如何通过节点或叶子中的标签折叠系统发育树中的分支?

时间:2015-12-21 20:39:58

标签: r plot phylogeny ape-phylo

我为一个蛋白质家族建立了一个系统发育树,可以将其分成不同的组,按照受体类型或反应类型对每个蛋白质进行分类。树中的节点被标记为受体类型。

在系统发育树中,我可以看到属于相同组或受体类型的蛋白质在同一分支中聚集在一起。所以我想要折叠这些具有共同标签的分支,并按给定的关键字列表对它们进行分组。

命令是这样的:

./ collapse_tree_by_label -f phylogenetic_tree.newick -l list_of_labels_to_collapse.txt -o collapsed_tree.eps(或pdf)

我的list_of_labels_to_collapse.txt会是这样的: 一个 乙 C d

我的纽约树将是这样的: (A_1:0.05,A_2:0.03,A_3:0.2,A_4:0.1):0.9;(((B_1:0.05,B_2:0.02,B_3:0.04):0.6,(C_1:0.6,C_2:0.08):0.7) :0.5,(D_1:0.3,D_2:0.4,D_3:0.5,D_4:0.7,D_5:0.4):1.2)

没有折叠的输出图像是这样的: http://i.stack.imgur.com/pHkoQ.png

输出图像折叠应该是这样的(collapsed_tree.eps): http://i.stack.imgur.com/TLXd0.png

三角形的宽度应代表分支长度,三角形的高度必须代表分支中的节点数。

我一直在和#34; ape"在R.中我能够绘制一个系统发育树,但我仍然无法弄清楚如何通过标签中的关键字来折叠分支:

require("ape")

这将加载树:

cat("((A_1:0.05,A_2:0.03,A_3:0.2,A_4:0.1):0.9,(((B_1:0.05,B_2:0.02,B_3:0.04):0.6,(C_1:0.6,C_2:0.08):0.7):0.5,(D_1:0.3,D_2:0.4,D_3:0.5,D_4:0.7,D_5:0.4):1.2):0.5);", file = "ex.tre", sep = "\n")
tree.test <- read.tree("ex.tre")

这里应该是要崩溃的代码

这将绘制树:

plot(tree.test)

4 个答案:

答案 0 :(得分:4)

存储在R中的树已将提示存储为多边形。这只是用三角形代表多边形来绘制树的问题。

ape中没有任何功能可以做到这一点,我知道,但是如果你把绘图功能弄得一点点就可以把它拉下来

# Step 1: make edges for descendent nodes invisible in plot:
groups <- c("A", "B", "C", "D")
group_edges <- numeric(0)
for(group in groups){
  group_edges <- c(group_edges,getMRCA(tree.test,tree.test$tip.label[grepl(group, tree.test$tip.label)]))
}
edge.width <- rep(1, nrow(tree.test$edge))
edge.width[tree.test$edge[,1] %in% group_edges ] <- 0


# Step 2: plot the tree with the hidden edges
plot(tree.test, show.tip.label = F, edge.width = edge.width)

# Step 3: add triangles
add_polytomy_triangle <- function(phy, group){
  root <- length(phy$tip.label)+1
  group_node_labels <- phy$tip.label[grepl(group, phy$tip.label)]
  group_nodes <- which(phy$tip.label %in% group_node_labels)
  group_mrca <- getMRCA(phy,group_nodes)

  tip_coord1 <- c(dist.nodes(phy)[root, group_nodes[1]], group_nodes[1])
  tip_coord2 <- c(dist.nodes(phy)[root, group_nodes[1]], group_nodes[length(group_nodes)])
  node_coord <- c(dist.nodes(phy)[root, group_mrca], mean(c(tip_coord1[2], tip_coord2[2])))

  xcoords <- c(tip_coord1[1], tip_coord2[1], node_coord[1])
  ycoords <- c(tip_coord1[2], tip_coord2[2], node_coord[2])
  polygon(xcoords, ycoords)
}

然后你只需循环遍历组添加三角形

for(group in groups){
  add_polytomy_triangle(tree.test, group)
}

答案 1 :(得分:1)

我认为剧本终于做了我想做的事。 从@CactusWoman提供的答案中,我稍微更改了代码,因此脚本将尝试找到代表与我的搜索模式匹配的最大分支的MRCA。这解决了不合并非多分支分支或折叠整个树的问题,因为一个匹配节点错误地在正确分支之外。

此外,我添加了一个参数,表示给定分支中模式丰度比的限制,因此我们可以选择和折叠/分组至少90%的提示与搜索模式匹配的分支,例如。

library(geiger)
library(phylobase)
library(ape)

