I am making a simple 'guess a number between one and ten' game. I have used some basic error handling and am printing the number generated by the random module for testing purposes.
However I would like to know if there is a less verbose way to write this.
This is the code:
import random
while True:
"""Variable declaration"""
number_of_attempts = 1
number = random.randrange (1,11)
print (number)
print("Time to play a guessing game! Muhahaha...")
"""Error handling and main game code/while loop"""
while True:
try:
guess = int(input("Guess a number between one and ten."))
except ValueError:
print("Input a whole number between one and ten silly!")
continue
if guess >= 1 and guess <= 10:
pass
else:
print("Input a number between one and ten silly!")
continue
if guess == number:
print("You were successful and it took you", number_of_attempts, "attempts!!!")
break
else:
print("Try again!")
number_of_attempts = number_of_attempts +1
"""Game Exit/Restart"""
play_again = input("Would you like to play again, y/n?")
if "y" in play_again or "yes" in play_again:
continue
else:
break
Thanks,
Ben
if guess >= 1 and guess <= 10:
Can be written as:
if 1 <= guess <= 10:
Also, your first conditional can simply be written as:
if not 1 <= guess <= 10:
print("Input a number between one and ten silly!")
continue
But this can also be put inside the try bit, saving you from writing continue twice:
try:
guess = int(input("Guess a number between one and ten."))
if not 1 <= guess <= 10:
print("Input a number between one and ten silly!")
continue
except ValueError:
print("Input a whole number between one and ten silly!")
continue
Finally your last conditional can simply be:
if play_again not in ('y', 'yes'):
break
The continue isn't needed.
You may also want to wrap this all up into a function as well, to get rid of those infinite while loops and to prevent you from using continue and break so much.
Why not put the actual conditions on the while loops so you don't have to hunt for breaks to understand the loops? It would make your code clearer and smaller.
if guess == number:
print("You were successful and it took you", number_of_attempts, "attempts!!!")
break
For instance if you put guess == number as the while loop conditional then the print would be the first thing after the loop. Initialize guess to -1 so it always works the first time. The play again if statement could also disappear into the loop conditional as well.
Related
I am using The Big Book of Small Python projects to increase my skills in python, and on the very first project, which is making a simple logic game, On the first try, the code goes all the way however if you get it wrong you it won't run properly.
Here is the code and a description of the game, the while loop with chances is supposed to run for the whole game, until you run out of chances, the second while loop is supposed to run in case user enters below or more than length three for the number
import re
import random
#In Bagels, a deductive logic game, you
#must guess a secret three-digit number
#based on clues. The game offers one of
#the following hints in response to your guess:
#“Pico” when your guess has a correct digit in the
#wrong place, “Fermi” when your guess has a correct
#digit in the correct place, and “Bagels” if your guess
#has no correct digits. You have 10 tries to guess the
#secret number.
choice_of_nums=['123','345','674','887','356','487','916']
random_three_num=random.choices(choice_of_nums)
count_bagel=0
count_fermi=0
Chances=10
while Chances!=0:
guess = input(f'Guess the three digit number! You have {Chances} to guess! ')
while len(guess)!=3:
guess=input('You must choose a three digit number! Try again! ')
for i in range(0,len(random_three_num)):
if guess==random_three_num:
print('YOU WIN! Well done')
break
elif guess[i] not in random_three_num:
count_bagel+=1
if count_bagel==len(random_three_num):
print('Bagels')
Chances=Chances-1
elif guess[i]==random_three_num[i]:
count_fermi+=1
Chances=Chances-1
print('Fermi')
elif guess in random_three_num:
print('Pico')
import random
choice_of_nums = ['123', '345', '674', '887', '356', '487', '916']
random_three_num = random.choice(choice_of_nums)
count_bagel = 0
count_fermi = 0
chances = 10
while chances > 0:
guess = input(f'Guess the three digit number! You have {chances} to guess! ')
while len(guess) != 3:
guess = input('You must choose a three digit number! Try again! ')
while not guess.isdigit():
guess = input('You must choose integer values! ')
number_is_present = any(number in guess for number in random_three_num)
if guess == random_three_num:
print('YOU WIN! Well done')
chances = 1 # combined w/ the last line, chances will become = 0
elif not number_is_present:
print('Bagel')
else:
index_is_right = False
for i in range(len(guess)):
if guess[i] == random_three_num[i]:
index_is_right = True
if index_is_right:
print('Fermi')
else:
print('Pico')
chances -= 1
(06/28/22) added chances = 1 if the guess is right, so to exit the while loop
random.choices returns a list
you don't need the re module
use snake case as suggested in PEP8
The break after print('YOU WIN! Well done') exits the for loop not the while loop. Put Chances = 0 before the break:
if guess==random_three_num:
print('YOU WIN! Well done')
Chances = 0
break
You should never check a while loop with a condition like x != 0. Always use <= or >=. The reason being, that if somehow the number zero is skipped and you end up at -1 then the loop will still exit.
