r - Understanding list behaviour -

i feel have understanding of data.frames , how work, aspects of lists confusing me.

here reproducible data start:

 list_a <- structure(list(`one` = structure(list(      words = c("a", "b","c", "d", "e", "f")), .names = "words", class = "data.frame", row.names = c(na,-6l)),       `two` = structure(list(words = c("a","s","t","z")), .names = "words", class = "data.frame", row.names = c(na, -4l))),      .names = c("one", "two")) 

this gives us:

list_a $one   words 1     2     b 3     c 4     d 5     e 6     f  $two   words 1     2     s 3     t 4     z 

now want loop through list return of results in data.frames.

list <- list()  for(i in list_a){list <- append(list, list_a$i$words)} 

this produces no results in list. neither does:

for(i in list_a){list <- append(list, list_a[[i]]$words)} error in list_a[[i]] : invalid subscript type 'list' 

i thought perhaps reason first loop didn't work was using list_a$i$words without defining correct names. tried:

for(i in names(list_a)){list <- append(list, list_a$i$words)} 

this still gives me list of length 0.

so not understand why attempts tried didnt give results expected, not know why using subscripts gave me error , figured out correct syntax:

for(i in list_a){list2 <- append(list2, i$words)}

however not know why works when using names method did not?

the arguments for expression in r consists of:

• lhs, iterator take each value of rhs
• in, language keyword
• rhs, vector, length of defines number of iterations occur.

when set first loop, rhs length 2 vector of type "list". on lhs have i 1 column data frame. asked $ extract "i" list_a, evaluated null. in second loop, rhs length 2 vector of type "character". same thing occurred.

$ not evaluate index. use [[ instead , answer expect in second loop.

# initialize list <- list() # loop (i in names(list_a)) {     list <- append(list, list_a[[i]]$words) } list # [[1]] # [1] "a" # # [[2]] # [1] "b" # ... 

as mentioned roland, appending expensive in r, each iteration creates new copy of object. here 1 alternative try:

# create data frame using of list_a,  # coerce character vector # coerce list as.list(unname(unlist(do.call(what = "rbind", args = list_a)))) 

note "data.frame" objects "list" objects "data.frame" class attribute applied. see same behaviour when working data.frames , $ unevaluated names lists. try this:

# print mtcars data.frame mtcars # set class attribute null class(mtcars) <- null # mtcars list :-) mtcars 

edit: $ , [[ operators, means functions can used in special way. can use them normal functions too, passing arguments in round brackets.

# $ function `$`(list_a, "one") #   words # 1     # 2     b # ... 

the behaviour of these functions different. [[ expects object interprets. $ expects element name tries find.

i <- "one" # $ function, there no element "i" `$`(list_a, i) # null # [[ function, , element "one" present `[[`(list_a, i) #   words # 1     # 2     b # ...