是否有可能从类似左边的分支系统树中生成0/1
字符矩阵,如右下方所示。矩阵中的1
表示存在将进化枝团结在一起的共享字符。
这段代码会生成漂亮的随机树,但我不知道从哪里开始将结果转换为字符矩阵。
library(ape) # Other package solutions are acceptable
forest <- rmtree(N = 2, n = 10, br = NULL)
plot(forest)
为清楚起见,我可以使用以下代码生成随机矩阵,然后绘制树。
library(ape)
library(phangorn)
ntaxa <- 10
nchar <- ntaxa - 1
char_mat <- array(0, dim = c(ntaxa, ntaxa - 1))
for (i in 1:nchar) {
char_mat[,i] <- replace(char_mat[,i], seq(1, (ntaxa+1)-i), 1)
}
char_mat <- char_mat[sample.int(nrow(char_mat)), # Shuffle rows
sample.int(ncol(char_mat))] # and cols
# Ensure all branch lengths > 0
dist_mat <- dist.gene(char_mat) + 0.5
upgma_tree <- upgma(dist_mat)
plot.phylo(upgma_tree, "phylo")
我想要的是生成随机树,然后从这些树中创建矩阵。 This solution的矩阵类型不正确。
为清楚起见进行编辑:我正在生成二进制字符矩阵,学生可以使用它们使用简单的简约来绘制系统树。 1
字符表示将分类单元合并为进化枝的同源性。因此,所有行必须共享一个字符(一列中所有行上的1
),并且某些字符必须仅由两个分类单元共享。 (我打折autapomorphies。)
示例:
答案 0 :(得分:2)
您可以直接看一下rTraitDisc
中的ape
函数:
library(ape)
## You'll need to simulate branch length!
forest <- rmtree(N = 2, n = 10)
## Generate on equal rate model character
(one_character <- rTraitDisc(forest[[1]], type = "ER", states = c(0,1)))
# t10 t7 t5 t9 t1 t4 t2 t8 t3 t6
# 0 0 0 1 0 0 0 0 0 0
# Levels: 0 1
## Generate a matrix of ten characters
(replicate(10, rTraitDisc(forest[[1]], type = "ER", states = c(0,1))))
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# t10 "0" "0" "0" "0" "1" "0" "0" "0" "0" "0"
# t7 "0" "0" "0" "0" "1" "0" "0" "0" "0" "0"
# t5 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t9 "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"
# t1 "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"
# t4 "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"
# t2 "0" "0" "1" "0" "0" "0" "0" "0" "0" "0"
# t8 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t3 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t6 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
要将其应用于多棵树,最好的方法是创建一个lapply函数,如下所示:
## Lapply wrapper function
generate.characters <- function(tree) {
return(replicate(10, rTraitDisc(tree, type = "ER", states = c(0,1))))
}
## Generate 10 character matrices for each tree
lapply(forest, generate.characters)
# [[1]]
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# t10 "0" "0" "0" "1" "0" "0" "0" "0" "0" "0"
# t7 "0" "0" "0" "1" "0" "0" "0" "0" "0" "0"
# t5 "0" "0" "0" "1" "0" "0" "0" "0" "0" "0"
# t9 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t1 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t4 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t2 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t8 "0" "0" "0" "1" "0" "1" "0" "0" "0" "1"
# t3 "0" "0" "0" "0" "0" "1" "0" "0" "0" "0"
# t6 "0" "0" "0" "0" "0" "1" "0" "0" "0" "0"
# [[2]]
# [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
# t7 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t9 "1" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t5 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t2 "0" "0" "0" "0" "0" "0" "0" "0" "0" "0"
# t4 "0" "1" "0" "0" "1" "0" "0" "0" "0" "0"
# t6 "0" "1" "0" "0" "1" "0" "0" "0" "0" "0"
# t10 "0" "1" "1" "0" "1" "1" "0" "0" "0" "1"
# t8 "0" "1" "1" "0" "1" "0" "0" "0" "0" "0"
# t3 "0" "1" "0" "0" "0" "0" "0" "0" "0" "0"
# t1 "0" "1" "0" "0" "0" "0" "0" "0" "0" "0"
另一种选择是使用sim.morpho
包中的dispRity
。该函数重用了rTraitDisc
函数,但是实现了更多模型,并且降低了作为采样分布的速率。它还允许字符看起来更“逼真”,而无需太多不变的数据,并确保生成的字符“看起来”像真实的形态字符(例如具有适当数量的同质体等)。
library(dispRity)
## You're first tree
tree <- forest[[1]]
## Setting up the parameters
my_rates = c(rgamma, rate = 10, shape = 5)
my_substitutions = c(runif, 2, 2)
## HKY binary (15*50)
matrixHKY <- sim.morpho(tree, characters = 50, model = "HKY",
rates = my_rates, substitution = my_substitutions)
## Mk matrix (15*50) (for Mkv models)
matrixMk <- sim.morpho(tree, characters = 50, model = "ER", rates = my_rates)
## Mk invariant matrix (15*50) (for Mk models)
matrixMk <- sim.morpho(tree, characters = 50, model = "ER", rates = my_rates,
invariant = FALSE)
## MIXED model invariant matrix (15*50)
matrixMixed <- sim.morpho(tree, characters = 50, model = "MIXED",
rates = my_rates, substitution = my_substitutions, invariant = FALSE,
verbose = TRUE)
我建议您阅读sim.morpho
函数,以获取有关模型工作方式的适当参考,或阅读dispRity package manual中的相关部分。
答案 1 :(得分:0)
我想出了如何使用phangorn包中的Descendants
来制作矩阵。我仍然必须使用合适的节点标签对其进行调整,以匹配原始问题中的示例矩阵,但是框架在那里。
library(ape)
library(phangorn)
ntaxa <- 8
nchar <- ntaxa - 1
tree <- rtree(ntaxa, br = NULL)
# Gets descendants, but removes the first ntaxa elements,
# which are the individual tips
desc <- phangorn::Descendants(tree)[-seq(1, ntaxa)]
char_mat <- array(0, dim = c(ntaxa, nchar))
for (i in 1:nchar) {
char_mat[,i] <- replace(char_mat[,i], y <- desc[[i]], 1)
}
rownames(char_mat) <- tree$tip.label
char_mat
#> [,1] [,2] [,3] [,4] [,5] [,6] [,7]
#> t6 1 1 0 0 0 0 0
#> t3 1 1 1 0 0 0 0
#> t7 1 1 1 1 0 0 0
#> t2 1 1 1 1 1 0 0
#> t5 1 1 1 1 1 0 0
#> t1 1 0 0 0 0 1 1
#> t8 1 0 0 0 0 1 1
#> t4 1 0 0 0 0 1 0
plot(tree)
由reprex package(v0.2.1)于2019-01-28创建