Merging multiple pairs of data-frame in R-CodePudding

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)

CodePudding user response：

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="")))))
}

CodePudding user response：

*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