anvd anvd - 6 months ago 22
Python Question

Group list of tuples by item

I have this list as example:

[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]


Now i want to group by the id, so I will use
itemgetter(0)
:

import operator, itertools
from decimal import *
test=[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

for _k, data in itertools.groupby(test, operator.itemgetter(0)):
print list(data)


I don't know why but I am getting this wrong output:

[(148, Decimal('3.0'))]
[(325, Decimal('3.0'))]
[(148, Decimal('2.0'))]
[(183, Decimal('1.0'))]
[(308, Decimal('1.0'))]
[(530, Decimal('1.0'))]
[(594, Decimal('1.0'))]
[(686, Decimal('1.0'))]
[(756, Decimal('1.0'))]
[(806, Decimal('1.0'))]


As you can see the output is not grouped by id. However the code above works fine if I use
itemgetter(1)
. The output is grouped by decimal val.

[(148, Decimal('3.0')), (325, Decimal('3.0'))]
[(148, Decimal('2.0'))]
[(183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]


What I am missing here?

Answer

You would first need to sort the data for the groupby to work, it groups consecutive elements based on the key you provide:

import operator, itertools
from decimal import *
test=[(148, Decimal('3.0')), (325, Decimal('3.0')), (148, Decimal('2.0')), (183, Decimal('1.0')), (308, Decimal('1.0')), (530, Decimal('1.0')), (594, Decimal('1.0')), (686, Decimal('1.0')), (756, Decimal('1.0')), (806, Decimal('1.0'))]

for _k, data in itertools.groupby(sorted(test), operator.itemgetter(0)):
    print list(data)

But you would be better using a dict to group to avoid an unnecessary O(n log n) sort:

from collections import defaultdict

d = defaultdict(list)

for t in test:
    d[t[0]].append(t)

for v in d.values():
    print(v)

Both would give you the same groupings, just not necessarily in the same order.