I was writing this code and I can't seem to get it to work correctly.
There's no Syntax errors so I'm clear on that.
It just is not giving me the correct output I want.
Here's the code :
This is the out put I get when x = 5 :
Big!
Done!
This is the output I get when x = 1
Small!
Done!
This is what I get when x = anything other than 1 or 5
Done!
What I want it to do is when x = anything between 1-5 to output Small! then Done!
Or if it is between 5-infinity to output Big! then Done! but if the number is not 1 or 5 it just outputs Done!
What changes should I do to my code?
As pointed by a user in the previous answer, what you need to implement is an if-else ladder and use logical operator for the case when your output is specifically either 1 OR 5
x=6 # You can replace this by a user defined input using input()
if x==5 or x==1:
print("Done!")
elif x<5:
print("Small!")
print("Done!")
elif x>5:
print("Big!")
print("Done!")
else:
print("Enter a valid number!")
I checked with various test cases like 1, 5 , numbers between 1 and 5 and numbers greater than 5 and all seem to work fine.
Here's a sample output
x = 5
if x in range(6):
print('Small')
elif x >=6 :
print('Big')
print('Done')
Try this. The range function checks if the number is in between 0 and 6 i.e 0 to 5. Anything lesser than 0 will be ignored
The problem with your code is that you're just checking two conditions if x== 5 and if x == 1. The print statements will be executed only if this condition is satisfied.
Cheers
You can create a if-else ladder to achieve this
def determine_size(inp):
if inp == 1 or inp == 5:
print("Done")
elif 0 <= inp < 5:
print("Small")
print("Done")
elif inp > 6:
print("Big")
print("Done")
else:
print("Negative Input")
>>> determine_size(0)
Small
Done
>>> determine_size(100)
Big
Done
>>> determine_size(60)
Big
Done
>>>
>>> determine_size(3)
Small
Done
>>> determine_size(4)
Small
Done
You can also play around with the if-else statements per your objectives
x = 40
# at first check if x is greater than 1 and less than 5.
# Only then it is between 1 and 5.
if x >=1 and x<=5:
print('Small!')
# Now chek if x is greater than 5
elif x>5:
print('Big!')
print('Done!')
if you meant the above mentioned scenario mentioned in comment
try:
x=int(input("enter a number: "))
if x >=0 or x<=5:
print("Small")
print("Done")
elif x>=6:
print("big")
print("Done")
except ValueError:
print("Done")
here the block of code is written in try block, why i have written in try block if someone enter something which is not number then program will not crash and raise a exception.
if you are not familiar with elif syntax, we can have simple construction like this :-
try:
x=int(input("enter a number: "))
if x >=0 and x<=5:
print("Small")
print("Done")
else:
print("big")
print("Done")
except ValueError:
print("Done")
Here all I have done is to change the logical operator from or to and.
I am new to python and programming and having difficulty breaking out of a while loop that calls a few functions. I have tried a variety of different options but they all end the same, well they don't end it just keeps running. I am only on here because I have really researched and tried for a long time to fix this. Below is some of the code where you can see there might be confusion with a function. I am not posting the full program just the last part. Thank you for any assistance while I learn. This is also my first time posting, I've been using stackoverflow for the past year dabbling.
def main():
choice = input('''Hello,
Would you like to do basic math or return an average?
Please select 1 for basic math and 2 for average and 3 to quit:
''')
if choice == '1':
print(performCalculation())
elif choice == '2':
print(calculateAverage())
elif choice == '3':
print(main())
j = 0
k = 0
while j < 3:
print(main())
while k == 3:
break
print('All Done!')
Simply change
j = 0
k = 0
while j < 3:
print(main())
while k == 3:
break
print('All Done!')
to
j = 0
while j < 3:
print(main())
j += 1
print('All Done!')
The reason your while loop never breaks is because you have while j < 3:, but you never change the value of j, so if it was smaller to begin with, it will forever be smaller.
