Question

我有更大的数据集如下：

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（small）和loctab（big）之间的qi值并创建一个newtable。

我尝试了以下但未提供正确答案。

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

预期输出

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

Answer 1

您正在寻找某种对merge()的电话：

## 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

Answer 2

merge(loctab,ex1,by.x=c('vloc','qi'),by.y=c('loc','qi'))

但如果你将列'vloc'和'loc'命名为同样的东西，你可以说：

merge(loctab,ex1)

子集数据集通过r匹配

2 个答案: