做一个“模糊”和非模糊，多对一与 data.table 合并
[英]Doing a "fuzzy" and non-fuzzy, many to 1 merge with data.table
问题描述
Lets assume I have two databases dfA and dfB .
假设我有两个数据库dfA和dfB 。 One has individual observations and one has country level data (which is applicable to multiple observations which are from the same year and country) For each of these databases I have created a key called matchcode.一个有个人观察，一个有国家级数据（适用于来自同一年份和国家的多个观察）对于这些数据库中的每一个，我都创建了一个名为 matchcode 的键。 This matchcode is a combination of a country code and a year.此匹配代码是国家/地区代码和年份的组合。
dfA <- read.table(
text = "A B C D E F G 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 <- read.table(
text = "A B C D H I J 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 2011 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
)
library(data.table)
setDT(dfA)
setDT(dfB)
Mostly when I merge these datasets I simply do:大多数情况下，当我合并这些数据集时，我只是这样做：
dfA<- merge(dfA, dfB, by= "matchcode", all.x = TRUE, allow.cartesian=FALSE)
The problem is that sometimes the years do not completely match.问题是有时年份并不完全匹配。 So I tried:所以我试过：
dfA <- dfA[dfB, on = .(iso, year), roll = "nearest", nomatch = 0]
But this reduces the amount of observations to 11.但这将观察量减少到 11。
# A tibble: 11 x 18
A B C D E F G iso year matchcode K L M N O P Q i.matchcode
<int> <int> <int> <int> <int> <int> <int> <fct> <int> <fct> <int> <int> <int> <int> <int> <int> <int> <fct>
1 0 1 1 1 0 1 0 NLD 2009 NLD2010 0 1 1 1 0 1 0 NLD2009
2 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 0 0 1 0 1 NLD2014
3 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 0 0 1 0 1 NLD2014
4 0 0 0 1 1 0 0 AUS 2011 AUS2010 0 0 0 1 1 0 0 AUS2011
5 1 0 1 0 0 1 0 AUS 2007 AUS2006 1 0 1 0 0 1 0 AUS2007
6 0 1 0 1 0 1 1 USA 2007 USA2008 0 1 0 1 0 1 1 USA2007
7 0 0 1 0 0 0 1 USA 2011 USA2010 0 0 1 0 0 0 1 USA2010
8 0 1 0 1 0 0 0 USA 2013 USA2012 0 1 0 1 0 0 0 USA2013
9 1 0 1 0 0 1 0 BLG 2007 BLG2008 1 0 1 0 0 1 0 BLG2007
10 0 1 0 1 1 0 1 BEL 2009 BEL2008 0 1 0 1 1 0 1 BEL2009
11 1 0 1 0 0 1 0 BEL 2012 BEL2010 1 0 1 0 0 1 0 BEL2012
The preferred output would be as follows:首选输出如下：
# A B C D E F G iso year matchcodeA H I J matchcodeB
# 1: 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 1 NLD2014
# 2: 0 0 0 1 1 0 0 AUS 2011 AUS2010 1 0 0 AUS2011
# 3: 1 0 1 0 0 1 0 AUS 2007 AUS2006 0 1 0 AUS2007
# 4: 0 0 1 0 0 0 1 USA 2011 USA2010 0 0 1 USA2010
# 5: 0 1 0 1 0 0 0 USA 2013 USA2012 0 0 0 USA2013
# 6: 0 1 0 1 1 0 1 BEL 2009 BEL2008 1 0 1 BEL2009
# 7: 0 1 1 1 0 1 0 NLD 2009 NLD2010 0 1 0 NLD2009
# 8: 0 1 0 1 0 1 1 USA 2007 USA2008 0 1 1 USA2007
# 9: 0 1 0 1 0 0 0 USA 2011 USA2012 0 0 1 USA2010
#10: 1 0 1 0 0 1 0 BEL 2009 BEL2010 1 0 1 BEL2009
#11: 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 1 NLD2014
#12: 0 0 0 1 1 0 0 AUS 2011 AUS2010 1 0 0 AUS2011
#13: 1 0 1 0 0 1 0 AUS 2007 AUS2006 0 1 0 AUS2007
#14: 0 0 1 0 0 0 1 USA 2011 USA2010 0 0 1 USA2010
#15: 0 1 0 1 0 0 0 USA 2013 USA2012 0 0 0 USA2013
#16: 0 1 0 1 1 0 1 BEL 2009 BEL2008 1 0 1 BEL2009
#17: 0 1 1 1 0 1 0 NLD 2009 NLD2010 0 1 0 NLD2009
#18: 0 1 0 1 0 1 1 USA 2007 USA2008 0 1 1 USA2007
#19: 0 1 0 1 0 0 0 USA 2011 USA2012 0 0 1 USA2010
#20: 1 0 1 0 0 1 0 BEL 2009 BEL2010 1 0 1 BEL2009
Additional Sources:其他来源：
1 个解决方案
解决方案1
2 已采纳 2019-01-04 12:40:38