Also, k will never equal to 3, and even if it will, the break statement within that while loop will only make that while loop terminate, not the main one.
You have several basic mistakes here. I'll go in order of your program execution.
You aren't incrementing your loop variables.
You've started your loop variables with j and k (why not i and j?) set to zero, and are looped based on the value of these variables. Since they are never incremented, your loop never hits an exit condition.
You can increment the variables with j += 1, and k += 1 at the end of the respective loops.
Your loops are "unpythonic"
These would typically be written as my example below. You don't need to declare i separately, or increment it here. Python handles that for you.
for i in range(0, 3):
...
You're printing a function that doesn't return anything.
Your main function doesn't have a return value, so calling print(main()) is nonsense. You can replace this with simply main() unless you change main() to have some kind of return "foo" statement.
Can someone explain why this causes an infinite loop even tho the continue from what I have been reading should "skip" the iteration
x = 0
while x < 50:
if x == 33:
print("I hit 33")
continue
else:
pass
print(x)
x+=1
The continue command restarts the innermost loop at the condition.
That means after x reaches 33, x += 1 will never execute because you will be hitting the continue and going back to the while line without running the rest of the code block.
x will forever be 33 so you will have a infinite loop.
You are skipping the increment that happens at the end of the while loop when you call continue. The following will automatically increment if you want to keep the continue statement:
for x in range(50):
if x == 33:
print("I hit 33")
continue
else:
print(x)
Otherwise, delete the continue.
I think you are confusing break and continue.
continue will skip to next iteration in innermost loop
break will leave innermost loop
continue goes to the next iteration. You want break which exits the loop. See:
for i in range(10):
if i == 5:
continue
if i == 8:
break
print(i)
outputs:
0
1
2
3
4
6
7
I'm guessing the code you're trying to get at is as follows, which will print out each integer from 0 - 50 (exclusive), except it will print "I hit 33" for the integer 33.
x = 0
while x < 50:
if x == 33:
print("I hit 33")
else:
print(x)
x += 1
You really don't need the continue or pass in this instance. continue continues with the next cycle of the nearest enclosing loop. pass is usually only used as a placeholder when a block is expecting a statement but you're not ready to use the statement.
I just learned about break and return in Python.
In a toy code that I wrote to get familiar with the two statements, I got stuck in a loop, but I don't know why. Here is my code:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
I'm new to programming, any suggestions will be appreciated.
With the for-else construct you only enter the else block if the for loop does not break, which your for loop always does because i inevitably becomes 3 with your range generator. Your infinite while loop is therefore never able to reach the return statement, which is only in the said else block.
nvm I'm super wrong here
First of all, when you define a function in Python, any code that belongs in the function should be in the same indentation block. With this in mind, your code would look like this:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
The next problem I see is that your else statement isn't correctly formatted with an if statement. If you mean for it to go on the 2nd if statement, your code would look like this:
def break_return():
while True:
for i in range(5):
if i < 2:
print(i)
if i == 3:
break
else:
print('i = ', i)
return 343
break_return()
This is only formatting. But in this example, the code would only run once because it immediately returns and exits the function.
I think this may be a better example of using both break and return:
def break_return(value):
for i in range(5):
print(i)
if i == 3:
break #This exits the for loop
if i == 4:
print("This won't print!")
#Won't print because the loop "breaks" before i ever becomes 4
return value * 2 #Returns the input value x 2
print(break_return(30)) #Display the return value of break_return()
This demonstrates how break exits a for loop and how return can return a value from the function.
The output of the code above is:
0 #Value of i
1 #Value of i
2 #Value of i
3 #Value of i
60 #The value returned by the function
Glad to hear you're learning Python! It's a lot of fun, and super useful.
I've noticed the following code is legal in Python. My question is why? Is there a specific reason?
n = 5
while n != 0:
print n
n -= 1
else:
print "what the..."
