Replace duplicate items from list while keeping the first occurrence

Question

I have a list lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]

I'm expecting the following output:

out = [1,"","",2,"","","",3,"","","","",4,"","","","","","","",""]

I want to keep the first occurrence of the item and replace all other occurrences of the same item with empty strings.

I tried the following approach.

`def splrep(lst):
    from collections import Counter
    C = Counter(lst)
    flst = [ [k,]*v for k,v in C.items()]
    nl = []
    for i in flst:
        nl1 = []
        for j,k in enumerate(i):
            nl1.append(j)
        nl.append(nl1)

    ng = list(zip(flst, nl))
    for i,j in ng:
        j.pop(0)
    for i,j in ng:
        for k in j:
            i[k] = ''
    final = [i for [i,j] in ng]
    fin = [i for j in final for i in j]
    return fin`

But I'm looking for some simpler or better approaches.

Answer 1

Use itertools.groupby, quite appropriate for grouping consecutively duplicate values.

from itertools import groupby
[v for k, g in groupby(lst) for v in [k] + [""] * (len(list(g))-1)]
# [1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']

If your list values are not consecutive, you may sort them first.