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

Question

I have a huge list of tuples in this format. The second field of the each tuple is the category field.

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

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

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

What is the most efficient way to break it down into sub-lists of the same category ( A, B, C .,etc)?

Answer 1

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))