Counting number of occurrences of a single digit in a list

[DeaD_EyE]

If the data is very big (does not fit in memory or does not fit on your hard drive,
you often have with frameworks a kind of generator or "magic" iterator, which
loads new data in chunks.

from collections import namedtuple


CountIndexResult = namedtuple('CountIndexResult', 'count index')


def count(iterable_or_gen, value):
    first_index = None
    count = 0
    for index, element in enumerate(iterable_or_gen):
        if element == value:
            if first_index is None:
                first_index = index
            count +=1
    return CountIndexResult(count, first_index)


count([2,2,2,2,2,2,1,1,1,1,1,1,2,2], 1)

Output:
CountIndexResult(count=6, index=6)

If your tuple/list has lesser than one million elements, you can use the methods count and index.

[python_newbie09]

If the data is very big (does not fit in memory or does not fit on your hard drive,
you often have with frameworks a kind of generator or "magic" iterator, which
loads new data in chunks.

from collections import namedtuple


CountIndexResult = namedtuple('CountIndexResult', 'count index')


def count(iterable_or_gen, value):
    first_index = None
    count = 0
    for index, element in enumerate(iterable_or_gen):
        if element == value:
            if first_index is None:
                first_index = index
            count +=1
    return CountIndexResult(count, first_index)


count([2,2,2,2,2,2,1,1,1,1,1,1,2,2], 1)

Output:
CountIndexResult(count=6, index=6)

If your tuple/list has lesser than one million elements, you can use the methods count and index.

Thank you. I will try this as well.

[perfringo]

If I understand the task at hand correctly then groupby can be used:

>>> data = [0,1,1,1,1,2,2,2,85,85,85,80,80,80,0,1,1,1,2,2,2,85,95,80,80,80]
>>> [[*group][0][0] for match, group in groupby(enumerate(data), key=lambda digit: digit[1] == 85) if match == True]
[8, 21]   # index of first 85 in group

EDIT:
If groups are large then indices can be obtained without unpacking next(group)[0]