How can I adjust the repetition of for loops?

Question

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..

Any ideas and suggestions are accepted here.

Answer 1

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']

Use

import itertools as it
my_string = '1234'
my_list = it.permutations(my_string, len(my_string))
with open('your_file.txt', 'w') as f:
    for item in my_list:
        f.write("%s\n" % item)

if you want to save the result to a file. If you print a very long result to console, the console usually starts deleting the old lines.

Answer 2

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.

Answer 3

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.

Answer 4

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}")

Hope this helps :)