Many beginners accidentally stumble on this syntax when they try to put an if/else block inside of a while or for loop, and don't indent the else properly. The solution is to make sure the else block lines up with the if, assuming that it was your intent to pair them. This question explains why it didn't cause a syntax error, and what the resulting code means. See also I'm getting an IndentationError. How do I fix it?, for the cases where there is a syntax error reported.
The else clause is only executed when your while condition becomes false. If you break out of the loop, or if an exception is raised, it won't be executed.
One way to think about it is as an if/else construct with respect to the condition:
if condition:
handle_true()
else:
handle_false()
is analogous to the looping construct:
while condition:
handle_true()
else:
# condition is false now, handle and go on with the rest of the program
handle_false()
An example might be along the lines of:
while value < threshold:
if not process_acceptable_value(value):
# something went wrong, exit the loop; don't pass go, don't collect 200
break
value = update(value)
else:
# value >= threshold; pass go, collect 200
handle_threshold_reached()
The else clause is executed if you exit a block normally, by hitting the loop condition or falling off the bottom of a try block. It is not executed if you break or return out of a block, or raise an exception. It works for not only while and for loops, but also try blocks.
You typically find it in places where normally you would exit a loop early, and running off the end of the loop is an unexpected/unusual occasion. For example, if you're looping through a list looking for a value:
for value in values:
if value == 5:
print "Found it!"
break
else:
print "Nowhere to be found. :-("
Allow me to give an example on why to use this else-clause. But:
my point is now better explained in Leo’s answer
I use a for- instead of a while-loop, but else works similar (executes unless break was encountered)
there are better ways to do this (e.g. wrapping it into a function or raising an exception)
Breaking out of multiple levels of looping
Here is how it works: the outer loop has a break at the end, so it would only be executed once. However, if the inner loop completes (finds no divisor), then it reaches the else statement and the outer break is never reached. This way, a break in the inner loop will break out of both loops, rather than just one.
for k in [2, 3, 5, 7, 11, 13, 17, 25]:
for m in range(2, 10):
if k == m:
continue
print 'trying %s %% %s' % (k, m)
if k % m == 0:
print 'found a divisor: %d %% %d; breaking out of loop' % (k, m)
break
else:
continue
print 'breaking another level of loop'
break
else:
print 'no divisor could be found!'
The else-clause is executed when the while-condition evaluates to false.
From the documentation:
The while statement is used for repeated execution as long as an expression is true:
while_stmt ::= "while" expression ":" suite
["else" ":" suite]
This repeatedly tests the expression and, if it is true, executes the first suite; if the expression is false (which may be the first time it is tested) the suite of the else clause, if present, is executed and the loop terminates.
A break statement executed in the first suite terminates the loop without executing the else clause’s suite. A continue statement executed in the first suite skips the rest of the suite and goes back to testing the expression.
The else clause is only executed when the while-condition becomes false.
Here are some examples:
Example 1: Initially the condition is false, so else-clause is executed.
i = 99999999
while i < 5:
print(i)
i += 1
else:
print('this')
OUTPUT:
this
Example 2: The while-condition i < 5 never became false because i == 3 breaks the loop, so else-clause was not executed.
i = 0
while i < 5:
print(i)
if i == 3:
break
i += 1
else:
print('this')
OUTPUT:
0
1
2
3
Example 3: The while-condition i < 5 became false when i was 5, so else-clause was executed.
i = 0
while i < 5:
print(i)
i += 1
else:
print('this')
OUTPUT:
0
1
2
3
4
this
My answer will focus on WHEN we can use while/for-else.
At the first glance, it seems there is no different when using
while CONDITION:
EXPRESSIONS
print 'ELSE'
print 'The next statement'
and
while CONDITION:
EXPRESSIONS
else:
print 'ELSE'
print 'The next statement'
Because the print 'ELSE' statement seems always executed in both cases (both when the while loop finished or not run).
Then, it's only different when the statement print 'ELSE' will not be executed.
