[R] add a new column to data frame

[Phil Spector]

Here are two ways:

dat0 = subset(dat,time==0)

one = as.data.frame(as.table(by(dat0,dat0$id,
                     function(x)as.character(x$mode)[which.max(x$y)])))
names(one) = c('id','type')
merge(dat,one)

two = sapply(split(dat0,dat0$id),
              function(x)as.character(x$mode)[which.max(x$y)])
merge(dat,data.frame(id=names(two),type=two))


 					- Phil Spector
 					 Statistical Computing Facility
 					 Department of Statistics
 					 UC Berkeley
 					 spector at stat.berkeley.edu




On Fri, 1 Oct 2010, array chip wrote:

> Hi, I am wondering if anyone can propose a simple/best way to do the following:
>
> Let's say I have a data frame
>
> dat <-
> cbind(expand.grid(mode=c('right','left'),time=0:3,id=c('p1','p2','p3')),y=c(3,5,rep(4,6),6,2,rep(3,6),4,4,rep(2,6)))
>
>
> dat
>    mode time id y
> 1  right    0 p1 3
> 2   left    0 p1 5
> 3  right    1 p1 4
> 4   left    1 p1 4
> 5  right    2 p1 4
> 6   left    2 p1 4
> 7  right    0 p2 6
> 8   left    0 p2 2
> 9  right    1 p2 3
> 10  left    1 p2 3
> 11 right    2 p2 3
> 12  left    2 p2 3
> 13 right    0 p3 4
> 14  left    0 p3 4
> 15 right    1 p3 2
> 16  left    1 p3 2
> 17 right    2 p3 2
> 18  left    2 p3 2
>
> Now I want to add a new column "type" to this data frame with values of either
> "left" or "right" for each "id" based on the following logic:
>
> For each "id", the value of "type" column is the value of "mode" for which the
> value of "y" is the maximum of "y" based on time=0. For example for id=p1, the
> value of "type" is "left" because for the 2 "y" values (3 & 5) based on time=0,
> mode=left has the bigger "y" value (5). But if the 2 "y" values are the same for
> "mode", then let type=right.
>
> In the end the new data frame is:
>
>    mode time id y  type
> 1  right    0 p1 3  left
> 2   left    0 p1 5  left
> 3  right    1 p1 4  left
> 4   left    1 p1 4  left
> 5  right    2 p1 4  left
> 6   left    2 p1 4  left
> 7  right    0 p2 6 right
> 8   left    0 p2 2 right
> 9  right    1 p2 3 right
> 10  left    1 p2 3 right
> 11 right    2 p2 3 right
> 12  left    2 p2 3 right
> 13 right    0 p3 4 right
> 14  left    0 p3 4 right
> 15 right    1 p3 2 right
> 16  left    1 p3 2 right
> 17 right    2 p3 2 right
> 18  left    2 p3 2 right
>
> Any suggestions would be appreciated.
>
> John
>
>
>
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>
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