Finding count of tuples with same first and third item in list of tuples

Question

I have a list of tuples each with three items :

z = [(1, 4, 2015), (1, 11, 2015), (1, 18, 2015), (1, 25, 2015), (2, 1, 2015), (2, 8, 2015), (2, 15, 2015), (2, 22, 2015), (3, 1, 2015), (3, 8, 2015), (3, 15, 2015), (3, 22, 2015), (3, 29, 2015), (4, 5, 2015), (4, 12, 2015), (4, 19, 2015), (4, 26, 2015), (5, 3, 2015), (5, 10, 2015), (5, 17, 2015), (5, 24, 2015), (5, 31, 2015), (6, 7, 2015), (6, 14, 2015), (6, 21, 2015), (6, 28, 2015), (7, 5, 2015), (7, 12, 2015), (7, 19, 2015), (7, 26, 2015), (8, 2, 2015), (8, 9, 2015), (8, 16, 2015), (8, 23, 2015), (8, 30, 2015), (9, 6, 2015), (9, 13, 2015), (9, 20, 2015), (9, 27, 2015), (10, 4, 2015), (10, 11, 2015), (10, 18, 2015), (10, 25, 2015), (11, 1, 2015), (11, 8, 2015), (11, 15, 2015), (11, 22, 2015), (11, 29, 2015), (12, 6, 2015), (12, 13, 2015), (12, 20, 2015), (12, 27, 2015), (1, 3, 2016), (1, 10, 2016), (1, 17, 2016), (1, 24, 2016), (1, 31, 2016)]

I want to find number of tuples in the list with same first and third items, like with first item 1 and third item 2015, there are 4 tuples; with first item 2 and third item 2015, there are 4 tuples.

I tried :

for tup in z:
    a=tup[0]
    b=tup[2]
    print(len(set({a:b})))

It doesn't give desired result. How to do it?

Answer 1

using standard python's itertools.groupby:

from itertools import groupby

for grp, elmts in groupby(z, lambda x: (x[0], x[2])):
    print(grp, len(list(elmts)))

Edit:

an even nicer solution by using operator.itemgetter instead of lambda:

from operator import itemgetter
from itertools import groupby

for grp, elmts in groupby(z, itemgetter(0, 2)):
    print(grp, len(list(elmts)))

Output:

(1, 2015) 4
(2, 2015) 4
(3, 2015) 5
(4, 2015) 4
(5, 2015) 5
(6, 2015) 4
(7, 2015) 4
(8, 2015) 5
(9, 2015) 4
(10, 2015) 4
(11, 2015) 5
(12, 2015) 4
(1, 2016) 5

Answer 2

Using collections.Counter with operator.itemgetter:

from collections import Counter
from operator import itemgetter

res = Counter(map(itemgetter(0, 2), z))

print(res)

Counter({(1, 2015): 4,
         (1, 2016): 5,
         (2, 2015): 4,
         (3, 2015): 5,
         (4, 2015): 4,
         (5, 2015): 5,
         (6, 2015): 4,
         (7, 2015): 4,
         (8, 2015): 5,
         (9, 2015): 4,
         (10, 2015): 4,
         (11, 2015): 5,
         (12, 2015): 4})