Couldn't your check if guess==random_three_num: be done before the for loop? Then the break statement would actually break the while loop. Now it's only breaking the for loop. This is one reason that could lead to a infinite loop.
Your second to last line elif guess in random_three_num: should probably be elif guess[1] in random_three_num:.
Chances=Chances-1 could probably be outside the for loop also, as the number of chances should decreasing only one per guess. Currently the number of chances decreases up to 3 times during the for loop (every time you hit 'Fermi'). This could lead to issue described in "1."
Python beginner here. Practicing user input control.
Trying to make user input loop to the beginning if anything but a whole number between 1 and 10 is used. Been trying for hours, tried using Try and Except commands but couldn't do it correctly. What am i doing wrong? Thank you.
Edit:
Thank you very much for your help everyone, however the problem is still not solved (but very close!) I'm trying to figure out how to loop back to the beginning if anything BUT a whole number is typed. Agent Biscuit (above) gave a great answer for floating numbers, but any word or letter that is typed still produces an error. I´m trying to understand how to loop when anything random (except whole numbers between 1 and 10) is typed. None of the above examples produced corrcct results. Thank you for your help
while True:
print("Enter a number between 1 and 10")
number = int(input())
if (number > 0) and (number < 10):
print("Thank you, the end.")
break
else number != (> 0 and < 10):
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
I have identified some problems.
First, the input statement you are using would just raise an error if a float value is entered, because the int at the start requires all elements of the input to be a number, and . is not a number.
Second; your else statement. else is just left as else:, and takes no arguments or parameters afterwards.
Now, how to check if the number is not whole? Try this:
while True:
print("Enter a number between 1 and 10")
number = float(input())
if (number > 0) and (number < 10) and (round(number)==number):
print("Thank you, the end.")
break
else:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
This accepts a float value, but only accepts it if it is equal to a whole number, hence the (round(number)==number).
Hope that answers your question.
First of all, you can't use a condition in a else statement. Also, you need to use or operator instead of and if one of the conditions is acceptable.
So, your code needs to be like this
while True:
print("Enter a number between 1 and 10")
number = int(input())
if (number > 0) and (number < 10):
print("Thank you, the end.")
break
elif number < 0 or number >10:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
Thanks to ack (above) for pointing me to a useful link. By studying another thread, I found the solution. It may not be perfect code, but it works 100%:
while True:
try:
print("Enter a number between 1 and 10")
number = float(input())
if (number > 0) and (number < 10) and (round(number)==number):
print("Thank you, the end.")
break
else:
print("\n")
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
print("\n")
continue
except ValueError:
print("It has to be a whole number between 1 and 10.")
print("Please try again:")
print("\n")
So, I'm working on Hand Cricket script & i want to stop "for" loop from iterating when a user's choice of number & CPU's choice of number are equal & if it's unequal it should keep iterating until it doesnt reach the final range's value
for i in range(1,7):
print("Ball %d"%i)
user_choice =int(input("Enter a number between 1 to 6 --> "))
cpu_choice=random.randint(1,6)
if user_choice < 7:
print("CPU picked --> ",cpu_choice)
run=cpu_choice+run
if user_choice == cpu_choice:
print("User is OUT!!")
run -= cpu_choice
print("Runs = %d \t Over = %d.%d\n"%(run,i//6,i%6))
break
print("Runs = %d \t Over = %d.%d\n"%(run,i//6,i%6))
else:
print("\nWRONG CHOICE!! %d ball is cancelled.\n"%i)
break
I might be missing something in your question, but it looks like you've already got it. break will force the for loop to exit. So if you wrap your break-statement in a conditional gate (if statement), you can set the criteria that must be met for the break.
