我有两个data.tables。如果dfB中的dfA对应于year 一个，我想将第二个data.table dfA与第一个year的行合并dfB中的前一年。
作为示例，dfB的第一行将与dfA的第一行合并，因为2009年的dfB年在{{1 }}，2010年。
dfA我想尝试：
library(data.table)
dfA <- fread("
A B C D E F G Z iso year matchcode
1 0 1 1 1 0 1 0 NLD 2010 NLD2010
2 1 0 0 0 1 0 1 NLD 2014 NLD2014
3 0 0 0 1 1 0 0 AUS 2010 AUS2010
4 1 0 1 0 0 1 0 AUS 2006 AUS2006
5 0 1 0 1 0 1 1 USA 2008 USA2008
6 0 0 1 0 0 0 1 USA 2010 USA2010
7 0 1 0 1 0 0 0 USA 2012 USA2012
8 1 0 1 0 0 1 0 BLG 2008 BLG2008
9 0 1 0 1 1 0 1 BEL 2008 BEL2008
10 1 0 1 0 0 1 0 BEL 2010 BEL2010
11 0 1 1 1 0 1 0 NLD 2010 NLD2010
12 1 0 0 0 1 0 1 NLD 2014 NLD2014
13 0 0 0 1 1 0 0 AUS 2010 AUS2010
14 1 0 1 0 0 1 0 AUS 2006 AUS2006
15 0 1 0 1 0 1 1 USA 2008 USA2008
16 0 0 1 0 0 0 1 USA 2010 USA2010
17 0 1 0 1 0 0 0 USA 2012 USA2012
18 1 0 1 0 0 1 0 BLG 2008 BLG2008
19 0 1 0 1 1 0 1 BEL 2008 BEL2008
20 1 0 1 0 0 1 0 BEL 2010 BEL2010",
header = TRUE)
dfB <- fread("
A B C D H I J K iso year matchcode
1 0 1 1 1 0 1 0 NLD 2009 NLD2009
2 1 0 0 0 1 0 1 NLD 2014 NLD2014
3 0 0 0 1 1 0 0 AUS 2011 AUS2011
4 1 0 1 0 0 1 0 AUS 2007 AUS2007
5 0 1 0 1 0 1 1 USA 2007 USA2007
6 0 0 1 0 0 0 1 USA 2010 USA2010
7 0 1 0 1 0 0 0 USA 2013 USA2013
8 1 0 1 0 0 1 0 BLG 2007 BLG2007
9 0 1 0 1 1 0 1 BEL 2009 BEL2009
10 1 0 1 0 0 1 0 BEL 2012 BEL2012",
header = TRUE)
但这会导致年份匹配，这不是我想要的。
我相信dfA <- merge(dfA , dfB, on =.(iso, year == year-1), all.x = TRUE, allow.cartesian=FALSE)
也会尝试找到最接近的匹配项。
我应该如何编写此合并？
期望的输出：
roll答案 0 :(得分：1)
有点混乱，但是请尝试：
dfB[dfA[,c(.SD,.(year1=year-1))],
on=.(A,B,C,D,iso,year == year1)]
A B C D H I J K iso year matchcode E F G Z i.year i.matchcode
1: 1 0 1 1 1 0 1 0 NLD 2009 NLD2009 1 0 1 0 2010 NLD2010
2: 2 1 0 0 NA NA NA NA NLD 2013 <NA> 0 1 0 1 2014 NLD2014
3: 3 0 0 0 NA NA NA NA AUS 2009 <NA> 1 1 0 0 2010 AUS2010
4: 4 1 0 1 NA NA NA NA AUS 2005 <NA> 0 0 1 0 2006 AUS2006
5: 5 0 1 0 1 0 1 1 USA 2007 USA2007 1 0 1 1 2008 USA2008
6: 6 0 0 1 NA NA NA NA USA 2009 <NA> 0 0 0 1 2010 USA2010
7: 7 0 1 0 NA NA NA NA USA 2011 <NA> 1 0 0 0 2012 USA2012
8: 8 1 0 1 0 0 1 0 BLG 2007 BLG2007 0 0 1 0 2008 BLG2008
9: 9 0 1 0 NA NA NA NA BEL 2007 <NA> 1 1 0 1 2008 BEL2008
10: 10 1 0 1 NA NA NA NA BEL 2009 <NA> 0 0 1 0 2010 BEL2010
11: 11 0 1 1 NA NA NA NA NLD 2009 <NA> 1 0 1 0 2010 NLD2010
12: 12 1 0 0 NA NA NA NA NLD 2013 <NA> 0 1 0 1 2014 NLD2014
