How do check the contents of a List using an if statement in python

Question

I am new to python, I have written the following code but it will not reduce the score if B is entered, it increases the score by 1 if How, Hi etc are entered but not it 8 is entered and does not reduce the score if B is entered. Can anyone help.

grid=[8,"B","How", "Hi", "Hey", "How are you doing", "How's it going", "How", "Hello"]

def Play(): 
 count=0 
 score=0 
 while (count<11):
 I=input("Enter your guess")
 if I in grid:
 score+=1
 print("Your score is:", score)
 elif I in grid == "B":
 score=0
 print("Your score is:",score)
 else:
 print("I don't understand")
 count+=1

• roganjosh: elif I in grid == "B": doesn't make sense. Take B out of grid and put it in a different list. Otherwise you need elif I in grid and I == 'B': or something similar, which is counter-intuitive as a setup.
• roganjosh: I in grid returns a bool. You're asking "If I is in grid..." (which returns True or False) "is equal to the letter B, do this". True and False won't be equal to the letter B.

Answer 1

input() stores input as a string. Therefore, if the user enters 8, it will be stored as "8". "8" is not the same as 8, so if I in grid will evaluate to False. To solve this, you can change 8 to "8" in the list, or add elif I == "8":

elif I in grid == "B": will never evaluate as true. This is because you are checking (I in grid) == ("B"). I in grid will in always be True or False, so you are comparing "B" with a boolean, which is always False.

An easier way to accomplish this is to change 8 to "8" in `grid:

grid=["8","B","How", "Hi", "Hey", "How are you doing", "How's it going", "How", "Hello"]

And check if I == "B" in the first if statement:

if I in grid:
 if I == "B":
 score=0
 else:
 score+=1
 print("Your score is:",score)
else:
 print("I don't understand")

• roganjosh: There was one thing we both haven't commented on, which might be worth noting. if I in grid: is True even for "B" so elif I in grid == "B": is never even evaluated anyway in the OP's approach since it breaks out at the first True evaluation.