打印列表中的元素，直到使用列表推导找到一个元素

Question

so i was doing this exercise from w3schools

28. Write a Python program to print all even numbers from a given numbers list in the same order and stop the printing if any numbers that come after 237 in the sequence.

and since i read about list comprehension i wanted to put it into practice, and i came to this

numbers = [ 233,12,59,213,69,923,30,10,420,237,432,
           233,98,912,5,61]

print([x for i,x in enumerate(numbers) if i in 
       [i for i,x in enumerate(numbers[:numbers.index(237)])] and x % 2 == 0])

it works but is this the proper way of doing it? is it very ugly?

Answer 1

Use enumerate and the % operator:

numbers = [ 233,12,59,213,69,923,30,10,420,237,432,
           233,98,912,5,61]

print([num for idx, num in enumerate(numbers) if num % 2 == 0 and idx < numbers.index(237)])

Output:

[12, 30, 10, 420]

Answer 2

end = numbers.index(237)
print([x for i,x in enumerate(numbers) if x%2==0 and i<end])

You could try this. Replace end with it's value to implement it in a single line.

Output:

[12, 30, 10, 420]

Answer 3

[x for x in numbers[:numbers.index(237)] if x%2==0]