问题
I would like to know if there is any way to optimize the distance calculation process below. I left a small example below, however I am working with a spreadsheet with more than 6000 rows, and it takes considerable time to calculate the variable d. It would be possible to somehow adjust this to have the same results, but in an optimized way.
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$cluster <- clusters
time<-toc()
1.54 sec elapsed
d
1 2 3 4 5 6 7 8
2 0.00
3 11075.61 11075.61
4 11075.61 11075.61 0.00
5 11075.61 11075.61 0.00 0.00
6 11075.61 11075.61 0.00 0.00 0.00
7 11075.61 11075.61 0.00 0.00 0.00 0.00
8 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00
9 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
10 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
11 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
12 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
13 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
14 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
15 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
16 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
17 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
18 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
19 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
9 10 11 12 13 14 15 16
2
3
4
5
6
7
8
9
10 0.00
11 10183.02 10183.02
12 10183.02 10183.02 0.00
13 10183.02 10183.02 0.00 0.00
14 10183.02 10183.02 0.00 0.00 0.00
15 10183.02 10183.02 0.00 0.00 0.00 0.00
16 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02
17 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
18 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
19 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
17 18
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18 0.00
19 0.00 0.00
Comparation
> df$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
回答1:
@Jose Perhaps not as sound mathematically (in terms of the clustering) but (generally) a better measure of great circle distances (Vincenty's formulae). And ~8 times faster to achieve (what I think is your desired result) - (just using your sample data).
# 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)
Libraries and sample data:
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()
来源：https://stackoverflow.com/questions/62055976/distance-calculation-optimization-in-r