It's when there is a breakinside the code block under while
In [17]: i = 0
In [18]: while i < 5:
print i
if i == 2:
break
i = i +1
else:
print 'ELSE'
print 'The next statement'
....:
0
1
2
The next statement
If differ to:
In [19]: i = 0
In [20]: while i < 5:
print i
if i == 2:
break
i = i +1
print 'ELSE'
print 'The next statement'
....:
0
1
2
ELSE
The next statement
return is not in this category, because it does the same effect for two above cases.
exception raise also does not cause difference, because when it raises, where the next code will be executed is in exception handler (except block), the code in else clause or right after the while clause will not be executed.
I know this is old question but...
As Raymond Hettinger said, it should be called while/no_break instead of while/else.
I find it easy to understeand if you look at this snippet.
n = 5
while n > 0:
print n
n -= 1
if n == 2:
break
if n == 0:
print n
Now instead of checking condition after while loop we can swap it with else and get rid of that check.
n = 5
while n > 0:
print n
n -= 1
if n == 2:
break
else: # read it as "no_break"
print n
I always read it as while/no_break to understand the code and that syntax makes much more sense to me.
thing = 'hay'
while thing:
if thing == 'needle':
print('I found it!!') # wrap up for break
break
thing = haystack.next()
else:
print('I did not find it.') # wrap up for no-break
The possibly unfortunately named else-clause is your place to wrap up from loop-exhaustion without break.
You can get by without it if
you break with return or raise → the entire code after the call or try is your no-break place
you set a default before while (e.g. found = False)
but it might hide bugs the else-clause knows to avoid
If you use a multi-break with non-trivial wrap-up, you should use a simple assignment before break, an else-clause assignment for no-break, and an if-elif-else or match-case to avoid repeating non-trival break handling code.
Note: the same applies to for thing in haystack:
Else is executed if while loop did not break.
I kinda like to think of it with a 'runner' metaphor.
The "else" is like crossing the finish line, irrelevant of whether you started at the beginning or end of the track. "else" is only not executed if you break somewhere in between.
runner_at = 0 # or 10 makes no difference, if unlucky_sector is not 0-10
unlucky_sector = 6
while runner_at < 10:
print("Runner at: ", runner_at)
if runner_at == unlucky_sector:
print("Runner fell and broke his foot. Will not reach finish.")
break
runner_at += 1
else:
print("Runner has finished the race!") # Not executed if runner broke his foot.
Main use cases is using this breaking out of nested loops or if you want to run some statements only if loop didn't break somewhere (think of breaking being an unusual situation).
For example, the following is a mechanism on how to break out of an inner loop without using variables or try/catch:
for i in [1,2,3]:
for j in ['a', 'unlucky', 'c']:
print(i, j)
if j == 'unlucky':
break
else:
continue # Only executed if inner loop didn't break.
break # This is only reached if inner loop 'breaked' out since continue didn't run.
print("Finished")
# 1 a
# 1 b
# Finished
The else: statement is executed when and only when the while loop no longer meets its condition (in your example, when n != 0 is false).
So the output would be this:
5
4
3
2
1
what the...
Suppose you've to search an element x in a single linked list
def search(self, x):
position = 1
p =self.start
while p is not None:
if p.info == x:
print(x, " is at position ", position)
return True
position += 1
p = p.link
else:
print(x, "not found in list")
return False
So if while conditions fails else will execute, hope it helps!
The better use of 'while: else:' construction in Python should be if no loop is executed in 'while' then the 'else' statement is executed. The way it works today doesn't make sense because you can use the code below with the same results...
n = 5
while n != 0:
print n
n -= 1
print "what the..."
As far as I know the main reason for adding else to loops in any language is in cases when the iterator is not on in your control. Imagine the iterator is on a server and you just give it a signal to fetch the next 100 records of data. You want the loop to go on as long as the length of the data received is 100. If it is less, you need it to go one more times and then end it. There are many other situations where you have no control over the last iteration. Having the option to add an else in these cases makes everything much easier.