Replace duplicate items from list while keeping the first occurrence

user_6017391 0
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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.

duplicates list python
7 Answers

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.

user_49090870

Using a simple iteration

Ex:

lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]
dup_check = set()
result = []
for i in lst:
    if i not in dup_check:
        result.append(i)
        dup_check.add(i)
    else:
        result.append("")
print(result)

Output:

[1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']
user_5323120

Try this simpler function:

def fill_blank(l):
    filled = []
    last = l[0]
    for i in l:
        if i != last:
            last = i
        filled.append(last)
    return filled

Use it as such:

>>> lst = [1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4]
>>> out = fill_blank(lst)
>>> print(out)
[1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']
user_106382320

Do this if you don't want to use set or itertools: 

lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]

new_lst = []

for i in lst:
    if i in new_lst:
        new_lst.append("")
    else:
        new_lst.append(i)

print(new_lst)

Output: 

[1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']
user_53495420

Short solution (one-liner) without any modules usage is as follows:

lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]
out = ["" if (i in lst[:inx]) else i for inx,i in enumerate(lst)]
print(out)

output

[1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']
user_107859750

You can try below approach.. 

lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]
fin=[]
for each in lst:
   if each not in fin:
       fin.append(each)
   else:
       fin.append("")
print(fin)
user_40245740

You can store the first occurences in a dictionary with dict.setdefault(), then keep the first found numbers depending if they exist in this dictionary:

lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]

first_indices = {}
for i, x in enumerate(lst):
    first_indices.setdefault(x, i)

result = [x if first_indices[x] == i else "" for i, x in enumerate(lst)]

print(result)
# [1, '', '', 2, '', '', '', 3, '', '', '', '', 4, '', '', '', '', '', '', '', '']
user_52389150
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