r - 如何按组连接两个数据框？

Question

我有一个数据框 (DF)，其中每个 CompanyID 都有 2006 年和 2007 年在那里工作的董事以及 2 个关于他们的信息(性别和年龄)。

DF <- 
CompanyID  Name    Country  ISIN     Director_2006 Gender_2006 Yearold_2006  Director_2007 Gender_2007 Yearold_2007   
25830      BANKxxx Austria  AT000504 11734844255        M            54        11734844255        M           55           
25830      BANKxxx Austria  AT000504 187836811559       F            45        5524344997         F           NA           
25830      BANKxxx Austria  AT000504 5524344997         F            NA        5524354997         M           39           
25830      BANKxxx Austria  AT000504 5524354997         M            38        5742347684         M           38           
25830      BANKxxx Austria  AT000504 6613115791         M            41        40160443378        M           30           
12339      BANKyyy Belgium  AT034003 9855321789         M            44        9855321789         M           45           
12339      BANKyyy Belgium  AT034003 277520199          M            NA        23779351           F           34

我有第二个数据框 (DF2)，其中每个 DirectorID(第一列)都有不同年份(第二列)的经验年数(第三列)。

DF2 <- 
  DirectorID     Year     YearsExperience
  11734844255    2006        0.4
  11734844255    2007        1.4
  187836811559   2006        1.5  
  5524344997     2006        2.4
  5524344997     2007        3.4
  5524354997     2006        1.8
  5524354997     2007        2.8  
  5742347684     2007        3.5
  40160443378    2007        4.3
  9855321789     2005        2.6
  9855321789     2006        3.6
  9855321789     2007        4.6
  277520199      2006        1.6
  23779351       2007        3.2
  55443322       2005        2.5
  55443322       2006        3.5

我想加入两个数据框的信息，创建一个新列，其中包含每家公司的每位董事在这两年(2006 年和 2007 年)的经验年限，即 Experience_2006 和 Experience_2007 列。

因此，我的预期输出如下:

DF_final <- 
 CompanyID   Name    Country ISIN       Director_2006  Gender_2006 YearBirth_2006  Experience_2006  Director_2007 Gender_2007 YearBirth_2007 Experience_2007  
 25830      BANKxxx  Austria  AT000504  11734844255          M        54                 0.4         11734844255      M           55                 1.4
 25830      BANKxxx  Austria  AT000504  187836811559         F        45                 1.5         5524344997       F           NA                 3.4
 25830      BANKxxx  Austria  AT000504  5524344997           F        NA                 2.4         5524354997       M           39                 2.8
 25830      BANKxxx  Austria  AT000504  5524354997           M        38                 1.8         5742347684       M           38                 3.5
 25830      BANKxxx  Austria  AT000504  6613115791           M        41                 NA          40160443378      M           30                 4.3
 12339      BANKyyy  Belgium  AT034003  9855321789           M        44                 3.6         9855321789       M           45                 4.6
 12339      BANKyyy  Belgium  AT034003  277520199            M        NA                 1.6         23779351         F           34                 3.2

拜托，有人可以告诉我吗？谢谢。

数据

DF <- read.table(text = 
               "CompanyID   Name    Country ISIN     Director_2006  Gender_2006 YearBirth_2006  Director_2007 Gender_2007 YearBirth_2007   
             25830      BANKxxx     Austria  AT000504  11734844255     M        54              11734844255     M           55           
             25830      BANKxxx     Austria  AT000504  187836811559    F        45              5524344997      F           NA           
             25830      BANKxxx     Austria  AT000504    5524344997    F        NA              5524354997      M           39           
             25830      BANKxxx     Austria  AT000504    5524354997    M        38              5742347684      M           38           
             25830      BANKxxx     Austria  AT000504    6613115791    M        41              40160443378     M           30           
             12339      BANKyyy     Belgium  AT034003    9855321789    M        44              9855321789      M           45           
             12339      BANKyyy     Belgium  AT034003     277520199    M        NA                23779351      F           34",
             header = T, stringsAsFactors = F)

DF2 <- read.table(text =
            "DirectorID     Year     YearsExperience
             11734844255    2006        0.4
             11734844255    2007        1.4
             187836811559   2006        1.5  
             5524344997     2006        2.4
             5524344997     2007        3.4
             5524354997     2006        1.8
             5524354997     2007        2.8  
             5742347684     2007        3.5
             40160443378    2007        4.3
             9855321789     2005        2.6
             9855321789     2006        3.6
             9855321789     2007        4.6
             277520199      2006        1.6
             23779351       2007        3.2
             55443322       2005        2.5
             55443322       2006        3.5",
            header = T, stringsAsFactors = F)

Answer 1

为了完成，我使用了 dplyr 和 'tidyr' 并与其他函数进行了基准测试。

更新:我在没有使用过滤器和选择函数 myfun4() 的情况下制作了另一个版本的@Jimbou 答案。这是我的基准测试中最快的加入。拉尔夫的答案现在排在第二位。我的初始版本 (myfun3()) 排在第三位。

 microbenchmark::microbenchmark(myfun1(),myfun2(),myfun3(),myfun4())
Unit: milliseconds
     expr     min       lq      mean   median       uq     max neval
 myfun1() 23.1527 28.36865 31.322275 31.53225 33.69430 52.7319   100
 myfun2()  5.2549  5.78445  8.241408  8.25995  9.63870 14.4018   100
 myfun3()  7.9534 10.15115 11.976498 11.40415 13.66255 20.9362   100
 myfun4()  2.9676  3.40105  5.032863  4.87115  5.56065 19.0217   100

函数代码:

myfun4<-function(){
colnames(DF2)[1]='Director_2007'
DF_final<-left_join(DF,DF2[DF2$Year==2006,-2],by='Director_2007') %>% 
              left_join(DF2[DF2$Year==2007,-2],by='Director_2007')
n=dim(DF_final)[2]
colnames(DF_final)[(n-1):n]=paste0('YearsExperience_',2006:2007)
}

myfun3<-function(){
DF2_spread<-tidyr::spread(DF2,Year,YearsExperience)[,-2]
colnames(DF2_spread)=c('Director_2007',paste0('Experience_',colnames(df2_spread)[2:3]))
DF_final<-dplyr::left_join(DF,DF2_spread,by='Director_2007')
}

myfun2<-function() {
  DF1 <- reshape(DF, direction = "long", varying = names(DF)[5:10], sep = "_", timevar = "Year")
DF3 <- merge(DF1, DF2, all.x = TRUE, by.x = c("Director" , "Year"), by.y = c("DirectorID", "Year"))
DF_final<-reshape(DF3, direction = "wide", v.names = names(DF3)[c(1,7,8,10)], timevar = "Year", sep = "_")
}

myfun1<-function(){
  DF %>% 
  left_join(DF2 %>% 
              filter(Year == 2006) %>% 
              select(DirectorID,YearsExperience_2016=YearsExperience), 
            by=c("Director_2006" =  "DirectorID")) %>% 
  left_join(DF2 %>% 
              filter(Year == 2007) %>% 
              select(DirectorID,YearsExperience_2017=YearsExperience), 
            by=c("Director_2007" =  "DirectorID")) 
}