3

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，但是我正在处理一个超过 6000 行的电子表格，计算变量 d 需要相当长的时间。有可能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
4

1 回答 1

1

@Jose 也许在数学上没有那么合理（就聚类而言），但（通常）可以更好地测量大圆距离（文森蒂公式）。并且实现速度快约 8 倍（我认为这是您想要的结果）-（仅使用您的示例数据）。

# 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()
于 2020-05-29T04:16:34.370 回答