Question

模拟数据：

set.seed(1)
df1 <- data.frame(country=c("US", "UK"),
                  year=c(2000, 2003))
df2 <- data.frame(country=rep(c("US", "UK"), 10),
                  year=rep(2000:2009, 2),
                  myvar=rnorm(20))

df1包含感兴趣的国家/地区年份。我希望获得此国家/地区的myvar值以及之前和之后的3年。

换句话说，合并是基于df2$country==df1$country和df2$year > df1$year - 3 & df2$year < df1$year + 3

的条件完成的

编辑：我的（工作，不优雅）解决方案是填充df1以创建我感兴趣的所有国家/地区年份，然后以常规方式与df2合并。

library(plyr)
ddply(df1, c("country", "year"), 
  function(df) data.frame(rep(df$country, 7), (df$year-3):(df$year+3)))

产生

   country year rep.df.country..7. X.df.year...3...df.year...3.
1       UK 2003                 UK                         2000
2       UK 2003                 UK                         2001
3       UK 2003                 UK                         2002
4       UK 2003                 UK                         2003
5       UK 2003                 UK                         2004
6       UK 2003                 UK                         2005
7       UK 2003                 UK                         2006
8       US 2000                 US                         1997
9       US 2000                 US                         1998
10      US 2000                 US                         1999
11      US 2000                 US                         2000
12      US 2000                 US                         2001
13      US 2000                 US                         2002
14      US 2000                 US                         2003

Answer 1

合并在哪里？这听起来像是一个子集问题，除非我误解了这个问题（正如我常常所做的那样）

set.seed(1)
df1 <- data.frame(country=c("US", "UK"),
                  year=c(2000, 2003))
df2 <- data.frame(country=rep(c("US", "UK"), 10),
                  year=rep(2000:2009, 2),
                  myvar=rnorm(20))


f <- lapply(df1$country, function(x) {
  tmp <- df2[df2$country == x, ]
  tmp[abs(tmp$year - df1[df1$country == x, 'year']) <= 3, ]
})


do.call(rbind, f)

#    country year       myvar
# 1       US 2000 -0.62645381
# 3       US 2002 -0.83562861
# 11      US 2000  1.51178117
# 13      US 2002 -0.62124058
# 2       UK 2001  0.18364332
# 4       UK 2003  1.59528080
# 6       UK 2005 -0.82046838
# 12      UK 2001  0.38984324
# 14      UK 2003 -2.21469989
# 16      UK 2005 -0.04493361

修改

set.seed(1)
df1 <- data.frame(country=c("US", "UK"),
                  year=c(2000, 2003, 2009, 2009))
df2 <- data.frame(country=rep(c("US", "UK"), 10),
                  year=rep(2000:2009, 2),
                  myvar=rnorm(20))

f <- lapply(seq_len(nrow(df1)), function(x) {
  y <- df1[x, 'country']
  tmp <- df2[df2$country == y, ]
  tmp[abs(tmp$year - df1[x, 'year']) <= 3, ]
})


do.call(rbind, f)

#    country year       myvar
# 1       US 2000 -0.62645381
# 3       US 2002 -0.83562861
# 11      US 2000  1.51178117
# 13      US 2002 -0.62124058
# 2       UK 2001  0.18364332
# 4       UK 2003  1.59528080
# 6       UK 2005 -0.82046838
# 12      UK 2001  0.38984324
# 14      UK 2003 -2.21469989
# 16      UK 2005 -0.04493361
# 7       US 2006  0.48742905
# 9       US 2008  0.57578135
# 17      US 2006 -0.01619026
# 19      US 2008  0.82122120
# 8       UK 2007  0.73832471
# 10      UK 2009 -0.30538839
# 18      UK 2007  0.94383621
# 20      UK 2009  0.59390132

Answer 2

在data.table中使用foverlaps的试用

set.seed(1)
df1 <- data.frame(country=c("US", "UK"),
                  year=c(2000, 2003, 2009, 2009))
df2 <- data.frame(country=rep(c("US", "UK"), 10),
                  year=rep(2000:2009, 2),
                  myvar=rnorm(20))
library(data.table)
setDT(df1); setDT(df2) # convert to data table
df1[, c("start", "end") := list(year-2, year+2)]
setkey(df1, country, start, end)
setkey(df2[, year2:=year], country, year, year2)
foverlaps(df1, df2, type="any")[,4:7:=NULL][]
    country year       myvar
 1:      UK 2001  0.18364332
 2:      UK 2001  0.38984324
 3:      UK 2003  1.59528080
 4:      UK 2003 -2.21469989
 5:      UK 2005 -0.82046838
 6:      UK 2005 -0.04493361
 7:      UK 2007  0.73832471
 8:      UK 2007  0.94383621
 9:      UK 2009 -0.30538839
10:      UK 2009  0.59390132
11:      US 2000 -0.62645381
12:      US 2000  1.51178117
13:      US 2002 -0.83562861
14:      US 2002 -0.62124058
15:      US 2008  0.57578135
16:      US 2008  0.82122120

Answer 3

使用data.table

的简单解决方案

library(data.table) # v1.9.7 (devel version)
# go here for install instructions
# https://github.com/Rdatatable/data.table/wiki/Installation    

# convert datasets into data.table
  setDT(df1)
  setDT(df2)


# create conditional columns in df1
  df1[, yearplus3  :=  year +3 ][, yearminus3 := year - 3 ]

# merge
    output <- df1[df2, on = .(country = country ,                # condition 1
                              yearminus3 < year,                 # condition 2
                              yearplus3  > year), nomatch = 0 ,  # condition 3
                  .(country, year,  myvar )]  # indicate columns in the output


output 
 >   country year       myvar
 >1:      US 2000 -0.62645381
 >2:      UK 2003  0.18364332
 >3:      US 2000 -0.83562861
 >4:      UK 2003  1.59528080
 >5:      UK 2003 -0.82046838
 >6:      US 2000  1.51178117
 >7:      UK 2003  0.38984324
 >8:      US 2000 -0.62124058

PS。请注意，截至今天（2016年5月12日），参数on =仍处于data.table的开发版本中

根据条件合并两个数据框，其中一列匹配

3 个答案: