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Is there a way to make while True loop to show randomly in Python?

I'm really new in Coding, with Python.
I was trying to make a Vocabulary exercise program for a Language that i am learning it right now. So the concept is, if the word "abhängen" is shown at the Console, i have to write "von" which is the right word to come after that word, which is "abhängen". And the program will show if its right or wrong, and loops the input to get the right answer.
But since there are tons of vocabulary, i have to make same loop over and over again just by using while True and changing a,b,c for the variables and the word between "". Is there a way to make it shorter maybe by using list or something?
And if its possible, can i somehow make the order of the questions randomly? Since this code always shows the first question as abhängen and second as abrechnen.
Sorry if this was some kind of dumb question to ask, have nowhere to ask haha
have a nice day guys
while True:
a = input("abhängen ")
if a == "von":
print("You're right")
break
else:
print("Wrong")
while True:
c = input("abrechnen ")
if c == "mit":
print("You're right")
break
else:
print("Wrong")

It looks like a great example to learn usage of lists in Python.
You can use list of tuples and random module for this task. See random.shuffle documentation here
words_and_answers = [("abhängen ", "von"), ("abrechnen ", "mit")]
random.shuffle(words_and_answers)
for input_word, correct_answer in words_and_answers:
while True:
user_response = input(input_word)
if user_response == correct_answer:
print("You're right")
break
else:
print("Wrong")

I think a possible solution is to use a dictionary in which the keys are the words "abhängen", "abrechnen", ecc... And the values are the correct words that come after. So you can write a thing like this:
vocabulary = {
"abhängen" : "von",
"abrechnen" : "mit"
}
for i in vocabulary:
a = input(i + " ")
if a == vocabulary[i]:
print("You are right")
else:
print("Wrong")
Note that in this case the loop is not infinite! This method allows you to have more than one word correct for each key
For the randomly access to the keys, you can make a list with the keys and use the random.shuffle() method, like this:
import random
keys = list(vocabulary.keys())
random.shuffle(keys)
And then your loop will be:
for i in keys:
...

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)

Why am I receiving "incorrect" output in this ifelse statement?

I am new to python and trying to learn by doing small projects.
I am trying to write a program that displays the names of the four properties and
asks the user to identify the property that is not a railroad. The user should be informed if the selection is correct or not.
properties = "Reading,","Pennsylvania","B & O","Short Line"
question = str(input("Which is not a railroad?")) **Short Line**
if properties == "Short Line":
print("correct")
else:
print("incorrect")
However, my final output shows as "incorrect", what am i doing wrong?
The four railroad properties
are Reading, Pennsylvania,
B & O, and Short Line.
Which is not a railroad? Short Line
Correct.
Short Line is a bus company.

Couple things I see with this code you have posted.
First, not sure if you actually have **Short Line** in your actual code but if you are trying to comment use # That way it won't be interpreted at run time.
Second as mentioned in other answers you are checking against properties which is pulling in your array. You should be checking against your input which is stored at question.
properties = "Reading,","Pennsylvania","B & O","Short Line"
question = str(input("Which is not a railroad?")) # **Short Line**
if question == "Short Line": # replaced properties with question
print("correct")
else:
print("incorrect")
print(properties)
print(question)
I find that when I am having troubles understanding why something is not working I throw some print statements in to see what the variables are doing.

You may want to catch the user in a loop, otherwise you would have to constantly have to run the code to find the correct answer(unless that is the desired behavior, then you can leave it as you have it). Also, be aware that you may want to uppercase or lowercase because a user may provide the answer as "Short line" (lower case "L"), and the code will return as incorrect. Of course, that depends on what you will accept as an answer.
Sample
print ("Reading,Pennsylvania,B & O, or Short Line. Which is not a railroad?")
user_input = input("Please provide an answer: ")
# != the loop will close once the user inputs short line in any form
# The upper.() will convert a user_input string to all caps
while user_input.upper() != "SHORT LINE":
print ("Incorrect, Please try again.")
user_input = input("Which one is not a railroad? ")
print ("Correct")

