I am using python 2.6.6
I am simply trying to restart the program based on user input from the very beginning.
thanks
import random
import time
print "You may press q to quit at any time"
print "You have an amount chances"
guess = 5
while True:
chance = random.choice(['heads','tails'])
person = raw_input(" heads or tails: ")
print "*You have fliped the coin"
time.sleep(1)
if person == 'q':
print " Nooo!"
if person == 'q':
break
if person == chance:
print "correct"
elif person != chance:
print "Incorrect"
guess -=1
if guess == 0:
a = raw_input(" Play again? ")
if a == 'n':
break
if a == 'y':
continue
#Figure out how to restart program
I am confused about the continue statement.
Because if I use continue I never get the option of "play again" after the first time I enter 'y'.
Use a continue statement at the point which you want the loop to be restarted. Like you are using break for breaking from the loop, the continue statement will restart the loop.
Not based on your question, but how to use continue:
while True:
choice = raw_input('What do you want? ')
if choice == 'restart':
continue
else:
break
print 'Break!'
Also:
choice = 'restart';
while choice == 'restart':
choice = raw_input('What do you want? ')
print 'Break!'
Output :
What do you want? restart
What do you want? break
Break!
I recommend:
Factoring your code into functions; it makes it a lot more readable
Using helpful variable names
Not consuming your constants (after the first time through your code, how do you know how many guesses to start with?)
.
import random
import time
GUESSES = 5
def playGame():
remaining = GUESSES
correct = 0
while remaining>0:
hiddenValue = random.choice(('heads','tails'))
person = raw_input('Heads or Tails?').lower()
if person in ('q','quit','e','exit','bye'):
print('Quitter!')
break
elif hiddenValue=='heads' and person in ('h','head','heads'):
print('Correct!')
correct += 1
elif hiddenValue=='tails' and person in ('t','tail','tails'):
print('Correct!')
correct += 1
else:
print('Nope, sorry...')
remaining -= 1
print('You got {0} correct (out of {1})\n'.format(correct, correct+GUESSES-remaining))
def main():
print("You may press q to quit at any time")
print("You have {0} chances".format(GUESSES))
while True:
playGame()
again = raw_input('Play again? (Y/n)').lower()
if again in ('n','no','q','quit','e','exit','bye'):
break
You need to use random.seed to initialize the random number generator. If you call it with the same value each time, the values from random.choice will repeat themselves.
After you enter 'y', guess == 0 will never be True.
Related
I am making a program that generates a random number and asks you to guess the number out of the range 1-100. Once you put in a number, it will generate a response based on the number. In this case, it is Too high, Too low, Correct, or Quit too soon if the input is 0, which ends the program(simplified, but basically the same thing).
It counts the number of attempts based on how many times you had to do the input function, and it uses a while loop to keep asking for the number until you get it correct. The problem that I am facing is that I have to make it break out of the while loop once the guess is either equal to the random number or 0. This normally isn't an issue, because you could use sys.exit() or some other function, but according to the instructions I can't use break, quit, exit, sys.exit, or continue. The problem is most of the solutions I've found for breaking the while loop implement break, sys.exit, or something similar and I can't use those. I used sys.exit() as a placeholder, though, so that it would run the rest of the code, but now I need to figure out a way to break the loop without using it. This is my code:
import random
import sys
def main():
global attempts
attempts = 0
guess(attempts)
keep_playing(attempts)
def guess(attempts):
number = random.randint(1,100)
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
while guess != 0:
if guess != number:
if guess < number:
print("Too low, try again")
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
elif guess > number:
print("Too high, try again")
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
else:
print()
print("Congratulations! You guessed the right number!")
print("There were", attempts,"attempts")
print()
#Ask if they want to play again
sys.exit()#<---- using sys.exit as a placeholder currently
else:
print()
print("You quit too early")
print("The number was ",number,sep='')
#Ask if they want to play again
sys.exit()#<----- using sys.exit as a placeholder currently
def keep_playing(attempts):
keep_playing = 'y'
if keep_playing == 'y' or keep_playing == 'n':
if keep_playing == 'y':
guess(attempts)
keep_playing = input("Another game (y to continue)? ")
elif keep_playing == 'n':
print()
print("You quit too early")
print("Number of attempts", attempts)
main()
If anyone has any suggestions or solutions for how to fix this, please let me know.
