left_join是理想的选择。

您的数据

df <- data.frame(
  Driver_name =  c("Rick","Julie","Denver","Johny","Cassandra","Phillip"),Driver_level = c("A","C","D","A","B","B"),Driver_speed = c(96,91,89,94,88,99),Competitor=    c("Yes","Yes","No","No"),Comp_name=     c("Julie","Rick","Johnny","NA","NA"),Comp_level=    c("B",Comp_speed=    c("???","???",Race_day=      c(165,165,72,92,65),Lap_number=    c(9,9,12,8,4),Humidity=      c(33,33,55),Temperature=   c(28,28,20,28)
)

我们加载了dplyr包

#install.packages("dplyr") #if you don't have it
library(dplyr)

让我们摆脱当前具有“ ???”的Comp_speed列值。

df <- df %>% select(-Comp_speed)

让我们创建一个仅包含名称和速度的第二个数据帧，然后即时将Driver_speed重命名为Comp_speed。

df2 <- df %>% 
  select(Driver_name,Comp_speed = Driver_speed)

现在我们可以将left_join数据帧df到df2。 Comp_name中的df与Driver_name中的df2匹配

df_updated <- df %>% 
  left_join(df2,by = c("Comp_name" = "Driver_name"))
#> Warning: Column `Comp_name`/`Driver_name` joining factors with different
#> levels,coercing to character vector

这是结果数据框df_updated

df_updated
#>   Driver_name Driver_level Driver_speed Competitor Comp_name Comp_level
#> 1        Rick            A           96        Yes     Julie          B
#> 2       Julie            C           91        Yes      Rick          B
#> 3      Denver            D           89        Yes    Johnny          D
#> 4       Johny            A           94        Yes    Denver          A
#> 5   Cassandra            B           88         No        NA         NA
#> 6     Phillip            B           99         No        NA         NA
#>   Race_day Lap_number Humidity Temperature Comp_speed
#> 1      165          9       33          28         91
#> 2      165          9       33          28         96
#> 3       72         12       88          12         NA
#> 4       72         12       88          12         89
#> 5       92          8       12          20         NA
#> 6       65          4       55          28         NA 

更新：

随着OP的提出，这对于不止一次赛车的赛车手来说并不牢固（我的疏忽）。

（从数据中）假设Race_day和Lap_number变量足以区分每个头对头种族，我们只需将它们保留在df2数据框中。然后在我们的left_join中加入这些列名称。这就是它的样子。

df2 <- df %>% 
  select(Driver_name,Comp_speed = Driver_speed,Race_day,Lap_number)

df_updated <- df %>% 
  left_join(df2,by = c("Comp_name" = "Driver_name","Race_day","Lap_number"))
#> Warning: Column `Comp_name`/`Driver_name` joining factors with different
#> levels,coercing to character vector

我们需要将df留给自己。
！names（df）％in％c（“ Comp_speed”）从第一个数据帧x中删除变量Comp_speed。

df [，c（“ Driver_name”，“ Driver_speed”）]]仅在第二个数据帧y中包含变量Driver_name和Driver_speed。

总而言之，x中的Comp_name与y中的Driver_name匹配，而y中的Driver_speed被报告为Driver_speed.y（Driver_speed.y，因为df中已经存在Driver_speed，在连接后将其更改为Driver_speed.x ）：

df <- merge(x=df[,!names(df)%in%c("Comp_speed")],y=df[,c("Driver_name","Driver_speed")],by.x="Comp_name",by.y="Driver_name",all.x=TRUE)

现在，我们只需要将“ Driver_speed.x”，“ Driver_speed.y”的名称更改为“ Driver_speed”，“ Comp_speed”：

library("data.table")
setnames(df,c("Driver_speed.x","Driver_speed.y"),c("Driver_speed","Comp_speed"))

我想<form> <input name="a.one"> <input name="a.two"> <input name="a.three"> <input name="b.one"> <input name="b.two"> <input name="b.three"> <input name="c.one"> <input name="c.two"> <input name="c.three"> </form> 可以满足您的需求