當前結果返回與Python返回值
[['jackson', ['21', '22'], ['20','24']], ['barkley', ['30', '32'], ['33', '34']]]
我要來分隔值，使
a = jackson, barkley
b = (jackson, 21), (jackson, 22), (jackson, 23), (jackson 24), (barkley, 30)..etc
什麼是最好的方法是什麼？謝謝。
>>> d = [['jackson', ['21', '22'], ['20','24']], ['barkley', ['30', '32'], ['33', '34']]]
>>> [(i[0],z) for i in d for j in i[1:] for z in j]
[('jackson', '21'), ('jackson', '22'), ('jackson', '20'), ('jackson', '24'), ('barkley', '30'), ('barkley', '32'), ('barkley', '33'), ('barkley', '34')]
你只是遍歷它，然後遍歷子列表... OK沒問題：P
我愛列表解析！它看起來很精英。 – ozgur 2015-02-10 00:55:34
所以你嘗試過這麼遠嗎？ – dragon2fly 2015-02-10 00:44:36
@mgilson是的。對不起，錯字.. – dangerous93 2015-02-10 00:47:04
提示：使用for循環。 – l00kitsjake 2015-02-10 00:54:48