合并列表的数据框并获取数据框的名称作为列

Question

I merged all data frames from a list in just one data frame.

The dataframes inside the list are called

TAI
NAM
HEE 

and each data frame looks like this

Yr-M   Compound1 Compound 2
2015-01   0.002    0.15
2015-02   0.004    0.02
2015-03   0.01     0.09

when I merge all dataframes with meanall<-do.call(rbind, meaneach) I get

         Yr-M  Compound1  Compound2
TAI.1   2015-01   0.002    0.15
TAI.2   2015-02   0.004    0.02
TAI.3   2015-03   0.01     0.09
  .
  .
  .
NAM.1   2015-01   0.03     0.4
NAM.2   2015-02   0.001    0.005

I would like to get a column with the names of the list and not as rownames (like above), and without the numbers (TAI.1, TAI.2...), I just want the name TAI

So that I get this:

 List    Yr-M  Compound1  Compound2
  TAI   2015-01   0.002    0.15
  TAI   2015-02   0.004    0.02
  TAI   2015-03   0.01     0.09
  .
  .
  .
  NAM   2015-01   0.03     0.4
  NAM   2015-02   0.001    0.005

How can I do this?

Answer 1

Rownames you can not have duplicates. So the best thing to do is convert the rownames to a column and then use gsub to remove the .[0-9], ie

df <- do.call(rbind, your_list)
df$list_id <- gsub('\\..*', '', rownames(df))

Note that you can use dplyr or data.table version of rbinding a list which have the option of including the list names as a column, ie

dplyr::bind_rows(your_list, .id = 'list_id')
data.table::rbindlist(your_list, idcol = 'list_id')

Answer 2

You can add an additional column with the listnames after merging the three lists via do.call :

nameColumn <- data.frame(listName = c(rep(c('TAI','NAM','HEE'),
                                          c(length(TAI),length(NAM),length(HEE) ))
meanall <- cbind(meanall, nameColumn)

If you want the nameColumn to be the first column, just switch arguments in cbind to

meanall <- cbind(nameColumn, meanall)