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,在这里我想得到三个数据( top-k =3),它可以是(a,b,c或其他点),它到其他点的距离最大/top-k最大不同。
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)我坚持,如何获得距离的最大值,并弹出到max_dist
预期输出:
[(9, 8, 3),(2, 15, 1),(5, 3, 6)] #I just choose these as random, I don't know the exact result例如:
第一个子集: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])第二个子集: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])第二个子集比第一个子集更好,因为第二个子集的总距离值更大,我想得到最大距离是所有组合中的最大值的子集(top-k=3)。
https://stackoverflow.com/questions/47587246
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