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whenever i try to use else and elif i get that error .
x=4
if x>0:
print("positive")
elif x=4:
print("equal")
else:
print("negative")
Msg Error :
File " '<'stdin'>' ", line 1
elif x=4 :
^
SyntaxError: invalid syntax
File "'<'stdin'>'", line 1
else:
^
SyntaxError: invalid syntax
= is an assignment operator.
== is a comparison operator.
You need to use the comparison operator in the elif statement like so:
x = 4
if x > 0:
print("positive")
elif x == 4:
print("equal")
else:
print("negative")
That is because you are using assignment operator = instead of equality operator ==.
x=4
if x>0:
print("positive")
elif x==4:
print("equal")
else:
print("negative")
Related
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def factorial(n):
if(n == 1):
fact = 1
return fact
else:
fact = a * factorial(n - 1)
return fact
a = int(input("Enter a number to find its factorial: ")
if (a > 0):
print("The factorial of", a, "is:", fact)
else:
print("Enter a positive value to find its factorial")
In the above code it tells me that - NameError name 'fact' is not defined .
Lines starting from a = int... should be outside your function. Once that is done all you need is to add fact = factorial(a).
Find the correct logic below.
def recur_factorial(n):
if n == 1:
return n
else:
return n*recur_factorial(n-1)
num = 7
# check if the number is negative
if num < 0:
print("Sorry, factorial does not exist for negative numbers")
elif num == 0:
print("The factorial of 0 is 1")
else:
print("The factorial of", num, "is", recur_factorial(num))
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New to this. My autochecker says that line 3 is incorrect.
score = input("Enter Score: ")
s = float(score)
if s >= 0.9
("A")
elif s >= 0.8
print ("B")
elif s >=0.7
print ("C")
elif s >=0.6
print ("D")
elif s < 0.5
print ("F")
else
print ("Enter number")
You need a colon after each if/elif/else line.
You need to add print before "A".
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I'm getting a syntax error on my 2nd while loop. Can't figure out why, any help appreciated :)
#intro
print("Welcome to my prime number detector.")
print("Provide an integer and I will determine if it is prime.")
#again loop
again = "Y"
while again == "Y":
num = (int(input("Enter an integer"))
#check for valid input
while num < 1:
num = (int(input("Enter an integer"))
#test for prime
for d in range(2,num):
if (num % d) == 0:
print(num,"is not prime.")
else:
print(num,"is prime.")
#ask again
again = intput("Do you want to play again? (Y/N)")
You are missing a closing parenthesis ) in the two of your following lines. The correct line of code is
num = (int(input("Enter an integer")))
Also, as sheepez mentioned below, your outer brackets are redundant. You can simply use
num = int(input("Enter an integer"))
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Closed 6 years ago.
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This is what I've input:
def greater_less_equal_5(answer):
if 6 > 5:
return 1
elif 4 < 5:
return -1
else:
return 0
print greater_less_equal_5(4)
print greater_less_equal_5(5)
print greater_less_equal_5(6)
and gave me this note
Oops, try again. It looks like your function output 1 instead of -1
when answer is 3. Make sure you filled in the if and elif statements
correctly!
and this is what came up in the upper right display:
>1
>1
>1
>None
No matter how I change around the numbers and the >/< I've even tried == and != it still outputs 1 1 1 None.
I've searched around for any possible tips and seen others stuck on the same problem as me and when I tried their solves I then get:
def greater_less_equal_5(answer):
if > 5:
return 1
elif < 5:
return -1
else:
return 0
print greater_less_equal_5(4)
print greater_less_equal_5(5)
print greater_less_equal_5(6)
and the output is:
File "python", line 2
if > 5:
^
SyntaxError: invalid syntax
Is this test rigged to always output a failed result to make me pay for pro and ask for their help?
And the hint given for this is:
Make sure the if and elif statements end with colons :
Your code should look something like:
if EXPRESSION:
do something
elif OTHER EXPRESSION:
do something
else:
do something
Am I just missing something horribly basic?
You are indeed missing something basic - namely, that the output of your function doesn't depend on answer at all. No matter what you feed in as answer, because 6 > 5 is always True, it will always return the result of that case.
What you need is
def greater_less_equal_5(answer):
if answer > 5:
return 1
elif answer < 5:
return -1
elif answer == 5:
return 0
You're missing 'answer' variable for expression, which you pass into your function
def greater_less_equal_5(answer):
if answer > 5:
return 1
elif answer < 5:
return -1
else:
return 0
print greater_less_equal_5(4)
print greater_less_equal_5(5)
print greater_less_equal_5(6)
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Closed 9 years ago.
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Here's my code;
def save(name):
if x['fname'] == 'ply.json':
save1(name)
elif x['fname'] not 'ply.json':
write_data({'fname':'ply.json', 'name':'Karatepig'}, 'ply.json')
I get an error stating that I have this syntax error:
File "<stdin>", line 4
elif x['fname'] not 'ply.json':
^
What am I doing wrong?
something not something is not a valid expression. If you want to test if it is not equal, use !=:
elif x['fname'] != 'ply.json':
However, since this is the exact opposite of the preceding if test, just use else here:
if x['fname'] == 'ply.json':
save1(name)
else:
write_data({'fname':'ply.json', 'name':'Karatepig'}, 'ply.json')
You need to use != to test inequality, like this:
elif x['fname'] != 'ply.json':
But why use elif?
def save(name):
if x['fname'] == 'ply.json':
save1(name)
else:
write_data({'fname':'ply.json', 'name':'Karatepig'}, 'ply.json)
You need to use != for "not equal":
elif x['fname'] != 'ply.json':