Index Lists For Specific Repeating Element
Solution 1:
This should work:
main_list= [True, True, False, False, True, True, True, True, True, False]
start_true=-1last_added=-1true_index= []
fori,valueinenumerate(main_list):if value:ifstart_true==-1:start_true=ielse:ifstart_true!=last_added:true_index.append(start_true)last_added=start_trueelse:start_true=-1print(true_index)
Also if you want the code to detect consecutive Trues including a single True here is a version that does that:
main_list= [True, False, False]
start_true=-1last_added=-1true_index= []
fori,valueinenumerate(main_list):if value:ifstart_true==-1:start_true=iifstart_true!=last_added:true_index.append(start_true)last_added=start_trueelse:start_true=-1print(true_index)
Solution 2:
The following is a much shorter and solution, and might be preferable -
main_list = [True, True, False, False, True, True, True, True, True, False]
defmy_diff(my_list):
return [1if my_list[0] else0] + [y - x for x, y inzip(my_list[:-1], my_list[1:])]
solution = [i for i, x inenumerate(my_diff(main_list)) if x == 1]
print(solution)
# [0, 4]
Explanation:
I'd personally solve this by using np.diff
, and simply searching for the "transitions" (from False
to True
). But seeing as numpy
is out of scope, I've just implemented a simple diff function, and the beginning of a sequence of True
s is the same as having a difference of 1
between two consecutive elements.
And to make sure the first element isn't missed, if it's True
then it's by definition the beginning of a sequence, and so we plant a 1
in place. Otherwise we don't :-).
To sum things up - look for "element - prev-element" being equal 1.
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