集群的SD(k均值)

问题描述 投票:0回答:1

我试图找出如何让我的集群的SD获得我的k-means聚类分析。我做了k-means并获得了几个输出,其中一个是“中心”,我认为是手段。我需要所有这些中心的标准偏差来呈现我的数据,我不知道,如何获得它们?

#kmeans
resultspoorT0t <- kmeans(poor_T0v, 3)
resultspoorT0t[["centers"]]

       ALH      BCF      LIN       VAP       VCL      VSL
1 5.130483 12.66909 40.14618  69.78680 146.97313 55.51221
2 3.098673 10.11618 34.38605  29.20927  69.74657 22.70321
3 7.212529 12.98836 41.71680 111.67745 229.73901 92.12502

我尝试了简单的sd()function,但这使得一个SD,我需要SD用于每个群集的每个参数

#SD
sd(resultspoorT0t$cluster, na.rm = FALSE)
[1] 0.758434
r cluster-computing k-means standard-deviation
1个回答
0
投票

我们假设你想要一个简单的循环SD。因此,您需要计算从群集到该群集中心的每个点的距离。它是欧几里德距离sqrt(sum((x_mean - x)** 2 +(y_mean - y)** 2 ...))。然后,您可以计算每个群集的距离SD。代码是:

# Some fake data
set.seed(2222)
df <- matrix(rnorm(6 * 50), 50)
colnames(df) <- letters[1:6]
df <- as.data.frame(df)
k_res <- kmeans(df, 3)

# SD = sd of points distances from cluster center
clusters <- k_res$cluster
centers <- k_res$centers


res_sd <- NULL
for (cl in c(unique(clusters))){
    df_part <- df[clusters == cl, ]

    # Calculate Euclidian distance between 
    # each point (row) and cluster center.
    dist <- sqrt(rowSums((df_part - centers[cl, ]) ** 2))

    # Calculate SD for each column (i.e. SD along each axis)
    sd_s <- apply(df_part - centers[cl, ], 2, sd)
    names(sd_s) <- paste("sd_", colnames(df_part), sep = "")

    res_part <- c(cluster = cl, total_sd = sd(dist), sd_s)
    res_sd <- rbind(res_sd, res_part)
}

res_sd <- as.data.frame(res_sd)
rownames(res_sd) <- res_sd$cluster
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