我们可以使用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

更新

对于更新的问题，可以使用reshape2中的melt或data.table

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

或者使用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

基准测试

以下是使用所公布的四种解决方案的一些基准测试

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

两个软件包中的melt函数在性能上只有细微的差别

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

我们可以使用非常高效的data.table包，

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

此外，data.table::melt()也可以完成这项工作（类似于@akrun的reshape2解决方案），即

data.table::melt(mylist)

我们已经得到了data.table和tidyverse解决方案（它们非常高效），但为了完整起见，这里使用的是基本R方法

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也可以替换为do.call

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