在保留第一次出现的同时替换列表中的重复项

Question

我有一份清单 lst = [1,1,1,2,2,2,2,3,3,3,3,3,4,4,4,4,4,4,4,4,4]

我期待以下输出:

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

我想保留第一次出现的项目,并用空字符串替换同一项目的所有其他出现次数.

我尝试了以下方法.

`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`

但我正在寻找一些更简单或更好的方法.

Answer 1

使用itertools.groupby,非常适合分组连续重复的值.

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

如果列表值不连续,您可以先对它们进行排序.