achieved by setting a generate number?

myPass = []
print("Calculate started..")
for a in string.digits:
    for b in string.digits:
        for c in string.digits:
            for d in string.digits:
                for e in string.digits:
                    for f in string.digits:
                        for g in string.digits:
                            for h in string.digits:
                                myPass.append(a   b   c   d   e   f   g   h)

print("Calculate finish..")

For example, I want to have a function that performs the above process by just setting a number. This is how I can adjust the number of strings:

def Generate(lettersCount):
    print("Generate for loops for 12 times..")  # for e.g.
    print("12 letters passwords calculated..")  # for e.g.

Generate(12) # 12 for loop's generated..

Please help me and any help is appreciated.

CodePudding user response：

Do you want something like this?

import itertools as it
my_string = '1234'
s = it.permutations(my_string, len(my_string))
print([x for x in s])

Output: [('1', '2', '3', '4'), ('1', '2', '4', '3'), ('1', '3', '2', '4'), ('1', '3', '4', '2'), ('1', '4', '2', '3'), ('1', '4', '3', '2'), ('2', '1', '3', '4'), ('2', '1', '4', '3'), ('2', '3', '1', '4'), ('2', '3', '4', '1'), ('2', '4', '1', '3'), ('2', '4', '3', '1'), ('3', '1', '2', '4'), ('3', '1', '4', '2'), ('3', '2', '1', '4'), ('3', '2', '4', '1'), ('3', '4', '1', '2'), ('3', '4', '2', '1'), ('4', '1', '2', '3'), ('4', '1', '3', '2'), ('4', '2', '1', '3'), ('4', '2', '3', '1'), ('4', '3', '1', '2'), ('4', '3', '2', '1')]

Edit: Use print(["".join(x) for x in s]) if you want to add to get strings. Output: ['1234', '1243', '1324', '1342', '1423', '1432', '2134', '2143', '2314', '2341', '2413', '2431', '3124', '3142', '3214', '3241', '3412', '3421', '4123', '4132', '4213', '4231', '4312', '4321']

CodePudding user response：

You can make the recursive function like the following.

class PasswordGenerator():
    def __init__(self):
        self.password_list = []

    def generate_password(self, len, added_string=""):
        if len == 0:
            self.password_list.append(added_string)
        else:
            for i in string.digits:
                self.generate_rand_with_for(len - 1, i   added_string)

Then you can use this class to get the list of passwords.

password_gen = PasswordGenerator()
password_gen.generate_password(12)
print(password_gen.password_list)

Or you can implement this using the python generator.

import string
from random import choices

def generate_random_string(len):
    while True:
        yield ''.join(choices(string.ascii_letters   string.digits, k = len))

gen = generate_random_string(12)

Then you can get one string from this generator at any time.

print(next(gen))

Or you can get any number of passwords like the following

number_of_passwords = 100000
for index, item in enumerate(gen_loop):
    print(item)
    if index == number_of_passwords:
        break

Hope it could help.

CodePudding user response：

In python there is a function called ord(). This function returns the unicode value of a character. Digits from 0 to 9 are also characters. We can view the unicode values of the characters from '0' to '9' as follows...

for c in ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']:
print('Character : ', c, ' and Unicode value : ', ord(c))

You will ge a out put like this...

Character :  0  and Unicode value :  48
Character :  1  and Unicode value :  49
Character :  2  and Unicode value :  50
Character :  3  and Unicode value :  51
Character :  4  and Unicode value :  52
Character :  5  and Unicode value :  53
Character :  6  and Unicode value :  54
Character :  7  and Unicode value :  55
Character :  8  and Unicode value :  56
Character :  9  and Unicode value :  57

There is a function in the module "random" called "randint()"

random.randint(a, b)

Return a random integer N such that a <= N <= b. Alias for randrange(a, b 1).

Now considering that your paawrod will only contain digits from '0' to '9', you can solve your problem with the code below ()...

def passwordGenerator(password_length):
    password = ''
    for length in range(password_length):
        password  = chr(random.randint(48, 57))
    return password

print(passwordGenerator(12))

Few examples of the generated passwords are given below...

852501224302
501575191222
271006502875
914595005843

The function chr() in python returns the string representation from the unicode value.

CodePudding user response：

Are you trying to make a password generator ?

That can be accomplished with the random module and a single for loop

all_symbols = "0123456789ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz"
import random

def password_gen():
    return ''.join(random.choice(all_symbols)for i in range(15))
    
password = password_gen()

    
print(f"Secure password - {password}")