我想知道是否有任何方法可以优化下面的距离计算过程。我在下面留下了一个小例子，但是我正在使用一个超过 6000 行的电子表格，计算变量 d 需要相当长的时间。可以以某种方式调整它以获得相同的结果，但以优化的方式。
library(rdist)
library(tictoc)
library(geosphere)
time<-tic()
df<-structure(list(Industries=c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19), Latitude = c(-23.8, -23.8, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9,
+ + -23.9, -23.9, -23.9, -23.9, -23.9), Longitude = c(-49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.7,
+ + -49.7, -49.7, -49.7, -49.7, -49.6, -49.6, -49.6, -49.6)), class = "data.frame", row.names = c(NA, -19L))
k=3
#clusters
coordinates<-df[c("Latitude","Longitude")]
d<-as.dist(distm(coordinates[,2:1]))
fit.average<-hclust(d,method="average")
clusters<-cutree(fit.average, k)
nclusters<-matrix(table(clusters))
df$cluster <- clusters
time<-toc()
1.54 sec elapsed
d
1 2 3 4 5 6 7 8
2 0.00
3 11075.61 11075.61
4 11075.61 11075.61 0.00
5 11075.61 11075.61 0.00 0.00
6 11075.61 11075.61 0.00 0.00 0.00
7 11075.61 11075.61 0.00 0.00 0.00 0.00
8 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00
9 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
10 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
11 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
12 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
13 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
14 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
15 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
16 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
17 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
18 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
19 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
9 10 11 12 13 14 15 16
2
3
4
5
6
7
8
9
10 0.00
11 10183.02 10183.02
12 10183.02 10183.02 0.00
13 10183.02 10183.02 0.00 0.00
14 10183.02 10183.02 0.00 0.00 0.00
15 10183.02 10183.02 0.00 0.00 0.00 0.00
16 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02
17 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
18 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
19 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
17 18
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18 0.00
19 0.00 0.00
比较
> df$cluster <- clusters
> df
Industries Latitude Longitude cluster
1 1 -23.8 -49.6 1
2 2 -23.8 -49.6 1
3 3 -23.9 -49.6 2
4 4 -23.9 -49.6 2
5 5 -23.9 -49.6 2
6 6 -23.9 -49.6 2
7 7 -23.9 -49.6 2
8 8 -23.9 -49.6 2
9 9 -23.9 -49.6 2
10 10 -23.9 -49.6 2
11 11 -23.9 -49.7 3
12 12 -23.9 -49.7 3
13 13 -23.9 -49.7 3
14 14 -23.9 -49.7 3
15 15 -23.9 -49.7 3
16 16 -23.9 -49.6 2
17 17 -23.9 -49.6 2
18 18 -23.9 -49.6 2
19 19 -23.9 -49.6 2
> clustered_df
Industries Latitude Longitude cluster Dist Cluster
1 11 -23.9 -49.7 3 0.00 1
2 12 -23.9 -49.7 3 0.00 1
3 13 -23.9 -49.7 3 0.00 1
4 14 -23.9 -49.7 3 0.00 1
5 15 -23.9 -49.7 3 0.00 1
6 3 -23.9 -49.6 2 10183.02 2
7 4 -23.9 -49.6 2 0.00 2
8 5 -23.9 -49.6 2 0.00 2
9 6 -23.9 -49.6 2 0.00 2
10 7 -23.9 -49.6 2 0.00 2
11 8 -23.9 -49.6 2 0.00 2
12 9 -23.9 -49.6 2 0.00 2
13 10 -23.9 -49.6 2 0.00 2
14 16 -23.9 -49.6 2 0.00 2
15 17 -23.9 -49.6 2 0.00 2
16 18 -23.9 -49.6 2 0.00 2
17 19 -23.9 -49.6 2 0.00 2
18 1 -23.8 -49.6 1 11075.61 3
19 2 -23.8 -49.6 1 0.00 3
@Jose 也许在数学上不那么合理(就聚类而言(，但(通常(是大圆距离的更好度量(文森蒂公式(。实现速度快 ~8 倍(我认为是您想要的结果( - (仅使用您的样本数据(。
# Order the dataframe by Lon and Lat: ordered_df => data.frame
ordered_df <-
df %>%
arrange(., Longitude, Latitude)
# Scalar valued at how many clusters we are expecting => integer vector
k = 3
# Matrix of co-ordinates: coordinates => matrix
coordinates <-
ordered_df %>%
select(Longitude, Latitude) %>%
as.matrix()
# Generate great circle distances between points and Long-Lat Matrix: d => data.frame
d <- data.frame(Dist = c(0, distVincentyEllipsoid(coordinates)))
# Segment the distances into groups: cluster => factor
d$Cluster <- factor(cumsum(d$Dist > (quantile(d$Dist, 1/k))) + 1)
# Merge with base data: clustered_df => data.frame
clustered_df <- cbind(ordered_df, d)
库和示例数据：
library(geosphere)
library(dplyr)
df <- structure(list(Industries=c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19),
Latitude = c(-23.8, -23.8, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9, -23.9),
Longitude = c(-49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.6, -49.7,-49.7, -49.7, -49.7, -49.7, -49.6, -49.6, -49.6, -49.6)),
class = "data.frame", row.names = c(NA, -19L))
start_time <- Sys.time()