I'm writing a simple For loop in Python. Is there a way to break the loop without using the 'break' command. I would think that by setting count = 10 that the exit condition would be met and the loop would stop. But that doesn't seem to be the case.
NOTE: Part of the challenge is to use the FOR loop, not the WHILE loop.
import random
guess_number = 0
count = 0
rand_number = 0
rand_number = random.randint(0, 10)
print("The guessed number is", rand_number)
for count in range(0, 5):
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print("You guessed it!")
count = 10
else:
print("Try again...")
count += 1
I'm new to programming, so I'm just getting my feet wet. I could use a 'break' but I'm trying figure out why the loop isn't ending when you enter the guessed number correctly.
The for loop that you have here is not quite the same as what you see in other programming languages such as Java and C. range(0,5) generates a list, and the for loop iterates through it. There is no condition being checked at each iteration of the loop. Thus, you can reassign the loop variable to your heart's desire, but at the next iteration it will simply be set to whatever value comes next in the list.
It really wouldn't make sense for this to work anyway, as you can iterate through an arbitrary list. What if your list was, instead of range(0,5), something like [1, 3, -77, 'Word', 12, 'Hello']? There would be no way to reassign the variable in a way that makes sense for breaking the loop.
I can think of three reasonable ways to break from the loop:
Use the break statement. This keeps your code clean and easy to understand
Surround the loop in a try-except block and raise an exception. This would not be appropriate for the example you've shown here, but it is a way that you can break out of one (or more!) for loops.
Put the code into a function and use a return statement to break out. This also allows you to break out of more than one for loop.
One additional way (at least in Python 2.7) that you can break from the loop is to use an existing list and then modify it during iteration. Note that this is a very bad way to it, but it works. I'm not sure that this will this example will work in Python 3.x, but it works in Python 2.7:
iterlist = [1,2,3,4]
for i in iterlist:
doSomething(i)
if i == 2:
iterlist[:] = []
If you have doSomething print out i, it will only print out 1 and 2, then exits the loop with no error. Again, this is a bad way to do it.
You can use while:
times = 5
guessed = False
while times and not guessed:
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print("You guessed it!")
guessed = True
else:
print("Try again...")
times -= 1
For loops in Python work like this.
You have an iterable object (such as a list or a tuple) and then you look at each element in the iterable, storing the current value in a specified variable
That is why
for i in [0, 1, 2, 3]:
print item
and
for j in range(4):
print alist[j]
work exactly the same. i and j are your storage variables while [0, 1, 2, 3] and range(4) are your respective iterables. range(4) returns the list [0, 1, 2, 3] making it identical to the first example.
In your example you try to assign your storage variable count to some new number (which would work in some languages). In python however count would just be reassigned to the next variable in the range and continue on. If you want to break out of a loop
Use break. This is the most pythonic way
Make a function and return a value in the middle (I'm not sure if this is what you'd want to do with your specific program)
Use a try/except block and raise an Exception although this would be inappropriate
As a side note, you may want to consider using xrange() if you'll always/often be breaking out of your list early.
The advantage of xrange() over range() is minimal ... except when ...
all of the range’s elements are never used (such as when the loop is
usually terminated with break)
As pointed out in the comments below, xrange only applies in python 2.x. In python 3 all ranges function like xrange
In Python the for loop means "for each item do this". To end this loop early you need to use break. while loops work against a predicate value. Use them when you want to do something until your test is false. For instance:
tries = 0
max_count = 5
guessed = False
while not guessed and tries < max_count:
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print("You guessed it!")
guessed = True
else:
print("Try again...")
tries += 1
What #Rob Watts said: Python for loops don't work like Java or C for loops. To be a little more explicit...
The C "equivalent" would be:
for (count=0; count<5; count++) {
/* do stuff */
if (want_to_exit)
count=10;
}
... and this would work because the value of count gets checked (count<5) before the start of every iteration of the loop.
In Python, range(5) creates a list [0, 1, 2, 3, 4] and then using for iterates over the elements of this list, copying them into the count variable one by one and handing them off to the loop body. The Python for loop doesn't "care" if you modify the loop variable in the body.
Python's for loop is actually a lot more flexible than the C for loop because of this.
What you probably want is to use break and to avoid assigning to the count variable.
See the following, I've edited it with some comments:
import random
guess_number = 0
count = 0
rand_number = 0
rand_number = random.randint(0, 10)
print("The guessed number is", rand_number)
# for count in range(0, 5): instead of count, use a throwaway name
for _ in range(0, 5): # in Python 2, xrange is the range style iterator
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print("You guessed it!")
