Is this correct lambda expression to find a even or odd number

Question

number = int(input("Enter a number >> "))
odd_even = lambda : "Even Number" if number % 2 == 0 else "Odd Number"

print(odd_even())

Answer 1

Why don't you put your function into Code Playground and test it?

Answer 2

🄼🄼🄷🄺
Yes the lambda expression is correct.
But I would say it is better if you pass 'number' as a parameter to the lambda function:

https://code.sololearn.com/cPT7YbpRZfzO/?ref=app

Answer 3

Four downvotes? For my answer? I thought my comment to be the most natural recommendation, and there was definitely no harm intended!

Normally you'd put a code here after you tried it but it puts out awkward results or gives errors and you don't understand what's wrong. Then you'd ask: 'What's wrong with this?'

But 'is this lambda expression correct?' is easily to be answered by just feeding numbers and check if they get evaluated correctly.

Answer 4

n=int(input())
odd_even=lambda n:"Even number" if(n%2==0) else "Odd number"
print(odd_even(n))