This question already has answers here:
Why is "None" printed after my function's output?
(7 answers)
Closed 5 years ago.
I'm trying to create a function that uses a while loop to count up from one to a number given by a user. The code executes as I intend it to but returns None at the end. How do I get rid of the None? Here's the code.
def printFunction(n):
i = 1
while i <= n:
print(i)
i+=1
print (printFunction(int(input())))
You can use this code to prevent none, tough its just the last line changed
def printFunction(n):
i = 1
while i <= n:
print(i)
i+=1
printFunction(int(input()))
In the last line you were using
print(printFunction(int(input()))) which was getting you None after printing the results.
Instead just use printFunction(int(input())). This will not print None. You can also use a message to ask user like printFunction(int(input("Enter a number"))). Since there is noting getting returned you no need to use print.
Related
This question already has answers here:
Why is "None" printed after my function's output?
(7 answers)
Closed 3 years ago.
I tried to make a program to determine whether a number is odd or not, I get the following output
def is_odd(n):
if n % 2 != 0:
print(True)
else:
print(False)
if I put
print(is_odd(1))
I get:
True
None
Whatever number I choose, I always get
None
at the end.
You have to return something to remove this None . You can also modify the code by returning the True or false and print it in the main
This question already has answers here:
Return multiple values over time
(2 answers)
Closed 5 years ago.
I've written a function in python to see If i know what I'm doing but unfortunately I'm not. I thought my defined arg variable will hold a number and multiply each of the numbers which *args variable holds and print it all at a time. But, it is giving me the first result only.
This is what I've written:
def res_info(arg,*args):
for var in args:
return var*arg
print(res_info(2,70,60,50))
Having:
140
Expected:
140
120
100
I don't understand why I'm not getting all the results. Thanks for taking a look into it.
You are on the right path. The problem you had was due to your use of the return statement. Use yield instead.
>>> def res_info(arg,*args):
for var in args:
yield var*arg
>>> list(res_info(2,70,60,50))
=> [140, 120, 100]
So, what was happening was, even though your logic was correct, since there was a return statement in the loop, your loop hence was never fully executed and your program scope would come out on the first iteration of the loop itself (ie, when var = 70).
Using yield, solved the problem as it returns a Generator object after all calculations, and hence does not exit the loop like return does.
def res_info(arg,*args):
result = []
for var in args:
result.append(var*arg)
return result
print(res_info(2,70,60,50))
This question already has answers here:
Assignment Condition in Python While Loop
(5 answers)
Closed 5 years ago.
I am learning python. I tried to the following code but it did not work.
How can I print inputs from a user repeatedly in python?
while ( value = input() ):
print(value)
Thank you very much.
Assignments within loops don't fly in python. Unlike other languages like C/Java, the assignment (=) is not an operator with a return value.
You will need something along the lines of:
while True:
value = input()
print(value)
while true is an infinite loop, therefore it will always take an input and print the output. value stores the value of a user input, and print prints that value after. This will always repeat.
while True:
value = input()
print(value)
Use this code
while 1:
print(input())
If you want to stop taking inputs, use a break with or without condition.
This question already has answers here:
How do I get a result (output) from a function? How can I use the result later?
(4 answers)
Closed 5 years ago.
I am having issues with values being returned exactly as they are passed to a method despite being modified inside the method.
def test(value):
value+=1
return value
value = 0
while True:
test(value)
print(value)
This stripped-down example simply returns zero every time instead of increasing integers as one might expect. Why is this, and how can I fix it?
I am not asking how the return statement works/what is does, just why my value isn't updating.
You need to assign the return'd value back
value = test(value)
Use this :-
def test(value):
value+=1
return value
value = 0
while True:
print(test(value))
You weren't using the returned value and using the test(value) inside the print statement saves you from creating another variable.
This question already has answers here:
calculate length of list recursively [duplicate]
(2 answers)
Closed 7 years ago.
So, my goal is to count the number of arguments in a list using recursion. However, as I'm only to use one argument in the function call, I don't know how to solve it without a second "count" argument.
So far I have this, where I accumulate the 1's together.
def countElements(a):
if a==[]:
return []
else:
return [1] + countElements(a[1:])
def main():
a=[3,2,5,3]
print(countElements(a))
main()
Instead of returning an empty list in the base case, return 0, and instead of [1] + countElements(a[1:]), return 1 + countElements(a[1:]). That way, you're just keeping the running count instead of getting a list back.