Question

Consider the “string” ( treat it as an array of digits )

0 0 1 8 8 8 1 0

The RLE ( “groupby” ) is:

[(0,2), (1, 1), (8,3), (1, 1), (0, 1)]

We then enrich the above RLE with the sum of the run lengths of the previous elements.

Hence, the enriched version of the above becomes:

[(0, (0,2)), (0+2, (1, 1)), (0+2+1, (8,3)), (0+1+2+3, (1, 1)), (0+1+2+3+1, (0, 1))]

The “string” split on 1:

0 0 , 8 8 8 , 0

RLE split on 1

[(0,2)] , [(8,3)] , [(0, 1)]

The “string” split on 8:

0 0 1 , , , 1 0

RLE split on 8

[(0,2), (1, 1)] , , , [(1, 1), (0, 1)]

Note : In my examples, I have cited the “RLE split on Z” lists without enriching them. This would not be so. I left them out to reduce clutter. For example, the “RLE split on 1” should really be treated as:

[(0, (0,2))] , [(0+2+1, (8,3))] , [(0+1+2+3+1, (0, 1)]

How can I achieve this “RLE split” on Z ( = 1, 8; in this case )

It’s fine to leave out empty arrays ( after split ).

Perhaps a clever list comp.? ( it seems a little easier to solve with a for loop with an append nested within )

Answer 1

Just for show the way how, I strongly advise you not to use this

“elegant” ugly way:

>>> data
[0, 0, 1, 8, 8, 8, 4, 4, 1, 0]
>>> def fromDataToSplitRLE(dat,n):
    RLE=[(k,len(tuple(g))) for k,g in itertools.groupby(dat)]
    tmp=tuple(zip(*RLE))
    return [list(g) for k,g in itertools.groupby((zip((sum(tmp[1][:i]) for i in range(len(tmp[1]))) ,(zip(*tmp)))),lambda x:x[1][0]!=n) if k]

>>> fromDataToSplitRLE(data,1)
[[(0, (0, 2))], [(3, (8, 3)), (6, (4, 2))], [(9, (0, 1))]]