stack the data frames on top of each other and return one single data frame

Question

I'm trying to write a function (stackDataInList) to stack the data frame together, but having trouble to do so, some help is very appreciated

>datList
[[1]]
ID value
1 a 0.9101742
2 b 0.3841854
3 c 1.6821761
[[2]]
ID value
1 d -0.6357365
2 e -0.4616447
[[3]]
ID value
1 f 1.4322822
2 g -0.6506964
3 h -0.2073807
4 i -0.3928079
[[4]]
ID value
1 j -0.3199929
2 k -0.2791133
3 l 0.4941883

dataList looks like it

So basically I need to create a function, and when I will call it I'll get results below:

>stackDataInList(datList[1])
ID value
1 a 0.9101742
2 b 0.3841854
3 c 1.6821761
OR:
>stackDataInList(datList[c(1,3,4)])
ID value
1 a 0.9101742
2 b 0.3841854
3 c 1.6821761
4 f 1.4322822
5 g -0.6506964
6 h -0.2073807
7 i -0.3928079
8 j -0.3199929
9 k -0.2791133
10 l 0.4941883

`do.call('rbind', datList)` probably should do the work, or you can use `dplyr::bind_rows(datList)` — PKumar, Jun 23 '19 at 05:21

Katia · Answer 1 · 2019-06-23T05:40:15.397

2

There are a number of ways you can do this. The fastest way (especially for large data frames) is to use rbindlist() function for the data.table package:

library(data.table)
datList <- list( data.frame(ID=c("a","b","c"), value=c(0.9101742,0.3841854, 1.6821761)),
             data.frame(ID=c("d","e"), value=c(-0.6357365,-0.4616447)),
             data.frame(ID=c("f","g","e"), value=c(1.4322822,-0.6506964,-0.2073807)))


rbindlist(datList)
#   ID      value
#1:  a  0.9101742
#2:  b  0.3841854
#3:  c  1.6821761
#4:  d -0.6357365
#5:  e -0.4616447
...

Or subsetting the list:

rbindlist(datList[c(1,3)])
#    ID      value
# 1:  a  0.9101742
# 2:  b  0.3841854
# 3:  c  1.6821761
# 4:  f  1.4322822
# 5:  g -0.6506964
# 6:  e -0.2073807

edited Jun 23 '19 at 05:40

answered Jun 23 '19 at 05:29

Katia

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@BonaguidiLab It is not clear what is the meaning of the argument `a` in your function (you list it as a second argument, but then you are using it as a local variable in your function). In addition to this your approach to start with an empty list and then use `rbind()` function to add new rows will be very slow. It is still not clear why the above answer does not work for you. Are you trying to implement this function yourself? – Katia Jun 23 '19 at 06:13

jay.sf · Answer 2 · 2019-06-23T17:09:52.650

As @PKumar pointed out in comments, in base R we may use do.call(rbind). Just put it into a function.

stackDataInList <- function(x) {
  return(do.call(rbind, x))
}

stackDataInList(L[1])  
#   ID     value
# 1  a 0.9101742
# 2  b 0.3841854
# 3  c 1.6821761

stackDataInList(L[c(1, 3, 4)])  
#    ID      value
# 1   a  0.9101742
# 2   b  0.3841854
# 3   c  1.6821761
# 4   f  1.4322822
# 5   g -0.6506964
# 6   h -0.2073807
# 7   i -0.3928079
# 8   j -0.3199929
# 9   k -0.2791133
# 10  l  0.4941883

Or use stack and mapply:

stackDataInList2 <- function(x) setNames(cbind.data.frame(stack(mapply(`[`, x, 1))[[1]],
                                                          stack(mapply(`[`, x, 2))[[1]]), 
                                         names(x[[1]]))

stackDataInList2(L[1])  
# ID     value
# 1  a 0.9101742
# 2  b 0.3841854
# 3  c 1.6821761
stackDataInList2(L[c(1, 3, 4)])  
# ID      value
# 1   a  0.9101742
# 2   b  0.3841854
# 3   c  1.6821761
# 4   f  1.4322822
# 5   g -0.6506964
# 6   h -0.2073807
# 7   i -0.3928079
# 8   j -0.3199929
# 9   k -0.2791133
# 10  l  0.4941883

Also Reduce is an option, found here

stackDataInList3 <- function(x) Reduce(rbind, x)
stackDataInList3(L[1])  
#   ID     value
# 1  a 0.9101742
# 2  b 0.3841854
# 3  c 1.6821761
stackDataInList3(L[c(1, 3, 4)]) 
#    ID      value
# 1   a  0.9101742
# 2   b  0.3841854
# 3   c  1.6821761
# 4   f  1.4322822
# 5   g -0.6506964
# 6   h -0.2073807
# 7   i -0.3928079
# 8   j -0.3199929
# 9   k -0.2791133
# 10  l  0.4941883

Data

L <- list(structure(list(ID = c("a", "b", "c"), value = c(0.9101742, 
0.3841854, 1.6821761)), row.names = c(NA, -3L), class = "data.frame"), 
    structure(list(ID = c("d", "e"), value = c(-0.6357365, -0.4616447
    )), row.names = c(NA, -2L), class = "data.frame"), structure(list(
        ID = c("f", "g", "h", "i"), value = c(1.4322822, -0.6506964, 
        -0.2073807, -0.3928079)), row.names = c(NA, -4L), class = "data.frame"), 
    structure(list(ID = c("j", "k", "l"), value = c(-0.3199929, 
    -0.2791133, 0.4941883)), row.names = c(NA, -3L), class = "data.frame"))

Do you know what I can use instead of ~do.call? Maybe ~apply, ~tapply function? — Bonaguidi Lab, Jun 23 '19 at 15:07
@BonaguidiLab To what purpose, what's your problem with `do.call`? — jay.sf, Jun 23 '19 at 15:11
Yes do.call works perfectly, but I need to provide a alternative solution, so in search for that.. — Bonaguidi Lab, Jun 23 '19 at 15:16
@BonaguidiLab I see no error, your data may differ from mine, add `dput(datList)` to your question. — jay.sf, Jun 23 '19 at 16:20

stack the data frames on top of each other and return one single data frame

2 Answers2