I am a beginner in Python and from what I understand, the continue statement in Python returns the control to the beginning of the while loop.
guesses = [0]
while True:
# we can copy the code from above to take an input
guess = int(input("I'm thinking of a number between 1 and 100.\n What is your guess? "))
if guess < 1 or guess > 100:
print('OUT OF BOUNDS! Please try again: ')
continue
# here we compare the player's guess to our number
if guess == num:
print(f'CONGRATULATIONS, YOU GUESSED IT IN ONLY {len(guesses)} GUESSES!!')
break
# if guess is incorrect, add guess to the list
guesses.append(guess)
# when testing the first guess, guesses[-2]==0, which evaluates to False
# and brings us down to the second section
if guesses[-2]:
if abs(num-guess) < abs(num-guesses[-2]):
print('WARMER!')
else:
print('COLDER!')
else:
if abs(num-guess) <= 10:
print('WARM!')
else:
print('COLD!')
Above is the code for the game called 'guess the number from 1 - 100'.
The first if statement where guess < 1 or guess > 100, it will print "Out of bounds!" and then continue which loops to the top of the code and asks for the user's input again.
But for the 3rd if statement where if guesses[-2]:, it does not require continue for neither if nor else.
Sorry if you do not understand what I am asking. But essentially, I want to know why 'continue' statement is not required after print('WARMER!), print('COLDER!'), print('WARM!') and print('COLD!').
guesses = [0]
while True:
# we can copy the code from above to take an input
guess = int(input("I'm thinking of a number between 1 and 100.\n What is your guess? "))
if guess < 1 or guess > 100:
print('OUT OF BOUNDS! Please try again: ')
continue
# here we compare the player's guess to our number
if guess == num:
print(f'CONGRATULATIONS, YOU GUESSED IT IN ONLY {len(guesses)} GUESSES!!')
break
# if guess is incorrect, add guess to the list
guesses.append(guess)
# when testing the first guess, guesses[-2]==0, which evaluates to False
# and brings us down to the second section
if guesses[-2]:
if abs(num-guess) < abs(num-guesses[-2]):
print('WARMER!')
**continue**
else:
print('COLDER!')
**continue**
else:
if abs(num-guess) <= 10:
print('WARM!')
**continue**
else:
print('COLD!')
**continue**
Note that all four of these print statements are the last of their execution branch.
Meaning, if guesses[-2] evaluates as true, than the else part won't be executed at all. Then, if abs(num-guess) < abs(num-guesses[-2]) evaluates to true, again - its else won't be executed. It means that for this execution branch the print('WARMER!') is the last statement of the loop and hence continue is not needed.
Same logic applies to all other 3 print statements.
In the case "guess < 1 or guess > 100" you want to skip the rest of the loop since you dont want to append the guess to the guesses list. Therefore you use continue.
The 'continue' statement is not required after print('WARMER!) etc because you dont need to skip any code afterwards. In this case because there wont be any lines executed after the print statement anyway.
First of all, if/else doesn't require continue. It's optional.
From what I understand you want to know why you used continue in the first if/else and not in the last if/else.
In the first one, after executing if statement it will transfer the control to while True: but for the last if/else statement it is not needed as the control will automatically go to while True: as it is the end of the loop and even if you had used continue it wouldn't have made a difference.
For further references
If you want to know why the continue is not always in if or else, this is due to the fact that when the operation arrives (true), the loop is exited and the code continues to run.
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."
I am just learning the basics of Python and created a number guessing game. I want the user to be able to guess the number as many times as possible until they guess correctly. I did this through a while loop but the code, "else guess == a:" near the end is giving me a syntax error. I am confused because the while loop ensures that the input guess is an integer by the if statement,
if guess.isdigit():
guess = int(guess)
Please help
import random
a = random.randint(1,10)
print("this is a number guessing game")
question_one = input("Would you like to play? Yes or No?:")
if question_one == "Yes":
print("Let's go!")
else:
print("That sucks!")
exit()
guess = None
while guess != a:
guess = (input("Alright, guess a number from 1-10"))
if guess.isdigit():
guess = int(guess)
if guess > a:
guess = int(input("Guess lower!"))
elif guess < a:
guess = int(input("Guess higher!"))
else guess == a:
print("you got it!")
else doesn't let you define a condition. else will execute if all other conditionals return false. You should change that last else to elif. Or you can simply leave out the conditional guess == a all together. If it is not greater than or less than, the only other thing it can be is equal to.
If you have the last else, the code inside the else will be executed if any other condition goes false. So I think if you change the last else with an elif we work properly.
An else statement contains the block of code that executes if the conditional expression in all your if or elif statements is false. Hence in your case you're supposed to use the elif statement instead of else. See the following:
elif guess == a:
print("you got it!")
How do I make a specific line of code execute only once inside a while loop?
I want the line:
"Hello %s, please enter your guess: " %p1" to run only once and not every time the player guesses wrong.
Is there are command or function I can use or do I have to structure the whole game differently? Is there a simple fix to the program in this form?
import random
number = random.randint(1,9)
p1 = input("Please enter your name: ")
count = 0
guess = 0
while guess != number and guess != "exit":
guess = input("Hello %s, please enter your guess: " % p1)
if guess == "exit":
break
guess = int(guess)
count += 1
if guess == number:
print("Correct! It Took you only", count, "tries. :)")
break
elif guess > number:
print("Too high. Try again.")
elif guess < number:
print("Too low. Try again.")