#functions
find_best_mrca <- function(phy, group, threshold){

     group_matches <- phy$tip.label[grepl(group, phy$tip.label, ignore.case=TRUE)]
     group_mrca <- getMRCA(phy,phy$tip.label[grepl(group, phy$tip.label, ignore.case=TRUE)])
     group_leaves <- tips(phy, group_mrca)
     match_ratio <- length(group_matches)/length(group_leaves)

      if( match_ratio < threshold){

           #start searching for children nodes that have more than 95% of descendants matching to the search pattern
           mrca_children <- descendants(as(phy,"phylo4"), group_mrca, type="all")
           i <- 1
           new_ratios <- NULL
           nleaves <- NULL
           names(mrca_children) <- NULL

           for(new_mrca in mrca_children){
                child_leaves <- tips(tree.test, new_mrca)
                child_matches <- grep(group, child_leaves, ignore.case=TRUE)
                new_ratios[i] <- length(child_matches)/length(child_leaves)
                nleaves[i] <- length(tips(phy, new_mrca))
                i <- i+1
           }



           match_result <- data.frame(mrca_children, new_ratios, nleaves)


           match_result_sorted <- match_result[order(-match_result$nleaves,match_result$new_ratios),]
           found <- numeric(0);

           print(match_result_sorted)

           for(line in 1:nrow(match_result_sorted)){
                 if(match_result_sorted$ new_ratios[line]>=threshold){
                     return(match_result_sorted$mrca_children[line])
                     found <- 1
                 }

           }

           if(found==0){return(found)}
      }else{return(group_mrca)}




}

add_triangle <- function(phy, group,phylo_plot){

     group_node_labels <- phy$tip.label[grepl(group, phy$tip.label)]
     group_mrca <- getMRCA(phy,group_node_labels)
     group_nodes <- descendants(as(tree.test,"phylo4"), group_mrca, type="tips")
     names(group_nodes) <- NULL

     x<-phylo_plot$xx
     y<-phylo_plot$yy


     x1 <- max(x[group_nodes])
     x2 <-max(x[group_nodes])
     x3 <- x[group_mrca]

     y1 <- min(y[group_nodes])
     y2 <- max(y[group_nodes])
     y3 <-  y[group_mrca]

     xcoords <- c(x1,x2,x3)
     ycoords <- c(y1,y2,y3)

     polygon(xcoords, ycoords)

     return(c(x2,y3))

}



#main

  cat("((A_1:0.05,E_2:0.03,A_3:0.2,A_4:0.1,A_5:0.1,A_6:0.1,A_7:0.35,A_8:0.4,A_9:01,A_10:0.2):0.9,((((B_1:0.05,B_2:0.05):0.5,B_3:0.02,B_4:0.04):0.6,(C_1:0.6,C_2:0.08):0.7):0.5,(D_1:0.3,D_2:0.4,D_3:0.5,D_4:0.7,D_5:0.4):1.2):0.5);", file = "ex.tre", sep = "\n")
tree.test <- read.tree("ex.tre")


# Step 1: Find the best MRCA that matches to the keywords or search patten

groups <- c("A", "B|C", "D")
group_labels <- groups

group_edges <- numeric(0)
edge.width <- rep(1, nrow(tree.test$edge))
count <- 1


for(group in groups){

    best_mrca <- find_best_mrca(tree.test, group, 0.90)

    group_leaves <- tips(tree.test, best_mrca)

    groups[count] <- paste(group_leaves, collapse="|")
    group_edges <- c(group_edges,best_mrca)

    #Step2: Remove the edges of the branches that will be collapsed, so they become invisible
    edge.width[tree.test$edge[,1] %in% c(group_edges[count],descendants(as(tree.test,"phylo4"), group_edges[count], type="all")) ] <- 0
    count = count +1

}


#Step 3: plot the tree hiding the branches that will be collapsed/grouped

last_plot.phylo <- plot(tree.test, show.tip.label = F, edge.width = edge.width)

#And save a copy of the plot so we can extract the xy coordinates of the nodes
#To get the x & y coordinates of a plotted tree created using plot.phylo
#or plotTree, we can steal from inside tiplabels:
last_phylo_plot<-get("last_plot.phylo",envir=.PlotPhyloEnv)

#Step 4: Add triangles and labels to the collapsed nodes
for(i in 1:length(groups)){

  text_coords <- add_triangle(tree.test, groups[i],last_phylo_plot)

  text(text_coords[1],text_coords[2],labels=group_labels[i], pos=4)

}

答案 2 :(得分:1)

我也一直在寻找这种工具,不仅仅是为了折叠分类组,而是为了根据数值支持值折叠内部节点。

ape包中的di2multi函数可以将节点折叠为多边形,但它目前只能通过分支长度阈值来实现。 这是一个粗略的适应,允许通过节点支持值阈值折叠(默认阈值= 0.5)。