# count = 10 # instead of this, you want to break
break
else:
print("Try again...")
# count += 1 also not needed
As others have stated the Python for loop is more like a a traditional foreach loop in the sense that it iterates over a collection of items, without checking a condition. As long as there is something in the collection Python will take them, and if you reassign the loop variable the loop won't know or care.
For what you are doing, consider using the for ... break ... else syntax as it is more "Pythonic":
for count in range(0, 5):
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print("You guessed it!")
break
else:
print("Try again...")
else:
print "You didn't get it."
As your question states NOTE: Part of the challenge is to use the FOR loop, not the WHILE loop and you don't want to use break, you can put it in a function and return when the correct number is guessed to break the loop.
import random
def main():
guess_number = 0
count = 0
rand_number = 0
rand_number = random.randint(0, 10)
print("The guessed number is", rand_number)
for count in range(0, 5):
guess_number = int(input("Enter any number between 0 - 10: "))
if guess_number == rand_number:
print ("You guessed it!")
return
else:
print("Try again...")
Related
import sys
ok = True
count = 0
while ok == True:
try:
a = int(input("What is a? Put number pls "))
ok = False
count = count + 1
except ValueError:
print("no u")
if (count >= 3):
sys.exit()
What I'm trying to do here is get the user to input an integer. If the user inputted an integer or the loop has ran 3 times, the program stops. I have tried putting the if statement both in the try and except statements, it still keeps on looping.
I have also tried this:
ok = True
count = 0
while ok == True:
try:
a = int(input("What is a? Put number pls "))
ok = False
count = count + 1
except ValueError:
print("no u")
if (count >= 3):
break
It still doesn't break out of the loop.
I have inputted numbers, breaks out of the loop as what I expected. I then tried to input random letters and hoped that the loop stops after 3 times, it still continues to loop until I have inputted a number. Happens to both versions.
Here's an approach which you might not have thought of:
count = 0
while count < 3:
try:
a = int(input("What is a? Put number pls "))
break
except ValueError:
print("no u")
count = count + 1
There is no need for ok because you can just use a break statement if you want to end the loop. What you really want to be testing for is if count is less than three, and incrementing count every time the user gets it wrong.
There is no reason to increment count if the loop is about to end, so you want to increment count whenever there is a ValueError (in the except-block) and the loop is about to start again.
I would recommend a slightly expanded version of michaels solution, that takes advantage of the while/else mechanism to determine when no good input was provided
count = 0
while count < 3:
try:
a = int(input("What is a? Put number pls "))
break
except ValueError:
print("no u")
count = count + 1
else: # is triggered if the loop completes without a break
print("No Break... used up all tries with bad inputs")
sys.exit(-1)
print(f"You entered a number {a}")
I am new to python and I am taking a summer online class to learn python.
Unfortunately, our professor doesn't really do much. We had an assignment that wanted us to make a program (python) which asks the user for a number and it determines whether that number is even or odd. The program needs to keep asking the user for the input until the user hit zero. Well, I actually turned in a code that doesn't work and I got a 100% on my assignment. Needless to say our professor is lazy and really doesn't help much. For my own knowledge I want to know the correct way to do this!!! Here is what I have/had. I am so embarrassed because I know if probably very easy!
counter = 1
num = 1
while num != 0:
counter = counter + 1
num=int(input("Enter number:"))
while num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)
There are a couple of problems with your code:
You never use counter, even though you defined it.
You have an unnecessary nested while loop. If the number the user inputs is even, then you're code will continue to run the while loop forever.
You are incorrectly using the else clause that is available with while loops. See this post for more details.
Here is the corrected code:
while True:
# Get a number from the user.
number = int(input('enter a number: '))
# If the number is zero, then break from the while loop
# so the program can end.
if number == 0:
break
# Test if the number given is even. If so, let the
# user know the number was even.
if number % 2 == 0:
print('The number', number, 'is even')
# Otherwise, we know the number is odd. Let the user know this.
else:
print('The number', number, 'is odd')
Note that I opted above to use an infinite loop, test if the user input is zero inside of the loop, and then break, rather than testing for this condition in the loop head. In my opinion this is cleaner, but both are functionally equivalent.
You already have the part that continues until the user quits with a 0 entry. Inside that loop, all you need is a simple if:
while num != 0:
num=int(input("Enter number:"))
if num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)
I left out the counter increment; I'm not sure why that's in the program, since you never use it.
Use input() and If its only number specific input you can use int(input()) or use an If/else statement to check
Your code wasn't indented and you need to use if condition with else and not while
counter = 1
num = 1
while num != 0:
counter = counter + 1
num = int(input("Enter number:"))
if num % 2 == 0:
print ("Even", num)
else:
print ("Odd", num)
Sample Run
Enter number:1
Odd 1
Enter number:2
Even 2
Enter number:3
Odd 3
Enter number:4
Even 4
Enter number:5
Odd 5
Enter number:6
Even 6
Enter number:0
Even 0
I have previously studied Visual Basic for Applications and am slowly getting up to speed with python this week. As I am a new programmer, please bear with me. I understand most of the concepts so far that I've encountered but currently am at a brick wall.
I've written a few functions to help me code a number guessing game. The user enters a 4 digit number. If it matches the programs generated one (I've coded this already) a Y is appended to the output list. If not, an N.
EG. I enter 4567, number is 4568. Output printed from the list is YYYN.
import random
def A():
digit = random.randint(0, 9)
return digit
def B():
numList = list()
for counter in range(0,4):
numList.append(A())
return numList
def X():
output = []
number = input("Please enter the first 4 digit number: ")
number2= B()
for i in range(0, len(number)):
if number[i] == number2[i]:
results.append("Y")
else:
results.append("N")
print(output)
X()
I've coded all this however theres a few things it lacks:
A loop. I don't know how I can loop it so I can get it to ask again. I only want the person to be able to guess 5 times. I'm imagining some sort of for loop with a counter like "From counter 1-5, when I reach 5 I end" but uncertain how to program this.
I've coded a standalone validation code snippet but don't know how I could integrate this in the loop, so for instance if someone entered 444a it should say that this is not a valid entry and let them try again. I made an attempt at this below.
while myNumber.isnumeric() == True and len(myNumber) == 4:
for i in range(0, 4)):
if myNumber[i] == progsNumber[i]:
outputList.append("Y")
else:
outputList.append("N")
Made some good attempts at trying to work this out but struggling to patch it all together. Is anyone able to show me some direction into getting this all together to form a working program? I hope these core elements that I've coded might help you help me!
To answer both your questions:
Loops, luckily, are easy. To loop over some code five times you can set tries = 5, then do while tries > 0: and somewhere inside the loop do a tries -= 1.
If you want to get out of the loop ahead of time (when the user answered correctly), you can simply use the break keyword to "break" out of the loop. You could also, if you'd prefer, set tries = 0 so loop doesn't continue iterating.
You'd probably want to put your validation inside the loop in an if (with the same statements as the while loop you tried). Only check if the input is valid and otherwise continue to stop with the current iteration of your loop and continue on to the next one (restart the while).
So in code:
answer = [random.randint(0, 9) for i in range(4)]
tries = 5
while tries > 0:
number = input("Please enter the first 4 digit number: ")
if not number.isnumeric() or not len(number) == len(answer):
print('Invalid input!')
continue
out = ''
for i in range(len(answer)):
out += 'Y' if int(number[i]) == answer[i] else 'N'
if out == 'Y' * len(answer):
print('Good job!')
break
tries -= 1
print(out)
else:
print('Aww, you failed')
I also added an else after the while for when tries reaches zero to catch a failure (see the Python docs or maybe this SO answer)
First, in my code i'm asked to enter a value from user and the program should place it in the correct orderly position without using the built-in python sort().
My completed code can do this without repetitively inserting the same element, only with the break command. Now i once i remove the break command the number i enter outputs in the list 3 times instead of 1.
Note: we aren't allowed to use the break statement in this Python course
Currently my code looks like this:
#List of numbers
myList = [1,9,16,24]
#Enter a number to be placed in the list
num = int(input("Enter a number: "))
#For loop to check the placement of entered number
#to compare it and see if it is less than or
#equal to the each and every element in myList
for x in range(0, len(myList)-1):
if num < myList[x]:
myList.insert(x,num)
#break
elif num > myList[x]:
myList.append(num)
#break
print(myList)
ex. output:
[-1,-1,-1,1,9,16,24]
Simple. Have an external condition that you can test at each insertion attempt. If you insert the item, set the condition to false. Then, outside of the loop, if the condition is still true you know the item goes at the end. No breaks involved.
myList = [1,9,16,24]
num = int(input("Enter a number: "))
condition = True
for index, x in enumerate(myList):
if condition and num < x:
myList.insert(index, num)
condition = False
if condition:
myList.append(num)
print(myList)
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.