Split a list of tuples into sub-lists of the same tuple field

Use itertools.groupby:

import itertools
import operator

data=[(1, 'A', 'foo'),
    (2, 'A', 'bar'),
    (100, 'A', 'foo-bar'),

    ('xx', 'B', 'foobar'),
    ('yy', 'B', 'foo'),

    (1000, 'C', 'py'),
    (200, 'C', 'foo'),
    ]

for key,group in itertools.groupby(data,operator.itemgetter(1)):
    print(list(group))

yields

[(1, 'A', 'foo'), (2, 'A', 'bar'), (100, 'A', 'foo-bar')]
[('xx', 'B', 'foobar'), ('yy', 'B', 'foo')]
[(1000, 'C', 'py'), (200, 'C', 'foo')]

Or, to create one list with each group as a sublist, you could use a list comprehension:

[list(group) for key,group in itertools.groupby(data,operator.itemgetter(1))]

---

The second argument to itertools.groupby is a function which itertools.groupby applies to each item in data (the first argument). It is expected to return a key. itertools.groupby then groups together all contiguous items with the same key.

operator.itemgetter(1) picks off the second item in a sequence.

For example, if

row=(1, 'A', 'foo')

then

operator.itemgetter(1)(row)

equals 'A'.

---

As @eryksun points out in the comments, if the categories of the tuples appear in some random order, then you must sort data first before applying itertools.groupby. This is because itertools.groupy only collects contiguous items with the same key into groups.

To sort the tuples by category, use:

data2=sorted(data,key=operator.itemgetter(1))

collections.defaultdict

itertools.groupby requires the input to be sorted by the key field, otherwise you will have to sort first, incurring O(n log n) cost. For guaranteed O(n) time complexity, you can use a defaultdict of lists:

from collections import defaultdict

dd = defaultdict(list)
for item in data:
    dd[item[1]].append(item)

res = list(dd.values())

print(res)

[[(1, 'A', 'foo'), (2, 'A', 'bar'), (100, 'A', 'foo-bar')],
 [('xx', 'B', 'foobar'), ('yy', 'B', 'foo')],
 [(1000, 'C', 'py'), (200, 'C', 'foo')]]