How to use laply match to lookup value and append in each row?











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I have two data tables as below:



library(data.table)

x <- data.table(id = c(1,1,1,2,2,2,3,3,3,4,4,4), date = as.Date(c("2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03")))

y <- data.table(id=c(1,2,3,4),new_id=c(10,20,30,40))


As mentioned now I want to append the new_id column in the data table x and then later drop column id .



I can do this by 



merge(x,y,by="id")


But I wanted to try the lapply .



So I tried 



x[,new_id:=0]

nm <- c("new_id")

x[nm] <- lapply(nm, function(z) y[[z]][match(y$id, x$id)])


Also which method will be good if I have wide columns and more rows.



It does not matches the column it seems.



Also which method will be efficient if I have wide columns and more rows.



Any help is appreciated.






r dplyr data.table







share|improve this question




















  • 1




    If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
    – akrun
    Nov 8 at 5:24










  • I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
    – Frank
    Nov 8 at 11:06















up vote
1
down vote

favorite












I have two data tables as below:



library(data.table)

x <- data.table(id = c(1,1,1,2,2,2,3,3,3,4,4,4), date = as.Date(c("2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03")))

y <- data.table(id=c(1,2,3,4),new_id=c(10,20,30,40))


As mentioned now I want to append the new_id column in the data table x and then later drop column id .



I can do this by 



merge(x,y,by="id")


But I wanted to try the lapply .



So I tried 



x[,new_id:=0]

nm <- c("new_id")

x[nm] <- lapply(nm, function(z) y[[z]][match(y$id, x$id)])


Also which method will be good if I have wide columns and more rows.



It does not matches the column it seems.



Also which method will be efficient if I have wide columns and more rows.



Any help is appreciated.






r dplyr data.table







share|improve this question




















  • 1




    If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
    – akrun
    Nov 8 at 5:24










  • I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
    – Frank
    Nov 8 at 11:06













up vote
1
down vote

favorite









up vote
1
down vote

favorite











I have two data tables as below:



library(data.table)

x <- data.table(id = c(1,1,1,2,2,2,3,3,3,4,4,4), date = as.Date(c("2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03")))

y <- data.table(id=c(1,2,3,4),new_id=c(10,20,30,40))


As mentioned now I want to append the new_id column in the data table x and then later drop column id .



I can do this by 



merge(x,y,by="id")


But I wanted to try the lapply .



So I tried 



x[,new_id:=0]

nm <- c("new_id")

x[nm] <- lapply(nm, function(z) y[[z]][match(y$id, x$id)])


Also which method will be good if I have wide columns and more rows.



It does not matches the column it seems.



Also which method will be efficient if I have wide columns and more rows.



Any help is appreciated.






r dplyr data.table







share|improve this question















I have two data tables as below:



library(data.table)

x <- data.table(id = c(1,1,1,2,2,2,3,3,3,4,4,4), date = as.Date(c("2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03","2015-5-26","2015-6-15","2015-4-03")))

y <- data.table(id=c(1,2,3,4),new_id=c(10,20,30,40))


As mentioned now I want to append the new_id column in the data table x and then later drop column id .



I can do this by 



merge(x,y,by="id")


But I wanted to try the lapply .



So I tried 



x[,new_id:=0]

nm <- c("new_id")

x[nm] <- lapply(nm, function(z) y[[z]][match(y$id, x$id)])


Also which method will be good if I have wide columns and more rows.



It does not matches the column it seems.



Also which method will be efficient if I have wide columns and more rows.



Any help is appreciated.






r dplyr data.table




r dplyr data.table






share|improve this question















share|improve this question













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share|improve this question








edited Nov 8 at 4:39

























asked Nov 8 at 4:36









Ricky

6621323




6621323








  • 1




    If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
    – akrun
    Nov 8 at 5:24










  • I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
    – Frank
    Nov 8 at 11:06














  • 1




    If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
    – akrun
    Nov 8 at 5:24










  • I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
    – Frank
    Nov 8 at 11:06








1




1




If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
– akrun
Nov 8 at 5:24




If you are using match, then you don't need lapply i.e. y$new_id[match(x$id, y$id)]
– akrun
Nov 8 at 5:24












I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
– Frank
Nov 8 at 11:06




I guess you are just looking for an update join: x[, new_id := y[x, on=.(id), x.id]] the x.* prefix refers to the x table's columns in j of x[i,on=, j]
– Frank
Nov 8 at 11:06

















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