您可以尝试：

  merge（d，m，by.x =v2，by.y =v3）
 v2 v1 v4 
 1 A 1 a 
 2 A 7 a 
 3 B 4 b 
 4 B 5 b 
 5 C 3 c 
 6 C 6 c 
 7 E 2 e 
 8 E 8 e 
 

修改

这是另一种方法来保存订单：

  data.frame（v1 = d $ v1，v4 = m [比赛（d $ v2，m $ v3），2]）
 v1 v4 
 1 1 a 
 2 2 e 
 3 3 c 
 4 4 b 
 5 5 b 
 6 6 c 
 7 7 a 
 8 8 e 
 

Complicated title but here is a simple example of what I am trying to achieve:

d <- data.frame(v1 = c(1,2,3,4,5,6,7,8), 
                v2 = c("A","E","C","B","B","C","A","E"))

m <- data.frame(v3 = c("D","E","A","C","D","B"), 
                v4 = c("d","e","a","c","d","b"))

Values in d$v2 should be replaced by values in m$v4 by matching the values from d$v2 in m$v3

The resulting data frame d should look like:

v1    v4
1      a
2      e
3      c
4      b
5      b
6      c
7      a
8      e

I tried different stuff and the closest I came was: d$v2 <- m$v4[which(m$v3 %in% d$v2)]

I try to avoid any for-loops again! Must be possible :-) somehow... ;)

You could try:

merge(d,m, by.x="v2", by.y="v3")
  v2 v1 v4
1  A  1  a
2  A  7  a
3  B  4  b
4  B  5  b
5  C  3  c
6  C  6  c
7  E  2  e
8  E  8  e

Edit

Here is another approach, to preserve the order:

data.frame(v1=d$v1, v4=m[match(d$v2, m$v3), 2])
  v1 v4
1  1  a
2  2  e
3  3  c
4  4  b
5  5  b
6  6  c
7  7  a
8  8  e