[FIXED] Cycle through letters of the alphabet in a loop and assign values to them in Python?

Issue

Here the number of variables (which correspond to numfactors) are assigned manually to each letter of the alphabet. Now I could just make a very ugly if-then sequence for each possible letter, but it seems there has to be a better way. Current method:

indexes = list()
for i in range(numfactors):
    indexes.append("paths[" + str(curvelen*i) + ":" + str(curvelen*(i+1)) + "]")

A=eval(indexes[0])
B=eval(indexes[1])
C=eval(indexes[2])
D=eval(indexes[3])

Okay that's hardcoded, and 4 values are expected A, B, C, D. But the actual number could be 1 to 20 for example. Something like this:

for i in range(indexes):
# loop 1
   A=eval(indexes[i])
# loop 2
   B=eval(indexes[i])
# loop 3
   C=eval(indexes[i])
# ... to end

I think that summarizes the problem. Each loop goes through A, B, C, D, E, F, G, H, I, J, K, L, M, N, O, P etc. and assigns the letter-variable value to indexes[i]. Any help is appreciated!

Update: thanks to @Diego Torres Milano and the linked post, this is all done in a few lines of code:

import string

# create path indexing for each input curve
indexes = list()
for i in range(numfactors):
    indexes.append("paths[" + str(curvelen*i) + ":" + str(curvelen*(i+1)) + "]")

# make a dictionary from A to Z for variables
variables = dict(zip(string.ascii_uppercase, indexes))

# assign each key name as variables (A-Z) to the paths defined prior
for k, v in variables.items():
    exec(f"{k} = {v}")

Solution

You can create a dictionary with uppercase letters as key using

import string
dict(zip(string.ascii_uppercase, indexes))

Answered By - Diego Torres Milano