# [R] Read 2 rows in 1 dataframe for diff - longitudinal data

arun smartpink111 at yahoo.com
Tue Jun 4 19:18:35 CEST 2013

```
Hi,

By comparing some of the solutions:
set.seed(25)
subid<- sample(30:50,22e5,replace=TRUE)
set.seed(27)
year<- sample(1990:2012,22e5,replace=TRUE)
set.seed(35)
var1<- sample(c(1,3,5,7),22e5,replace=TRUE)
df2<- data.frame(subid,year,var1)
df2<- df2[order(df2\$subid,df2\$year),]
system.time(res<-subset(ddply(df2,.(subid),mutate,delta=c(FALSE,var1[-1]!=var1[-length(var1)])),delta)[,-4])
#  user  system elapsed
# 8.036   0.132   8.188

system.time(res2<-df2[ as.logical( ave( df2\$var1, df2\$subid, FUN=function(x) c( FALSE, x[-1] != x[-length(x)]) ) ), ])
#  user  system elapsed
# 1.220   0.000   1.222
system.time(res3<-df2[with(df2,unlist(tapply(var1,list(subid),FUN=function(x) c(FALSE,diff(x)!=0)),use.names=FALSE)),])
#  user  system elapsed
# 1.729   0.000   1.730
identical(res2,res3)
#[1] TRUE

row.names(res)<-1:nrow(res)
row.names(res2)<-1:nrow(res)
identical(res,res2)
#[1] TRUE

I found half an hour a bit too extreme by comparing the above numbers.

A.K.

David:

6     47 1999   1

should not be included in the output list because, we are trying
to detect changes within the subid's.  1999 was the first year for
subject 47 and changes have to be detected after that year - hence we
were using ddply to group. Your solution ran very fast as expected.

AK- I have a large dataset and your solution is taking too long -
as a matter of fact i had to kill it afte 1/2 hr on a 22K row dataset.

Thanks for the suggestions.

-ST

----- Original Message -----
From: David Winsemius <dwinsemius at comcast.net>
To: arun <smartpink111 at yahoo.com>
Cc: R help <r-help at r-project.org>
Sent: Tuesday, June 4, 2013 11:13 AM
Subject: Re: [R] Read 2 rows in 1 dataframe for diff - longitudinal data

On Jun 3, 2013, at 9:51 PM, arun wrote:

> If it is grouped by "subid" (that would be the difference in the number of changes)
>
> subset(ddply(df1,.(subid),mutate,delta=c(FALSE,var[-1]!=var[-length(var)])),delta)[,-4]
> #   subid year var
> #3     36 2003   3
> #7     47 2001   3
> #9     47 2005   1
> #10    47 2007   3
> A.K.

I'm not sure why the first one retruns integer values from the ave() call but the second version works:

> df1[ ave( df1\$var, df1\$subid, FUN=function(x) c( FALSE, x[-1] != x[-length(x)]) ), ]
subid year var
1      36 1999   1
1.1    36 1999   1
1.2    36 1999   1
1.3    36 1999   1

ave( df1\$var, df1\$subid, FUN=function(x) c( FALSE, x[-1] != x[-length(x)]))
[1] 0 0 1 0 0 0 1 0 1 1

Perhaps one of the single item groups sabotaged my simple function.

> df1[ as.logical( ave( df1\$var, df1\$subid, FUN=function(x) c( FALSE, x[-1] != x[-length(x)]) ) ), ]
subid year var
3     36 2003   3
7     47 2001   3
9     47 2005   1
10    47 2007   3

--
David.
>
>
> ----- Original Message -----
> From: David Winsemius <dwinsemius at comcast.net>
> To: arun <smartpink111 at yahoo.com>
> Cc: R help <r-help at r-project.org>
> Sent: Tuesday, June 4, 2013 12:37 AM
> Subject: Re: [R] Read 2 rows in 1 dataframe for diff - longitudinal data
>
>
> On Jun 3, 2013, at 7:10 PM, arun wrote:
>
>> Hi,
>> May be this helps:
>> res1<-df1[with(df1,unlist(tapply(var,list(subid),FUN=function(x) c(FALSE,diff(x)!=0)),use.names=FALSE)),]
>>   res1
>> #   subid year var
>> #3     36 2003   3
>> #7     47 2001   3
>> #9     47 2005   1
>> #10    47 2007   3
>> #or
>> library(plyr)
>>   subset(ddply(df1,.(subid),mutate,delta=c(FALSE,diff(var)!=0)),delta)[,-4]
>> #   subid year var
>> #3     36 2003   3
>> #7     47 2001   3
>> #9     47 2005   1
>> #10    47 2007   3
>> A.K.
>>
> It's pretty simple with logical indexing:
>
>> df1[ c(FALSE, df1\$var[-1]!=df1\$var[-length(df1\$var)]), ]
>    subid year var
> 3     36 2003   3
> 6     47 1999   1
> 7     47 2001   3
> 9     47 2005   1
> 10    47 2007   3
>
>
> When I count the number of changes in value of var is give me 5. Not sure why you are both leaving out row 6.
>
> --
> David.
>>
>>
>> I need to output a dataframe whenever var changes a value.
>>
>> df1 <- data.frame(subid=rep(c(36,47),each=5),year=rep(seq(1999,2007,2),2),var=c(1,1,3,3,3,1,3,3,1,3))
>>     subid year var
>> 1     36 1999   1
>> 2     36 2001   1
>> 3     36 2003   3
>> 4     36 2005   3
>> 5     36 2007   3
>> 6     47 1999   1
>> 7     47 2001   3
>> 8     47 2003   3
>> 9     47 2005   1
>> 10    47 2007   3
>>>
>>
>> I need:
>> 36 2003   3
>> 47 2001   3
>> 47 2005   1
>> 47 2007   3
>>
>> I am trying to use ddply over subid and use the diff function, but it is not working quiet right.
>>
>>> dd <- ddply(df1,.(subid),summarize,delta=diff(var) != 0)
>>> dd
>>    subid delta
>> 1    36 FALSE
>> 2    36  TRUE
>> 3    36 FALSE
>> 4    36 FALSE
>> 5    47  TRUE
>> 6    47 FALSE
>> 7    47  TRUE
>> 8    47  TRUE
>>
>> I would appreciate any help on this.
>> Thank You!
>> -ST
>>
>> ______________________________________________
>> R-help at r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> and provide commented, minimal, self-contained, reproducible code.
>
> David Winsemius
> Alameda, CA, USA
>

David Winsemius
Alameda, CA, USA

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