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我想将 hclust-dendrogram 从 R 导出到数据表中，以便随后将其导入另一个（“自制”）软件。str(unclass(fit))提供树状图的文本概述，但我正在寻找的实际上是一个数字表。我看过 Bioconductor ctc 包，但它产生的输出看起来有些神秘。我想要类似于这张表的东西：http ://stn.spotfire.com/spotfire_client_help/heat/heat_importing_exporting_dendrograms.htm 有没有办法从 R 中的 hclust 对象中得到它？

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2 回答 2

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如果有人也对树状图导出感兴趣，这是我的解决方案。很可能，它不是最好的，因为我最近才开始使用 R，但至少它有效。因此，欢迎提出有关如何改进代码的建议。

所以，ifhr是我的 hclust 对象并且df是我的数据，它的第一列包含一个从 0 开始的简单索引，行名是聚集项的名称：

# Retrieve the leaf order (row name and its position within the leaves)
leaf.order <- matrix(data=NA, ncol=2, nrow=nrow(df),
              dimnames=list(c(), c("row.num", "row.name")))
leaf.order[,2] <- hr$labels[hr$order]
for (i in 1:nrow(leaf.order)) {
   leaf.order[which(leaf.order[,2] %in% rownames(df[i,])),1] <- df[i,1]
}
leaf.order <- as.data.frame(leaf.order)

hr.merge <- hr$merge
n <- max(df[,1])

# Re-index all clustered leaves and nodes. First, all leaves are indexed starting from 0.
# Next, all nodes are indexed starting from max. index leave + 1.
for (i in 1:length(hr.merge)) {
  if (hr.merge[i]<0) {hr.merge[i] <- abs(hr.merge[i])-1}
  else { hr.merge[i] <- (hr.merge[i]+n) }
}
node.id <- c(0:length(hr.merge))

# Generate dendrogram matrix with node index in the first column.
dend <- matrix(data=NA, nrow=length(node.id), ncol=6,
           dimnames=list(c(0:(length(node.id)-1)),
              c("node.id", "parent.id", "pruning.level",
              "height", "leaf.order", "row.name")) )
dend[,1] <- c(0:((2*nrow(df))-2))  # Insert a leaf/node index

# Calculate parent ID for each leaf/node:
# 1) For each leaf/node index, find the corresponding row number within the merge-table.
# 2) Add the maximum leaf index to the row number as indexing the nodes starts after indexing all the leaves.
for (i in 1:(nrow(dend)-1)) {
  dend[i,2] <- row(hr.merge)[which(hr.merge %in% dend[i,1])]+n
}

# Generate table with indexing of all leaves (1st column) and inserting the corresponding row names into the 3rd column.
hr.order <- matrix(data=NA,
           nrow=length(hr$labels), ncol=3,
           dimnames=list(c(), c("order.number", "leaf.id", "row.name")))
hr.order[,1] <- c(0:(nrow(hr.order)-1))
hr.order[,3] <- t(hr$labels[hr$order])
hr.order <- data.frame(hr.order)
hr.order[,1] <- as.numeric(hr.order[,1])

# Assign the row name to each leaf.
dend <- as.data.frame(dend)
for (i in 1:nrow(df)) {
      dend[which(dend[,1] %in% df[i,1]),6] <- rownames(df[i,])
}

# Assign the position on the dendrogram (from left to right) to each leaf.
for (i in 1:nrow(hr.order)) {
      dend[which(dend[,6] %in% hr.order[i,3]),5] <- hr.order[i,1]-1
}

# Insert height for each node.
dend[c((n+2):nrow(dend)),4] <- hr$height

# All leaves get the highest possible pruning level
dend[which(dend[,1] <= n),3] <- nrow(hr.merge)

# The nodes get a decreasing index starting from the pruning level of the
# leaves minus 1 and up to 0

for (i in (n+2):nrow(dend)) {
   if ((dend[i,4] != dend[(i-1),4]) || is.na(dend[(i-1),4])){
        dend[i,3] <- dend[(i-1),3]-1}
      else { dend[i,3] <- dend[(i-1),3] }
}
dend[,3] <- dend[,3]-min(dend[,3])

dend <- dend[order(-node.id),]

# Write results table.
write.table(dend, file="path", sep=";", row.names=F)
于 2012-04-24T10:05:29.527 回答
1

有一个包与你想要的完全相反 - Labeltodendro ;-)

但说真的，你不能手动从hclust对象中提取元素（例如$merge, $height, $order）并从提取的元素中创建自定义表吗？

于 2012-04-10T16:46:28.887 回答