Question

情况是飙车...有时车手与竞争对手比赛，有时他们只是一个人比赛。驾驶员及其技能水平始终是完全随机的。比赛在第12圈结束后，每天进行一次比赛，持续10年。有数百个驱动程序。独立的观察员在比赛期间记录了数据，包括驾驶员的速度，但仅限于其中一名驾驶员！因此，数据丢失。这是数据的前6行：

    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)
    )

每行用于不同的驱动程序，但我需要填写数据以了解竞争对手的速度！我将手动输入速度，以说明其余数据集需要执行的操作。

    df_1 <- 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(91,       96,       94,       89,      "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)
    )

Answer 1

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",    "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

Answer 2

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"))

Answer 3

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

仅当满足多个条件时，才将不同列中的值匹配到行

3 个答案:

更新：