1

我有更大的数据集，如下所示：

l = 9
loc <- c(paste('Loc', 1:l, sep = ''))
 vloc <- c(rep(loc, each=2))
 qi <- c(  13, 12, 27, 20, 16, 18,  14, 17, 15, 22, 21, 26,12, 14, 11,
   18,  8, 24  )
    afreq <- c( 0.308, 0.4, 0.041, 0.5, 0.125, 0.5,0.139, 0.2, 0.219, 0.2,0.176, 
    0.33,0.358, 0.4, 0.274, 0.5, 0.173, 0.15)   
 loctab <- data.frame(vloc = vloc, qi = qi, afreq = afreq)
loctab 

   vloc     qi afreq
1  Loc1     13 0.308
2  Loc1     12 0.400
3  Loc2     27 0.041
4  Loc2     20 0.500
5  Loc3     16 0.125
6  Loc3     18 0.500
7  Loc4     14 0.139
8  Loc4     17 0.200
9  Loc5     15 0.219
10 Loc5     22 0.200
11 Loc6     21 0.176
12 Loc6     26 0.330
13 Loc7     12 0.358
14 Loc7     14 0.400
15 Loc8     11 0.274
16 Loc8     18 0.500
17 Loc9      8 0.173
18 Loc9     24 0.150

较小的数据集如下：

ex1 <-data.frame (loc, qi = c(13, 27, 16, 14, 15, 21, 12, 11,  8)
ex1

   loc qi
1 Loc1 13
2 Loc2 27
3 Loc3 16
4 Loc4 14
5 Loc5 15
6 Loc6 21
7 Loc7 12
8 Loc8 11
9 Loc9  8

对于每个 loc，我需要匹配 ex1（小）和 loctab（大）之间的 qi 值并创建一个新表。

我尝试了以下但没有提供正确的答案。

 nloct <- loctab[loctab$qi %in%  ex1$qit, ]

预期的输出是

nloct <- data.frame (loc, qi = c(13, 27, 16, 14, 15, 21, 12, 11,  8), 
afreq = c( 0.308,  0.041,  0.125, 0.139, 0.219,0.176, 0.358,  0.274,  0.173))
  loc qi afreq
1 Loc1 13 0.308
2 Loc2 27 0.041
3 Loc3 16 0.125
4 Loc4 14 0.139
5 Loc5 15 0.219
6 Loc6 21 0.176
7 Loc7 12 0.358
8 Loc8 11 0.274
9 Loc9  8 0.173
4

2 回答 2

3

您正在寻找某种呼叫merge()：

## e.g. :
## merge(ex1, loctab)[c(2,1,4)]
## OR
merge(ex1, loctab, by.x=c("loc", "qi"), by.y=c("vloc", "qi"))
   loc qi afreq
1 Loc1 13 0.308
2 Loc2 27 0.041
3 Loc3 16 0.125
4 Loc4 14 0.139
5 Loc5 15 0.219
6 Loc6 21 0.176
7 Loc7 12 0.358
8 Loc8 11 0.274
9 Loc9  8 0.173
于 2012-06-05T20:29:56.677 回答
0 
merge(loctab,ex1,by.x=c('vloc','qi'),by.y=c('loc','qi'))

但是，如果您将列 'vloc' 和 'loc' 命名为相同的东西，您可以说：

merge(loctab,ex1)
于 2012-06-05T20:34:14.220 回答