Hers is my (default) approach for a join like this, using data.table Hers 是我使用data.table进行这样的连接的（默认）方法
code代码
library( data.table )
#change the name of the matchcode-column
setnames(dfA, c("matchcode", "iso", "year"), c("matchcodeA", "isoA", "yearA"))
setnames(dfB, c("matchcode", "iso", "year"), c("matchcodeB", "isoB", "yearB"))
#store column-order for in the end
namesA <- as.character( names( dfA ) )
namesB <- as.character( setdiff( names(dfB), names(dfA) ) )
colorder <- c(namesA, namesB)
#create columns to join on
dfA[, `:=`(iso.join = isoA, year.join = yearA)]
dfB[, `:=`(iso.join = isoB, year.join = yearB)]
#perform left join
result <- dfB[dfA, on = c("iso.join", "year.join"),roll = "nearest" ]
#drop columns that are not needed
result[, grep("^i\\.", names(result)) := NULL ]
result[, grep("join$", names(result)) := NULL ]
#set column order
setcolorder(result, colorder)
result结果
# A B C D E F G isoA yearA matchcodeA H I J isoB yearB matchcodeB
# 1: 0 1 1 1 0 1 0 NLD 2010 NLD2010 0 1 0 NLD 2009 NLD2009
# 2: 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 1 NLD 2014 NLD2014
# 3: 0 0 0 1 1 0 0 AUS 2010 AUS2010 1 0 0 AUS 2011 AUS2011
# 4: 1 0 1 0 0 1 0 AUS 2006 AUS2006 0 1 0 AUS 2007 AUS2007
# 5: 0 1 0 1 0 1 1 USA 2008 USA2008 0 1 1 USA 2007 USA2007
# 6: 0 0 1 0 0 0 1 USA 2010 USA2010 0 0 1 USA 2011 USA2010
# 7: 0 0 1 0 0 0 0 USA 2012 USA2012 0 0 1 USA 2011 USA2010
# 8: 1 0 1 0 0 1 0 BLG 2008 BLG2008 0 1 0 BLG 2007 BLG2007
# 9: 0 1 0 1 1 0 1 BEL 2008 BEL2008 1 0 1 BEL 2009 BEL2009
# 10: 0 1 0 1 0 1 0 BEL 2010 BEL2010 1 0 1 BEL 2009 BEL2009
# 11: 0 1 1 1 0 1 0 NLD 2010 NLD2010 0 1 0 NLD 2009 NLD2009
# 12: 1 0 0 0 1 0 1 NLD 2014 NLD2014 1 0 1 NLD 2014 NLD2014
# 13: 0 0 0 1 1 0 0 AUS 2010 AUS2010 1 0 0 AUS 2011 AUS2011
# 14: 1 0 1 0 0 1 0 AUS 2006 AUS2006 0 1 0 AUS 2007 AUS2007
# 15: 0 1 0 1 0 1 1 USA 2008 USA2008 0 1 1 USA 2007 USA2007
# 16: 0 0 1 0 0 0 1 USA 2010 USA2010 0 0 1 USA 2011 USA2010
# 17: 0 0 1 0 0 0 0 USA 2012 USA2012 0 0 1 USA 2011 USA2010
# 18: 1 0 1 0 0 1 0 BLG 2008 BLG2008 0 1 0 BLG 2007 BLG2007
# 19: 0 1 0 1 1 0 1 BEL 2008 BEL2008 1 0 1 BEL 2009 BEL2009
# 20: 0 1 0 1 0 1 0 BEL 2010 BEL2010 1 0 1 BEL 2009 BEL2009
sample data样本数据
dfA <- fread(
"A B C D E F G iso year matchcode
0 1 1 1 0 1 0 NLD 2010 NLD2010
1 0 0 0 1 0 1 NLD 2014 NLD2014
0 0 0 1 1 0 0 AUS 2010 AUS2010
1 0 1 0 0 1 0 AUS 2006 AUS2006
0 1 0 1 0 1 1 USA 2008 USA2008
0 0 1 0 0 0 1 USA 2010 USA2010
0 1 0 1 0 0 0 USA 2012 USA2012
1 0 1 0 0 1 0 BLG 2008 BLG2008
0 1 0 1 1 0 1 BEL 2008 BEL2008
1 0 1 0 0 1 0 BEL 2010 BEL2010
0 1 1 1 0 1 0 NLD 2010 NLD2010
1 0 0 0 1 0 1 NLD 2014 NLD2014
0 0 0 1 1 0 0 AUS 2010 AUS2010
1 0 1 0 0 1 0 AUS 2006 AUS2006
0 1 0 1 0 1 1 USA 2008 USA2008
0 0 1 0 0 0 1 USA 2010 USA2010
0 1 0 1 0 0 0 USA 2012 USA2012
1 0 1 0 0 1 0 BLG 2008 BLG2008
0 1 0 1 1 0 1 BEL 2008 BEL2008
1 0 1 0 0 1 0 BEL 2010 BEL2010",
header = TRUE
)
dfB <- fread(
"A B C D H I J iso year matchcode
0 1 1 1 0 1 0 NLD 2009 NLD2009
1 0 0 0 1 0 1 NLD 2014 NLD2014
0 0 0 1 1 0 0 AUS 2011 AUS2011
1 0 1 0 0 1 0 AUS 2007 AUS2007
0 1 0 1 0 1 1 USA 2007 USA2007
0 0 1 0 0 0 1 USA 2011 USA2010
0 1 0 1 0 0 0 USA 2013 USA2013
1 0 1 0 0 1 0 BLG 2007 BLG2007
0 1 0 1 1 0 1 BEL 2009 BEL2009
1 0 1 0 0 1 0 BEL 2012 BEL2012",
header = TRUE
)
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