how to add add list element names as a variable in the output

Question

I have a list of data.frames looks like this:

$IB1Q
         V1          V2          V3          V4          V5          V6          V7          V8 
0.184011746 0.805070251 0.840738769 0.701748191 0.493042421 0.889981450 0.003934755 0.794561335 
         V9         V10         V11 
0.664718057 0.880357401 0.545795252 

$IBCA
       V1        V2        V3        V4        V5        V6        V7        V8        V9 
0.8719393 0.8177656 0.9205127 0.7200678 0.7456413 0.4643883 0.8006192 0.1430972 0.7622733 
      V10       V11 
0.4716331 0.8681162

and each element name (IB1Q, IBCA) conveys information. I want to row.bind them together and in my data.farme I want a variable for the list elemnt each observation originated in. So I want to get a data.frame with 2 columns (first column popID and second columns my proportions):

IB1Q   0.184011746
IB1Q   0.805070251
IB1Q   0.840738769
IB1Q   0.701748191
IB1Q   0.493042421
.
.
.
IBCA   0.4716331
IBCA   0.8681162

Data frames have only one row?

pogibas
– pogibas

2017-09-29 23:09:01 +00:00
Commented Sep 29, 2017 at 23:09 — pogibas
– pogibas, Commented Sep 29, 2017 at 23:09

r2evans · Accepted Answer · 2017-09-29 23:19:01Z

1

Here's one thought:

mylist <- list(
  IB1Q = as.data.frame(setNames(
    list(0.184011746, 0.805070251, 0.840738769, 0.701748191, 0.493042421, 0.889981450, 0.003934755, 0.794561335, 0.664718057, 0.880357401, 0.545795252),
    paste0("V", 1:11)
  )),
  IBCA = as.data.frame(setNames(
    list(0.8719393, 0.8177656, 0.9205127, 0.7200678, 0.7456413, 0.4643883, 0.8006192, 0.1430972, 0.7622733, 0.4716331, 0.8681162),
    paste0("V", 1:11)
  ))
)

do.call(rbind.data.frame,
        lapply(names(mylist), function(n) data.frame(K=n, V=unlist(mylist[[n]], use.names = FALSE)))
        )
#       K           V
# 1  IB1Q 0.184011746
# 2  IB1Q 0.805070251
# 3  IB1Q 0.840738769
# 4  IB1Q 0.701748191
# 5  IB1Q 0.493042421
# 6  IB1Q 0.889981450
# 7  IB1Q 0.003934755
# 8  IB1Q 0.794561335
# 9  IB1Q 0.664718057
# 10 IB1Q 0.880357401
# 11 IB1Q 0.545795252
# 12 IBCA 0.871939300
# 13 IBCA 0.817765600
# 14 IBCA 0.920512700
# 15 IBCA 0.720067800
# 16 IBCA 0.745641300
# 17 IBCA 0.464388300
# 18 IBCA 0.800619200
# 19 IBCA 0.143097200
# 20 IBCA 0.762273300
# 21 IBCA 0.471633100
# 22 IBCA 0.868116200

Since we're discarding the column names of the nested data.frames, I don't think the use of unlist is too harsh/lossy. (I use use.names=FALSE because I find row names to be unreliable in the long run ... feel free to keep them if you'd like.)

answered Sep 29, 2017 at 23:19

r2evans

167k8 gold badges92 silver badges176 bronze badges

Sign up to request clarification or add additional context in comments.

4 Comments

Anna1364 Over a year ago

I cannot do something like that, my list has more than 50 elements. For each element doing something like mylist <- list( IB1Q = as.data.frame(setNames( list(0.184011746, 0.805070251, 0.840738769, 0.701748191, 0.493042421, 0.889981450, 0.003934755, 0.794561335, 0.664718057, 0.880357401, 0.545795252), paste0("V", 1:11) is very time consumening

r2evans Over a year ago

But you don't need to do that ... since you did not provide data in a readily-reproducible form, that code is just to create a data object for the rest of code. You, the OP, get to skip that part and go to the gusto starting with do.call( and replacing mylist with whatever the name of your list of data.frames is.

Anna1364 Over a year ago

Yesss, thanks so much @r2evans. it did exactly what I wanted. Sorry I did not get your point first.

r2evans Over a year ago

(In the future, if you can provide code similar to my declaration of mylist within your question, it will make answering it faster and less likely to fall apart due to faulty assumptions. Better yet, if there is any doubt, provide similar code for desired output ... though there was no doubt in this instance.)

Rich Scriven · Accepted Answer · 2017-09-29 23:36:18Z

1

You don't need loops here. You can use rep() and unlist().

data.frame(
    popID = rep(names(mylist), lengths(mylist)),
    prop = unlist(mylist, use.names=FALSE) # use.names=FALSE for speed
)

lengths() is useful here because it gives us the number of columns per list element.

answered Sep 29, 2017 at 23:36

Rich Scriven

99.8k11 gold badges191 silver badges252 bronze badges

1 Comment

markdly Over a year ago

I like this answer better than mine. Very concise and clear (+1)

markdly · Accepted Answer · 2017-09-29 23:48:31Z

Here's yet another way it could be done (using mylist provided in @r2evans answer):

library(tidyverse)
bind_rows(mylist, .id = "popID") %>% 
  gather(key, value, -popID) %>% select(-key) %>% arrange(popID)
#>    popID       value
#> 1   IB1Q 0.184011746
#> 2   IB1Q 0.805070251
#> 3   IB1Q 0.840738769
#> 4   IB1Q 0.701748191
#> 5   IB1Q 0.493042421
#> 6   IB1Q 0.889981450
#> 7   IB1Q 0.003934755
#> 8   IB1Q 0.794561335
#> 9   IB1Q 0.664718057
#> 10  IB1Q 0.880357401
#> 11  IB1Q 0.545795252
#> 12  IBCA 0.871939300
#> 13  IBCA 0.817765600
#> 14  IBCA 0.920512700
#> 15  IBCA 0.720067800
#> 16  IBCA 0.745641300
#> 17  IBCA 0.464388300
#> 18  IBCA 0.800619200
#> 19  IBCA 0.143097200
#> 20  IBCA 0.762273300
#> 21  IBCA 0.471633100
#> 22  IBCA 0.868116200

Collectives™ on Stack Overflow

how to add add list element names as a variable in the output

3 Answers 3

4 Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

4 Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related