Try to implement this solution to your code:
is_playing = True
while is_playing:
if guess == 0:
is_playing = False
your code...
else:
if guess == number:
is_playing = False
your code...
else:
your code...
Does not use any break etc. and It does breaks out of your loop as the loop will continue only while is_playing is True. This way you will break out of the loop when the guess is 0 (your simple exit way) or when the number is guessed correctly. Hope that helps.
I am not a fan of global variables but here it's your code with my solution implemented:
import random
def main() -> None:
attempts = 0
global is_playing
is_playing = True
while is_playing:
guess(attempts)
keep_playing()
def guess(attempts: int) -> None:
number = random.randint(1,100)
print(number)
is_guessing = True
while is_guessing:
attempts += 1
guess = int(input("Enter a number between 1 and 100, or 0 to quit: "))
if guess == 0:
is_guessing = False
print("\nYou quit too early.")
print("The number was ", number,sep='')
else:
if guess == number:
is_guessing = False
print("\nCongratulations! You guessed the right number!")
print("There were", attempts, "attempts")
else:
if guess < number:
print("Too low, try again.")
elif guess > number:
print("Too high, try again.")
def keep_playing() -> None:
keep_playing = input('Do you want to play again? Y/N ')
if keep_playing.lower() == 'n':
global is_playing
is_playing = False
main()
TIP:
instead
"There were", attempts, "attempts"
do: f'There were {attempts} attempts.'
This question already has answers here:
Asking the user for input until they give a valid response
(22 answers)
Closed 1 year ago.
I have a y/n query at the end of the game which will restart the game or exit the game.
y and n work as they should, but if I enter any other letter even if it does print the "please try again message", it restarts the game anyway. I thought it should only do that if I put continue, but I haven't
I also want it to recognize y or n as upper case, is there a command in python that allows a letter to be recognized regardless of if its upper or lowercase?
play_again = input('Would you like to play again?(y/n) ')
if play_again == 'y': #loop restarts
print('Starting another game...')
continue
elif play_again == 'n': #loop is exited
print('Thank you for playing! You have now exited the game.')
break
else:
print ("I don't recognize that character, please select y or n")
I have provided the loop as requested, but the last time I did this I got in trouble:
number_of_stones = int(input('Select between 30-50 stones to start your game: ' )) #player1 chooses stones
if number_of_stones >= 30 and number_of_stones <= 50:
print('Starting the game...')
has_winner = False # Winner = False since there no winners (start of game)
while True:
for player in player_list:
if has_winner: # if there is a winner, exit loop
break
# display whose turn is it
print('\n{}\'s turn:'.format(player))
while True:
# if the player is computer use strategy that is mentioned in assignment 2
if player == 'Mr Computer':
remainder = number_of_stones%3
if remainder == 0:
stones_to_remove = 2
else:
stones_to_remove = 1
# if the player is not the computer
else:
stones_to_remove = int(input('Select between 1-3 stones to remove: '))
# assess the number of stones remaining and print the remainder to the screen
if stones_to_remove >= 1 and stones_to_remove <= 3:
number_of_stones -= stones_to_remove
print('Number of stones remaining:',number_of_stones)
if number_of_stones <= 0:
print('The winner is:',player,"!")
has_winner = True
break
else: # show error and let the user input again
print("I'm sorry, that number is outside the range. Please choose a number between 1 and 3")
if has_winner == True: # if the winner=true then exit loop
break
play_again = input('Would you like to play again?(y/n) ').lower() #only accepts lower case letter
if play_again == 'y': #loop restarts
os.system('cls') # clear the console
print('Starting another game...')
continue
elif play_again == 'n': #loop is exited
print('Thank you for playing! You have now exited the game.')
break
else:
print("I'm sorry, that number is outside the range. Please select between 30 and 50 stones.")
