使用 ward.D2 方法聚类 pheatmap 和 corrplot() 之间的差异

Question

我对我的数据进行了 Spearman 相关。然后我尝试使用 corrplot() 和 pheatmap() 的“ward.D2”方法对我的数据进行聚类和绘图。但是，两个图之间的变量顺序不同。

有人可以帮我澄清这一点，从而更正我的代码并创建两个具有相同聚类变量顺序的图吗？非常感谢。

#A sample of my dataset: 

dput(Data_corr)

structure(list(S_cHDLP = c(0.299999999999999, -2.78, 0.880000000000001, 
2.48, 2.15, 5.31, 3.02, 1.19, 2.1, -1.18, -0.34, 1.25, -3.25, 
-3.16, 0.19, -0.100000000000001, -2.16, -0.220000000000001, 0.77, 
-2.12), H7P = c(-0.18, -0.48, -0.13, -0.21, 0.07, 0.64, -0.13, 
-0.1, 0.12, -0.22, 0.09, -0.0399999999999999, -1.56, 0.39, 0.58, 
-0.49, 0.2, 0.13, 0.11, 0.06), H6P = c(0, 0, 0, 0.16, -0.23, 
0, 0, 0, -0.26, -0.28, 0.06, -0.17, 1.16, -0.12, -0.32, -0.17, 
0.38, 0.05, 0.01, 0), H5P = c(0, 0.84, 0.47, 1.21, 0.01, 0.21, 
1.36, 0.2, -0.12, 0.93, -1.01, 0, -0.58, -0.97, -1, 0.97, -0.89, 
0.35, -0.59, -0.12), H4P = c(-0.12, -1.27, -0.18, 0.25, 1.02, 
1.26, -0.62, -0.16, 0.25, -0.01, 0.44, 0.17, 0.19, 0.97, 2.35, 
0.3, -0.18, 0.03, 0.0899999999999999, 0.38), H3P = c(-0.31, 0.39, 
0.13, 0.29, 0, 0.02, -0.07, 0, 0, -0.32, 0, -0.79, 0, -0.53, 
-0.71, -0.2, 2.08, 0.86, 0, 0), H2P = c(-1.28, -0.619999999999999, 
-1.07, 1.96, 0.15, 4.92, 1.55, 3, -0.459999999999999, -0.56, 
1.12, 3.44, -1.48, -1.27, 1.45, 0.609999999999999, -1.59, -1.57, 
2.04, 2.03), H1P = c(1.58, -2.15, 1.96, 0.51, 2, 0.37, 1.47, 
-1.83, 2.56, -0.62, -1.46, -2.19, -1.77, -1.9, -1.25, -0.73, 
-0.57, 1.35, -1.28, -4.14), TRLZ_TRL = c(4.61, 1.49, -2.71, 1.54, 
-5.46, 2.18, 3.48, 12.83, 7.51, 7.74, -8.38, -0.729999999999997, 
6.11, -19.74, -0.869999999999997, -1.82, -1.57000000000001, 0.609999999999999, 
-14.79, -18.65), LDLZ = c(-0.0599999999999987, -0.400000000000002, 
-0.289999999999999, -1.2, -0.479999999999997, -0.59, -1.29, -0.0599999999999987, 
0.210000000000001, -1.58, 1.97, 0.0800000000000018, -1, 1.95, 
1.41, 0.00999999999999801, 0.430000000000003, -0.289999999999999, 
0.68, 0.52), HDLZ = c(-0.200000000000001, -0.200000000000001, 
-0.0700000000000003, 0, -0.0200000000000014, -0.0199999999999996, 
-0.0399999999999991, -0.119999999999999, -0.0900000000000016, 
-0.0500000000000007, -0.15, -0.16, -0.640000000000001, 0.42, 
0.16, -0.130000000000001, 0.15, 0.41, -0.0300000000000011, 0.18
)), class = "data.frame", row.names = c(NA, -20L))

library(pheatmap)
library(corrplot)


CorMethod <- "spearman"

CorMatrix <- cor(Data_corr, method=CorMethod, use="pairwise.complete.obs")


## 1st Plot

Plot3<-pheatmap(CorMatrix, cluster_cols=T, cluster_rows=T, cutree_rows = 3, angle_col=45, fontsize_col=5, fontsize_row = 7, treeheight_col=0, clustering_method="ward.D2")



#2nd Plot

Plot8 <-corrplot(CorMatrix, method="square", type="lower", order="hclust", hclust.method="ward.D2", tl.pos="ld", tl.cex = 0.5, tl.col="black", tl.srt=45)

Answer 1

您可以使用 pheatmap 给出的相同顺序创建一个 corrplot，如下所示：

#2nd Plot
library(RColorBrewer)
ord <- Plot3$tree_row$order
ReordCorMatrix <- CorMatrix[ord, ord]
Plot8 <-corrplot(ReordCorMatrix, method="square", type="lower", order="original", 
                 hclust.method="ward.D2", 
                 tl.pos="ld", tl.cex = 0.5, tl.col="black", tl.srt=45, 
                 col=colorRampPalette(rev(brewer.pal(n = 7, name="RdYlBu")))(100))

使用 ward.D2 方法聚类 pheatmap 和 corrplot() 之间的差异

Clustering differences between pheatmap and corrplot() using ward.D2 method

r

r-corrplot

pheatmap