計算操作的輸出的兩個dataframes之間有條件

Question

假設我有兩個dataframes：

set.seed(123) 
df1<-data.frame(rep=rep(1:4,each=360),parc=rep(1:40,each=36),trat=rep(sample(1:10),each=36),tree=rep(1:36,40),med=1,dap_prev=rnorm(1440, mean = 12)) 
df2<-data.frame(med=rep(1:18,each=10),trat=rep(sample(1:10)),b0=rnorm(180),b1=rnorm(180)) 

在df2我需要時，他們符合標準df1$med == df2$med和df1$trat == df2$trat檢索的df2$b0和df2$b1值。然後在df1中創建一個新列，其產品爲df2$b0 + df2$b1 * df1$dap_prev。

我試着用下面這個命令，當然它沒有工作：

df1$ddap_cm <- df2$b0[df2$med == df1$med & df2$trat == df.1$trat] + df2$b1[df1$med == df2$med & df1$trat == df2$trat] * df1$dap_prev 

Answer 1

我不知道我得到了你的權利，所以我建議這兩個步驟的答案

set.seed(123) 
df1<-data.frame(rep=rep(1:4,each=360),parc=rep(1:40,each=36),trat=rep(sample(1:10),each=36),tree=rep(1:36,40),med=1,dap_prev=rnorm(1440, mean = 12)) 
df2<-data.frame(med=rep(1:18,each=10),trat=rep(sample(1:10)),b0=rnorm(180),b1=rnorm(180)) 

### Answer 
library(dplyr) 
df1_v2 <- left_join(df1, df2, by = c("med", "trat")) # matching with "med" and "trat" 
df1_v2$ddap_cm <- df1_v2$b0 + (df1_v2$b1*df1_v2$dap_prev) # new variable 
str(df1_v2)