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How to code options for a CYOA game(in Python)?

really new to python and stack overflow, and I'm currently trying out a choose your own adventure game. I'm already a bit familiar with if-else statements but I need a little help in a specific mechanic I want in the game. Basically, the character will have a certain set of choices so like:
print('[a]Wash the Dishes')
print('[b]Feed the Dog')
print('[c]Brush Teeth')
input = choice('What should I do? ')
So how could I make it that once the character inputs their choice, it'll show the same set of options, but now removing the one already selected?

game_choice = {'a':"Wash the Dishes",'b':"Feed the Dog",'c':"Brush Teeth"}
# printing the choices to user
for key in game_choice:
print(f"[{key}]: {game_choice[key]}")
# Iteratively asking user for input
while(len(game_choice)>0):
choice = input('What should I do? ')
if choice in game_choice:
print(game_choice[choice])
game_choice.pop(choice)
else:
print("You Entered wrong choice, Please try again")
Above we created a Dict() to store all the options and values in key pair groups.
You can use a similar approach to do build your game.

choices = {'a': 'Wash the Dished', 'b': 'Feed the Dog', 'c': 'Brush teeth'}
def print_choices():
for key in choices:
print(f'[{key}]: {choices[key]}')
# by using f'' you can use Curly Braces to call variables like what i did here
# instead of doing .format or 'text' + variable + 'text' or ...
def choice(txt):
# your code
input_value = input(txt)
choices.pop(input_value, None)
# None here causes the dict to NOT raise an error if the key doesn't exist in the dictionary
print_choices()
input_value = choice('What should I do? ')
This answer is based on considering you already know what a dictionary is in python.
Also if you wanted to have a message shown to the player if their input is not VALID you could do this instead:
def choice(txt):
# your code
input_value = input(txt)
if input_value not in choices:
print("Invalid Input")
else:
choices.pop(input_value)

Issues with lists? Error checking not working

I am relatively new to python, and I just started learning how to use classes. This is the first program I've made where I've tried to integrate them, but I'm coming up with a small issue I can't seem to fix, and I think it has to do with lists. The code is as follows:
(The topic is getting the user to choose what type of seat to purchase).
class SeatBooking:
def __init__(self, seat):
self.seat = seat
possible_types = []
possible_types.extend(["Low_Economy", "Standard_Economy", "High_Economy",
"Business", "First", "Residence"])
possible_types = " ".join(possible_types)
while True:
if self.seat not in possible_types:
print("Sorry, but this is not a valid answer. Please try again!")
self.seat = str(input("What type of ticket would you like? The possible types are: {} "
.format(possible_types)))
else:
print("You have chosen to book a {} ticket.".format(self.seat))
confirmation = str(input("Please confirm with 'Yes' or 'No': ")).lower()
if confirmation == "yes":
print("Excellent decision! Ready to continue")
print("=" * 170)
break
elif confirmation == "no":
self.seat = str(input("What type of ticket would you like? The possible types are: {} "
.format(possible_types)))
else:
print("That doesn't seem to be a valid answer.")
Here is the main file (to execute the different classes I'll make):
import type_seat
# Choose the seat to book
print("=" * 170)
print("Welcome to Etihad! This program can help you organize your flight, payments and usage of miles!")
possible_types = []
possible_types.extend(["Low_Economy", "Standard_Economy", "High_Economy",
"Business", "First", "Residence"])
possible_types = " ".join(possible_types)
seat_type = str(input("What type of ticket would you like? The possible types are: {}. "
.format(possible_types)))
type_seat.SeatBooking(seat_type)
The problem I have is that I seem to be able to enter certain letters and it doesn't count them as an error even though they're not one of the available seats. For example, when I enter the letters "h" or "s", my error checking part of the code doesn't respond to it, but when I enter the letter "b" or random words like "try" it does. It doesn't seem to be completely random though, and it seems to only happen with letters or parts of the first 3 'items' in the possible_types[] list. However, I haven't tested this fully. This is why I thought it had something to do with lists, so if anyone knows what's causing this, I'd really appreciate it if they could help me resolve this and perhaps help me from repeating this mistake in the future!
Note, for the lists I am using .join, but I also tried str().

