Using 'Not equal to' with if statement in python

Asked Aug 10 '21 at 05:47

Active Aug 10 '21 at 05:51

Viewed 31 times

for the following code I am getting an error which says "ValueError: invalid literal for int() with base 10: '{'". Is something wrong in the way I compare 'i' with '{' , '}' and ','

c={1,2,3,4,5}
LE=[]  #to store even numbers
LO=[]   #to store odd numbers
for i in c:
    if(i != '{' or i != ',' or i != '}'):
        if(int(i)%2==0):
            LE.append(i)
        else:
            LO.append(i)
    else:
        continue

print(LE)            
print(LO)

edited Aug 10 '21 at 05:51

asked Aug 10 '21 at 05:47

adf_hunter

let the input be {1,2,3,4,5}...still the error comes. – adf_hunter Aug 10 '21 at 05:48
@Guy thats why I have used != with if condition – adf_hunter Aug 10 '21 at 05:50
1

Please provide a [mcve], we shouldn't have to perform any input at all if that is not an actual part of the problem. Also, `LE` and `LO` are probably unnecessary as well. – Ulrich Eckhardt Aug 10 '21 at 05:50
1

`or` should be `and` – Barmar Aug 10 '21 at 05:50
`,` is not equal to `{`, so the condition is true and you try to convert it to `int`. – Barmar Aug 10 '21 at 05:51
@Barmar I just want that i should not be equal to '{', '} or ','....can you tell why what I have written in the if condition is wrong – adf_hunter Aug 10 '21 at 05:53
Read the linked question above, it explains it all. – Barmar Aug 10 '21 at 05:54
You can also write `if i not in "{,}":` – Barmar Aug 10 '21 at 05:54
if(int(i)%2==0): has a problem If you want to convert a char into its ASCII value, you need to use **ord** function. Changing to the following if (ord(i)%2 == 0): may help. I think there is still some logic issue with the program. – Ravi S Aug 10 '21 at 06:06

Using 'Not equal to' with if statement in python

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