在 R 中合并多对数据帧

Question

I am looking to merge multiple pairs of data frames. For example I have following data frames Con_2014, Con_2015, Con_2016..... and Income_2014, Income_2015, Income_2016. So I want to merge data for each year like the following but at once.

Con_Income_2014 <- merge(Income_2014,Con_2014)
Con_Income_2015 <- merge(Income_2015,Con_2015)
Con_Income_2016 <- merge(Income_2016,Cons_2016)`

I tried the following way but it does not work.

for (i in 1: length(years)){
  assign(paste("Cons_Income", as.character(years[i]),sep =""),merge(paste("Income_",as.character(years[i]),sep=""),paste("Cons_",as.character(years[i]),sep="")))}

It runs but it give me just name of each pairs. Its merging strings(the name of the file but not the data)

Answer 1

It's not the prettiest thing, but it works:

Income_2014 = data.frame(id = 1:5, x = runif(5,-1,1))
Income_2015 = data.frame(id = 6:10, x = runif(5,-1,1))
Income_2016 = data.frame(id = 11:15, x = runif(5,-1,1))
Cons_2014 = data.frame(id = 1:5, y = runif(5,2,3))
Cons_2015 = data.frame(id = 6:10, y = runif(5,2,3))
Cons_2016 = data.frame(id = 11:15, y = runif(5,2,3))
years <- 2014:2016
for (i in 1: length(years)){
  assign(paste("Cons_Income", as.character(years[i]),sep =""),
    merge(eval(parse(text=paste("Income_",as.character(years[i]),sep=""))),                  
          eval(parse(text=paste("Cons_",as.character(years[i]),sep="")))))
}

Answer 2

*Observation. I believe the last object should be con_... and not cons_... (a typo)

We do not neet for loops. If all years are associated with a pair of dataframes, and the naming is consistent, we can use mget(ls(pattern =....)) to create two lists of dataframes, then do the pairwise merge with mapply :

mapply(merge,
       mget(ls(pattern = "Income_20\\d{2}")),
       mget(ls(pattern = "Con_20\\d{2}")),
       SIMPLIFY = FALSE) %>%
    setNames(paste0('income_con_', 2014:2016))

Data created by @DaveArmstrong:

set.seed(1)

Income_2014 = data.frame(id = 1:5, x = runif(5,-1,1))
Income_2015 = data.frame(id = 6:10, x = runif(5,-1,1))
Income_2016 = data.frame(id = 11:15, x = runif(5,-1,1))
Cons_2014 = data.frame(id = 1:5, y = runif(5,2,3))
Cons_2015 = data.frame(id = 6:10, y = runif(5,2,3))
Cons_2016 = data.frame(id = 11:15, y = runif(5,2,3))

Output:

$income_con_2014
  id          x        y
1  1 -0.4689827 2.497699
2  2 -0.2557522 2.717619
3  3  0.1457067 2.991906
4  4  0.8164156 2.380035
5  5 -0.5966361 2.777445

$income_con_2015
  id          x        y
1  6  0.7967794 2.934705
2  7  0.8893505 2.212143
3  8  0.3215956 2.651674
4  9  0.2582281 2.125555
5 10 -0.8764275 2.267221

$income_con_2016
  id          x        y
1 11 -0.5880509 2.386114
2 12 -0.6468865 2.013390
3 13  0.3740457 2.382388
4 14 -0.2317926 2.869691
5 15  0.5396828 2.340349