non-recursive unlist lists of tuples [duplicate]

Question

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'), ()]

I was wondering is there a one-liner to unlist this non-recursively such that the remaining elements are all tuples.

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')]

Thank you so much for guidance.

Philip Tzou · Accepted Answer · 2016-07-28 00:09:47Z

0

Use itertools.chain.

from itertools import chain

def unlist(nlist):
    return list(chain(*[[n] if isinstance(n, tuple) else n for n in nlist]))

answered Jul 28, 2016 at 0:09

Philip Tzou

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