Hi Elk,
I tried to test with another case where coconut is in different position in
the sublist.
mango <- list( y=c(4,5,3,2,1,8), bw=c(4,18,12,3,4,9),
nn=paste0("a",1:6),data.frame(coconut=1:6))
banana <- list(data.frame(coconut=c(1,2,18,16,15)), y=c(4,5,9,2,1),
bw=c(4,18,22,3,4), nn=paste0("a",1:5))
pineapple <- list(data.frame(coconut=c(4,6,9)), y=c(8,24,12),
bw=c(14,31,36), nn=paste0("a",1:3))
list5<-list(mango,banana,pineapple)
#Your code
fun<-max # or min, median etc
sapply(lapply(list5,"[[",1),fun)
[1] 8 18 9
# Here, first number should be 6, instead it did the max of y from mango.
#Then, I tried second solution of yours
sapply(lapply(lapply(list5,"[[",1),"[[","coconut"),fun)
#Error in FUN(X[[1L]], ...) : subscript out of bounds
#Third solution
sapply(lapply(list5,function(x)x[[1]][["coconut"]]),fun)
#Error in x[[1]][["coconut"]] : subscript out of bounds
#I tried with another way to come up with the result (not as elegant)
b3<-do.call("c",list5)
names(b3)[4]<-"mango"
names(b3)[5]<-"banana"
names(b3)[9]<-"pineapple"
b4<-data.frame(key=names(b3),value=unlist(lapply(b3,FUN=function(x)
max(x))))
key1<-c("mango","banana","pineapple")
b5<-subset(b4, b4$key%in% key1)
b5
# key value
#4 mango 6
#5 banana 18
#9 pineapple 9
b6<-within(b5,{value<-as.numeric(as.character(value))})
max(b6$value)
#[1] 18
A.K.
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