You don't have a list, you are testing characters against one long string:
possible_types = " ".join(possible_types)
The letters h and s are in that string (in the words High_Economy and Business, respectively), but the sequence try doesn't appear anywhere in the string.
If you only wanted to allow whole words to match, you'd need to leave possbile_types a list, or ideally convert it to a set (as sets allow for fast membership testing). You can define the list, no need for list.extend() here:
possible_types = ["Low_Economy", "Standard_Economy", "High_Economy",
"Business", "First", "Residence"]
or make it a set by using {...}:
possible_types = {"Low_Economy", "Standard_Economy", "High_Economy",
"Business", "First", "Residence"}
Do not join this into a string, just test directly against the object:
if self.seat not in possible_types:
If you still need to show the values to a user in an error message, join the values then, or store the str.join() result in a different variable for that purpose.
Note that you shouldn't deal with user input validation in the class __init__ method. Leave user interaction to a separate piece of code, and create instances of your class after you validated. That way you can easily swap out user interfaces without having to adjust all your data objects too.

possible_types = " ".join(possible_types)
Above statement will create one string as "Low_Economy Standard_Economy High_Economy Business First Residence".
Now you are doing
if self.seat not in possible_types:
This will check for a particular character in the string present or not. In your case you are finding 'h' which is present and 'try' which isn't.
Your program will work if you remove this statement
possible_types = " ".join(possible_types)

NameError when programming in python

I made a program in python that is supposed to accept a name as user input. It will then check if the name given is contained inside a string that is already given and if it is then the program will print out the telephone next to that name. My code is as follows:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
check=str(input("Give a name: "))
for check in tilefwnikos_katalogos:
if check=="Christoforos":
arxi=check.find("Christoforos")
elif check=="Eirini":
arxi=check.find("Eirini")
elif check=="Costas":
arxi=check.find("Costas")
elif check=="George":
arxi=check.find("George")
elif check=="Panayiotis":
arxi=check.find("Panayiotis")
elif check=="Katerina":
arxi=check.find("Katerina")
s=check.find(" ",arxi)
arxi=s
y=check.find(":",arxi)
telos=y
apotelesma=tilefwnikos_katalogos[arxi+1:telos]
print(apotelesma)
But when I try to run it, I input the name and then the following message pops up:
Traceback (most recent call last):
File "C:\Users\Sotiris\Desktop\test.py", line 16, in <module> s=check.find(" ",arxi)
NameError: name 'arxi' is not defined
What am I doing wrong?

You're getting your error because arxi isn't getting defined in the first place when then name the user gave is not present on your list.You can fix that by simply adding an unconditional else case to your if/else if bundle as pointed in the comments. But the very way you tackled this problem is faulty, storing data like this in a string is a bad idea, you want to use a dictionary:
phone_catalog = {'Christoforos': 99111111, 'Eirini': 99556677, 'Costas': 99222222, 'George':99333333, 'Panayiotis':99444444, 'Katerina': 96543217}
Also check isn't a very clear variable name, maybe you should try using something better like:
user_name = str(input("Give a name: "))
And now you can do your if/elif condition but replacing it for using dictionary logic and making sure you have a final else, like such:
if user_name in phone_catalog:
print(phone_catalog[user_name])
else:
print("Unknown user")
See how the dictionary made your life much easier and your code cleaner here? Read more on Python Data Structures.

so there are a few things you have overlooked / not going as expected, the first of which is how iterating over strings in python works:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
for check in tilefwnikos_katalogos:
print(check)
#print(repr(check)) #this shows it as you would write it in code ('HI' instead of just HI)
so check can never be equal to any of the things you are checking it against, and without an else statement the variable arxi is never defined. I'm assuming you meant to use the check from the user input instead of the one in the loop but I'm not sure you need the loop at all:
tilefwnikos_katalogos = "Christoforos 99111111: Eirini 99556677: Costas 99222222: George 99333333: Panayiotis 99444444: Katerina 96543217"
check=str(input("Give a name: ")) #the str() isn't really necessary, it is already a str.
if check=="Christoforos":
arxi=check.find("Christoforos")
elif check=="Eirini":
arxi=check.find("Eirini")
elif check=="Costas":
arxi=check.find("Costas")
elif check=="George":
arxi=check.find("George")
elif check=="Panayiotis":
arxi=check.find("Panayiotis")
elif check=="Katerina":
arxi=check.find("Katerina")
else: raise NotImplementedError("need a case where input is invalid")
s=check.find(" ",arxi)
arxi=s
y=check.find(":",arxi)
telos=y
apotelesma=tilefwnikos_katalogos[arxi+1:telos]
print(apotelesma)
but you could also just see if check is a substring of tilefwnikos_katalogos and deal with other conditions:
if check.isalpha() and check in tilefwnikos_katalogos:
# ^ ^ see if check is within the string
# ^ make sure the input is all letters, don't want to accept number as input
arxi=check.find(check)
else:
raise NotImplementedError("need a case where input is invalid")
although this would make an input of C and t give Cristoforos' number since it retrieves the first occurrence of the letter. An alternative approach which includes the loop (but not calling the variable check!) would be to split up the string into a list:
tilefwnikos_katalogos = "..."
check = input(...)
for entry in tilefwnikos_katalogos.split(":"):
name, number = entry.strip().split(" ")
if check == name:
apotelesma=number
break
else:
raise NotImplementedError("need a case where input is invalid")
although if you are going to parse the string anyway and you may use the data more then once it would be even better to pack the data into a dict like #BernardMeurer suggested:
data = {}
for entry in tilefwnikos_katalogos.split(":"):
name, number = entry.strip().split(" ")
data[name] = number #maybe use int(number)?
if check in data:
apotelesma = data[check]
else:
raise NotImplementedError("need a case where input is invalid")

Nested Loop 'If'' Statement Won't Print Value of Tuple

Current assignment is building a basic text adventure. I'm having trouble with the following code. The current assignment uses only functions, and that is the way the rules of the assignment state it must be done.
def make_selections(response):
repeat = True
while repeat == True:
selection = raw_input('-> ')
for i, v in enumerate(response):
i +=1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
if selection == i:
print v[1]
else:
print "There's an error man, what are you doing?!?!?"
firstResponse = 'You chose option one.'
secondResponse = 'You chose option two.'
thirdResponse = 'You chose option three.'
responses = [(0, firstResponse), (1, secondResponse),( 0, thirdResponse)]
make_selections(responses)
My intention in that code is to make it so if the user selects a 1, it will return firstResponse, if the user selects 2 it will return secondResponse, etc.
I am basically just bug testing the code to make sure it produces the appropriate response, hence the "Error man..." string, but for some reason it just loops through the error message without printing the appropriate response string. Why is this?
I know that this code is enumerating the list of tuples and I can call them properly, as I can change the code to the following and get the expected output:
for i, v in enumerate(response):
i += 1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
print i, v
Also, two quick asides before anyone asks:
I know there is currently no way to get out of this while loop. I'm just making sure each part of my code works before I move on to the next part. Which brings me to the point of the tuples.
When I get the code working, a 0 will produce the response message and loop again, asking the user to make a different selection, whereas a 1 will produce the appropriate response, break out of the loop, and move on to the next 'room' in the story... this way I can have as many 'rooms' for as long of a story as I want, the player does not have to 'die' each time they make an incorrect selection, and each 'room' can have any arbitrary amount of options and possible responses to choose from and I don't need to keep writing separate loops for each room.

