nlist = [[('4698874', '0', '58'), ('838286', '58', '310', '1.01')],('2588097', '368', '179', '1.01'), ('2746740', '547', '342', '1.44'),('3873988', '889', '259', '1.01'), ('808046', '1148', '236', '1.01'), ('2588498', '1384', '158', '1.01'), ('2492893', '1542', '196', '1.02'), ('2168413', '1738', '165', '1.02'), ('1778345', '1903', '448', '1.07'), ('2989691', '2351', '194', '0.99'), [('4698875', '2545', '256'), ('2985955', '2801', '257', '1.54')], [('4698876', '3058', '177'), ('1728736', '3235', '270', '0.96')], ('2615446', '3505', '172', '0.93'),[('4698877', '3677', '177'), ('4698878', '3854', '144'), ('515524', '3998', '134', '1.10')], [('4698879', '4132', '172'), ('4698880', '4304', '98'), ('2444241', '4402', '146', '1.04')], ('4698881', '4548', '-1', '1.00'), ()]
我想知道是否存在一个非递归的单行列表,以便其余的元素都是元组。
nlist = [('4698874', '0', '58'), ('838286', '58', '310', '1.01'),('2588097', '368', '179', '1.01'), ('2746740', '547', '342', '1.44'),('3873988', '889', '259', '1.01'), ('808046', '1148', '236', '1.01'), ('2588498', '1384', '158', '1.01'), ('2492893', '1542', '196', '1.02'), ('2168413', '1738', '165', '1.02'), ('1778345', '1903', '448', '1.07'), ('2989691', '2351', '194', '0.99'), ('4698875', '2545', '256'), ('2985955', '2801', '257', '1.54'), ('4698876', '3058', '177'), ('1728736', '3235', '270', '0.96')], ('2615446', '3505', '172', '0.93'),('4698877', '3677', '177'), ('4698878', '3854', '144'), ('515524', '3998', '134', '1.10'), ('4698879', '4132', '172'), ('4698880', '4304', '98'), ('2444241', '4402', '146', '1.04'), ('4698881', '4548', '-1', '1.00')]
非常感谢您的指导。
答案 0 :(得分:0)
from itertools import chain
def unlist(nlist):
return list(chain(*[[n] if isinstance(n, tuple) else n for n in nlist]))