deleting an object in a loop that runs through the range of the list??

Question

I have a list composed of [start position, stop position, [sample names with those positions]]

My goal is to remove the duplicates with exact start and stop positions and just add the extra sample to the sample names section. The problem I'm encountering is that when I delete from the list, I end up with an out of range error, because it's not recalculating the len(list) within the loops.

for g in range (len(list)) :

    for n in range(len(list)):
    #compares the start and stop position of one line to the start and stop of another line 
          if (list[g][0]==list[n+1][0] and list[g][1]==[n+1][1])
          #adds new sample numbers to first start and stop entry with duplication
          labels1=list[g][2]
          labels2=list[n+1][2]
          labels=labels1+labels2
          list[g][2]=labels
    #now delete the extra line
          del list[n+1]

Answer 1

I not sure I understand what you want, but it might be this:

from collections import defaultdict
d = defaultdict(list)
for start, stop, samples in L1:
    d[start, stop].extend(samples)
L2 = [[start, stop, samples] for (start, stop), samples in d.items()]

Which will take L1:

L1 = [ [1, 5, ["a", "b", "c"]], [3, 4, ["d", "e"]], [1, 5, ["f"]] ]

and make L2:

L2 = [ [1, 5, ["a", "b", "c", "f"]], [3, 4, ["d", "e"]] ]

Please note that this does not guarantee the same order of the elements in L2 as in L1, but from the looks of your question, that doesn't matter.

Answer 2

Your loops should not be for loops, they should be while loop with an increment step. I guess you can just manually check the condition within your for loop (continue if it's not met), but a while loop makes more sense, imo.

Answer 3

I've just put together a nice little list comprehension that does pretty much what you did, except without the nasty del s.

from functools import reduce
from operator import add
from itertools import groupby

data = [
    [1, 1, [2, 3, 4]],
    [1, 1, [5, 7, 8]],
    [1, 3, [2, 8, 5]],
    [2, 3, [1, 7, 9]],
    [2, 3, [3, 8, 5]],
]

data.sort()
print(
    [[key[0], key[1], reduce(add, (i[2] for i in iterator))]
     for key, iterator in groupby(data, lambda item: item[:2])
    ]
)

Answer 4

Here is truppo's answer, re-written to preserve the order of entries from L1. It has a few other small changes, such as using a plain dict instead of a defaultdict, and using explicit tuples instead of packing and unpacking them on the fly.

L1 = [ [1, 5, ["a", "b", "c"]], [3, 4, ["d", "e"]], [1, 5, ["f"]] ]


d = {}
oplist = []  # order-preserving list

for start, stop, samples in L1:
    tup = (start, stop)  # make a tuple out of start/stop pair
    if tup in d:
        d[tup].extend(samples)
    else:
        d[tup] = samples
        oplist.append(tup)

L2 = [[tup[0], tup[1], d[tup]] for tup in oplist]

print L2
# prints: [[1, 5, ['a', 'b', 'c', 'f']], [3, 4, ['d', 'e']]]