我有一个数据框（DF），我在每个CompanyID上都有2006年和2007年在那里工作的Director以及2个有关它们的信息（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），每个DirectorID（fisrt专栏）都有不同年份的经验年限（第三栏）（第二栏）。
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，在两年（2006年和2007年）创建一个新的专栏，其中包含每个公司的每位董事的经验年数，即专栏Experience_2006和Experience_2007。
因此，我的预期输出看起来像：
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？感谢。
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)
答案 0 :(得分：2)
为了完成起见，我使用了<img src="<?php echo get_the_post_thumbnail_url($loop->post->ID); ?>" class="img-responsive" alt=""/>
和＆＃39; tidyr＆＃39;并与其他功能进行了基准测试。
更新我在不使用过滤器并选择函数dplyr的情况下创建了@ Jimbou答案的另一个版本。这是我的基准测试中加速最快的。拉尔夫的答案现在排在第二位。我的初始版本（myfun4()）排在第三位。
myfun3()功能代码：
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
答案 1 :(得分：1)
你可以尝试
library(tidyverse)
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"))
CompanyID Name Country ISIN Director_2006 Gender_2006 YearBirth_2006 Director_2007 Gender_2007
1 25830 BANKxxx Austria AT000504 11734844255 M 54 11734844255 M
2 25830 BANKxxx Austria AT000504 187836811559 F 45 5524344997 F
3 25830 BANKxxx Austria AT000504 5524344997 F NA 5524354997 M
4 25830 BANKxxx Austria AT000504 5524354997 M 38 5742347684 M
5 25830 BANKxxx Austria AT000504 6613115791 M 41 40160443378 M
6 12339 BANKyyy Belgium AT034003 9855321789 M 44 9855321789 M
7 12339 BANKyyy Belgium AT034003 277520199 M NA 23779351 F
YearBirth_2007 YearsExperience_2016 YearsExperience_2017
1 55 0.4 1.4
2 NA 1.5 3.4
3 39 2.4 2.8
4 38 1.8 3.5
5 30 NA 4.3
6 45 3.6 4.6
7 34 1.6 3.2
答案 2 :(得分：1)
使用基本R函数：
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"))
reshape(DF3, direction = "wide", v.names = names(DF3)[c(1,7,8,10)], timevar = "Year", sep = "_")
#> CompanyID Name Country ISIN id Director_2007 Gender_2007
#> 1 12339 BANKyyy Belgium AT034003 7 23779351 F
#> 3 25830 BANKxxx Austria AT000504 3 5524354997 M
#> 4 25830 BANKxxx Austria AT000504 2 5524344997 F
#> 5 25830 BANKxxx Austria AT000504 4 5742347684 M
#> 8 25830 BANKxxx Austria AT000504 5 40160443378 M
#> 9 12339 BANKyyy Belgium AT034003 6 9855321789 M
#> 11 25830 BANKxxx Austria AT000504 1 11734844255 M
#> YearBirth_2007 YearsExperience_2007 Director_2006 Gender_2006
#> 1 34 3.2 277520199 M
#> 3 39 2.8 5524344997 F
#> 4 NA 3.4 187836811559 F
#> 5 38 3.5 5524354997 M
#> 8 30 4.3 6613115791 M
#> 9 45 4.6 9855321789 M
#> 11 55 1.4 11734844255 M
#> YearBirth_2006 YearsExperience_2006
#> 1 NA 1.6
#> 3 NA 2.4
#> 4 45 1.5
#> 5 38 1.8
#> 8 41 NA
#> 9 44 3.6
#> 11 54 0.4