2016-10-09 30 views
1

我有以下代碼將列表SampleDf和sampleDF2綁定在一起,然後將兩個字符向量綁定到它上面。我想要做的是創建一個函數,在這個函數中我可以傳遞另一個sampleDF3,Recipe3和成分列表,並將它們同樣地綁定在一起。有沒有一種簡單的方法可以像lapply或do.call一樣執行此操作?我的最終目標是能夠將函數傳遞給sampleDF,Recipe和成分列表,並將它們全部綁定在一起,類似於下面的示例。rbind和cbind一起列出

Code: 
try1<-cbind(cbind(RecipeName<-c("Recipe1","Recipe2"),ingredients<-c("","Beans")) 
      ,rbind(
        SampleDf 
       ,sampleDf2 

       ) 
      ) 


Data: 

dput(SampleDf) 
structure(c(45.8490717149901, 75.6532220962743, 49.4757541141121, 
21.7923657299986, 153.255016847245), .Dim = c(1L, 5L), .Dimnames = list(
    "Test set", c("ME", "RMSE", "MAE", "MPE", "MAPE"))) 

dput(sampleDf2) 
structure(c(-1.39930351254246, 65.1992541962796, 46.5664097914753, 
-364.369685854671, 412.539393211685), .Dim = c(1L, 5L), .Dimnames = list(
    "Test set", c("ME", "RMSE", "MAE", "MPE", "MAPE"))) 

dput(sampleDf3) 
structure(c(0, 65.1992541962796, 1, 
-364.369685854671, 10), .Dim = c(1L, 5L), .Dimnames = list(
    "Test set", c("ME", "RMSE", "MAE", "MPE", "MAPE"))) 
+0

如果你看一下輸出'try1'它是所有字符列。我想這不是你想要的,對嗎?將'sample'數據集放入'list'中,轉換爲'data.frame',然後使用'do.call(rbind'即'res < - do.call(rbind,lapply(mget(ls(pattern = (?i)SampleDf \\ d *「)),as.data.frame))' – akrun

回答

1

你可以做到以下幾點:

require(dplyr) 
bind_all <- function(rows, cols){ 
    rows <- lapply(rows, as.data.frame) 
    cols <- vapply(cols, as.data.frame, list(1)) 
    bind_cols(bind_rows(rows), cols) 
} 

bind_all(list(SampleDf, sampleDf2), 
     list(RecipeName=c("Recipe1","Recipe2"),ingredients=c("","Beans"))) 

它給你:

  ME  RMSE  MAE  MPE  MAPE RecipeName ingredients 
1 45.849072 75.65322 49.47575 21.79237 153.2550 Recipe1    
2 -1.399304 65.19925 46.56641 -364.36969 412.5394 Recipe2  Beans