New to stackoverflow, and new to python (python-3). Currently learning on edx.org and ran into the following error.
I created a function that checks a user-input str against the answer str and returns True or False.
When testing the function, I created a while loop to stop at the 3rd unsuccessful attempt. However, whenever there is an unsuccessful attempt, the function prints the error message twice when it should only print it once.
I fixed the error by storing the returning Bool value of the function into a variable rather than calling the function directly in the if condition within the while loop. However, I would like to understand the logic behind the error message printing twice. Here is the original code that prints the error message twice :
def letter_guess(letter, guess):
if len(guess) == 1 and guess.isalpha() and guess < letter:
print(guess,"is lower than the answer. Try again.\n")
return False
elif len(guess) == 1 and guess.isalpha() and guess > letter:
print(guess,"is higher than the answer. Try again.\n")
return False
elif len(guess) == 1 and guess.isalpha() and guess == letter:
print("Correct answer!")
return True
else:
print("Please only enter one alphabet for the letter. Try again.\n")
return False
answer2 = "m"
guess2 = input("Please enter a single alphabet : ")
i = 0
while i < 3:
if letter_guess(answer2, guess2):
break
elif letter_guess(answer2, guess2) == False and i == 2:
print("You have reached 3 guesses. Game over.")
break
else:
i += 1
guess2 = input("Please guess again : ")
You want to call input() inside the while loop:
# ...
answer2 = "m"
i = 0
while i < 3:
guess2 = input("Please enter a single alphabet : ")
# ...
Otherwise the user doesn't have a chance to change their answer, guess2 never changes and they get the same error message multiple times.
You call the function twice, in first if and in elif, with the same wrong guess. You fixed it right calling only once and storing the return value.
I try to explain it better: the function is always called by the first if, to evaluate its condition; if return value is false, is called again to evaluate the elif condition, with same arguments as before.
Related
When I run the code, the 3rd guess gives me an output of High/Low. On the 3rd try, I don't need it to tell me if I'm high or low. How can I fix the problem with out using "while" loops or "range." We haven't covered these two keywords yet.
python
print("You have 3 tries to guess the letter.")
letter = "F"
tries = 0
# [ ] create letter_guess() function, call the function to test
def letter_guess(tries):
if not tries == 3:
guess = input("Guess a letter: ")
tries = tries + 1
check = check_guess (guess,letter)
if check == True:
print('Winner')
else:
letter_guess(tries)
else:
print ("GAME OVER!")
pass
# def check_guess(guess,letter)
def check_guess (guess, letter):
#if else to see if correct letter
if letter == guess.upper():
print ("correct")
return True
elif letter < guess.upper():
print ("You are wrong, guess lower.")
return False
elif letter > guess.upper():
print ("You are wrong, guess higher.")
return False
else:
print("Invalid response!")
return False
letter_guess(tries)
One way to get result similar to a loop is to use recursion. But if you have not done loops yet, you almost certainly have not done recursion. So just use straightforward code.
The tricky part is that you need to ask for a guess, input the guess, and check for a correct guess three times, once for each guess. However, you need to give feedback on whether the guess is high or low only two times. Therefore, you cannot put those actions in the same function. Just split them into separate functions, and handle each guess in your main routine. No need to count guesses--the position in the main routine makes that clear.
"""Guess-a-letter game."""
def get_guess(letter):
"""Get a guess an note if it is correct. If correct, return None.
Otherwise, return the guess."""
guess = input("Guess a letter: ").upper()
if guess == letter:
print("Correct: You win!")
return None
else:
return guess
def give_feedback(guess, letter):
"""Give feedback on a wrong guess."""
if letter < guess:
print("You are wrong, guess lower.")
else:
print("You are wrong, guess higher.")
def letter_guess():
# Store the letter for the user to guess.
letter = "F"
# Introduce the game.
print("You have 3 tries to guess the letter.")
# Handle the first guess.
guess = get_guess(letter)
if guess is None:
return # Success!
give_feedback(guess, letter)
# Handle the second guess.
guess = get_guess(letter)
if guess is None:
return # Success!
give_feedback(guess, letter)
# Handle the third and last guess.
guess = get_guess(letter)
if guess is None:
return # Success!
print("You were wrong three times. GAME OVER!")
letter_guess()
I'm making a simple guessing game in python and was trying to create an "Invalid entry" message for when the user enters in any input that is not an integer.
I have tried to use just 'int' in an if statement to address all integers, but that is not working.
I know that I have the syntax wrong. I'm just not sure what the correct syntax to do it would be.
import random
play = True
while play:
count = 1
hidden = random.randrange(1,5)
guess = int(input("Guess a number between 1 and 5:"))
if guess != int
guess = int(input("Invalid Entry. Please enter an Integer between 1 and 5:"))
while guess != hidden:
count+=1
if guess > hidden + 10:
print("your guess is to high!")
elif guess < hidden -10:
print("your too low!")
elif guess > hidden:
print("your really warm, but still to high!")
elif guess < hidden:
print("your really warm, but still to low")
print("You have guessed incorrectly, Try again!. \n")
#reset the guess variable and make another guess
guess = int(input("Guess a number between 1 and 5:"))
print("Nice!!! Your guess was correct!\n you got the correct number in" , count , "tries.")
count = 1
playagain = str(input("Do you want to play again?\nType yes or no: "))
if playagain == "no" or "n" or "N" or "no thank you":
play = False
elif playagain == "yes" or "y" or "Y" or "YES" or "yes":
play = True
else: playagain != "yes" or "y" or "Y" or "YES" or "yes" "no" or "n" or "N" or "no thank you"
playagain = str(input("Invalid Entry. Please Type yes or no: "))
This is the error that I'm getting. There may be some other mistakes in my code as well.
File "comrandomguess.py", line 18
if guess != int
^
SyntaxError: invalid syntax
If you really want to verify that the user entry is an int, you want to keep the input in string form. Then write a small function to test the input. Here, I'll use a list comprehension and the string join and isdigit methods, to ensure the user has only entered digits 0-9 in the string, i.e. then this function returns True (else False) (*modified as per Jack Taylor comment below, also for s = '' case):
def testForInt(s):
if s:
try:
_ = s.encode('ascii')
except UnicodeEncodeError:
return False
test = ''.join([x for x in s if x.isdigit()])
return (test == s)
else:
return False
If you want to sandbox the user entirely, wrap it in a loop like this:
acceptable = False
while not acceptable:
entry = input("Enter an int: ")
if testForInt(entry):
entry = int(entry)
acceptable = True
else:
print("Invalid Entry")
If you want a simpler version with no function call(see Jack Taylor comment), this works too:
acceptable = False
while not acceptable:
entry = input("Enter an int: ")
try:
entry = int(entry)
acceptable = True
except ValueError as e:
print(f"Failed due to {str(e)}")
Now you've got what you know is an int, with no worries. This kind of approach to verifying user entry saves many headaches if consistently implemented. See SQL injection etc.
I always use this method to check if something is not an integer:
Python 3
if not round(guess) == guess: print("Do Stuff")
Python 2
if not round(guess) == guess: print "Do Stuff"
You need to do something like this:
play = True
while play:
guess = input("Guess a number between 1 and 5: ")
try:
number = int(guess)
except ValueError:
print("You need to input an integer.")
continue
if number < 1 or number > 5:
print("You need to input an integer between 1 and 5.")
continue
# ...
print("Your number was: " + guess)
play = False
When you first use input(), you get a string back. If you try to turn that string into an integer straight away by doing int(input()), and if the player types a string like "abcd", then Python will raise an exception.
>>> int(input("Guess a number: "))
Guess a number: abcd
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: invalid literal for int() with base 10: 'abcd'
To avoid this, you have to handle the exception by doing int(guess) inside a try/except block.
The continue statement skips back to the start of the while loop, so if you use it you can get away with only having to ask for input once.
Parse the user input as string to avoid ValueError.
guess = input("Guess a number between 1 and 5: ")
while not guess.isdigit() or int(guess) > 5 or int(guess) < 1:
guess = input("Invalid Entry. Please enter an Integer between 1 and 5: ")
guess = int(guess)
Above code ensures that user input is a positive integer and between 1 and 5. Next, convert the user input to integer for further use.
Additionally, if you want to check the data type of a python object/variable then use the isinstance method. Example:
a = 2
isinstance(a, int)
Output:
>>> True
This question already has an answer here:
Python "if" statement not working
(1 answer)
Closed 4 years ago.
I'm learning Python for fun at the moment, and it went all well until now. I'm trying to extend the "Guess the Word"-Game, for example being able to let the Player choose a Word by himself (when 2 People play, 1 chooses the Word, the other guesses) I bet that my Mistake is obvious to me as soon as you point it out, but I'm gonna ask anyway. Well, here is the Code. I put in the entire Program, even tough only the top part should matter. I just put in the rest because it isn't much and maybe you guys can understand it better then.
print("Do you wish to set the Word yourself, or let the program choose?")
user_input = input("1 for own Input - 0 for Program-Input")
if user_input == 1:
Keyword = input("Type in the Word you want to use!")
else:
Keyword = "castle"
word = list(Keyword)
length = len(word)
right = list ("_" * length)
used_letters = list()
finished = False
while finished == False:
guess = input("Guess a Letter!")
if guess not in Keyword:
print("This letter is not in the word. Sorry...")
for letter in word:
if letter == guess:
index = word.index(guess)
right[index] = guess
word[index] = "_"
if guess in used_letters[0:100]:
print("You already used that letter before!")
else:
used_letters.append(guess)
list.sort(used_letters)
print(right)
print("Used letters:")
print(used_letters)
if list(Keyword) == right:
print("You win!")
finished = True
input('Press ENTER to exit')
My problem is, I wanna add the Function to be able to choose if you want to set a Word yourself, or use the word the Program has, defined as "Keyword". But no matter what I input, it always starts with "Guess a Letter" instead of skipping down to where the program sets the Keyword. Thank you in advance for your answers! :)
There's 2 issues with your code.
