CODEWITHSUNDEEP
python
Aadit S. Bagga
Aadit S. Bagga

Reputation: 65

How can i solve this error in my python code?

def almost_there(n): 
    list=[x for x in range(90,111)]
    list_1=[i for i in range(190,211)]
    if n in list or list_1: 
        return True 
    else:
        return False
print(almost_there(1))
>>> True

Why is it giving true even when the value is 1?

Upvotes: 0

Views: 70

Answers (3)

Tsubasa
Tsubasa

Reputation: 1429

Try this one.

def almost_there(n): 
    list_1 = [x for x in range(90,111)]
    list_2 = [i for i in range(190,211)]
    
    if n in list_1 or n in list_2: 
        return True 
    else:
        return False

print(almost_there(1))
# False

Check if n is either in list_1 or list_2. If you do if n in list_1 or list_2 this will only check if n is in list_1 and not list_2.
One more thing don't use list as your variable name. That will override the list() function.

Upvotes: 2

George Bou
George Bou

Reputation: 588

The problem is in you if statement.

if n in list or list_1: evaluates to: 1) if n in list (false) or 2) if list_1 (true)

Now the second argument is always going to be True because it's a populated list.

The correct way to write your if statement should be: if (n in list or n in list_1)

This evaluates to if 1) n in list (false) or 2) n in list_1 (false)

And thus you 'll get the expected results.

Upvotes: 0

SWater
SWater

Reputation: 443

Because of the condition: if n in list or list_1:

When you have or condition, both part of the condition should be False for the condition to be False

In your case:
n is not in list is False , but list_1 is not empty and is True.

So you have if False or True: -- and this returns True (the condition is fulfilled)

Upvotes: 2

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