Number of events in python array

Question

I have the following array:

a = [0,0,0,1,1,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0]

Every time I have a "1" or a series of them (sequential), this is one event. I need to get in Python how many events my array has. Thus, in this case we will have 5 events (this is 5 times 1 or its sequence). I need to count such events in order to get:

b = [5]

thank

Answer 1

You can use itertools.groupby(it does exactly what you want - groups consecutive elements) and count all groups starting with1

In [1]: from itertools import groupby

In [2]: a = [0,0,0,1,1,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0]

In [3]: len([k for k, _ in groupby(a) if k == 1])
Out[3]: 5

---

what if I wanted to add a condition that an event is given as long as there are 2 or more between them.

, groupby key:

from itertools import groupby


class GrouperFn:
    def __init__(self):
        self.prev = None

    def __call__(self, n):
        assert n is not None, 'n must not be None'

        if self.prev is None:
            self.prev = n
            return n

        if self.prev == 1:
            self.prev = n
            return 1

        self.prev = n
        return n


def count_events(events):
    return len([k for k, _ in groupby(events, GrouperFn()) if k == 1])


def run_tests(tests):
    for e, a in tests:
        c = count_events(e)
        assert c == a, 'failed for {}, expected {}, given {}'.format(e, a, c)

    print('All tests passed')


def main():
    run_tests([
        ([0, 1, 1, 1, 0], 1),
        ([], 0),
        ([1], 1),
        ([0], 0),
        ([0, 0, 0], 0),
        ([1, 1, 0, 1, 1], 1),
        ([0, 1, 1, 0, 1, 1, 0], 1),
        ([1, 0, 1, 1, 0, 1, 1, 0, 0, 1], 2),
        ([1, 1, 0, 0, 1, 1], 2),
        ([0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0], 4)
    ])


if __name__ == "__main__":
    main()

- 0 1, . ( 1), ( 0)

, , , [1, 1, 0, 0] [[1, 1, 0], [0]].

Answer 2

( ):

a = [0,0,0,1,1,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0]
events = (a[0] + a[-1] + sum(a[i] != a[i-1] for i in range(1, len(a)))) / 2
print events

, :)

Answer 3

def num_events(list):
    total = 0
    in_event = 0
    for current in list:
        if current and not in_event:
            total += 1
        in_event = current
    return total

, 1 0, . , 1 .

Answer 4

a = [0,0,0,1,1,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0]

flag = 0
cnt = 0
for j in (a):
    if( j == 1):
        if(flag == 0):
            cnt += 1
        flag = 1
    elif (j == 0):
        flag = 0


print cnt                   

Answer 5

:

sum([(a[i] - a[i-1])>0 for i in range(1, len(a))])

Answer 6

Me - , -)

def get_number_of_events(event_list):
    num_of_events = 0
    for index, value in enumerate(event_list):
        if 0 in (value, index):
            continue
        elif event_list[index-1] == 0:
            num_of_events += 1
    return num_of_events

Answer 7

To replace multiple occurrences with one occurrence, add each item to the list, unless the last added item is equal to the current one, and complete the count.

def count_series(seq):
    seq = iter(seq)
    result = [next(seq, None)]
    for item in seq:
        if result[-1] != item:
            result.append(item)
    return result

a = [0,0,0,1,1,1,0,0,1,0,0,0,1,1,0,0,0,0,0,0,1,1,0,1,1,1,0,0,0]

count_series(a).count(1)
5