Question

简化示例，示例数据：

data_f <- data.frame(id=rep(1:10),cat=rep("non",10),nam=paste(rep(1:10),rep(1:10),letters[1:10],sep=""))
data_f[,2] <- c("one","","","two","","tree","","four","","")

data_d <- data.frame(nam1=paste(rep(1:10),rep(1:10),letters[1:10],sep=""))

期望的结果：

      nam1  new.name
1     11 a       one   
2     22 b       one
3     33 c       one
4     44 d       two
5     55 e       two
6     66 f       tree 
7     77 g       tree
8     88 h       four
9     99 i       four
10    1010 j     four

单词：如果列nam的任何条目与nam1中的data_d匹配，则从cat获取该类别的data_f列将它放在nam1的相应值旁边。请注意，nam1和nam的条目随机分散，不按示例顺序排列。

我未能找到有效的解决方案。

Answer 1

这是一种可能性：

merge(data_d, 
      transform(data_f[-1], cat = ave(cat, cumsum(cat != ""), FUN = function(x) x[1])), 
      by.x = "nam1", by.y = "nam", sort = FALSE)

结果：

    nam1  cat
1    11a  one
2    22b  one
3    33c  one
4    44d  two
5    55e  two
6    66f tree
7    77g tree
8    88h four
9    99i four
10 1010j four

重命名R中另一列的列条目

1 个答案: