我试图实现K-means但代码有问题，因为我没有得到结果。
这是欧几里德距离计算：
euclid <- function(p1, p2) {
dm <- matrix(NA, nrow=dim(p1)[1], ncol=dim(p2)[1])
for(i in 1:nrow(p2)) {
dm[,i] <- sqrt(rowSums(t(t(p1)-p2[i,])^2))
}
dm
}
这是K-mean算法。
K_means <- function(dt, c, itr) {
clusterHistory <- vector(itr, mode="list")
centerHistory <- vector(itr, mode="list")
for(i in 1:itr) {
distsToCenters <- euclid(dt,c)
clusters <- apply(distsToCenters, 1, which.min)
centers <- apply(dt, 2, tapply, clusters, mean)
# Saving history
clusterHistory[[i]] <- clusters
centerHistory[[i]] <- c
}
list(clusters=clusterHistory, c = centerHistory)
}
这些是我的中心：
c <- data.frame(Label = c("A","C","F","M"), X = c(3,2.1,3,5), Y =c(6.1,5.0,5.0,5.0))
这些是我的全部观点：
dt2 <- data.frame(Label = c("A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T"), X = c(3.0,4.0,2.1,4.0,7.0,3.0,6.1,7.0,4.0,3.0,6.2,7.0,5.0,3.5,2.5,3.5,5.5,6.0,0.5,0.8), Y =c(6.1,2.0,5.0,6.0,3.0,5.0,4.0,2.0,1.5,2.0,2.0,3.0,5.0,4.5,6.0,5.5,4.5,1.0,1.5,1.2))
显示结果：
res <- K_means(dt2[,2:3], c[,2:3], 4)
res
我想在适当的群集中放置适当的标签，但它不起作用。
答案 0 :(得分：2)
BTW如果假设这是一个自学问题，假设R中有kmeans函数。如果是这样，datacamp（第1章免费）上有关于kmeans的良好课程
问题似乎与你的euclid功能有关，特别是这一行
dm[,i] <- sqrt(rowSums(t(t(p1)-p2[i,])^2))
这为dm
提供了以下结果 [,1] [,2] [,3] [,4]
[1,] 0 0.9 0.0 2.0
[2,] 0 1.1 1.1 1.1
[3,] 0 0.9 0.0 2.0
[4,] 0 1.1 1.1 1.1
[5,] 0 0.9 0.0 2.0
[6,] 0 1.1 1.1 1.1
[7,] 0 0.9 0.0 2.0
[8,] 0 1.1 1.1 1.1
[9,] 0 0.9 0.0 2.0
[10,] 0 1.1 1.1 1.1
[11,] 0 0.9 0.0 2.0
[12,] 0 1.1 1.1 1.1
[13,] 0 0.9 0.0 2.0
[14,] 0 1.1 1.1 1.1
[15,] 0 0.9 0.0 2.0
[16,] 0 1.1 1.1 1.1
[17,] 0 0.9 0.0 2.0
[18,] 0 1.1 1.1 1.1
[19,] 0 0.9 0.0 2.0
[20,] 0 1.1 1.1 1.1
我替换了这一行
euclid <- function(p1, p2) {
dm <- matrix(NA, nrow=dim(p1)[1], ncol=dim(p2)[1])
for(i in 1:nrow(p2)) {
dm[,i] <- sqrt(rowSums(sapply(1:ncol(p1), function(c) {(p1[,c] - p2[i,c])^2})))
}
dm
}
这使用sapply迭代每列。
这给出了
[,1] [,2] [,3] [,4]
[1,] 0.000000 1.421267 1.1000000 2.2825424
[2,] 4.220190 3.551056 3.1622777 3.1622777
[3,] 1.421267 0.000000 0.9000000 2.9000000
[4,] 1.004988 2.147091 1.4142136 1.4142136
[5,] 5.060632 5.292447 4.4721360 2.8284271
[6,] 1.100000 0.900000 0.0000000 2.0000000
[7,] 3.744329 4.123106 3.2572995 1.4866069
[8,] 5.728001 5.745433 5.0000000 3.6055513
[9,] 4.707441 3.982462 3.6400549 3.6400549
[10,] 4.100000 3.132092 3.0000000 3.6055513
[11,] 5.200961 5.080354 4.3863424 3.2310989
[12,] 5.060632 5.292447 4.4721360 2.8284271
[13,] 2.282542 2.900000 2.0000000 0.0000000
[14,] 1.676305 1.486607 0.7071068 1.5811388
[15,] 0.509902 1.077033 1.1180340 2.6925824
[16,] 0.781025 1.486607 0.7071068 1.5811388
[17,] 2.968164 3.436568 2.5495098 0.7071068
