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我是粒子羣優化的新手。我閱讀了關於基於PSO和K-means的聚類的研究論文，但是我沒有找到相同的工作示例。任何形式的幫助都非常感謝。提前致謝！基於PSO和K-means的文本文檔聚類R
我想在R中使用PSO和K-means進行文本文檔聚類。我的基本思想是，首先PSO會給我聚類質心的優化值，然後我必須使用羣集質心的優化值PSO作爲k-均值的初始聚類質心以獲得文檔簇。
下面是描述我迄今爲止所做的工作的代碼！
#Import library
library(pdist)
library(hydroPSO)
#Create matrix and suppose it is our document term matrix which we get after
the cleaning of corpus
（在我的實際數據我有一個951個條款，即暗淡（DTM）= 20 * 951 20個文檔）
matri <- matrix(data = seq(1, 20, 1), nrow = 4, ncol = 7, byrow = TRUE)
matri
[,1] [,2] [,3] [,4] [,5] [,6] [,7]
[1,] 1 2 3 4 5 6 7
[2,] 8 9 10 11 12 13 14
[3,] 15 16 17 18 19 20 1
[4,] 2 3 4 5 6 7 8
#Initially select first and second row as centroids
cj <- matri[1:2,]
#Calculate Euclidean Distance of each data point from centroids
vm <- as.data.frame(t(as.matrix(pdist(matri, cj))))
vm
V1 V2 V3 V4
1 0.00000 18.52026 34.81379 2.645751
2 18.52026 0.00000 21.51744 15.874508
#Create binary matrix S in which 1 means Instance Ii is allocated to the cluster Cj otherwise 0.
S <- matrix(data = NA, nrow = nrow(vm), ncol = ncol(vm))
for(i in 1:nrow(vm)){
for(j in 1:ncol(vm)){
cd <- which.min(vm[, j])
ifelse(cd==i, S[i,j] <-1, S[i,j] <-0)
}
}
S
[,1] [,2] [,3] [,4]
[1,] 1 0 0 1
[2,] 0 1 1 0
#Apply `hydroPSO()` to get optimised values of centroids.
set.seed(5486)
D <- 4 # Dimension
lower <- rep(0, D)
upper <- rep(10, D)
m_s <- matrix(data = NA, nrow = nrow(S), ncol = ncol(matri))
Fn= function(y) { #Objective Function which has to be minimised
for(j in 1:ncol(matri)){
for(i in 1:nrow(matri)){
for(k in 1:nrow(y)){
for(l in 1:ncol(y)){
m_s[k,] <- colSums(matri[y[k,]==1,])/sum(y[k,])
}
}
}
}
sm <- sum(m_s)/ nrow(S)
return(sm)
}
hh1 <- hydroPSO(S,fn=Fn, lower=lower, upper=upper,
control=list(write2disk=FALSE, npart=3))
但上面hydroPSO()功能無法正常工作。它給錯誤錯誤1：nrow（y）：參數的長度爲0。我搜索了它，但沒有得到任何解決方案，這對我有用。
我也做了一些改變，我的目標函數和這個時候hydroPSO（）工作，但我猜測不正確。我將我的初始質心矩陣作爲尺寸爲2 * 7的參數傳遞，但函數僅返回1 * 7個優化值。我沒有得到它的理由。
set.seed(5486)
D <- 7# Dimension
lower <- rep(0, D)
upper <- rep(10, D)
Fn = function(x){
vm <- as.data.frame(t(as.matrix(pdist(matri, x))))
S <- matrix(data = NA, nrow = nrow(vm), ncol = ncol(vm))
for(i in 1:nrow(vm)){
for(j in 1:ncol(vm)){
cd <- which.min(vm[, j])
ifelse(cd==i, S[i,j] <-1, S[i,j] <-0)
}
}
m_s <- matrix(data = NA, nrow = nrow(S), ncol = ncol(matri))
for(j in 1:ncol(matri)){
for(i in 1:nrow(matri)){
for(k in 1:nrow(S)){
for(l in 1:ncol(S)){
m_s[k,] <- colSums(matri[S[k,]==1,])/sum(S[k,])
}
}
}
}
sm <- sum(m_s)/ nrow(S)
return(sm)
}
hh1 <- hydroPSO(cj,fn=Fn, lower=lower, upper=upper,
control=list(write2disk=FALSE, npart=2, K=2))
上述函數的輸出。
## $par
## Param1 Param2 Param3 Param4 Param5 Param6 Param7
## 8.6996174 2.1952303 5.6903588 0.4471795 3.7103161 1.6605425 8.2717574
##
## $value
## [1] 61.5
##
## $best.particle
## [1] 1
##
## $counts
## function.calls iterations regroupings
## 2000 1000 0
##
## $convergence
## [1] 3
##
## $message
## [1] "Maximum number of iterations reached"
我想我以錯誤的方式將參數傳遞給hydroPSO()。請糾正我在哪裏做錯了。
非常感謝！