R中的距离计算优化 [英] Distance calculation optimization in R
本文介绍了R中的距离计算优化的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！
问题描述
我想知道下面是否有任何方法可以优化距离计算过程.我在下面留下了一个小示例，但是我正在处理包含6000行以上的电子表格，并且计算变量d花费了大量时间.可以通过某种方式将其调整为具有相同结果，但是以一种优化的方式.
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$cluster <- clusters
time<-toc()
1.54 sec elapsed
d
1 2 3 4 5 6 7 8
2 0.00
3 11075.61 11075.61
4 11075.61 11075.61 0.00
5 11075.61 11075.61 0.00 0.00
6 11075.61 11075.61 0.00 0.00 0.00
7 11075.61 11075.61 0.00 0.00 0.00 0.00
8 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00
9 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
10 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
11 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
12 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
13 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
14 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
15 15048.01 15048.01 10183.02 10183.02 10183.02 10183.02 10183.02 10183.02
16 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
17 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
18 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
19 11075.61 11075.61 0.00 0.00 0.00 0.00 0.00 0.00
9 10 11 12 13 14 15 16
2
3
4
5
6
7
8
9
10 0.00
11 10183.02 10183.02
12 10183.02 10183.02 0.00
13 10183.02 10183.02 0.00 0.00
14 10183.02 10183.02 0.00 0.00 0.00
15 10183.02 10183.02 0.00 0.00 0.00 0.00
16 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02
17 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
18 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
19 0.00 0.00 10183.02 10183.02 10183.02 10183.02 10183.02 0.00
17 18
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18 0.00
19 0.00 0.00
比较
> df$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
推荐答案
@Jose也许在数学上(就聚类而言)听起来不那么合理，但(通常)可以更好地度量大圆距离(Vincenty公式).大约要快8倍(我认为这是您想要的结果)-(仅使用示例数据即可).
@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)
库和样本数据:
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()
这篇关于R中的距离计算优化的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
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