You haven't added necessary information ,i.e. rest of the loop , try add that .
For case insensitivity , just use the following
if play_again == "n" || play_again == "N":
.
guys! I'm new here. Nice to meet you!
I have the following problem. My random number generator always do exactly the same number. I have a guess why it's happening, but I'm not sure. Can you, please, explain what I'm doing wrong and just write some words about my code at all? Thanks!
import random
def begin():
r_num = random.randint(1, 10)
p_num = int(input('Enter your number: '))
ch = (r_num, p_num)
if ch[0] == ch[1]:
print(ch[0])
print("You've won! Excellent!")
play_again = input("Do you want to play again? y/n")
if play_again == 'y' or play_again == 'Y':
begin()
elif play_again == 'n' or play_again == 'N':
print('Goodbye!')
elif ch[1] < ch[0]:
print(ch[0])
print("Your number is lower than it must be!")
begin()
elif ch[1] > ch[0]:
print(ch[0])
print("Your number is higher than it must be!")
begin()
def start():
gen = input('Print generate to begin playing! \n')
if gen == 'generate':
print('Success!')
begin()
else:
print('Fail!')
start()
print('Welcome to my first Python game! Guess random generated number from 1 to 10!')
start()
The code works perfectly fine and so you genuinely must just have had incredibly good/bad luck (depending on which way you think about it). Try running the code in a different IDE - it's the only thing I can think of that may be causing the issue. Try running the script again?
I am attempting to create a loop which a user can stop at the end of the program. I've tried various solutions, none of which have worked, all I have managed to do is create the loop but I can't seem to end it. I only recently started learning Python and I would be grateful if someone could enlighten me on this issue.
def main():
while True:
NoChild = int(0)
NoAdult = int(0)
NoDays = int(0)
AdultCost = int(0)
ChildCost = int(0)
FinalCost = int(0)
print ("Welcome to Superslides!")
print ("The theme park with the biggest water slide in Europe.")
NoAdult = int(raw_input("How many adults are there?"))
NoChild = int(raw_input("How many children are there?"))
NoDays = int(raw_input("How many days will you be at the theme park?"))
WeekDay = (raw_input("Will you be attending the park on a weekday? (Yes/No)"))
if WeekDay == "Yes":
AdultCost = NoAdult * 5
elif WeekDay == "No":
AdultCost = NoAdult * 10
ChildCost = NoChild * 5
FinalCost = (AdultCost + ChildCost)*NoDays
print ("Order Summary")
print("Number of Adults: ",NoAdult,"Cost: ",AdultCost)
print("Number of Children: ",NoChild,"Cost: ",ChildCost)
print("Your final total is:",FinalCost)
print("Have a nice day at SuperSlides!")
again = raw_input("Would you like to process another customer? (Yes/No)")
if again =="No":
print("Goodbye!")
return
elif again =="Yes":
print("Next Customer.")
else:
print("You should enter either Yes or No.")
if __name__=="__main__":
main()
You can change the return to break and it will exit the while loop
if again =="No":
print("Goodbye!")
break
Instead of this:
while True:
You should use this:
again = True
while again:
...
usrIn = raw_input("Would you like to process another customer? y/n")
if usrIn == 'y':
again = True
else
again = False
I just made it default to False, but you can always just make it ask the user for a new input if they don't enter y or n.
I checked your code with python 3.5 and it worked after I changed the raw_input to input, since input in 3.5 is the raw_input of 2.7. Since you're using print() as a function, you should have an import of the print function from future package in your import section. I can't see no import section in your script.
What exactly doesn't work?
Additionally: It's a good habit to end a command line application by exiting with an exit code instead of breaking and ending. So you would have to
import sys
in the import section of your python script and when checking for ending the program by the user, do a
if again == "No":
print("Good Bye")
sys.exit(0)
This gives you the opportunity in case of an error to exit with a different exit code.