You can create a flag variable, e. g.
print_username = True
before the while loop. Inside the loop uncheck it after loop's first iteration:
if print_username:
guess = input("Hello %s, please enter your guess: " % p1)
print_username = False
else:
guess = input("Try a new guess:")
You have to ask for a new guess on every iteration - else the code will loop either endlessly (after first wrong guess) or finish immediately.
To change up the message you can use a ternary (aka: inline if statement) inside your print to make it conditional:
# [start identical]
while guess != number and guess != "exit":
guess = input("Hello {}, please enter your guess: ".format(p1) if count == 0
else "Try again: ")
# [rest identical]
See Does Python have a ternary conditional operator?
The ternary checks the count variable that you increment and prints one message if it is 0 and on consecutive runs the other text (because count is no longer 0).
You might want to switch to more modern forms of string formatting as well: str.format - works for 2.7 as well
A way to execute an instruction only x times in a while loop could be to implement a counter, and add an if condition that checks if the counter < x before executing the instruction.
You should ask for the username outside of the loop and request input at the beginning of the loop.
Inside the loop you create output at the end and request input on the next iteration. The same would work for the first iteration: create output (outside of the loop) and then request input (first thing inside the loop)
I'm making a number guessing game and can someone tell me why it wont work properly?
import random
import time
time.time()
count=0
a = random.randint(0,100)
b = int(input("What number will you guess?"))
while b != a:
if b > a:
print("TOO BIG")
count = count + 1
elif b < a:
print("TOO SMALL")
count = count + 1
else b == a:
print("YOU WIN")
count = count + 1
time=time.time()
print("You took",count,"tries")
print("It took you",time,"second")
The reason this isn't working properly is because you're calling b = input outside of your while loop. As it stands, the user will be asked a single time what their guess is, then the loop will just move infinitely, because b is never being modified.
This loop will accomplish more what you want:
a = random.randint(0,100)
b = -1
while b != a:
b = int(input("What number will you guess?"))
A few notes, though:
Firstly, as Lee Daniel Crocker points out, the user will actually never see "YOU WIN", because you've structured your if-elif-else statement incorrectly. else by definition cannot have a condition - it exists purely in exclusion to all other conditionals in the same block. Additionally, your else statement is the opposite of your while condition. When that else becomes true, the loop exits. You'll need to handle printing "YOU WIN" somewhere else.
Second, you're not validating the user's input in any way - if they enter 'a', the program will crash, because you can't cast 'a' to an int. Either add an exception handler (for ValueError) or using isdigit() on the string, then casting it.
Third, you're not using time.time() correctly - you need to subtract the time at which the user wins from the time at which they started, then represent that value, which is in seconds, in some meaningful way. As it stands you're telling each player that they took the number of seconds since the beginning of the UNIX epoch to complete the game.
Also, for usability reasons, you should probably provide the user some way to break out - a string like "QUIT" - because as it stands, the only way to restart/quit is to either close the application or KeyboardInterrupt.
You need to accept input in the loop. Move the line b = int(input("What number will you guess?")) within the loop.
I'm having some trouble with breaking out of these loops:
done = False
while not done:
while True:
print("Hello driver. You are travelling at 100km/h. Please enter the current time:")
starttime = input("")
try:
stime = int(starttime)
break
except ValueError:
print("Please enter a number!")
x = len(starttime)
while True:
if x < 4:
print("Your input time is smaller than 4-digits. Please enter a proper time.")
break
if x > 4:
print("Your input time is greater than 4-digits. Please enter a proper time.")
break
else:
break
It recognizes whether the number is < 4 or > 4 but even when the number inputted is 4-digits long it returns to the start of the program rather than continues to the next segment of code, which isn't here.
You obviously want to use the variable done as a flag. So you have to set it just before your last break (when you are done).
...
else:
done = 1
break
The reason it "returns to the beginning of the program" is because you've nested while loops inside a while loop. The break statement is very simple: it ends the (for or while) loop the program is currently executing. This has no bearing on anything outside the scope of that specific loop. Calling break inside your nested loop will inevitably end up at the same point.
If what you want is to end all execution within any particular block of code, regardless of how deeply you're nested (and what you're encountering is a symptom of the issues with deeply-nested code), you should move that code into a separate function. At that point you can use return to end the entire method.
Here's an example:
def breakNestedWhile():
while (True):
while (True):
print("This only prints once.")
return
All of this is secondary to the fact that there's no real reason for you to be doing things the way you are above - it's almost never a good idea to nest while loops, you have two while loops with the same condition, which seems pointless, and you've got a boolean flag, done, which you never bother to use. If you'd actually set done to True in your nested whiles, the parent while loop won't execute after you break.
input() can take an optional prompt string. I've tried to clean up the flow a bit here, I hope it's helpful as a reference.
x = 0
print("Hello driver. You are travelling at 100km/h.")
while x != 4:
starttime = input("Please enter the current time: ")
try:
stime = int(starttime)
x = len(starttime)
if x != 4:
print("You input ({}) digits, 4-digits are required. Please enter a proper time.".format(x))
except ValueError:
print("Please enter a number!")