Hi,
May be this helps:
dat2<-read.table(text="
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
",sep="",header=TRUE)
dat2$flag<-unlist(lapply(split(dat2,dat2$idr),function(x)
rep(ifelse(any(apply(x,1,function(x) x[2]<=5 &
x[3]==0)),1,0),nrow(x))),use.names=FALSE)
dat2
# idr schyear year flag
#1 1 4 -1 1
#2 1 5 0 1
#3 1 6 1 1
#4 1 7 2 1
#5 2 9 0 0
#6 2 10 1 0
#7 2 11 2 0
A.K.
----- Original Message -----
From: Christopher Desjardins <cddesjard...@gmail.com>
To: jim holtman <jholt...@gmail.com>
Cc: r-help@r-project.org
Sent: Saturday, November 3, 2012 7:09 PM
Subject: Re: [R] Replacing NAs in long format
I have a similar sort of follow up and I bet I could reuse some of this
code but I'm not sure how.
Let's say I want to create a flag that will be equal to 1 if schyear < = 5
and year = 0 for a given idr. For example
> dat
idr schyear year
1 4 -1
1 5 0
1 6 1
1 7 2
2 9 0
2 10 1
2 11 2
How could I make the data look like this?
idr schyear year flag
1 4 -1 1
1 5 0 1
1 6 1 1
1 7 2 1
2 9 0 0
2 10 1 0
2 11 2 0
I am not sure how to end up not getting both 0s and 1s for the 'flag'
variable for an idr. For example,
dat$flag = ifelse(schyear <= 5 & year ==0, 1, 0)
Does not work because it will create:
idr schyear year flag
1 4 -1 0
1 5 0 1
1 6 1 0
1 7 2 0
2 9 0 0
2 10 1 0
2 11 2 0
And thus flag changes for an idr. Which it shouldn't.
Thanks,
Chris
On Sat, Nov 3, 2012 at 5:50 PM, Christopher Desjardins <
cddesjard...@gmail.com> wrote:
> Hi Jim,
> Thank you so much. That does exactly what I want.
> Chris
>
>
> On Sat, Nov 3, 2012 at 1:30 PM, jim holtman <jholt...@gmail.com> wrote:
>
>> > x <- read.table(text = "idr schyear year
>> + 1 8 0
>> + 1 9 1
>> + 1 10 NA
>> + 2 4 NA
>> + 2 5 -1
>> + 2 6 0
>> + 2 7 1
>> + 2 8 2
>> + 2 9 3
>> + 2 10 4
>> + 2 11 NA
>> + 2 12 6
>> + 3 4 NA
>> + 3 5 -2
>> + 3 6 -1
>> + 3 7 0
>> + 3 8 1
>> + 3 9 2
>> + 3 10 3
>> + 3 11 NA", header = TRUE)
>> > # you did not specify if there might be multiple contiguous NAs,
>> > # so there are a lot of checks to be made
>> > x.l <- lapply(split(x, x$idr), function(.idr){
>> + # check for all NAs -- just return indeterminate state
>> + if (sum(is.na(.idr$year)) == nrow(.idr)) return(.idr)
>> + # repeat until all NAs have been fixed; takes care of contiguous
>> ones
>> + while (any(is.na(.idr$year))){
>> + # find all the NAs
>> + for (i in which(is.na(.idr$year))){
>> + if ((i == 1L) && (!is.na(.idr$year[i + 1L]))){
>> + .idr$year[i] <- .idr$year[i + 1L] - 1
>> + } else if ((i > 1L) && (!is.na(.idr$year[i - 1L]))){
>> + .idr$year[i] <- .idr$year[i - 1L] + 1
>> + } else if ((i < nrow(.idr)) && (!is.na(.idr$year[i +
>> 1L]))){
>> + .idr$year[i] <- .idr$year[i + 1L] -1
>> + }
>> + }
>> + }
>> + return(.idr)
>> + })
>> > do.call(rbind, x.l)
>> idr schyear year
>> 1.1 1 8 0
>> 1.2 1 9 1
>> 1.3 1 10 2
>> 2.4 2 4 -2
>> 2.5 2 5 -1
>> 2.6 2 6 0
>> 2.7 2 7 1
>> 2.8 2 8 2
>> 2.9 2 9 3
>> 2.10 2 10 4
>> 2.11 2 11 5
>> 2.12 2 12 6
>> 3.13 3 4 -3
>> 3.14 3 5 -2
>> 3.15 3 6 -1
>> 3.16 3 7 0
>> 3.17 3 8 1
>> 3.18 3 9 2
>> 3.19 3 10 3
>> 3.20 3 11 4
>> >
>> >
>>
>>
>> On Sat, Nov 3, 2012 at 1:14 PM, Christopher Desjardins
>> <cddesjard...@gmail.com> wrote:
>> > Hi,
>> > I have the following data:
>> >
>> >> data[1:20,c(1,2,20)]
>> > idr schyear year
>> > 1 8 0
>> > 1 9 1
>> > 1 10 NA
>> > 2 4 NA
>> > 2 5 -1
>> > 2 6 0
>> > 2 7 1
>> > 2 8 2
>> > 2 9 3
>> > 2 10 4
>> > 2 11 NA
>> > 2 12 6
>> > 3 4 NA
>> > 3 5 -2
>> > 3 6 -1
>> > 3 7 0
>> > 3 8 1
>> > 3 9 2
>> > 3 10 3
>> > 3 11 NA
>> >
>> > What I want to do is replace the NAs in the year variable with the
>> > following:
>> >
>> > idr schyear year
>> > 1 8 0
>> > 1 9 1
>> > 1 10 2
>> > 2 4 -2
>> > 2 5 -1
>> > 2 6 0
>> > 2 7 1
>> > 2 8 2
>> > 2 9 3
>> > 2 10 4
>> > 2 11 5
>> > 2 12 6
>> > 3 4 -3
>> > 3 5 -2
>> > 3 6 -1
>> > 3 7 0
>> > 3 8 1
>> > 3 9 2
>> > 3 10 3
>> > 3 11 4
>> >
>> > I have no idea how to do this. What it needs to do is make sure that for
>> > each subject (idr) that it either adds a 1 if it is preceded by a value
>> in
>> > year or subtracts a 1 if it comes before a year value.
>> >
>> > Does that make sense? I could do this in Excel but I am at a loss for
>> how
>> > to do this in R. Please reply to me as well as the list if you respond.
>> >
>> > Thanks!
>> > Chris
>> >
>> > [[alternative HTML version deleted]]
>> >
>> > ______________________________________________
>> > R-help@r-project.org mailing list
>> > https://stat.ethz.ch/mailman/listinfo/r-help
>> > PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> > and provide commented, minimal, self-contained, reproducible code.
>>
>>
>>
>> --
>> Jim Holtman
>> Data Munger Guru
>>
>> What is the problem that you are trying to solve?
>> Tell me what you want to do, not how you want to do it.
>>
>
>
[[alternative HTML version deleted]]
______________________________________________
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
______________________________________________
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
