Stack a named Date list to data.frame

Question

I am doing a school project of data cleaning with tidyverse package. Now I get a list output from purrr::map() like this:

(mylist <- list(A = as.Date(sample(1e3:1e4, 4), origin = "1960-01-01"),
                B = as.Date(sample(1e3:1e4, 2), origin = "1960-01-01"),
                C = as.Date(sample(1e3:1e4, 3), origin = "1960-01-01")))


$A
[1] "1970-06-12" "1984-05-28" "1967-06-28" "1982-12-14"

$B
[1] "1966-02-04" "1967-02-21"

$C
[1] "1977-07-19" "1968-03-11" "1964-02-13"

I want to stack them to:

df <- data.frame(Value = reduce(mylist, c))
df$Class <- rep(names(mylist), sapply(mylist, length))
df

       Value Class
1 1970-06-12     A
2 1984-05-28     A
3 1967-06-28     A
4 1982-12-14     A
5 1966-02-04     B
6 1967-02-21     B
7 1977-07-19     C
8 1968-03-11     C
9 1964-02-13     C

Note1: The length of every cell is different and the values in the list are Date class actually.
Note2: stack(mylist) doesn't work in Date list.

Are there any methods to achieve it efficiently with functions in tidyverse or other packages?

I edit my question. `stack()` doesn't work in Date list. – Darren Tsai Jan 04 '19 at 09:43 — Darren Tsai, Jan 04 '19 at 09:43

akrun · Accepted Answer · 2019-01-04T18:41:30.850

We can use stack

stack(mylist)
#   values ind
#1      1   A
#2      2   A
#3      3   A
#4      4   B
#5      5   B
#6      6   B
#7      7   B
#8      8   C
#9      9   C

Update

For the updated question, either melt from reshape2 or data.table can be used

library(reshape2)
melt(mylist)
#        value L1
#1 3658-09-23  A
#2 2390-06-01  A
#3 2744-01-09  A
#4 2432-02-21  A
#5 4077-11-13  B
#6 4022-11-13  B
#7 3923-11-19  C
#8 2836-08-20  C
#9 3411-01-23  C

Or using tidyverse

library(tidyverse)
enframe(mylist) %>% 
     unnest
# A tibble: 9 x 2
#  name  value     
#  <chr> <date>    
#1 A     3658-09-23
#2 A     2390-06-01
#3 A     2744-01-09
#4 A     2432-02-21
#5 B     4077-11-13
#6 B     4022-11-13
#7 C     3923-11-19
#8 C     2836-08-20
#9 C     3411-01-23

Benchmarks

Here are some benchmarks using the four solutions posted

mylist2 <- rep(mylist, 1e4) #slightly bigger list

system.time({data.table::melt(mylist2)}) #Sotos solution
#user  system elapsed 
#  3.432   0.025   3.436 
system.time({reshape2::melt(mylist2)})
#user  system elapsed 
#  3.461   0.021   3.472

The melt functions from both packages have only slight difference in performance

system.time({rbindlist(lapply(mylist2, as.data.table), idcol = names(mylist2))})
# user  system elapsed 
#  2.889   0.160   3.029 
system.time({enframe(mylist2) %>% 
     unnest})
# user  system elapsed 
#  0.149   0.004   0.152 

system.time({data.frame(value = reduce(mylist2, c), 
        class = rep(names(mylist2), lengths(mylist2)))})
#   user  system elapsed 
#  14.714   8.890  23.550 



library(microbenchmark)

microbenchmark(
ak1 = reshape2::melt(mylist2),
ak2 = enframe(mylist2) %>% 
     unnest,
st1 = data.table::melt(mylist2),
st2 = rbindlist(lapply(mylist2, as.data.table), idcol = names(mylist2)),

unit = 'relative', times = 10L)
#Unit: relative
# expr      min       lq     mean   median       uq      max neval cld
#  ak1 26.51333 24.38488 21.52345 23.65577 19.36023 18.71291    10   c
#  ak2  1.00000  1.00000  1.00000  1.00000  1.00000  1.00000    10 a  
#  st1 25.34465 24.56934 21.97353 26.27053 19.23078 17.32802    10   c
#  st2 16.19268 16.96023 14.28353 16.25288 12.69549 11.58340    10  b

Sorry, at first I think the solution of a numeric list is equivalent to a Date list. Thanks for your edit. — Darren Tsai, Jan 04 '19 at 09:49
@DarrenTsai I think the attributes of `Date` created issue in `stack`. I updated with one more solution — akrun, Jan 04 '19 at 09:50
I think the `stack` solution should be removed here, since it is not working for date list. — Maël, Sep 02 '22 at 08:55

score 3 · Answer 2 · answered Jan 04 '19 at 09:48

We can use data.table package which is very efficient,

library(data.table)

rbindlist(lapply(mylist, as.data.table), idcol = names(mylist))
#   A         V1
#1: A 4394-02-08
#2: A 4580-05-16
#3: A 2476-01-24
#4: A 2928-11-03
#5: B 4652-12-02
#6: B 3758-02-20
#7: C 2331-09-07
#8: C 3092-02-15
#9: C 3494-03-07

Additionally, data.table::melt() will also do the job (similar to @akrun's reshape2 solution), i.e.

data.table::melt(mylist)

Ronak Shah · Answer 3 · 2019-01-04T10:40:06.377

We have already got data.table and tidyverse solution (which are highly efficient) but just for completeness sake here is a base R approach

data.frame(value = Reduce(c, mylist), class = rep(names(mylist), lengths(mylist)))


#       value class
#1 1983-04-14     A
#2 1979-01-15     A
#3 1977-08-22     A
#4 1974-06-12     A
#5 1975-07-10     B
#6 1980-02-08     B
#7 1986-11-29     C
#8 1984-03-31     C
#9 1985-03-24     C

Reduce can also be replaced with do.call

data.frame(value = do.call(c, mylist), class = rep(names(mylist), lengths(mylist)))

Stack a named Date list to data.frame

3 Answers3

Update

Benchmarks

Linked

Related