Using break in a list comprehension

Question

How can I break a list comprehension based on a condition, for instance when the number 412 is found?

Code:

numbers = [951, 402, 984, 651, 360, 69, 408, 319, 601, 485, 980, 507, 725, 547, 544,
           615, 83, 165, 141, 501, 263, 617, 865, 575, 219, 390, 984, 592, 236, 105, 942, 941,
           386, 462, 47, 418, 907, 344, 236, 375, 823, 566, 597, 978, 328, 615, 953, 345, 399,
           162, 758, 219, 918, 237, 412, 566, 826, 248, 866, 950, 626, 949, 687, 217, 815, 67,
           104, 58, 512, 24, 892, 894, 767, 553, 81, 379, 843, 831, 445, 742, 717, 958, 609, 842,
           451, 688, 753, 854, 685, 93, 857, 440, 380, 126, 721, 328, 753, 470, 743, 527]

even = [n for n in numbers if 0 == n % 2]

So functionally, it would be something you can infer this is supposed to do:

even = [n for n in numbers if 0 == n % 2 and break if n == 412]

I really prefer:

• a one-liner
• no other fancy libraries like itertools, "pure python" if possible (read: the solution should not use any import statement or similar)

Answer 1

even = [n for n in numbers[:None if 412 not in numbers else numbers.index(412)] if not n % 2] 

Just took F.J.'s code above and added a ternary to check if 412 is in the list. Still a 'one liner' and will work even if 412 is not in the list.