Question

我试图剥离其data.frame结构的两个数据帧，提取每个data.frame中的元素，并将数据帧中提取的数据合并到一个data.frame中。这应该导致data.frame由两列作为向量组成。请参阅下面的输出（以粗体标记）。

问题：输出包含多个data.frame元素，而不是包含输入数据框中矢量的单个data.frame。

每个数据框都有一个向量。

[EDIT ^ v回应评论。]

到目前为止，我尝试了as()和unlist()的各种组合无济于事......

我正在尝试使用内置R函数和向量化来解决此问题（不使用plyr和loops：Merge several data.frames into one data.frame with a loop，Merge many data frames from csv files，Recombining a list of Data.frames into a single data frame ）

可重现的代码：我无法复制错误，但这是我希望我的代码能够正常工作的方式：

df1<-data.frame<-c(1, 2, 3)
df2<-data.frame<-c(2, 4, 6)

output<-cbind(df1, df2)
print(output)       #Returns a data.frame
str(output)         #                     of vectors
                    #In my case however, a data.frame returns data.frames)

返回：

       df1 df2
[1,]   1   2
[2,]   2   4
[3,]   3   6

现实：

readmultiple <- function(directory = "bigdata") {

    ....


    ....
    ....
        output <- cbind.data.frame(filename, readmultiplesum) 
        # This is probably where things go wrong
        return(output)
    }
    output <- lapply(filenames, complete.cases.sum)
    assign("Global.output", output, envir = .GlobalEnv) 
    # There is probably a better way to do this too

    if (firstoutput == 1) {
        Global.output <- merge(as(unlist(Global.output[1]), "vector"), 
                           as(unlist(output[1])), "vector") 
    # as, unlist... Not sure what's needed here
    } else {
        firstoutput <- 1
    }
    str(output)
    return(Global.output)
}

输出看起来像

[[1]]
   filename result 
          1         142 

[[2]]
   filename result
          1        521

[[3]]
   filename result
          1         324

但我希望它是

filename        result 

[1,]   filename[i]  142 

[2,]   filename[i]  521

[3,]   filename[i]  324

...其中filename [i]是文件名的索引。

str（输出）返回

List of 2400
 $ :'data.frame':       1 obs. of  2 variables:
  ..$ filename   : Factor w/ 1 level "bigdata/001.csv": 1
  ..$ sumrows: num 142
 $ :'data.frame':       1 obs. of  2 variables:
  ..$ filename   : Factor w/ 1 level "bigdata/001.csv": 1
  ..$ sumrows: num 521
 $ :'data.frame':       1 obs. of  2 variables:
  ..$ filename   : Factor w/ 1 level "bigdata/001.csv": 1
  ..$ sumrows: num 324
 $ :'data.frame':       1 obs. of  2 variables:
  ..$ filename   : Factor w/ 1 level "bigdata/001.csv": 1

.....

dput（head（output））返回

    list(structure(list(filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 142), .Names = c("filename", "sumrows"), row.names = c(NA, 
-1L), class = "data.frame"), structure(list(filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 521), .Names = c("filename", "sumrows"
), row.names = c(NA, -1L), class = "data.frame"), structure(list(
    filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 324), .Names = c("filename", "sumrows"), row.names = c(NA, 
-1L), class = "data.frame"), structure(list(filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 1896), .Names = c("filename", "sumrows"
), row.names = c(NA, -1L), class = "data.frame"), structure(list(
    filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 1608), .Names = c("filename", "sumrows"
), row.names = c(NA, -1L), class = "data.frame"), structure(list(
    filename = structure(1L, .Label = "bigdata/001.csv", class = "factor"), 
    sumrows = 912), .Names = c("filename", "sumrows"), row.names = c(NA, 
-1L), class = "data.frame"))

Answer 1

将列表更改为data.frame的一般技巧是使用do.call

ll <- list(c(filename=1 ,result=142 ),c(filename=2 ,result=521 ))
> do.call(rbind,ll)
     filename result
[1,]        1    142
[2,]        2    521

当我将此应用到您的列表中时，我得到：

do.call(rbind,ll)
         filename sumrows
1 bigdata/001.csv     142
2 bigdata/001.csv     521
3 bigdata/001.csv     324
4 bigdata/001.csv    1896
5 bigdata/001.csv    1608
6 bigdata/001.csv     912

不幸的是，你不确切知道什么是文件名[i]？

修改

此解决方案似乎适用于OP：

library(plyr) ldply(ll)

通常你可以使用：

ldply(ll,function(x){ ##you process the row x here } )

两个数据帧的合并导致R中的多个数据帧

1 个答案: