Question

我有2个看起来像这样的数据框

data_frame_1 <- data.frame(DATE = seq(as.Date("2016-01-01"),as.Date("2016-01-10"),by = "1 day"),
                           Att1 = c(1,3,4,5,NA,4,5,NA,NA,9),
                           Att2 = c(4,5,6,7,3,4,NA,7,2,NA)
                           )

> data_frame_1
         DATE Att1 Att2
1  2016-01-01    1    4
2  2016-01-02    3    5
3  2016-01-03    4    6
4  2016-01-04    5    7
5  2016-01-05   NA    3
6  2016-01-06    4    4
7  2016-01-07    5   NA
8  2016-01-08   NA    7
9  2016-01-09   NA    2
10 2016-01-10    9   NA


data_frame_2 <- data.frame(DATE = seq(as.Date("2016-01-01"),as.Date("2016-01-10"),by = "1 day"),
                           Att1 = c(3,3,21,5,8,4,5,11,5,9),
                           Att2 = c(23,7,9,13,3,4,9,7,2,12)
)

> data_frame_2
         DATE Att1 Att2
1  2016-01-01    1    4
2  2016-01-02    3    5
3  2016-01-03    4    6
4  2016-01-04    5    7
5  2016-01-05    3    3
6  2016-01-06    4    4
7  2016-01-07    5    9
8  2016-01-08   11    7
9  2016-01-09    5    2
10 2016-01-10    9   12

现在我想要数据frame_1中的每一列，其中日期大于2016-01-08，并且有NA它会被data_frame_2中的相应值所关联所以最终结果看起来像这个

> data_frame_1_mod
         DATE Att1 Att2
1  2016-01-01    1    4
2  2016-01-02    3    5
3  2016-01-03    4    6
4  2016-01-04    5    7
5  2016-01-05   NA    3
6  2016-01-06    4    4
7  2016-01-07    5   NA
8  2016-01-08   11    7
9  2016-01-09    5    2
10 2016-01-10    9   12

我能在R中实现这一目标的最快方式是什么？

修改

我试过这个方法

replace_func <- function(column,date,data1,data2){
  rel_frame <- data.frame(date = data1$DATE, data_1 = data1[,column], data_2 = data2[,column] )
  rel_frame$data_1_mod <- ifelse(rel_frame$date >= date & is.na(rel_frame$data_1),rel_frame$data_2,rel_frame$data_1)
  rel_frame <- rel_frame[c("date","data_1_mod")]
  colnames(rel_frame) <- c("DATE",column)
  return(rel_frame)
}


all_frames <- lapply(c("Att1", "Att2"), function(x) replace_func(x,as.Date("2016-01-08"),data_frame_1,data_frame_2))
data_frame_1_mod <- Reduce(function(x, y) merge(x, y, all.x=TRUE),all_frames )


> data_frame_1_mod
         DATE Att1 Att2
1  2016-01-01    1    4
2  2016-01-02    3    5
3  2016-01-03    4    6
4  2016-01-04    5    7
5  2016-01-05   NA    3
6  2016-01-06    4    4
7  2016-01-07    5   NA
8  2016-01-08   11    7
9  2016-01-09    5    2
10 2016-01-10    9   12

我想知道是否有更好的方法

Answer 1

这个怎么样：

 n <- which(data_frame_1$DATE > as.Date("2016-01-08") & (is.na(data_frame_1$Att1) | is.na(data_frame_1$Att2))

 n
 [1]  9 10

 for (i in n) {
      data_frame_1[i,] <- data_frame_2[i,]
 }

#result
> data_frame_1
         DATE Att1 Att2
1  2016-01-01    1    4
2  2016-01-02    3    5
3  2016-01-03    4    6
4  2016-01-04    5    7
5  2016-01-05   NA    3
6  2016-01-06    4    4
7  2016-01-07    5   NA
8  2016-01-08   NA    7
9  2016-01-09    5    2
10 2016-01-10    9   12
>

Answer 2

这是一个基于data.table的快速解决方案：

library(data.table)
## replace data.frame by data.table 
setDT(data_frame_1)
setDT(data_frame_2)
## since the number of columns to check can be big ,
## better to put your data in the long format 
dx1 <- melt(data_frame_1,id="DATE")
dx2 <- melt(data_frame_2,id="DATE")
## setkey for fast join 
setkey(dx1,DATE,variable)
setkey(dx2,DATE,variable)
## use tab2 as an index , and replace all miising values after a certain    dates by the corresponding values from tab2 , finally come back to the wide format using `dcast`
dcast(dx1[dx2][is.na(value) & DATE >= "2016-01-08",value:=i.value][,i.value:=NULL],
      DATE~variable)

给出了预期的结果：

          DATE Att1 Att2
 1: 2016-01-01    1    4
 2: 2016-01-02    3    5
 3: 2016-01-03    4    6
 4: 2016-01-04    5    7
 5: 2016-01-05   NA    3
 6: 2016-01-06    4    4
 7: 2016-01-07    5   NA
 8: 2016-01-08   11    7
 9: 2016-01-09    5    2
10: 2016-01-10    9   12

有条件地替换多个Dataframe的值R.

2 个答案: