使用plyr匹配两个数据帧中的列值

Question

DF

av  bv  tv   u   l value           s
30 120 360 330 210  6600 0.005238424 
35 125 360 325 200  6875 0.005028887
40 130 360 320 190  7150 0.004835468
45 135 360 315 180  7425 0.004656377
50 140 360 310 170  7700 0.004490078
55 145 360 305 160  7975 0.004335247
60 150 360 300 150  8250 0.004190739
65 155 360 295 140  8525 0.004055554
70 160 360 290 130  8800 0.003928818
75 165 360 285 120  9075 0.003809763
80 170 360 280 110  9350 0.003697711

dput(DF)

df<-structure(list(av = c(30, 35, 40, 45, 50, 55, 60, 65, 70, 75, 
80), bv = c(120, 125, 130, 135, 140, 145, 150, 155, 160, 165, 
170), tv = c(360, 360, 360, 360, 360, 360, 360, 360, 360, 360, 
360), u = c(330, 325, 320, 315, 310, 305, 300, 295, 290, 285, 
280), l = c(210, 200, 190, 180, 170, 160, 150, 140, 130, 120, 
110), value = c(6600, 6875, 7150, 7425, 7700, 7975, 8250, 8525, 
8800, 9075, 9350), s = c(0.005238424, 0.00502888704, 0.00483546830769231, 
0.00465637688888889, 0.00449007771428572, 0.00433524744827586, 
0.0041907392, 0.00405555406451613, 0.003928818, 0.00380976290909091, 
0.00369771105882353)), .Names = c("av", "bv", "tv", "u", "l", 
"value", "s"), row.names = c(1L, 13L, 25L, 37L, 49L, 61L, 73L, 
85L, 97L, 109L, 121L), class = "data.frame")

DF2

  av  bv  tv   u  l value
  30 120   0   0  0     0
  30 120  20   0  0     0
  30 120  40  10  0   550
  30 120  60  30  0  1650
  30 120 120  90  0  4950
  30 120 180 150 30  6600

dput(DF2)

df2<-structure(list(av = c(30, 30, 30, 30, 30, 30), bv = c(120, 120, 
120, 120, 120, 120), tv = c(0, 20, 40, 60, 120, 180), u = c(0, 
0, 10, 30, 90, 150), l = c(0, 0, 0, 0, 0, 30), value = c(0, 0,
550, 1650, 4950, 6600)), .Names = c("av", "bv", "tv", "u", "l", 
"value"), row.names = c(1L, 2602L, 5203L, 7804L, 10405L, 13006L
), class = "data.frame")

我想要做的就是将df$s值添加df到df2where df$bv == df2$bv.df2会有更多相同的bv值df,所以会有一些重复的s值.

我正在尝试以下方面

     newDF <- ddply(df2, .(bv,tv), summarise, s = df[df$bv %in% df2$bv,]$s)

虽然这对我不起作用,但也许是因为我并不真正理解这个函数中的变量参数.

实际上所有其他列都是任意的,但我想保持整个数据帧完好无损.

Answer 1

这会将'df'中相应的's'项目拉入匹配的'df2'行:

df2$s <- df$s[ match(df2$bv, df$bv)]
df2
#-----------------------
      av  bv  tv   u  l value           s
1     30 120   0   0  0     0 0.005238424
2602  30 120  20   0  0     0 0.005238424
5203  30 120  40  10  0   550 0.005238424
7804  30 120  60  30  0  1650 0.005238424
10405 30 120 120  90  0  4950 0.005238424
13006 30 120 180 150 30  6600 0.005238424

这比'subset()' - ting和'merge()' - 更有效.Oooops.我没有看到plyr部分.它也会比任何plyr方法快得多,但那就是因为'我是一个基础R家伙.如果你想用plyr做,那么这就是我认为你要求的:

> newDF <- ddply(df2, .(bv), summarise, s = df$s[match(df2$bv , df$bv)])
> newDF
   bv           s
1 120 0.005238424
2 120 0.005238424
3 120 0.005238424
4 120 0.005238424
5 120 0.005238424
6 120 0.005238424