Number of possible solutions for generating clusters.

[JojoSouza]

I have information of 29 properties (Points_properties), and I wanted to find out the number of possible solutions to generate clusters with at least 2 properties, i.e. I don't want cluster with only 1 property. I made code that shows there are 12 possible solutions, but I know because I used a value for k in the cutree function. Notice that in nclusters, I have 13 solutions, the first with 1 it is not considered, only the others that have at least 2 properties, that is, there are 12 possible solutions in total. So how do I find that there are 12 possible solutions, without using a number defined in cutree. If you have an easier way to find out, feel free to enter.

library(geosphere)

Points_properties<-structure(list(
Propertie=c(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29), 
Latitude = c(-24.781624, -24.775017, -24.769196, 
-24.761741, -24.752019, -24.748008, -24.737312, -24.744718, -24.751996, 
-24.724589, -24.8004, -24.796899, -24.795041, -24.780501, -24.763376, 
-24.801715, -24.728005, -24.737845, -24.743485, -24.742601, -24.766422, 
-24.767525, -24.775631, -24.792703, -24.790994, -24.787275, -24.795902, 
-24.785587, -24.787558), Longitude = c(-49.937369, 
-49.950576, -49.927608, -49.92762, -49.920608, -49.927707, -49.922095, 
-49.915438, -49.910843, -49.899478, -49.901775, -49.89364, -49.925657, 
-49.893193, -49.94081, -49.911967, -49.893358, -49.903904, -49.906435, 
-49.927951, -49.939603, -49.941541, -49.94455, -49.929797, -49.92141, 
-49.915141, -49.91042, -49.904772, -49.894034)), row.names = c(NA, -29L), class = c("tbl_df", "tbl", 
"data.frame"))



coordinates<-subset(Points_properties,select=c("Latitude","Longitude"))
d<-distm(coordinates[,2:1])
d<-as.dist(d)
fit.average<-hclust(d,method="average")

clusters<-cutree(fit.average, 13) 
nclusters<-matrix(table(clusters))
nclusters
> nclusters
      [,1]
 [1,]    1
 [2,]    2
 [3,]    2
 [4,]    3
 [5,]    3
 [6,]    2
 [7,]    2
 [8,]    3
 [9,]    2
[10,]    3
[11,]    2
[12,]    2
[13,]    2

[nirgrahamuk]

Points_properties <- data.frame(
  Propertie = c(1,2,3,4,5,6,7,8,9,10,11,12,13,
                14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29),
  Latitude = c(-24.781624,-24.775017,-24.769196,
               -24.761741,-24.752019,-24.748008,-24.737312,-24.744718,-24.751996,
               -24.724589,-24.8004,-24.796899,-24.795041,-24.780501,
               -24.763376,-24.801715,-24.728005,-24.737845,-24.743485,
               -24.742601,-24.766422,-24.767525,-24.775631,-24.792703,
               -24.790994,-24.787275,-24.795902,-24.785587,-24.787558),
  Longitude = c(-49.937369,-49.950576,-49.927608,
                -49.92762,-49.920608,-49.927707,-49.922095,-49.915438,-49.910843,
                -49.899478,-49.901775,-49.89364,-49.925657,-49.893193,
                -49.94081,-49.911967,-49.893358,-49.903904,-49.906435,
                -49.927951,-49.939603,-49.941541,-49.94455,-49.929797,-49.92141,
                -49.915141,-49.91042,-49.904772,-49.894034)
)
library(geosphere)
library(tidyverse)
coordinates<-subset(Points_properties,select=c("Latitude","Longitude"))
d<-distm(coordinates[,2:1])
d<-as.dist(d)
fit.average<-hclust(d,method="average")

(max_feasible_cut_groups <- floor(29/2))
full_results <- map_dfr(seq_len(max_feasible_cut_groups),
    ~{
      r1 <- cutree(fit.average,k= .x)
      ct <- table(r1)
      tibble(k=.x,
           cluster_label_for_each_node=list(r1),
           coverage_table = list(ct),
       problematic_unique_clusters= length(ct[which(ct<2)]))
      })

print(full_results,n=Inf)

(last_valid_run <- filter(full_results,
       problematic_unique_clusters==0) %>% slice_tail(n = 1))

last_valid_run$cluster_label_for_each_node
last_valid_run$coverage_table

[JojoSouza]

That's it @nirgrahamuk r, thank you so much for that. Could you please just make a comment exactly on what you are doing in max_feasible_cut_groups, full_results, last_valid_run, to make it more understandable to me, what was your idea. Thank you again!