Change this code snippet
if again =="No":
print("Goodbye!")
exit() #this will close the program
elif again =="Yes":
print("Next Customer.")
exit()#this will close the program
else:
print("You should enter either Yes or No.")
For menu-driven programming, how is the best way to write the Quit function, so that the Quit terminates the program only in one response.
Here is my code, please edit if possible:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=input(">>> ")
choice=choose.lower()
while choice!="q":
if choice=="v":
highScore()
main()
elif choice=="s":
setLimit()
main()
elif choice=="p":
game()
main()
else:
print("Invalid choice, please choose again")
print("\n")
print("Thank you for playing,",name,end="")
print(".")
When the program first execute and press "q", it quits. But after pressing another function, going back to main and press q, it repeats the main function.
Thanks for your help.
Put the menu and parsing in a loop. When the user wants to quit, use break to break out of the loop.
source
name = 'Studboy'
while True:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choice = raw_input(">>> ").lower().rstrip()
if choice=="q":
break
elif choice=="v":
highScore()
elif choice=="s":
setLimit()
elif choice=="p":
game()
else:
print("Invalid choice, please choose again\n")
print("Thank you for playing,",name)
print(".")
def Menu:
while True:
print("1. Create Record\n2. View Record\n3. Update Record\n4. Delete Record\n5. Search Record\n6. Exit")
MenuChoice=int(input("Enter your choice: "))
Menu=[CreateRecord,ViewRecord,UpdateRecord,DeleteRecord,SearchRecord,Exit]
Menu[MenuChoice-1]()
You're only getting input from the user once, before entering the loop. So if the first time they enter q, then it will quit. However, if they don't, it will keep following the case for whatever was entered, since it's not equal to q, and therefore won't break out of the loop.
You could factor out this code into a function:
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=input(">>> ")
choice=choose.lower()
And then call it both before entering the loop and then as the last thing the loop does before looping back around.
Edit in response to comment from OP:
The following code below, which implements the factoring out I had mentioned, works as I would expect in terms of quitting when q is typed.
It's been tweaked a bit from your version to work in Python 2.7 (raw_input vs. input), and also the name and end references were removed from the print so it would compile (I'm assuming those were defined elsewhere in your code). I also defined dummy versions of functions like game so that it would compile and reflect the calling behavior, which is what is being examined here.
def getChoice():
print("\nMenu\n(V)iew High Scores\n(P)lay Game\n(S)et Game Limits\n(Q)uit")
choose=raw_input(">>> ")
choice=choose.lower()
return choice
def game():
print "game"
def highScore():
print "highScore"
def main():
print "main"
def setLimit():
print "setLimit"
choice = getChoice()
while choice!="q":
if choice=="v":
highScore()
main()
elif choice=="s":
setLimit()
main()
elif choice=="p":
game()
main()
else:
print("Invalid choice, please choose again")
print("\n")
choice = getChoice()
print("Thank you for playing,")
This is a menu driven program for matrix addition and subtraction
def getchoice():
print('\n What do you want to perform:\n 1.Addition\n 2. Subtraction')
print('Choose between option 1,2 and 3')
cho = int(input('Enter your choice : '))
return cho
m = int(input('Enter the Number of row : '))
n = int(input('Enter the number of column : '))
matrix1 = []
matrix2 = []
print('Enter Value for 1st Matrix : ')
for i in range(m):
a = []
for j in range(n):
a.append(int(input()))
matrix1.append(a)
print('Enter Value for 2nd Matrix : ')
for i in range(m):
a = []
for j in range(n):
a.append(int(input()))
matrix2.append(a)
choice = getchoice()
while choice != 3:
matrix3 = []
if choice == 1:
for i in range(m):
a = []
for j in range(n):
a.append(matrix1[i][j] + matrix2[i][j])
matrix3.append(a)
for r in matrix3:
print(*r)
elif choice == 2:
for i in range(m):
a = []
for j in range(n):
a.append(matrix1[i][j] - matrix2[i][j])
matrix3.append(a)
for r in matrix3:
print(*r)
else:
print('Invalid Coice.Please Choose again.')
choice = getchoice()