将嵌套列表分组为元组

Question

I am trying to group nested lists in python (I already sorted the data). The final output should be comma separated tuples, one per line

Current nested list:

[['A123', '2012-01-01', 'estrogen'],
 ['A123', '2012-01-01', 'ibuprofen '],
 ['A123', '2014-01-01', 'oxaliplatin'],
 ['A123', '2014-01-01', 'penicilin']]

Desired outcome (comma separated tuple)

[(('A123', '2012-01-01', 'estrogen'),
  ('A123', '2012-01-01', 'ibuprofen ')),
 (('A123', '2014-01-01', 'oxaliplatin'),
  ('A123', '2014-01-01', 'penicilin'))]

I understand the sorting but don't quite know how to group the lists together into one. The final step I assume is the convert into tuples.

Answer 1

Since you already sorted your data, you can convert the lists to tuples and create a tuple (couple) of tuples 2 by 2:

l = [['A123', '2012-01-01', 'estrogen'],['A123', '2012-01-01', 'ibuprofen'],['A123', '2014-01-01', 'oxaliplatin'],
 ['A123', '2014-01-01', 'penicilin']]

result = [(tuple(l[i]),tuple(l[i+1])) for i in range(0,len(l),2)]
print(result)

variant using zip and sliced sub-lists:

result = [(tuple(u),tuple(v)) for u,v in zip(l[::2],l[1::2])]

you don't control the way the data is printed unless you use pprint module.

import pprint
pprint.pprint(result,width = 50)

Since I chose a width of 50, I get one element per line (it depends on the data, but it seems that it provides exactly the output mentioned in your question, horizontal alignment included)

[(('A123', '2012-01-01', 'estrogen'),
  ('A123', '2012-01-01', 'ibuprofen')),
 (('A123', '2014-01-01', 'oxaliplatin'),
  ('A123', '2014-01-01', 'penicilin'))]

Answer 2

This looks like a case for itertools.groupby :

>>> data = [['A123', '2012-01-01', 'estrogen'],
...  ['A123', '2012-01-01', 'ibuprofen '],
...  ['A123', '2014-01-01', 'oxaliplatin'],
...  ['A123', '2014-01-01', 'penicilin']]
>>> [tuple(grp) for key, grp in itertools.groupby(data, key=lambda x: x[:2])]
[(['A123', '2012-01-01', 'estrogen'], ['A123', '2012-01-01', 'ibuprofen ']),
 (['A123', '2014-01-01', 'oxaliplatin'], ['A123', '2014-01-01', 'penicilin'])]

This groups all items that have the same first two elements ( key=lambda x: x[:2] ) together in one tuple. Note that the data has to be sorted by the key for this to work, but according to you this is already the case.