R 中的 KNN 异常值检测

Question

我正在尝试运行使用加权 KNN 异常值分数执行异常值检测的脚本，但不断出现以下错误：

Error in apply(kNNdist(x = dat, k = k), 1, mean) :
dim(X) must have a positive length

我正在尝试运行的脚本如下。它是单个脚本块，但我在导致错误的脚本部分正上方添加了注释，即函数：

WKNN_Outlier <- apply(kNNdist(x=dat, k = k), 1, mean)

如果有人对无监督异常值检测有更好或更简单的想法，我会洗耳恭听（可以这么说...）

    library(dbscan)
    library(ggplot2)

    set.seed(0)

    x11 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 1 (x1 coordinate)
    x21 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 1 (x2 coordinate)
    x12 <- rnorm(n = 100, mean = 20, sd = 1) # Cluster 2 (x1 coordinate)
    x22 <- rnorm(n = 100, mean = 10, sd = 1) # Cluster 2 (x2 coordinate)
    x13 <- rnorm(n = 100, mean = 15, sd = 3) # Cluster 3 (x1 coordinate)
    x23 <- rnorm(n = 100, mean = 25, sd = 3) # Cluster 3 (x2 coordinate)
    x14 <- rnorm(n = 50, mean = 25, sd = 1)  # Cluster 4 (x1 coordinate)
    x24 <- rnorm(n = 50, mean = 25, sd = 1)  # Cluster 4 (x2 coordinate)

    dat <- data.frame(x1 = c(x11,x12,x13,x14), x2 = c(x21,x22,x23,x24))

    ( g0a <- ggplot() + geom_point(data=dat, mapping=aes(x=x1, y=x2), shape = 19) )

    k <- 4 # KNN parameter
    top_n <- 20 # No. of top outliers to be displayed

    KNN_Outlier <- kNNdist(x=dat, k = k)
    rank_KNN_Outlier <- order(x=KNN_Outlier, decreasing = TRUE)    # Sorting (descending)
    KNN_Result <- data.frame(ID = rank_KNN_Outlier, score = KNN_Outlier[rank_KNN_Outlier])

    head(KNN_Result, top_n)

    graph <- g0a +
      geom_point(data=dat[rank_KNN_Outlier[1:top_n],], mapping=aes(x=x1,y=x2), shape=19, 
      color="red", size=2) +
      geom_text(data=dat[rank_KNN_Outlier[1:top_n],],
      mapping=aes(x=(x1-0.5), y=x2, label=rank_KNN_Outlier[1:top_n]), size=2.5)

    graph

    ## Use KNNdist() to calculate the weighted KNN outlier score

    k <- 4       # KNN parameter
    top_n <- 20  # No. of top outliers to be displayed

下面的 WKNN_Outler 函数是导致错误的原因。据我所知，apply 函数应该没有任何问题，因为数据 (dat) 被转换为 data.frame，这应该可以防止错误，但没有。

    WKNN_Outlier <- apply(kNNdist(x=dat, k = k), 1, mean)  # Weighted KNN outlier score (mean)


    rank_WKNN_Outlier <- order(x=WKNN_Outlier, decreasing = TRUE) 
    WKNN_Result <- data.frame(ID = rank_WKNN_Outlier, score = WKNN_Outlier[rank_WKNN_Outlier])

    head(WKNN_Result, top_n)

    ge1 <- g0a +
      geom_point(data=dat[rank_WKNN_Outlier[1:top_n],], mapping=aes(x=x1,y=x2), shape=19, 
      color="red", size=2) +
      geom_text(data=dat[rank_WKNN_Outlier[1:top_n],],
      mapping=aes(x=(x1-0.5), y=x2, label=rank_WKNN_Outlier[1:top_n]), size=2.5)

    ge1

Answer 1

函数 kNNdist(x=dat, k = k) 生成向量而不是矩阵，这就是为什么当您尝试执行 apply 函数时它会告诉您 dim(X) must have a positive length （向量有一个 NULL昏暗）。

尝试：

WKNN_Outlier <- apply(kNNdist(x=dat, k = k, all=T), 1, mean)

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