[18,] 5.916925 5.586591 5.0000000 4.1231056
[19,] 5.235456 3.848376 4.3011626 5.7008771
[20,] 5.371220 4.016217 4.3908997 5.6639209
已编辑 - 标签
要获取群集标签，建议您将包含标签的整个data.frame传递到K_means函数，如下所示。
请注意，我还更改了代码，以便在后续迭代中使用新的中心。在您之前的代码中，您始终使用最初传递的中心。
K_means <- function(dt, c, itr) {
clusterHistory <- vector(itr, mode="list")
centerHistory <- vector(itr, mode="list")
dt_labels <- dt[,1]
dt_data <- dt[,-1]
c_labels <- c[,1]
c_data <- c[,-1]
for(i in 1:itr) {
distsToCenters <- euclid(dt_data,c_data)
clusters <- apply(distsToCenters, 1, which.min)
c_data <- apply(dt_data, 2, tapply, clusters, mean)
# Saving history
clusterHistory[[i]] <- data.frame(Label = dt_labels, cluster_label = c_labels[clusters] ,cluster_number = clusters)
centerHistory[[i]] <- data.frame(Label = c_labels, c_data)
}
list(clusters=clusterHistory, c = centerHistory)
}
更改调用以传递整个data.frame。我假设标签在第1列。
c <- data.frame(Label = c("A","C","F","M"), X = c(3,2.1,3,5), Y =c(6.1,5.0,5.0,5.0))
dt2 <- data.frame(Label = c("A","B","C","D","E","F","G","H","I","J","K","L","M","N","O","P","Q","R","S","T"), X = c(3.0,4.0,2.1,4.0,7.0,3.0,6.1,7.0,4.0,3.0,6.2,7.0,5.0,3.5,2.5,3.5,5.5,6.0,0.5,0.8), Y =c(6.1,2.0,5.0,6.0,3.0,5.0,4.0,2.0,1.5,2.0,2.0,3.0,5.0,4.5,6.0,5.5,4.5,1.0,1.5,1.2))
res <- K_means(dt2, c, 4)
res
这给出了以下结果
$clusters
$clusters[[1]]
Label cluster_label cluster_number
1 A A 1
2 B F 3
3 C C 2
4 D A 1
5 E M 4
6 F F 3
7 G M 4
8 H M 4
9 I F 3
10 J F 3
11 K M 4
12 L M 4
13 M M 4
14 N F 3
15 O A 1
16 P F 3
17 Q M 4
18 R M 4
19 S C 2
20 T C 2
$clusters[[2]]
Label cluster_label cluster_number
1 A A 1
2 B F 3
3 C A 1
4 D A 1
5 E M 4
6 F A 1
7 G M 4
8 H M 4
9 I F 3
10 J F 3
11 K M 4
12 L M 4
13 M A 1
14 N F 3
15 O A 1
16 P A 1
17 Q M 4
18 R M 4
19 S C 2
20 T C 2
$clusters[[3]]
Label cluster_label cluster_number
1 A A 1
2 B F 3
3 C A 1
4 D A 1
5 E M 4
6 F A 1
7 G M 4
8 H M 4
9 I F 3
10 J F 3
11 K M 4
12 L M 4
13 M A 1
14 N A 1
15 O A 1
16 P A 1
17 Q M 4
18 R M 4
19 S C 2
20 T C 2
$clusters[[4]]
Label cluster_label cluster_number
1 A A 1
2 B F 3
3 C A 1
4 D A 1
5 E M 4
6 F A 1
7 G M 4
8 H M 4
9 I F 3
10 J F 3
11 K M 4
12 L M 4
13 M A 1
14 N A 1
15 O A 1
16 P A 1
17 Q M 4
18 R M 4
19 S C 2
20 T C 2
$c
$c[[1]]
Label X Y
1 A 3.166667 6.033333
2 C 1.133333 2.566667
3 F 3.500000 3.416667
4 M 6.225000 3.062500
$c[[2]]
Label X Y
1 A 3.300 5.514286
2 C 0.650 1.350000
3 F 3.625 2.500000
4 M 6.400 2.785714
$c[[3]]
Label X Y
1 A 3.325000 5.387500
2 C 0.650000 1.350000
3 F 3.666667 1.833333
4 M 6.400000 2.785714
$c[[4]]
Label X Y
1 A 3.325000 5.387500
2 C 0.650000 1.350000
3 F 3.666667 1.833333
4 M 6.400000 2.785714