Answer by schot for Count and summation of positive and negative number sequences

Two different lazy solutions in Python, using the itertools module.

Using itertools.groupby (and accumulate)

from itertools import accumulate, groupbyresult = (    item    for _, group in groupby(x, key=lambda n: n < 0)    for item in enumerate(accumulate(group), 1))

Using itertools.accumulate with a custom accumulation function

from itertools import accumulatedef sign_count_sum(count_sum, value):    count, prev_sum = count_sum    same_sign = (prev_sum < 0) is (value < 0)    if same_sign:        return count + 1, prev_sum + value    else:        return 1, valueresult = accumulate(x, sign_count_sum, initial=(0, 0))next(result)  # needed to skip the initial (0, 0) item

The initial keyword argument was added in Python 3.8. In earlier versions you can use itertools.chain to prepend the (0,0)-tuple:

result = accumulate(chain([(0, 0)], x), sign_count_sum)

The output is as expected:

for (i, v), (c, s) in zip(enumerate(x), result):    print(f"{i:3} {v:7.3f} {c:3} {s:7.3f}")

  0  -0.010   1  -0.010  1   0.003   1   0.003  2  -0.002   1  -0.002  3   0.018   1   0.018  4   0.002   2   0.020  5   0.006   3   0.026  6  -0.012   1  -0.012  7   0.014   1   0.014  8  -0.017   1  -0.017  9  -0.007   2  -0.024 10   0.002   1   0.002 11   0.002   2   0.004 12  -0.004   1  -0.004 13   0.015   1   0.015 14   0.002   2   0.017 15  -0.001   1  -0.001 16  -0.008   2  -0.009 17   0.010   1   0.010 18  -0.018   1  -0.018 19   0.046   1   0.046