使用列表中的数据框:删除变量，添加新变量 [英] Working with dataframes in a list: Drop variables, add new ones

问题描述

使用两个数据框 df1 和 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

我想在每个数据帧中删除变量 a .接下来，我想从外部数据帧中添加一个具有每个数据帧的ID的变量，例如:

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

要删除不必要的var，我没有运气尝试过此操作:

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

也不知道如何添加ID.

Not sure how to add the id either.

我也尝试过，也许更合适:

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

我感到困惑的是，str(out[1])返回一个列表，str(out[[1]])返回一个数据帧.我认为这可能与它有关.

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.

推荐答案

或尝试以下操作:将ids提取到命名向量中，该向量将数据帧名称映射到id:

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

然后使用mapply两者(a)从每个数据帧中删除a列，以及(b)添加id列:

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

请注意，我们将df2id[ names(dats) ]传递给mapply －这确保了df2id中的数据帧与dats中的数据帧对齐".

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.

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