Python:进行重复计数时删除重复元素

Question

下面是 3D 点组的元组

f = [[10, 20, 0], 
    [40, 20, 30], 
    [20, 0, 30], 
    [10, 10, 0], 
    [30, 10, 10], 
    [20, 0, 30], 
    [20, 10, 20], 
    [10, 10, 0]]

每个点对应一个特定的数字(索引)，指示点的类型(假设)

ic=[1,2,3,2,1,3,2,3,1]

因此，前面的元组可以表示为

f = [[10, 20, 0, 1], 
    [40, 20, 30, 2], 
    [20, 0, 30, 3], 
    [10, 10, 0, 2], 
    [30, 10, 10, 1], 
    [20, 0, 30, 3], 
    [20, 10, 20, 2], 
    [10, 10, 0, 3],
    [20, 0, 30, 1]]

这是我的代码:

def selectUnique(x):
    count = Counter((i[0], i[1], i[2]) for i in x)
    out = [i for i in x if count[(i[0], i[1], i[2])] == 1]
    return out

selectUnique(f)

我想要得到的结果如下所示:

独特部分:

index       point         equivalent points    index
  0      [10, 20, 0, 1]           1             [1]
  1      [40, 20, 30, 2]          1             [2]
  2      [20, 0, 30, 3]           3             [3,3,1]
  3      [10, 10, 0, 2]           2             [2,3]
  4      [30, 10, 10, 1]          1             [1]
  6      [20, 10, 20, 2]          1             [2]

重复部分:

index       point         Duplicate or not
  5      [20, 0, 30, 3]       duplicate
  7      [10, 10, 0, 3]       duplicate
  8      [20, 0, 30, 1]       duplicate

我的代码旨在实现挑出重复点并指示其在列表中的索引的功能。另外，我还需要实现显示我的独特部分中有多少等价点以及这些等价点的索引的功能。

但是，我的代码无法运行，我该如何修改它？

Answer 1

uniq = []
dup = []
count = 0
for i, j, k  in f:
    if not [f.index([i,j,k]),i,j,k] in uniq:
        uniq.append([count,i,j,k])
    else:
        dup.append([count,i,j,k,"duplicate"])
    count += 1
uniq.extend(dup)
print uniq

[[0, 10, 20, 0], [1, 40, 20, 30], [2, 20, 0, 30], [3, 10, 10, 0], [4, 30, 10, 10], [6, 20, 10, 20], [5, 20, 0, 30, 'duplicate'], [7, 10, 10, 0, 'duplicate']]

for j in (uniq):
    j.append(ic[j[0]])
print uniq


 [[0, 10, 20, 0, 1], [1, 40, 20, 30, 2], [2, 20, 0, 30, 3], [3, 10, 10, 0, 2], [4, 30, 10, 10, 1], [6, 20, 10, 20, 2], [5, 20, 0, 30, 'duplicate', 3], [7, 10, 10, 0, 'duplicate', 3]]