在嵌套列表中查找和分组相同的值（浮点向量）

I have a list containing vectors (float x, y ,z values) and i'm trying group the same values together....

我有一个包含向量的列表(浮点数x,y,z值),我正在尝试将相同的值组合在一起....

Here is a working example on a single list:

以下是单个列表上的工作示例:

from collections import defaultdict

example_list = [1,5,1,4,2,5,5,4,5,7,2,1,7,9] 

d = defaultdict(list)

for item in example_list:
     d[item].append(item)

groupedlist = sorted(d[x] for x in d)

# Returns [[1, 1, 1], [2, 2], [4, 4], [5, 5, 5, 5], [7, 7], [9]]

I'm trying the achieve the same result for nested lists of 3D Vectors (X,Y,Z)...

我正在尝试为3D矢量(X,Y,Z)的嵌套列表实现相同的结果......

example_vectorlist = [[1,2,4], [1,2,3], [3,4,3], [1,3,2], [5,6,7], [3,4,3], [5,6,7]]

# Desired output = [[1,2,4],[[1,2,3],[1,2,3]],[1,3,2],[[3,4,3], [3,4,3]],[[5,6,7], [5,6,7]]

3 个解决方案

#1

Just make the keys for your defaultdict into tuples:

只需将defaultdict的键设置为元组:

from collections import defaultdict

example_list = [[1,2,4], [1,2,3], [3,4,3], [1,3,2], [5,7,1], [3,4,3], [5,6,1]]

d = defaultdict(list)

for item in example_list:
    d[tuple(item)].append(item)

groupedlist = sorted(d[x] for x in d)

The problem with just using the original "vectors" as keys to d is that lists aren't hashable; making tuples of them addresses this.

仅使用原始“向量”作为d的键的问题是列表不可清除;制作他们的元组可以解决这个问题。

#2

Without using defaultdict:

不使用defaultdict:

example_vectorlist = [[1,2,4], [1,2,3], [3,4,3], [1,3,2], [5,7,1],[3,4,3], [5,6,1]]
d = {}
for el in example_vectorlist:
    try:
        d[tuple(el)].append(el)
    except KeyError:
        d[tuple(el)] = [el]
print d.values()

#3

Your desired output does not reflect what you have as input, if you want to group common sublists you can use itertools.groupby, considering you want sorted output using a dict and then sorting makes less sense than just creating the groupings from the sorted list using groupby:

您想要的输出并不能反映出您输入的内容,如果您想对可以使用itertools.groupby的公共子列表进行分组,考虑到您希望使用dict排序输出,然后排序不仅仅是使用排序列表创建分组通过...分组:

from itertools import groupby

print([list(v) for _,v in groupby(sorted(example_vectorlist))])

Which for you original list outputs the same:

哪个为您原始列表输出相同:

example_list = [1,5,1,4,2,5,5,4,5,7,2,1,7,9]
print([list(v) for _,v in groupby(sorted(example_list))])
[[1, 1, 1], [2, 2], [4, 4], [5, 5, 5, 5], [7, 7], [9]]

#1