我是粒子群优化的新手。我阅读了关于基于PSO和K-means的聚类的研究论文,但我没有找到任何相同的工作实例。任何形式的帮助都非常感谢。提前致谢!
我想在R中使用PSO和K-means执行文本文档聚类。我有一个基本的想法,即第一个PSO会给我优化的聚类质心值,然后我必须使用那些优化的聚类质心值PSO作为k-means的初始集群质心来获取文档集群。
以下代码描述了我到目前为止所做的工作!
#Import library
library(pdist)
library(hydroPSO)
#Create matrix and suppose it is our document term matrix which we get after
the cleaning of corpus
(在我的实际数据中,我有20个文件,951个术语,即 dim(dtm)= 20 * 951 )
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))
但上述hydroPSO()
功能无效。它在1:nrow(y):长度为0的参数中给出错误错误。我搜索了它,但没有得到任何适合我的解决方案。
我也对我的目标函数进行了一些更改,这次hydroPSO()工作但我觉得不正确。我将初始质心矩阵作为参数传递,其尺寸为2 * 7,但该函数仅返回1 * 7优化值。我没理由。
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))
输出上述功能。
## $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"
我想我是以错误的方式将参数传递给hydroPSO()
。请纠正我在哪里做错了。
非常感谢!
答案 0 :(得分:0)
我没有将 cj 传递给hydroPSO()
,而是在我的第二种方法中使用了 as.vector(t(cj)),而且它对我来说很好。我得到了14个优化值