How to be able to retry an input for a specific key in dictionary in Python without skipping it

Question

I'm trying to make a script that will ask for 5 score input from the user. Each score corresponds to a specific dilution. The input are strings and only six "characters" are allowed to be entered and q to quit the script and show each dilution with their score that have been entered. The problem is that when a invalid "character" is used it skips the dilution completely and does not add it to the dictionary and what I want is to retry the input for the specific dilution.

dil = ["1:16", "1:32", "1:64", "1:128", "1:256"]
corr_input = ["+", "++-", "-", "+-", "-a", "A"]
scores = {}

for dil in dil:
    testscore = input("Enter score: ")
    try:
        if testscore in corr_input:
            scores[dil] = testscore
        elif testscore == "q":
            print("Done!")
            break
        else:
            print("Not a valid score!")
    except TypeError:
        print("Invalid input! Try again")
        break
print(scores)

What I get!

######### 
Enter score: +
Enter score: +
Enter score: v
Not a valid score!
Enter score: +
Enter score: +
{'1:16': '+', '1:32': '+', '1:128': '+', '1:256': '+'}`

This question is similar to: repeat an iteration of for loop. If you believe it’s different, please edit the question, make it clear how it’s different and/or how the answers on that question are not helpful for your problem. — Timur Shtatland
– Timur Shtatland, Commented Jun 26 at 21:03

Krishnan Shankar · Accepted Answer · 2025-06-26 20:45:32Z

2

The typical way of accomplishing this is with a nested loop. For every dilution, you want a sub-loop that runs until a valid score (or q) is entered.

dil = ["1:16", "1:32", "1:64", "1:128", "1:256"]
corr_input = ["+", "++-", "-", "+-", "-a", "A"]
scores = {}

for dil in dil:
    while dil not in scores:  # This loop will run until the dilution has a score
        testscore = input("Enter score: ")
        try:
            if testscore in corr_input:
                scores[dil] = testscore
            elif testscore == "q":
                print("Done!")
                exit()  # Use exit() to exit the program with 'q'
            else:
                print("Not a valid score! Try again")
        except TypeError:
            print("Invalid input! Try again")

print(scores)

Note that the above example uses exit() to break out of both loops at once, instantly ending the program. Alternatively, you will need to use a variable like to_exit to break out of the outer for loop, like so:

dil = ["1:16", "1:32", "1:64", "1:128", "1:256"]
corr_input = ["+", "++-", "-", "+-", "-a", "A"]
scores = {}

to_exit = False

for dil in dil:
    while dil not in scores:  # This loop will run until the dilution has a score
        testscore = input("Enter score: ")
        try:
            if testscore in corr_input:
                scores[dil] = testscore
            elif testscore == "q":
                print("Done!")
                to_exit = True
                break  # This will break out of the while loop but NOT the for loop
            else:
                print("Not a valid score! Try again")
        except TypeError:
            print("Invalid input! Try again")
    if to_exit:
        break  # This will break out of the for loop

print(scores)

edited Jun 26 at 20:45

answered Jun 26 at 20:39

Krishnan Shankar

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7 Comments

user30902982 Jun 27 at 9:26

I see now the the exit()/to_exit is what I was missing. On another forum someone said that I might not need the TypeError and I'm wondering if a ValueError might be better or something else. Or do you think TypeError is fine?

jackal Jun 27 at 9:27

This is unnecessarily complicated. There's no code in this answer that would cause a TypeError exception. Why have 16 lines of (significant) code when you only need 5? Also, for dil in dil: is not best practice. Sure, it works, but seriously?

jackal Jun 27 at 9:29

@user30902982 There is no need for any kind of exception handling here

user30902982 Jun 27 at 11:49

The for dil in dil was a experiment I forgot to take out of my code. The code is actually for d in dil. It was more of me not paying attention when I copied the code onto here. As for exception handling, I've removed it.

Kelly Bundy Jun 27 at 16:58

@Ramrab You don't need 5 lines. You can do it in 1:

scores = dict(zip(dils, (testscore for testscore in iter(lambda: input("Enter score: "), "q") if testscore in corr_input or print("Not a valid score!"))))

|

jackal · Accepted Answer · 2025-06-27 09:38:11Z

0

You just need a while loop that is limited by the number of available dil values and also account for user entering "q" (which would terminate the loop early).

PROMPT = "Enter score: "
INIT_SCORES = True
DIL = ["1:16", "1:32", "1:64", "1:128", "1:256"]
CORR_INPUT = {"+", "++-", "-", "+-", "-a", "A"}  # use a set here
# initialise the dictionary
# by doing this, if user enters "q" you'll easily be able
# to identify keys that do not have valid associated values
scores = {key: None for key in DIL} if INIT_SCORES else {}
dil = DIL.copy()

while dil and (testscore := input(PROMPT)) != "q":
    if testscore in CORR_INPUT:
        scores[dil.pop(0)] = testscore
    else:
        print("\aNot a valid score!")

print(scores)

Example 1:

Enter score: +
Enter score: +
Enter score: v
Not a valid score!
Enter score: +
Enter score: +
Enter score: +-
{'1:16': '+', '1:32': '+', '1:64': '+', '1:128': '+', '1:256': '+-'}

Example 2:

Enter score: ++-
Enter score: +
Enter score: q
{'1:16': '++-', '1:32': '+', '1:64': None, '1:128': None, '1:256': None}

Note:

This is destructive w.r.t the dil list. If you need that list for other purposes you'll need to take a copy and work with that

edited Jun 27 at 9:38

answered Jun 27 at 8:50

jackal

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3 Comments

user30902982 Jun 27 at 9:32

Thank you for the reply. I do need it later on for the second part which I will start on as soon as this part is complete. I was thinking of using the .item() as I need to iterate of the key-value pair till the last +/++- and print out that dil. But I'll take it one loop at a time!

jackal Jun 27 at 9:37

@user30902982 I will edit this answer to make it non-destructive. Also, your additional requirement cannot be satisfied unless you ensure that either "+" or "++-" has been input

user30902982 Jun 27 at 11:56

That is the idea, sometimes not all 5 dilutions will have a + or ++- score sometimes it's the first three of them and the fourth will be - or +- or -a as an example.

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