Split tuple / collection into instances of pattern

July 17, 2022

I have a tuple [but it can be any collection] that contains elements:

tple = (1.02, 'a', 'b', 1.02, 'a', 'b')

I’m trying to find a way to count the number of times the pattern (1.02, 'a', 'b') occurs within the tple. The pattern may not exist at all, in which case I would want to return 0.

Is there such a way?

>Solution :

One approach using itertools:

from itertools import tee


def wise(iterable):
    """Iterate over contiguous overlapping chunks"""
    a, b, c = tee(iterable, 3)
    next(b, None)
    next(c, None)
    next(c, None)
    return zip(a, b, c)


tple = (1.02, 'a', 'b', 1.02, 'a', 'b')
pattern = (1.02, 'a', 'b')

res = sum(i == pattern for i in wise(tple))
print(res)

Output

The function wise, is a generalization of itertools.pairwise. For the above example it returns something similar to:

[(1.02, 'a', 'b'), ('a', 'b', 1.02), ('b', 1.02, 'a'), (1.02, 'a', 'b')]

Note that by using itertools tple can be any collection. The expression:

res = sum(i == pattern for i in wise(tple))

is equivalent to the following for-loop:

res = 0
for i in wise(tple):
    if pattern == i:
        res += 1
print(res)

If you want to iterate in chunks of different lengths, use the following general wise function:

def wise(iterable, n=3):
    """Iterate over contiguous overlapping chunks"""
    its = tee(iterable, n)
    for i, it in zip(range(n), its):
        for _ in range(i):
            next(it)
    return zip(*its)