Prettied it up for you
print( "Reading, Pennsylvania, B & O, and Short Line. Which is not a railroad?" )
print("Which is not a railroad?")
answer = input()
if answer == "Short Line":
print("correct")
else:
print("incorrect")

Index Out Of Range When Artificially Limited

when I run this program, sometimes I receive an error.This error however is not possible as I am using an 8x8 grid and I limit the inputs so that they can only be numbers from 0-7, to obey the fact that list indexes start at 0.
The user must input coordinates (1-8),(A-H) and the program will check to see if those coordinates are correct, by systematically going through the CompShips list and repeatedly comparing those coordinates to ones given by the user. If the cords match, then a message will appear and a "Z" will change to an "X" on those coordinates, indicating a HIT. If the guess does not match, a "Z" will change to an "M" on those coordinates indicating a MISS.
CompShips=[[1,0],[1,1],[2,2],[2,3],[2,4],[3,0],[3,1],[3,2],[5,4],[5,5],[5,6],[5,7],[1,7],[2,7],[3,7],[4,7],[5,7]]
FRow1=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow2=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow3=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow4=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow5=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow6=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow7=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow8=["Z","Z","Z","Z","Z","Z","Z","Z",]
def PrintFireBoard():
print(Index)
print(FRow1)
print(FRow2)
print(FRow3)
print(FRow4)
print(FRow5)
print(FRow6)
print(FRow7)
print(FRow8)
FireBoard=[FRow1,FRow2,FRow3,FRow4,FRow5,FRow6,FRow7,FRow8]
while len(CompShips) !=0 or CompSuccess==17:
FireRow=input("Please Choose The Row That You Wish To Fire Upon (1-8) ")
FireIndex=input("Please Choose The Column That You Wish To Fire Upon (A-H) ")
#As Lists start at 0
FireRow=int(FireRow)-1
if FireIndex==("A"):
FireIndex=0
elif FireIndex==("B"):
FireIndex=1
elif FireIndex==("C"):
FireIndex=2
elif FireIndex==("D"):
FireIndex=3
elif FireIndex==("E"):
FireIndex=4
elif FireIndex==("F"):
FireIndex=5
elif FireIndex==("G"):
FireIndex=6
elif FireIndex==("H"):
FireIndex=7
Guess=[FireRow,FireIndex]
#Check To See If Correct
UserSuccess=0
for i in CompShips:
if Guess==i:
CompShips.remove(Guess)
UserSuccess=1
else:
pass
if UserSuccess==1:
print("HIT")
print(FireRow)
print(FireIndex)
FireBoard[[FireRow][FireIndex]]=("H")
PrintFireBoard()
else:
print("MISS")
print(FireRow)
print(FireIndex)
FireBoard[[FireRow][FireIndex]]=("M")
PrintFireBoard()
I receive the error:
IndexError: string index out of range

Looks like these two lines
FireBoard[[FireRow][FireIndex]]=("H")
FireBoard[[FireRow][FireIndex]]=("M")
should be
FireBoard[FireRow][FireIndex]="H"
FireBoard[FireRow][FireIndex]="M"
Explanation: In your old code, FireBoard[[FireRow][FireIndex]]=("H")
[FireRow][FireIndex] means, given a list [FireRow] (which contains just one element), get the FireIndex-th element. This is not what you're trying to do.
For example [3][0] returns 3, and [3][1] gives IndexError.
Take a look at How to define a two-dimensional array in Python
Also note that ("H") is the same as the string "H". There is no need to add parentheses.

