问题描述
在R中，我有一个由12个子列表组成的列表，每个列表本身都由5个子列表组成，如下所示:
In R, I have a list, composed of 12 sublists, each themselves composed of 5 subsublists as follow
在此示例中，我要为每个子列表提取信息"MSD".
In this example, I want to extract the info "MSD", for each of these sublists.
我可以使用
lapply(letters, '[[', "statistics")
效果很好.对于每个列表，它为我提供了子列表统计"中包含的所有值但是，我想将其下调一点，因为我对MSerror，Df等其他数据不感兴趣.仅MSD
That worked well. It gave me all the values contained in the sublist "statistics", for each listHowever, I want to go one level down of that, as I am not interested in the other data such as MSerror, Df, ..... Only MSD
我尝试了
lapply(letters, '[[', "statistics", "MSD")
和许多其他没有成功的人.
and many others without success.
如果我只想要第一个子列表，它将与
If I wanted only the first sublist, it will work with
letters[[1]][["statistics"]][["MSD"]]
但是，我必须这样做:
letters[[1]][["statistics"]][["MSD"]]
letters[[2]][["statistics"]][["MSD"]]
letters[[3]][["statistics"]][["MSD"]]
我想暂时避免.
感谢您的帮助.
推荐答案
我们可以使用lambda/匿名函数
We can use a lambda/anonymous function
lapply(letters, function(x) x[["statistics"]][["MSD"]])
此功能的好处在于，如果我们有多个嵌套元素，则不必调用 n lapply ，它应该更快
The benefit of this function is that if we have multiple nested elements, we don't have to call n lapply and should be faster
或使用 map
library(tidyverse)
map(letters, ~ .x[["statistics"]][["MSD"]])
此外，关于这样的说法，即如果列表中没有某些元素，
Also, regarding the claim that this wouldn't work if there are not some elements in the list,
set.seed(24)
lst1 <- replicate(3, list(statistics = list(MSD = rnorm(20))))
names(lst1)[2] <- "Hello"
这确实是行不通的.但是，它不适用于声称有效的解决方案.
It is true that it wouldn't work. However, it wouldn't work with the solution claimed to be work as well.
这篇关于列表的子集的子集的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！