使用风险自负,但它适用于我的根贝叶树。

di2multi4node <- function (phy, tol = 0.5) 
  # Adapted di2multi function from the ape package to plot polytomies
  # based on numeric node support values
  # (di2multi does this based on edge lengths)
  # Needs adjustment for unrooted trees as currently skips the first edge
{
  if (is.null(phy$edge.length)) 
    stop("the tree has no branch length")
  if (is.na(as.numeric(phy$node.label[2])))
    stop("node labels can't be converted to numeric values")
  if (is.null(phy$node.label))
    stop("the tree has no node labels")
  ind <- which(phy$edge[, 2] > length(phy$tip.label))[as.numeric(phy$node.label[2:length(phy$node.label)]) < tol]
  n <- length(ind)
  if (!n) 
    return(phy)
  foo <- function(ancestor, des2del) {
    wh <- which(phy$edge[, 1] == des2del)
    for (k in wh) {
      if (phy$edge[k, 2] %in% node2del) 
        foo(ancestor, phy$edge[k, 2])
      else phy$edge[k, 1] <<- ancestor
    }
  }
  node2del <- phy$edge[ind, 2]
  anc <- phy$edge[ind, 1]
  for (i in 1:n) {
    if (anc[i] %in% node2del) 
      next
    foo(anc[i], node2del[i])
  }
  phy$edge <- phy$edge[-ind, ]
  phy$edge.length <- phy$edge.length[-ind]
  phy$Nnode <- phy$Nnode - n
  sel <- phy$edge > min(node2del)
  for (i in which(sel)) phy$edge[i] <- phy$edge[i] - sum(node2del < 
                                                           phy$edge[i])
  if (!is.null(phy$node.label)) 
    phy$node.label <- phy$node.label[-(node2del - length(phy$tip.label))]
  phy
}

答案 3 :(得分:1)

这是我基于phytools::phylo.toBackbone函数的答案, 见http://blog.phytools.org/2013/09/even-more-on-plotting-subtrees-as.htmlhttp://blog.phytools.org/2013/10/finding-edge-lengths-of-all-terminal.html。首先,在代码末尾加载函数。

library(ape)
library(phytools)  #phylo.toBackbone
library(phangorn) 

cat("((A_1:0.05,E_2:0.03,A_3:0.2,A_4:0.1,A_5:0.1,A_6:0.1,A_7:0.35,A_8:0.4,A_9:01,A_10:0.2):0.9,((((B_1:0.05,B_2:0.05):0.5,B_3:0.02,B_4:0.04):0.6,(C_1:0.6,C_2:0.08):0.7):0.5,(D_1:0.3,D_2:0.4,D_3:0.5,D_4:0.7,D_5:0.4):1.2):0.5);", file = "ex.tre", sep = "\n")

phy <- read.tree("ex.tre")
groups <- c("A", "B|C", "D") 

backboneoftree<-makebackbone(groups,phy)
#   tip.label clade.label  N     depth
# 1       A_1           A 10 0.2481818
# 2       B_1         B|C  6 0.9400000
# 3       D_1           D  5 0.4600000

par(mfrow=c(2,2), mar=c(0,2,2,0) )
plot(phy, main="Complete tree" )
plot(backboneoftree)

makebackbone<-function(groupings,phy){ 
  listofspecies<-phy$tip.label
  listtopreserve<-list()
  lengthofclades<-list()
  meandistnode<-list()
  newedgelengths<-list()
  for (i in 1:length(groupings)){
    groupings<-groups
    bestmrca<-getMRCA(phy,grep(groupings[i], phy$tip.label) )
    mrcatips<-phy$tip.label[unlist(phangorn::Descendants(phy,bestmrca, type="tips") )]
    listtopreserve[i]<- mrcatips[1]
    meandistnode[i]<- mean(dist.nodes(phy)[unlist(lapply(mrcatips,  
    function(x) grep(x, phy$tip.label) ) ),bestmrca] )
    lengthofclades[i]<-length(mrcatips)
    provtree<-drop.tip(phy,mrcatips, trim.internal=F, subtree = T)
    n3<-length(provtree$tip.label)
    newedgelengths[i]<-setNames(provtree$edge.length[sapply(1:n3,function(x,y) 
      which(y==x),y=provtree$edge[,2])],
      provtree$tip.label)[provtree$tip.label[grep("tips",provtree$tip.label)] ]
  }  
  newtree<-drop.tip(phy,setdiff(listofspecies,unlist(listtopreserve)), 
                    trim.internal = T)
  n<-length(newtree$tip.label)
  newtree$edge.length[sapply(1:n,function(x,y) 
    which(y==x),y=newtree$edge[,2])]<-unlist(newedgelengths)+unlist(meandistnode)
  trans<-data.frame(tip.label=newtree$tip.label,clade.label=groupings,
                    N=unlist(lengthofclades), depth=unlist(meandistnode) )
  rownames(trans)<-NULL
  print(trans)
  backboneoftree<-phytools::phylo.toBackbone(newtree,trans)
  return(backboneoftree)
}

enter image description here

编辑:我没有试过这个,但它可能是另一个答案:“脚本和函数来转换树的尖端分支,即厚度或三角形,两者的宽度与某些参数相关(例如,分支的种类数量)(tip.branches.R)“ http://www.sysbot.biologie.uni-muenchen.de/en/people/cusimano/use_r.html http://www.sysbot.biologie.uni-muenchen.de/en/people/cusimano/tip.branches.R