Scanning list for 5 consecutive values greater than x

Question

I want to scan a large list for consecutive values that are greater than x. This example x is greater than 1.0.

For example,

my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9, 1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1., 1.6, 0.2, 0.5, 1.0, 1.1, 0.2]

I can subset this list by

for i in range(0, len(my_list)):
    subset = my_list[i:i+5]

so I get

[0.2, 0.1, 0.3, 1.1, 0.7]
[0.1, 0.3, 1.1, 0.7, 0.5]
[0.3, 1.1, 0.7, 0.5, 1.2]
[1.1, 0.7, 0.5, 1.2, 1.3]
[0.7, 0.5, 1.2, 1.3, 1.4]
[0.5, 1.2, 1.3, 1.4, 1.2]
[1.2, 1.3, 1.4, 1.2, 1.9] <-- values I want
[1.3, 1.4, 1.2, 1.9, 1.1] <-- values I want
[1.4, 1.2, 1.9, 1.1, 0.2]
[1.2, 1.9, 1.1, 0.2, 1.3]
[1.9, 1.1, 0.2, 1.3, 1.5]
[1.1, 0.2, 1.3, 1.5, 1.4]
[0.2, 1.3, 1.5, 1.4, 1.2]
[1.3, 1.5, 1.4, 1.2, 1.1] <-- values I want

What is the best way to do this?

Answer 1

Here's an itertools based approach that won't need any extra memory and returns results as a generator:

from itertools import tee, islice

def find_consecutive(the_list, threshold, count=5):
    my_iters = tee(the_list, count)
    for i, it in enumerate(my_iters):
        next(islice(it, i, i), None)
    return (f for f in zip(*my_iters) if all(x > threshold for x in f))

my_list = [0.2, 0.1, 0.3, 1.1, 0.7, 0.5, 1.2, 1.3, 1.4, 1.2, 1.9, 1.1, 0.2, 1.3, 1.5, 1.4, 1.2, 1.1, 0.2, 1.3, 0.1, 1.6, 0.2, 0.5, 1.0, 1.1, 0.2]
list(find_consecutive(my_list, 1.0))
# [(1.2, 1.3, 1.4, 1.2, 1.9),
# (1.3, 1.4, 1.2, 1.9, 1.1),
# (1.3, 1.5, 1.4, 1.2, 1.1)]

The function is parameterized by threshold and count so you can look for any N consecutive values. You could even factor out the condition by passing in a function for that instead of just a threshold value.