将R中两个或多个列表中的对应元素绑定

Question

I have 3 lists, each with 500 elements. Here for demonstrative purposes, I have 2 lists with 1 element each:

structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame")   

structure(list(timeseries = c(5, 6, 7), t = c(8, 9, 10)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame") 

My aim is to rbind the first element in list 1 with the first element in list 2 and 3. Then, the second element in list 1 with the second element in list 2 and 3. And so on.

In my example, I would end up with a list of this form

structure(list(timeseries = c(1,7,59,5, 6, 7), t = c(1,3,7,8, 9, 10)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame") 

How do I do this?

Thank you!

****EDIT*** Improved example of the intended outcome. I have a and b. I want to obtain C.

a<-list(structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"),
structure(list(timeseries = c(1, 7, 59), t = c(1, 3, 7)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"))


b<-list(structure(list(timeseries = c(2, 3, 5), t = c(2, 4, 6)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"),
        structure(list(timeseries = c(60, 70, 80), t = c(20, 30, 40)), .Names = c("timeseries", "t"), row.names = c(NA, 3L), class = "data.frame"))


c<-list(structure(list(timeseries = c(1, 7, 59, 2,3, 5), t = c(1, 3, 7, 2, 4, 6)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame"),
        structure(list(timeseries = c(1, 7, 59, 60, 70, 80), t = c(1, 3, 7, 20, 30, 40)), .Names = c("timeseries", "t"), row.names = c(NA, 6L), class = "data.frame"))

Answer 1

Assuming length of a and b is the same we can do

lapply(seq_along(a), function(x) rbind(a[[x]], b[[x]]))

#[[1]]
#  timeseries t
#1          1 1
#2          7 3
#3         59 7
#4          2 2
#5          3 4
#6          5 6

#[[2]]
#  timeseries  t
#1          1  1
#2          7  3
#3         59  7
#4         60 20
#5         70 30
#6         80 40

seq_along generates sequence from 1 to length of the object. If you do

seq_along(a) #you would get output as
#[1] 1 2

as length(a) is 2. So we rbind the dataframe one by one rbind(a[[1]], b[[1]]) first, then rbind(a[[2]], b[[2]]) and so on. lapply ensures the final output is a list.

Answer 2

Just try the map2 function :

purrr::map2(a,b,rbind) -> d
identical(c,d)
# [1] TRUE