12

Due to time constraints, I've decided to use data tables in my code instead of data frames, as they are much faster. However, I still want the functionality of data frames. I need to merge two data tables, conserving all values (like setting all=TRUE in merge).

Some example code:

> x1 = data.frame(index = 1:10)
> y1 = data.frame(index = c(2,4,6), weight = c(.2, .5, .3))
> x1
   index
1      1
2      2
3      3
4      4
5      5
6      6
7      7
8      8
9      9
10    10
> y1
  index weight
1     2    0.2
2     4    0.5
3     6    0.3

> merge(x,y, all=TRUE)
      index weight
 [1,]     1     NA
 [2,]     2      1
 [3,]     3     NA
 [4,]     4      2
 [5,]     5     NA
 [6,]     6      3
 [7,]     7     NA
 [8,]     8     NA
 [9,]     9     NA
[10,]    10     NA

Now can I do a similar thing with data tables? (The NA's don't necessarily have to stay, I change them to 0's anyways). 

> x2 = data.table(index = 1:10, key ="index")
> y2 = data.table(index = c(2,4,6), weight= c(.3,.5,.2))

I know you can merge, but I also know that there is a faster way.

4

2 に答える 2

8

したがって、外部キーのSQL結合の変換からRdata.table構文への続き

x2 = data.table(index = 1:10, key ="index")
y2 = data.table(index = c(2,4,6), weight= c(.3,.5,.2),key="index")
y2[J(x2$index)]
于 2012-07-11T15:37:27.083 に答える
1

私は次のような関数を使用します：

mergefast<-function(x,y,by.x,by.y,all) {
  x_dt<-data.table(x)
  y2<-y
  for (i in 1:length(by.y)) names(y2)[grep(by.y[i],names(y2))]<-by.x[i]
  y_dt<-data.table(y2)
  setkeyv(x_dt,by.x)
  setkeyv(y_dt,by.x)
  as.data.frame(merge(x_dt,y_dt,by=by.x,all=all))
}

あなたの例では次のように使用できます。

mergefast(x1,y1,by.x="index",by.y="index",all=T)

merge、 、 などbyの機能が少し不足していますがall.x、all.yこれらは簡単に組み込むことができます。

于 2013-12-18T16:20:18.113 に答える