python - How to nest 2 for loops into one if statement -

how put 2 loops inside if statement?

ships = [     ["aircraft carrier", 5],      ["battleship", 4],      ["submarine", 3],         ["destroyer", 3],      ["patrol boat", 2] ] ships_left = ["a","b","s","d","p"]  if [ship ship in ships_left name name in ships if name[0][0] == ship]:     print(name[0]) 

expected output:

aircraft carrier 

this because if both iterate once ship should equal "a" , name should ["aircraft carrier", 5], name[0][0] should "a".

how code independently iterate through both lists , branch on given statement associating 2 lists?

calculate ships left first, before using if test if there any:

left_names = [name name, size in ships if name[0] in ships_left] if left_names:     print(left_names[0]) 

by calculating ships left first, can re-use result both if test , print() function; otherwise you'd have make same calculation twice.

you don't need 2 loops; need loop on ships list , test each name against ships_left list. i'd make ships_left set, however, faster membership testing:

ships_left = {"a", "b", "s", "d", "p"} 

membership testing in list takes n steps (where n length of list), while in set membership testing takes constant time (o(1)). makes removing ship once it's been sunk (or placed on board) easy , fast too:

ships_left.remove(name[0]) 

you use next() function generator expression if need first match; avoids extracting names:

ship_left = next((name name, size in ships if name[0] in ships_left), none) if ship_left:     print(ship_left) 

demo:

>>> ships = [ ...     ["aircraft carrier", 5], ...     ["battleship", 4], ...     ["submarine", 3], ...     ["destroyer", 3], ...     ["patrol boat", 2] ... ] >>> ships_left = {"a", "b", "s", "d", "p"} >>> next((name name, size in ships if name[0] in ships_left), none) 'aircraft carrier' >>> ships_left.remove('a') >>> next((name name, size in ships if name[0] in ships_left), none) 'battleship' >>> ships_left.clear()  # remove ships >>> next((name name, size in ships if name[0] in ships_left), none) none true