使用 apply() 函数跨子列表工作 [英] Working across sub-lists with apply() functions
问题描述
我正在尝试引导 7 个人饮食项目的发生比例并计算 sd()
I am trying to the bootstrap the proportional occurrence of diet items for 7 individuals and calculate a sd()
假设菜单上有 9 个猎物.
Lets say there are 9 prey items on the menu.
Diet <- c("Beaver","Bird", "Bobcat","Coyote", "Deer", "Elk",
"Porcupine", "Raccoon", "SmMamm")
而且这些猎物被同一物种的 7 个不同个体吃掉
And that these prey items are eaten by 7 different individuals of the same species
Inds <- c("P01", "P02", "P03", "P04", "P05", "P06", "P07")
我的目标是引导每个人每个饮食项目的出现比例.
下面的循环为每个个体生成五种饮食(每种饮食包含 N = 20 次喂养)，这些饮食被替换采样.数据存储为个体列表，每个个体包含一个样本饮食列表.
My goal is the bootstrap the proportional occurrence of each diet item for each individual.
The loop below generates five diets for each individual (each diet containing N = 20 feedings) that were sampled with replacement. The data are stored as a list of the individuals, each of which contains a list of the sample diets.
BootIndDiet <- list()
IndTotboot <- list()
for(i in Inds){
for(j in 1:5){
BootIndDiet[[j]] <- prop.table(table(sample(Diet, 20 ,replace = T)))
}
IndTotboot[[i]] <- BootIndDiet
}
下面我已经包括了个体 P07 的前两种饮食作为循环结果的示例
Below I have included the first two diets of individual P07 as an example of the loop results
$P07
$P07[[1]]
Beaver Bird Bobcat Deer Elk
0.05 0.15 0.20 0.10 0.15
Porcupine Raccoon SmMamm
0.15 0.15 0.05
$P07[[2]]
Beaver Bird Bobcat Coyote Deer
0.15 0.10 0.20 0.05 0.05
Elk Porcupine Raccoon SmMamm
0.05 0.20 0.10 0.10
然后我想计算每个物种对每个个体的比例的 sd().模棱两可，对于每个个体 (P01 - P07)，我想要 sd() 在 5 种饮食中每个猎物物种的出现比例.
I then want to calculate the sd() of the proportional of each species for each individual. Equivocally, for each individual (P01 - P07) I want the sd() of the proportional occurrence of each prey species across the 5 diets.
当我上面的循环运行时，我怀疑有更好的方法(可能使用 boot() 函数)来避免列表......
While my loop above runs, I suspect there is a better way (possibly using the boot() function) that avoids lists...
虽然我在这里只为每个人包含了 5 个样本(引导程序)，但我希望生成 10000 个.
While I have only included 5 samples (bootstraps) for each individual here, I hope to generate 10000.
非常感谢有关不同策略或如何在子列表中应用 sd() 的建议.
Suggestions on a different strategy or how to apply sd() across sub-lists is greatly appreciated.
推荐答案
我会尝试以这种方式获取数组(而不是嵌套列表):
I'd try to obtain an array (instead of a nested list) in this way:
IndTotboot <-array(replicate(5*length(Inds),prop.table(table(sample(as.factor(Diet), 20 ,replace = T))),simplify=T), dim=c(length(Diet),5,length(Inds)), dimnames=list(Diet,NULL,Inds))
使用replicate，您可以执行给定次数的表达式并将结果存储为数组(如果可能).我在 Diet 之前添加了一个 as.factor 以确保表格跟踪每个饮食(即使是频率为 0 的饮食).
With replicate you can execute an expression a given number of times and store the result as an array (if possible). I added an as.factor before Diet to make sure that the table takes trace of every Diet (even the ones with a 0 frequency).
获得的IndTotboot对象是一个3维数组，其中第一个索引表示Diet，第二个索引表示引导复制，第三个索引表示Inds.从那里您可以以标准方式使用 apply.
The IndTotboot object obtained is a 3-dimensional array where the first index indicates the Diet, the second the bootstrap replications and the third the Inds. From there you can use apply in the standard way.
如果你尝试 str(IndTotboot) 你会得到:
If you try str(IndTotboot) you get:
> str(IndTotboot)
num [1:9, 1:5, 1:7] 0.1 0.15 0.15 0.1 0.1 0.1 0.15 0.05 0.1 0.15 ...
- attr(*, "dimnames")=List of 3
..$ : chr [1:9] "Beaver" "Bird" "Bobcat" "Coyote" ...
..$ : NULL
..$ : chr [1:7] "P01" "P02" "P03" "P04" ...
第一行是最重要的.它说 num [1:9, 1:5, 1:7]，这意味着一个 9x5x7 数组.其余表示dimnames，维度的名称，是一个列表.它们是矩阵的 rownames 和 colnames 的概括.
The first line is the most important. It says num [1:9, 1:5, 1:7], which means a 9x5x7 array. The rest indicates the dimnames, the names of the dimensions, which is a list. They are the generalization of the rownames and the colnames for a matrix.
现在，要获取每个 Diet 和 Inds 的 sd，您只需使用 apply:
Now, to obtain the sd for every Diet and Inds you just use apply:
apply(IndTotboot,MARGIN=c(1,3),sd)
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