Question

这是我的数据：

a = (9,5,3)
b = (5,3,6)
c = (1,6,6)
d = (2,5,0)
e = (9,8,3)
f = (7,3,6)
g = (2,15,1)
data = [a,b,c,d,e,f,g]

我有7个数据点，在这里我想获得三个数据（top-k = 3），它可以是（a，b，c或其他点）与其他点/ top-的最大距离k max多样化。

from scipy.spatial import distance
d = distance.euclidean(a,b)

k = 3
i = 1
distancelist = []
max_dist = []
while (i < k):
    for x in (data):
        for y in (data):
            dist = distance.euclidean(x,y)
            distancelist.append(dist)
            # stuck in here
            max_dist = #
    i = i+1
print(max_dist)

我卡住了，如何获得最大距离值，然后弹出max_dist

预期产出：

[(9, 8, 3),(2, 15, 1),(5, 3, 6)] #I just choose these as random, I don't know the exact result

例如：

第一子集：总距离18.987490074177131

# combination (a,b,c) or [(9,5,3),(5,3,6),(1,6,6)]
distance.euclidean(data[0], data[1]) + distance.euclidean(data[1], data[2]) + distance.euclidean(data[0], data[2])

第二子集：总距离20.000937912998413

# combination (a,b,d) or [(9,5,3),(5,3,6),(2,5,0)]
distance.euclidean(data[0], data[1]) + distance.euclidean(data[1], data[3]) + distance.euclidean(data[0], data[3])

第二个子集优于第一个子集，因为第二个子集具有更大的总距离值，我想得到子集（top-k = 3），其中最大距离是所有组合的最大值。

Answer 1

以下情况如何。

首先，将所有距离和点（x，y）放入max_dixdance。这里，所有对都由0.34×0.33+0.33×0.0.1089=0.148137生成，而不是双循环。

combinations

此代码几乎（不完全）等同于以下内容：

from scipy.spatial import distance
from itertools import combinations
max_dixdance = []
# for x, y in combinations(data, 2):
#     dis = distance.euclidean(x, y)
#     max_dixdance.append((dis, (x, y)))

## modified version
for xyz in combinations(data, 3):
    # print(list(xyz)) # verify all combinations appeared

    # calculate a sum of all piarwise distance
    dis = 0
    for xy in combinations(xyz, 2):
        # print(list(xy)) # verify all pairs appeared
        dis += distance.euclidean(*xy)
    max_dixdance.append((dis, tuple(xyz)))

然后，使用dis值对列表进行排序，并获取前3个元素。

## modified version - 2
for x, y, z in combinations(data, 3):
    xyz = (x, y, z)

    # calculate a sum of all piarwise distance
    dis = 0
    for x, y in combinations(xyz, 2):
        dis += distance.euclidean(x, y)
    max_dixdance.append((dis, xyz))

Answer 2

没有scipy使用max使用键功能的强力：

from itertools import combinations

def dist2(points):  # distance of 2 points
     return sum((a_ - b_)**2 for a_, b_ in zip(*points))**0.5

def dist3(points):  # sum of triangle sides for 3 points
    return sum(map(dist2, combinations(points, 2)))

>>> max(combinations(data, 3), key=dist3)
((2, 5, 0), (7, 3, 6), (2, 15, 1))

Answer 3

这是我对问题的理解，即每点得到前3个距离，即

#`cdist` will give the distance from every point to one another. 
mat = scipy.spatial.distance.cdist(data,data, metric='euclidean')
#             0          1          2          3          4          5          6
#0   0.000000   5.385165   8.602325   7.615773   3.000000   4.123106  12.369317
#1   5.385165   0.000000   5.000000   7.000000   7.071068   2.000000  13.341664
#2   8.602325   5.000000   0.000000   6.164414   8.774964   6.708204  10.344080
#3   7.615773   7.000000   6.164414   0.000000   8.185353   8.062258  10.049876
#4   3.000000   7.071068   8.774964   8.185353   0.000000   6.164414  10.099505
#5   4.123106   2.000000   6.708204   8.062258   6.164414   0.000000  13.928388
#6  12.369317  13.341664  10.344080  10.049876  10.099505  13.928388   0.000000

#this is for mapping
di = dict(zip(np.arange(7),list('abcdefg')))

#Get top three distances indices using argsort
max3 = mat.argsort(1)[:,-3:]
#map the indices with the names

max3_with_names = np.array(np.vectorize(di.get)(max3)).tolist()
# [['d', 'c', 'g'],
#  ['d', 'e', 'g'],
#  ['a', 'e', 'g'],
#  ['f', 'e', 'g'],
#  ['d', 'c', 'g'],
#  ['c', 'd', 'g'],
#  ['a', 'b', 'f']]

list(zip(list('abcdefg'),max3_with_names))
# [('a', ['d', 'c', 'g']),# d,c,g is the 3 points maximum distances with respect to a.
#  ('b', ['d', 'e', 'g']),
#  ('c', ['a', 'e', 'g']),
#  ('d', ['f', 'e', 'g']),
#  ('e', ['d', 'c', 'g']),
#  ('f', ['c', 'd', 'g']),
#  ('g', ['a', 'b', 'f'])]

找到最大总和距离

3 个答案: