仅当满足多个条件时，才将不同列中的值匹配到行

这是aleft_join的理想选择。

您的资料

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",    "Yes",    "Yes",    "No",        "No"),
  Comp_name=     c("Julie", "Rick",   "Johnny", "Denver", "NA",        "NA"),
  Comp_level=    c("B",     "B",      "D",      "A",      "NA",        "NA"),
  Comp_speed=    c("???",   "???",    "???",    "???",    "NA",        "NA"),
  Race_day=      c(165,      165,      72,       72,       92,          65),
  Lap_number=    c(9,        9,        12,       12,       8,           4),
  Humidity=      c(33,       33,       88,       88,       12,          55),
  Temperature=   c(28,       28,       12,       12,       20,          28)
)

我们加载dplyr包裹

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

让我们摆脱Comp_speed当前具有“ ???”的列 价值观。

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

让我们创建一个仅包含名称和速度的数据框，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left_join将df数据框更改为df2。Comp_nameindf与Driver_namein匹配df2

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）。

假设（从数据中）Race_day和Lap_number变量足以区分每个正面竞争，我们只需将它们保留在df2数据框中即可。然后在我们的列表中加入这些列名称left_join。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