Working with dataframes in a list: Drop variables, add new ones

data-management, data-manipulation, lapply, r

Solution

Or try this: Extract your `ids` into a named vector that maps the data-frame name to the id:

df2id <- ids$id
names(df2id) <- ids$df

> df2id
df1 df2 
id1 id2 
Levels: id1 id2

Then use `mapply` to both (a) drop the `a` column from each data-frame, and (b) add the `id` column:

> mapply( function(d,x) cbind( subset(d, select = -a),
+                              id = x),
+         dats, df2id[ names(dats) ] ,
+         SIMPLIFY=FALSE)
$df1
   b  id
1 12 id1
2 11 id1
3 13 id1

$df2
   b  id
1 12 id2
2 11 id2
3 13 id2

Note that we are passing `df2id[ names(dats) ]` to the `mapply` -- this ensures that the data-frames in `df2id` are "aligned" with the data-frames in `dats`.

Problem

Define a list dats with two dataframes, df1 and df2 ``` dats <- list( df1 = data.frame(a=sample(1:3), b = sample(11:13)), df2 = data.frame(a=sample(1:3), b = sample(11:13))) > dats $df1 a b 1 2 12 2 3 11 3 1 13 $df2 a b 1 3 13 2 2 11 3 1 12 ``` I would like to drop variable a in each data frame. Next I would like to add a variable with the id of each dataframe from an external dataframe, like: ``` ids <- data.frame(id=c("id1","id2"),df=c("df1","df2")) > ids id df 1 id1 df1 2 id2 df2 ``` To drop unnecessary vars I tried this without luck: ``` > dats <- lapply(dats, function(x) assign(x, x[,c("b")])) > Error in assign(x, x[, c("b")]) : invalid first argument ``` Not sure how to add the id either. I also tried, perhaps more appropriately: ``` > temp <- lapply(dats, function(x) subset(x[1], select=x[[1]]$b)) Error in x[[1]]$b : $ operator is invalid for atomic vectors ``` What I find confusing is that `str(out[1])` returns a list, `str(out[[1]])` returns a dataframe. I think that may have something to do with it.

Original source