[R] function similar to ddply? + calculations based on previous row
Ista Zahn
istazahn at gmail.com
Thu Feb 16 13:53:03 CET 2012
Hi,
On Wednesday, February 15, 2012 08:02:44 AM Nerak wrote:
> Hi all,
>
> I was wondering if there is a function kind of similar that splits a
> dataframe, applies a function to each row and returns in a data frame. I
> know ddply but this one isn’t useful in this situation.
Why not? Sounds like a description of exactly the thing ddply is designed to
do!
>
> I have a dataframe with values for each day (rows) for different objects
> (columns). I have values for several years. Now, I want to do calculations
> on only the data of that year. With the ddply function you can use as second
> argument Year to split the data frame into years. But the function you use
> is for that whole part so you get only one output value for that year (for
> example the sum all the values in that column belonging to that year).
Not true. ddply is designed to return a data.frame, not a single value. I
think if you read the JSS article (http://www.jstatsoft.org/v40/i01) you'll
find that ddply will do exactly what you want.
Best,
Ista
I
> want to calculate a new value for each day of that year (what would be
> possible with apply if you have only data for one year). I found another
> way to do this by using a for loop (y in Year[,]:Year[length(Year)] and
> test2.dataframe<-test.dataframe[which(year==y)] to select the rows
> belonging to that year on which I make to calculations. The problem is that
> for loops take a lot of time to run and I’m trying to avoid using them
> whenever possible. (Example almost reproducible script below)
>
> I’m also wondering if it’s possible to refer to a value of the row below
> from another data vector or data frame or …. The line I mean in the script
> below is this one (and is the one that is the course that the script doesn’t
> work because that n is not known):
>
> test.number$numberb[y-Year[1]+1]<-length(which(test.starty==1 &
> test.f[(n+1)]== 1 ))
>
> I want that for a certain row, the according value of test.starty (on the
> row with the same number (e.g. n) ) = 1 and the according value of the row
> below row n of test.f ==1. How can I do this without having to loop (which I
> want to learn to avoid as much as possible). I tried to search on
> Rhelpforum already and found:
> http://r.789695.n4.nabble.com/How-to-calc-ratios-base-on-current-and-previou
> s-row-td2341407.html My n+1 is based on the original value so there is
> should be a solution without looping but I don’t understand how I should
> index…
>
> I’ll illustrate what I mean with a loop to solve this kind (different
> script) of problems:
> test[1,]<-ifelse(AAA[1,]>1,1,0)
> for (t in 2:10)
> {
> test <- ifelse(AAA[t,]>1 & AAA[t-1,]==0,1,0)
> }
>
>
> Below you can see how I did it with the for loop and what I want to create:
>
>
> Year<-data.frame(Date=c(1980,1980,1980,1980,1981,1981,1981,1981,1982,1982,19
> 82,1982,1983,1983,1983,1983))
> test.b<-data.frame(C=c(0,0,0,0,5,2,0,0,0,15,12,10,6,0,0,0),B=c(0,0,0,0,9,6,
> 2,0,0,24,20,16,2,0,0,0),F=c(0,0,0,0,6,5,1,0,0,18,16,12,10,5,1,0))
> test.start<-data.frame(C=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0),B=c(0,0,0,0,1,0
> ,0,0,0,1,0,0,0,0,0,0),F=c(0,0,0,0,1,0,0,0,0,1,0,0,0,0,0,0))
>
>
>
> test.2b<-test.b>1
> test.number<-data.frame(c(1980:1983))
> for (l in 1:nrow(test.b))
> {
> for (y in 1980:1983)
> {
>
> test.f<-test.2b[which(Year == y),l]
> test.starty<-test.start[which(Year ==y),l]
> test.number$numberb[y-Year[1]+1]<-length(which(test.starty==1 &
> test.f[(n-1)]== 1 ))
>
> }
> test.number[,l+1]<-cbind(test.number$numberb)
>
> }
>
>
> If someone knows a way to get rid of the loops, let me know! Because I want
> to make this script as fast as possible for larger datasets. I'm trying to
> get through the apply family to find solutions but it's a hard issue.
>
> Many thanks in advance,
> Kind regards,
> Nerak
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/function-similar-to-ddply-calculations-based-
> on-previous-row-tp4390925p4390925.html Sent from the R help mailing list
> archive at Nabble.com.
>
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