Here is a much cleaner code!
CompShips=[[1,0],[1,1],[2,2],[2,3],
[2,4],[3,0],[3,1],[3,2],
[5,4],[5,5],[5,6],[5,7],
[1,7],[2,7],[3,7],[4,7],
[5,7]]
FRow=[["Z"]*8]*8 #1 More Pythonic
def PrintFireBoard():
#print(Index)
for i in range(0,8):
print(FRow[i])
FireBoard=FRow[:] #NOTE THIS ONE!!!
mydict = {}
for i,key in enumerate(["A","B","C","D","E","F","G","H"]): #2 More Pythonic
mydict[key] = i
while len(CompShips) !=0 or CompSuccess==17:
FireRow=input("Please Choose The Row That You Wish To Fire Upon (1-8) ")
FireIndex=input("Please Choose The Column That You Wish To Fire Upon (A-H) ")
FireRow=int(FireRow)-1
FireIndex = mydict[FireIndex]
Guess=[FireRow,FireIndex]
print(Guess)
UserSuccess=0
for i in CompShips:
if Guess==i:
CompShips.remove(Guess)
UserSuccess=1
else:
pass
if UserSuccess==1:
print("HIT")
print(FireRow,FireIndex)
FireBoard[FireRow][FireIndex]="H" #3 your problem here
PrintFireBoard()
else:
print("MISS")
print(FireRow,FireIndex)
FireBoard[FireRow][FireIndex]="M"
PrintFireBoard()
1) As explained in the comments that's just a more nicer way to create a list of lists!. Remember DRY principle! Do Not Repeat yourself!
2) Instead of having all that if else to convert the 'A' to 0. You can use a dictionary lookup instead!
3) Your problem seems to be here! correct this to FireBoard[FireRow][FireIndex]="H"
PS: NOTE THIS ONE!!!: I'm not just making FireBoard as an alias to FRow! I'm copying it into a FireBoard as a new list! There's a subtle difference read about it here. I'm doing this incase you don't want your original FRow list to be modified!

The indentation in your question was off. I think that all the code from
Guess=[FireRow,FireIndex]
until the end should be preceded by 4 spaces.
I've removed print(Index) since it was not defined.
To access FireBoard use:
FireBoard[FireRow][FireIndex]
Instead of
FireBoard[[FireRow][FireIndex]]
This should be working
CompShips=[[1,0],[1,1],[2,2],[2,3],[2,4],[3,0],[3,1],[3,2],[5,4],
[5,5],[5,6],[5,7],[1,7],[2,7],[3,7],[4,7],[5,7]]
FRow1=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow2=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow3=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow4=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow5=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow6=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow7=["Z","Z","Z","Z","Z","Z","Z","Z",]
FRow8=["Z","Z","Z","Z","Z","Z","Z","Z",]
def PrintFireBoard():
print(FRow1)
print(FRow2)
print(FRow3)
print(FRow4)
print(FRow5)
print(FRow6)
print(FRow7)
print(FRow8)
FireBoard=[FRow1,FRow2,FRow3,FRow4,FRow5,FRow6,FRow7,FRow8]
while len(CompShips) !=0 or CompSuccess==17:
FireRow=input("Please Choose The Row That You Wish To Fire Upon (1-8) ")
FireIndex=input("Please Choose The Column That You Wish To Fire Upon (A-H) ")
#As Lists start at 0
FireRow=int(FireRow)-1
if FireIndex==("A"):
FireIndex=0
elif FireIndex==("B"):
FireIndex=1
elif FireIndex==("C"):
FireIndex=2
elif FireIndex==("D"):
FireIndex=3
elif FireIndex==("E"):
FireIndex=4
elif FireIndex==("F"):
FireIndex=5
elif FireIndex==("G"):
FireIndex=6
elif FireIndex==("H"):
FireIndex=7
Guess=[FireRow,FireIndex]
#Check To See If Correct
UserSuccess=0
for i in CompShips:
if Guess==i:
CompShips.remove(Guess)
UserSuccess=1
else:
pass
if UserSuccess==1:
print("HIT")
print(FireRow)
print(FireIndex)
FireBoard[FireRow][FireIndex]=("H")
PrintFireBoard()
else:
print("MISS")
print(FireRow)
print(FireIndex)
FireBoard[FireRow][FireIndex]=("M")
PrintFireBoard()

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?!?!?"