问用于对地图和聚类上的点进行分组的算法必须具有相同的第三个要素总和

EN

k <- 3 # define how many clusters you want


#A really simple gleedy clustering algorithm, basically you start each list with an elemnt and add the next element to the lowest scoring list
clustering <- function(df,k){
clusters <- list()
for (r in 1:k) {
clusters[[r]] <- df[r,]
}
for (i in 4:nrow(df)){
  a = data.frame(sum(clusters[[1]]$expenses))
  for (j in 2:k) {
    a = rbind(a,sum(clusters[[j]]$expenses))
  }
  minimo = which.min(a[,1])
  clusters[[minimo]] <- rbind(clusters[[minimo]],df[i,])
}
return(clusters)
}

#calculate the difference between the lowest and highest list
distance <- function(){
  A <- clustering(df,k)
  for (k in 1:k) {
    start <- c(start,sum(A[[k]]$expenses))  
  }
  distance <- max(start) - min(start) 
  return(distance)
}


#repeat the process with a diferent starting point and save the clusters which has the lowest variance
max.distance = distance()
Clusters <- clustering(df,k)
for (i in 2:50) {
df <- slice(df, sample(1:n()))
g=distance()
if (max.distance>g) {
  max.distance <- distance()
  Clusters <- clustering(df,k)
}
}