如何增加 function 中的循环数？

Question

I have a code here

letters = ['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']
ret = ""
for i in range(26):
  for j in letters:
    ret += f"({letters[i]},{j})"
print(ret)

#Output: 
(a,a)(a,b)(a,c)(a,d)(a,e)(a,f)(a,g)(a,h)(a,i)(a,j)(a,k)(a,l)(a,m)(a,n)(a,o)(a,p)...(z,y)(z,z)

The problem I am trying to figure out is:

When I am trying to increase the number of permutations, for example from 2 digits to 3 digits, I have to add another loop. How do I increase the number of loops as I increase the number of digits, when I want to implement this as a function, can anyone help:(

---

This is what i did when i tried to increase the number of digits from 2 to 3:

for i in range(26):
  for k in range(26):
    for j in letters:
      ret += "(" + str(letters[i]) + "," + str(letters[k]) + "," + j + ")"
print(ret)

#Output
(a,a,a)(a,a,b)(a,a,c)(a,a,d)(a,a,e)(a,a,f)(a,a,g)(a,a,h)(a,a,i)(a,a,j)(a,a,k)(a,a,l)(a,a,m)...(z,z,z)

How do i implement this in a function?

Answer 1

You can achieve this using the itertools like this

import itertools

def perm_func(arr, x):
    ret_str = ""
    p for p in itertools.product(arr, repeat=x):
        ret_str += str(p)

    return ret_str