use*_*113 15 replace r pattern-matching dataframe
复杂的标题,但这是我想要实现的一个简单的例子:
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"))
Run Code Online (Sandbox Code Playgroud)
在价值d$v2应该由值来代替m$v4从值相匹配d$v2的m$v3
生成的数据框d应如下所示:
v1 v4
1 a
2 e
3 c
4 b
5 b
6 c
7 a
8 e
Run Code Online (Sandbox Code Playgroud)
我尝试了不同的东西,我最接近的是: d$v2 <- m$v4[which(m$v3 %in% d$v2)]
我试着再次避免任何for循环!必须是可能的:-)不知怎的......;)
joh*_*nes 18
你可以尝试:
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
Run Code Online (Sandbox Code Playgroud)
这是保持顺序的另一种方法:
data.frame(v1=d$v1, v4=m[match(d$v2, m$v3), 2])
v1 v4
1 1 a
2 2 e
3 3 c
4 4 b
5 5 b
6 6 c
7 7 a
8 8 e
Run Code Online (Sandbox Code Playgroud)
Esb*_*rdt 10
您可以使用标准的左连接.
加载数据:
d <- data.frame(v1 = c(1,2,3,4,5,6,7,8), v2 = c("A","E","C","B","B","C","A","E"), stringsAsFactors=F)
m <- data.frame(v3 = c("D","E","A","C","D","B"), v4 = c("d","e","a","c","d","b"), stringsAsFactors=F)
Run Code Online (Sandbox Code Playgroud)
更改列名,以便我可以按列"v2"加入
colnames(m) <- c("v2", "v4")
Run Code Online (Sandbox Code Playgroud)
左连接并维护data.frame d的顺序
library(dplyr)
left_join(d, m)
Run Code Online (Sandbox Code Playgroud)
输出:
v1 v2 v4
1 1 A a
2 2 E e
3 3 C c
4 4 B b
5 5 B b
6 6 C c
7 7 A a
8 8 E e
Run Code Online (Sandbox Code Playgroud)
小智 6
这将为您提供所需的输出:
d$v2 <- m$v4[match(d$v2, m$v3)]
Run Code Online (Sandbox Code Playgroud)
match 函数返回 m 矩阵的 v3 列中d$v2匹配值的位置。获得索引(通过 using match())后,m$v4使用这些索引访问元素以替换 d 矩阵第 v2 列中的元素。