Something like this comes to mind:
# Calculate the CPU's Random Integer ONCE
cpu_choice=random.randint(1,6)
# Iteratively Ask User for Input and Validate
for i in range(1,7):
# Capture Input from User
user_choice =int(input("Enter a number between 1 to 6 --> "))
# Verify Input in Specific Range
if user_choice not in range(1,7):
print("{} is not in the valid range. Try again.".format(user_choice))
else:
# Check if User Input Matches CPU's Selection
if user_choice == cpu_choice:
print("You've got the right number! The number was: {}".format(user_choice))
break # break out of the `for` loop!
# Not Correct Input from User
else:
print("{} is not the correct number. Try again.".format(user_choice))
Again, it seems like you've already come to this answer in a way. Are you asking something else instead?
I am trying to create a number guessing game in Python for a school project. I have made a basic game that will work fairly well, but I want to add in some exception handling in case the user enters something incorrectly. For example, this is a section of my code:
def Normal_Guess():
number = round(random.uniform(0.0, 100.0),2)
guess = ""
while guess != number:
try:
guess = float(input("Please guess a number between 0 and 100: "))
except ValueError or TypeError:
print ("That isn't even a number!")
if guess < number and guess >= 0:
print ("You need to guess higher!")
elif guess > number and guess <= 100:
print ("You need to guess lower!")
elif guess == number:
print("Congratulations! You guessed the number!")
elif guess < 0 or guess > 100:
print ("The number you guessed is not between 0 and 100")
else:
print("That isn't even a number!")
New_Game()
This works fine when the user enters a float or integer value as "guess", and the Try-Except clause I have seems to catch if the user enters anything but a number at first, but the program seems to also carry on to the "if" statements. I am getting a TypeError saying that "'<' not supported between instances of 'str' and 'float'".
I have tried encompassing the entire loop in a Try-Except clause, and that doesn't work. I have no clue what I am doing wrong. Any help is much appreciated.
First off, the way you are catching the exception is invalid. The value of the expression ValueError or TypeError is always just going to be ValueError because that is how short-circuiting works with two non-False arguments. To get both types of errors to trigger the block, use a tuple, like (ValueError, TypeError).
The problem is that even if an exception is caught in your code, it will continue on to the if block. You have four simple options to avoid this:
Use a continue statement in the except block to tell the loop to move on without processing the following if structure:
try:
guess = float(input("Please guess a number between 0 and 100: "))
except (ValueError, TypeError):
print ("That isn't even a number!")
continue
This is probably the cleanest and easiest of the four options.
Do not use an except block to respond to the error. Instead, rely on the fact that the value of guess is still "". For this to work, you will have to pre-initialize guess with every iteration of the loop instead of once outside the loop:
while guess != number:
guess = ""
try:
guess = float(input("Please guess a number between 0 and 100: "))
except (ValueError, TypeError):
pass
if guess == "":
print ("That isn't even a number!")
elif guess < number and guess >= 0:
...
Personally, I am not a fan of this approach because it requires an initialization in every loop. This is not bad, just not as clean as option #1.
A variation on this option is to check directly if guess is an instance of str. You can then initialize it to the user input, making the conversion operation cleaner:
while guess != number:
guess = input("Please guess a number between 0 and 100: ")
try:
guess = float(guess)
except (ValueError, TypeError):
pass
if isinstance(guess, str):
print ("That isn't even a number!")
elif guess < number and guess >= 0:
...
Use the else clause that is one of the possible elements of a try block. This clause gets executed only if no exception occurred:
try:
guess = float(input("Please guess a number between 0 and 100: "))
except (ValueError, TypeError):
print ("That isn't even a number!")
else:
if guess < number and guess >= 0:
...
While this option creates an added layer of indentation, it is a possibility worth keeping in mind for those cases where a plain continue won't work. This happens sometimes when you need to do additional processing for both error and non-error cases, before you branch.
Put the entire if block into the try block. This way it will only be executed if there is no error. This is my least favorite option because I like my try blocks to be as trimmed-down as possible to avoid catching exceptions I did not intend to. In Python, try is relatively less of a performance-killer than in a language like Java, so for your simple case, this is still an option:
try:
guess = float(input("Please guess a number between 0 and 100: "))
if guess < number and guess >= 0:
...
except (ValueError, TypeError):
print ("That isn't even a number!")
Try using an else statement.
Your except catch print, but let script continue running. It will continue to all of if statements even when the catch is hit. What you want to do is skip the main logic of your function when the except is hit. Use the ELSE clause of the try-catch-else-finally block.
import random
def Normal_Guess():
number = round(random.uniform(0.0, 100.0),2)
guess = ""
while guess != number:
try:
guess = float(input("Please guess a number between 0 and 100: "))
except (ValueError, TypeError):
print ("That isn't even a number!")
else:
if guess < number and guess >= 0:
print ("You need to guess higher!")
elif guess > number and guess <= 100:
print ("You need to guess lower!")
elif guess == number:
print("Congratulations! You guessed the number!")
elif guess < 0 or guess > 100:
print ("The number you guessed is not between 0 and 100")
else:
print("That isn't even a number!")
Normal_Guess()
I am learning Python on Codecademy, and I am supposed to give the user 3 guesses before showing "you lose". I think my code allows 3 entries, but the website shows "Oops, try again! Did you allow the user 3 guesses, or did you incorrectly detect a correct guess?" unless the user guesses correctly within 3 trials. Can someone tell me what's wrong?
from random import randrange
random_number = randrange(1, 10)
count = 0
# Start your game!
guess= int(raw_input("Please type your number here:"))
while count < 2:
if guess==random_number:
print "You win!"
break
else:
guess=int(raw_input("Please guess again:"))
count+=1
else:
print "You lose!"
print random_number
Your loop will indeed ask the user for three guesses. (As can be trivially seen by running the code—ignore those other answers telling you to change the loop condition, that's the wrong solution.)
The problem with your loop is a more subtle one: because of the way it's structured, the third guess is never tested! You can see this by setting random_number to a constant and guessing wrong twice, then right on the last try; you still lose.
Your best bet is to use a more straightforward loop structure, where the asking and the checking happens in the same iteration of the loop.
for attempt in xrange(3):
guess = int(raw_input("Please enter a number: "))
if guess == random_number:
print "You win!"
break
print "Wrong! Try again."
else:
print "You lose! The number was", random_number
If you want a different prompt on the second and subsequent guesses, try this:
prompt = "Please enter a number"
for attempt in xrange(3):
guess = int(raw_input(prompt + ": "))
if guess == random_number:
print "You win!"
break
prompt = "Wrong! Try again"
else:
print "You lose! The number was", random_number
You need while count <= 2. Your count starts at 0. Then it goes through the body of your loop once. Then it gets incremented to 1. Then it goes through your loop body another time. Finally, once it increments to 2, your while condition evaluates to false, and the loop body doesn't execute a third time.
Be careful with corner cases when you're setting up conditions. :)
The condition should be:
while count < 3:
To make it easier to understand, I suggest you start the counter in count = 1 and write the condition like this:
while count <= 3:
Now it's more clear that exactly 3 repetitions are allowed. But let's see why your code was wrong:
count starts at 0, and it's true that 0 < 2, so we enter the loop
At the first failed attempt, count gets incremented to 1, and it's true that 1 < 2 so we enter the loop once more
At the second failed attempt, count gets incremented to 2, and it's no longer true that 2 < 2 so we exit the loop
So you see, only two attempts were being considered.