The best way to clarify such doubts is to play with the REPL:

>>> range(10)
range(0, 10)
>>> list(range(10))
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> x = list(range(10))
>>> x[:3]
[0, 1, 2]
>>> x[:1]
[0]
>>> x[1:10]
[1, 2, 3, 4, 5, 6, 7, 8, 9]

You can have your function take a function as an argument to use as the predicate for building your intervals:

def indexscope(dlist, predicate):
    scope = []
    start = end = -1
    for i, v in enumerate(dlist):
        if predicate(v):
            if start == -1:
                start = end = i
                continue
            if end + 1 == i:
                end = i
            else:
                scope.append([start] if start == end else [start, end])
                start = end = i
    if start != -1: 
        scope.append([start] if start == end else [start, end])
    return scope

a = [12,11,5,7,2,21,32,13,6,42,1,8,9,0,32,38]

def less_than_10(n):
    return n < 10

print(indexscope(a, less_than_10))
print(indexscope(a, lambda x: x > 20))


[[2, 4], [8], [10, 13]]
[[5, 6], [9], [14, 15]]

with scipy:

import numpy as np
import scipy.ndimage as nd

def passing_ranges(a, predicate):
    return nd.find_objects(nd.label(predicate(a))[0])

The results are returned as slice objects, but that is to your advantage because you can use them to against your original np array:

small_a = [12,11,5,7,2,21,32,13,6,42,1,8,9,0,32,38]
small_np_array = np.array(small_a)

valid_ranges = passing_ranges(small_np_array, lambda n: n < 10)

for r in valid_ranges:
    print(r[0], small_np_array[r])

slice(2, 5, None) [5 7 2]
slice(8, 9, None) [6]
slice(10, 14, None) [1 8 9 0]

benchmarks

large_a = [12,11,5,7,2,21,32,13,6,42,1,8,9,0,32,38]*1000000
large_np_array = np.array(large_a)

%timeit passing_ranges(large_np_array, lambda x: x < 10)
1 loops, best of 3: 1.2 s per loop

%timeit indexscope(large_a, lambda n: n < 10)
1 loops, best of 3: 6.99 s per loop

Here is your answer, I even inline the predicate to remove a function call:

from itertools import groupby, count

def xibinke(a):
    l = [idx for idx,value in enumerate(a) if value<10]
    return [list(g) for _,g in groupby(l,key=lambda n,c=count():n-next(c))]

%timeit xibinke(large_a)
1 loops, best of 3: 14.6 s per loop

You could try something like this:

list1 = [0, 1, 2, 3, 4, 5]
for ln in range(1, len(list1)):
    list2 = [list1[n] for n in range(-ln, ln)] #magic line
    print(list2)

or you could also try this instead of the magic line:

list2 = list1[ln:] + list1[:ln]

Where ln is the number of elements you want from the end and the beginning of the list.