You put the entire block of code into the else statement. This means that if the if user_input == 1: block ever executed, you would only ask your user for a word and then the program would end because the else statement would be skipped.
You are using if user_input == 1: as your check and this will never be true because user inputs are always read in as strings. A string 1 will never equal the integer 1. This is why your program always skips to the else statement. You need to do if int(user_input) == 1:
Whenever you collect a user's input using the input function, it is a string, not int. this means you will have to either parse the value into an int or evaluate it with a string.
option 1: parsing to int:
user_input = int(input("1 for own Input - 0 for Program-Input"))
option 2: evaluating with string:
if user_input == "1":
input returns a string not a integer so it can never be equal to 1 instead it will be equal to "1".
Plus the code for the user guessing only runs when the program chooses the word so it needs to be unindented.
As a side note your code currently registered capital letters as being different from lower case, you can fix this by putting a .lower() after each input which will turn all capital letters into lowercase.
print("Do you wish to set the Word yourself, or let the program choose?: ")
user_input = input("1 for own Input - 0 for Program-Input")
if user_input == "1":
Keyword = input("Type in the Word you want to use: ").lower()
else:
Keyword = "castle"
word = list(Keyword)
length = len(word)
right = list ("_" * length)
used_letters = list()
finished = False
while finished == False:
guess = input("Guess a Letter: ").lower()
if guess not in Keyword:
print("This letter is not in the word. Sorry...")
for letter in word:
if letter == guess:
index = word.index(guess)
right[index] = guess
word[index] = "_"
if guess in used_letters[0:100]:
print("You already used that letter before!")
else:
used_letters.append(guess)
list.sort(used_letters)
print(right)
print("Used letters:")
print(used_letters)
if list(Keyword) == right:
print("You win!")
finished = True
input('Press ENTER to exit')
The first IF statement is being ignored and I have no idea what could cause this. I checked the indentation and everything seems fine.As you can see in the code it prints numberRolled but when I run it it justs ignores the first IF.`
import random
numberRolled = random.randint(1,6)
print numberRolled
while True:
userGuess = raw_input("Guess a number\n")
if userGuess == numberRolled:
print "You got it right!"
quitYN = raw_input("Would you like to play again?\n").lower()
if quitYN == "yes":
continue
else:
break
elif userGuess != numberRolled:
print "Wrong!"`
raw_input() returns a string, but random.randint() returns an int. This means that when doing userGuess == numberRolled you are comparing a string to an int (which returns False).
To fix this simply convert one of the variables to the correct type:
userGuess == str(numberRolled)
Take a look at this answer for more information about variable types and how to compare them in python.
The assignment was to make a guessing game where the parameter is the answer. At the end if the person gets it right, it prints a congratulatory statement and returns the number of tries it took, if they type in quit, it displays the answer and tries == -1. other than that, it keeps looping until they get the answer correct.
def guessNumber(num):
tries = 1
while tries > 0:
guess = input("What is your guess? ")
if guess == num:
print ("Correct! It took you" + str(tries)+ "tries. ")
return tries
elif guess == "quit":
tries == -1
print ("The correct answer was " + str(num) + ".")
return tries
else:
tries += 1
When i run it, no matter what i put in it just keeps asking me for my guess.
Since you called your variable num so I'm guessing it's a integer, you were checking equality between an integer and a string so it's never True. Try changing the num to str(num) when comparing, so:
def guessNumber(num):
tries = 1
while tries > 0:
guess = input("What is your guess? ")
if guess == str(num):
print ("Correct! It took you {0} tries. ".format(tries))
return tries
elif guess == "quit":
tries = -1
print ("The correct answer was {0}.".format(num))
return tries
else:
tries += 1
Is the code properly indented?
The body of the function you are defining is determined by the level of indentation.
In the example you pastes, as the line right after the def has less indentation, the body of the function is 'empty'.
Indenting code is important in python
Additionally, for assigning one value to a variable you have to use a single '=', so the:
tries == -1
should be
tries = -1
if you want to assign the -1 value to that variable.