13: 13 0 0 0 NA NA NA NA AUS 2009 <NA> 1 1 0 0 2010 AUS2010
14: 14 1 0 1 NA NA NA NA AUS 2005 <NA> 0 0 1 0 2006 AUS2006
15: 15 0 1 0 NA NA NA NA USA 2007 <NA> 1 0 1 1 2008 USA2008
16: 16 0 0 1 NA NA NA NA USA 2009 <NA> 0 0 0 1 2010 USA2010
17: 17 0 1 0 NA NA NA NA USA 2011 <NA> 1 0 0 0 2012 USA2012
18: 18 1 0 1 NA NA NA NA BLG 2007 <NA> 0 0 1 0 2008 BLG2008
19: 19 0 1 0 NA NA NA NA BEL 2007 <NA> 1 1 0 1 2008 BEL2008
20: 20 1 0 1 NA NA NA NA BEL 2009 <NA> 0 0 1 0 2010 BEL2010
答案 1 :(得分：1)
我不知道这是否是您想要的，但我希望它能对您有所帮助。有点复杂
df<-numeric()
for(i in 1:nrow(dfA)){
d<-numeric()
d<-which(dfA$iso[i]==dfB$iso & (dfA$year[i]-1)==dfB$year)
if(length(d)>0){
df<-rbind(df,c(dfA[i,1:8],dfB[d,5:8],dfB$year[d],dfA$iso[i],dfA$year[i],dfA$matchcode[i],dfB$matchcode[d]))
}else{
df<-rbind(df,c(dfA[i,1:8],rep(NA,5),dfA$iso[i],dfA$year[i],dfA$matchcode[i],NA))
}
}
colnames(df)[13:17]<-c("year_from_B", "iso", "year" , "matchcode1", "matchcode2")
df
这是输出
A B C D E F G Z H I J K year_from_B iso year matchcode1 matchcode2
[1,] 1 0 1 1 1 0 1 0 1 0 1 0 2009 "NLD" 2010 "NLD2010" "NLD2009"
[2,] 2 1 0 0 0 1 0 1 NA NA NA NA NA "NLD" 2014 "NLD2014" NA
[3,] 3 0 0 0 1 1 0 0 NA NA NA NA NA "AUS" 2010 "AUS2010" NA
[4,] 4 1 0 1 0 0 1 0 NA NA NA NA NA "AUS" 2006 "AUS2006" NA
[5,] 5 0 1 0 1 0 1 1 1 0 1 1 2007 "USA" 2008 "USA2008" "USA2007"
[6,] 6 0 0 1 0 0 0 1 NA NA NA NA NA "USA" 2010 "USA2010" NA
[7,] 7 0 1 0 1 0 0 0 NA NA NA NA NA "USA" 2012 "USA2012" NA
[8,] 8 1 0 1 0 0 1 0 0 0 1 0 2007 "BLG" 2008 "BLG2008" "BLG2007"
[9,] 9 0 1 0 1 1 0 1 NA NA NA NA NA "BEL" 2008 "BEL2008" NA
[10,] 10 1 0 1 0 0 1 0 1 1 0 1 2009 "BEL" 2010 "BEL2010" "BEL2009"
[11,] 11 0 1 1 1 0 1 0 1 0 1 0 2009 "NLD" 2010 "NLD2010" "NLD2009"
[12,] 12 1 0 0 0 1 0 1 NA NA NA NA NA "NLD" 2014 "NLD2014" NA
[13,] 13 0 0 0 1 1 0 0 NA NA NA NA NA "AUS" 2010 "AUS2010" NA
[14,] 14 1 0 1 0 0 1 0 NA NA NA NA NA "AUS" 2006 "AUS2006" NA
[15,] 15 0 1 0 1 0 1 1 1 0 1 1 2007 "USA" 2008 "USA2008" "USA2007"
[16,] 16 0 0 1 0 0 0 1 NA NA NA NA NA "USA" 2010 "USA2010" NA
[17,] 17 0 1 0 1 0 0 0 NA NA NA NA NA "USA" 2012 "USA2012" NA
[18,] 18 1 0 1 0 0 1 0 0 0 1 0 2007 "BLG" 2008 "BLG2008" "BLG2007"
[19,] 19 0 1 0 1 1 0 1 NA NA NA NA NA "BEL" 2008 "BEL2008" NA
[20,] 20 1 0 1 0 0 1 0 1 1 0 1 2009 "BEL" 2010 "BEL2010" "BEL2009"