Question

I am having issues trying to get a while loop to validate the users input, and then make sure the user doesn’t repeat values. Shown bellow are the two methods I have tried, but I can’t get my head around how to get them to work.

Method 1

def test():
my_list = ["", "", ""]
for i in range(3):
    while (my_list[i] != "one") and \
          (my_list[i] != "two") and \
          (my_list[i] != "three"):

        while (my_list[i] == my_list[0]) and \
              (my_list[i] == my_list[1]) and \
              (my_list[i] == my_list[2]):

            text = "Enter, one, two or three", i + 1, ":"
            try:
                my_list[i] = input(text)
            except KeyboardInterrupt:
                sys.exit()

print(my_list)

Method 2

def test2():

my_list= ["", "", ""]
while len(my_list)!=len(set(my_list)) == True:
    for c in range(4):
        while (my_list[i] != "one") and \
              (my_list[i] != "two") and \
              (my_list[i] != "three"):
            text = "Enter, one, two or three", c + 1, ":"
            try:
                my_list[c] = input(text)
            except KeyboardInterrupt:
                sys.exit()


print(my_list)

Answer 1

Why not simplify it a bit:

my_list = []
while len(data) < 3:
    data = input(text) # Valid for Python 3, use raw_input(text) in Python 2
    if data in ("one", "two", "three") and data not in my_list:
        my_list.append(data)

This is a more-or-less direct translation of your requirements:

1. I have a list
2. While the list is too small
   1. Get some data
   2. If the data is one of the valid values, and not already in the list
      1. Add it to the list

Using for x in range(y) in this case just adds complexity, because you don’t actually know how many times you want the loop to run.

There’s no reason to pre-fill the list with invalid values (my_list = ["", "", ""]), since lists can be resized with append.

Also, your sys.exit code is unnecessary and probably harmful. Just let the exception propagate and it will crash the program itself (unless you catch it, which you shouldn’t almost never do for a KeyboardInterrupt exception).

Note: Your first version would work if you combined the two while loops into one, it’s just unnecessarily complicated.