There are a few problems here.
First, there's no good reason to iterate through all the numbers just to see if one of them matches selection; you already know that will be true if 1 <= selection <= len(response), and you can then just do response[selection-1] to get the v. (If you know anything about dicts, you might be able to see an even more convenient way to write this whole thing… but if not, don't worry about it.)
But if you really want to do this exhaustive search, you shouldn't print out There is an error man after any mismatch, because then you're always going to print it at least twice. Instead, you want to only print it if all of them failed to match. You can do this by keeping track of a "matched" flag, or by using a break and an else: clause on your for loop, whichever seems simpler, but you have to do something. See break and continue Statements, and else Clauses on Loops in the tutorial for more details.
But the biggest problem is that raw_input returns a string, and there's no way a string is ever going to be equal to a number. For example, try '1' == 1 in your interactive interpreter, and it'll say False. So, what you need to do is convert the user's input into a number so you can compare it. You can do that like this:
try:
selection = int(selection)
except ValueError:
print "That's not a number!"
continue

Seems like this is a job for dictionaries in python. Not sure if your assignment allows this, but here's my code:
def make_selections(response):
selection = raw_input('-> ')
print response.get(selection, err_msg)
resp_dict = {
'1':'You chose option one.',
'2':'You chose option two.',
'3':'You chose option three.'
}
err_msg = 'Sorry, you must pick one of these choices: %s'%sorted(resp_dict.keys())
make_selections(resp_dict)

The problem is that you are comparing a string to an integer. Selection is raw input, so it comes in as a str. Convert it to an int and it will evaluate as you expect.
You can check the type of a variable by using type(var). For example, print type(selection) after you take the input will return type 'str'.
def make_selections(response):
repeat = True
while repeat == True:
selection = raw_input('-> ')
for i, v in enumerate(response):
i +=1 # adds 1 to the index to make list indices correlate to a regular 1,2,3 style list
if int(selection) == i:
print v[1]
else:
print "There's an error man, what are you doing?!?!?"

Dictionaries: How to get duplicate input by the user and count how many times it was entered?

names = []
print("Duplicate names won't be in the list!") # reminder
while True:
userInput = input("Enter a name: ")
if userInput == "": # condition to terminate loop
print("You've entered these names, duplicated names won't be included")
print(names)
break
elif userInput not in names: # condition to see if input is duplicated
names.append(userInput) # if satisfied, adds it to the list
else:
print("value exists!") # reminder about the value entered is duplicate
That's my code so far, I don't know how could I count how many times a specific duplicate value was entered, a Dictionary will solve it a lot of people say but I am not too familiar yet with it. In the output, When the input of names is complete, the console user should be able to search the list for
names by simply typing the name at the console. The program should output either “Not
found” or display the name and the number of times it was entered.
When the searching is complete, the console user should be able to delete names from the list
by typing the name. If a name is not found, the program should output “Not found”. If a name
is found then it should be removed from the list and the name is displayed along with a
“Deleted” message to the user e.g. “Deleted Ted”.
Thank you very much to people that would help!

You should use a Counter.
c = Counter() #Initialize the counter
print("Duplicate names won't be in the list!")
while True:
userInput = input("Enter a name: ") #input used to be raw_input in Python 2
if userInput == "":
print("You've entered these names, duplicated names won't be included")
print(c.keys())
break
elif userInput not in c:
c.update([userInput]) # if satisfied, adds it to the counter. You have to provide a list, hence the [ ]. If you provide a string, it will be broke down into characters.
else:
print("value exists!")
In order to show a specific count,
print c[word]
You can also see the most common terms, etc.
However your code forbids that a name be entered more than once, but if you want to count the number of times people enter your name, this is the way to go.

A dictionary consists of a key and a value, a key in a dictionary should always be unique, so unless you have a unique key with every value you use, it wouldn't support duplicate values and will not suit your purposes.