how i can solve this

Question

Fill in the blanks to print the first element of the list, if it contains an even number of elements.

list = [1, 2, 3, 4]
if (list) % 2 == 0
print(list[ ])

Answer 1

If I'm getting the question right, you want to print the first element if the list contains an even number..

list = [1,2,3,4]

Then you just have to iterate through the list

for i in list:
if i%2==0:
print(list[0])
break

The break is just there to save time.
Since you've gotten what u need, there's no need looping through all elements

Answer 2

Don't forget to to add semi colons ':'

Answer 3

#It will only work this way....

list = [1, 2, 3, 4]
if len(list) % 2 == 0:
print(list[0])

Answer 4

Try like this:-
list=[1,2,3,4]
if len(list)%2==0:
print(list[0])
maybe this would help you out 😊

Answer 5

list = [1, 2, 3, 4]
if len(list) % 2 == 0
print(list[0])

Answer 6

thanks Njeri am also a n00b in python, am kind of treating it like C

Answer 7

iterate through the list, and check for an even value
if len(list) %2==0
print(list[0])

Answer 8

✳AsterisK✳ The question asks to print the first element of the list if the list contains an even number of elements in it, not to print the even numbers in the list.

Also you have used i to iterate through the list and also as an index, but indices start at 0, not 1, so your code will check whether the next element in the list is even for a given element rather than the current one. It will produce an IndexError when it gets to 4, since list[4] is the fifth element of the list, which doesn't exist.

Answer 9

✳AsterisK✳ No problem.

Answer 10

Print(list[0])