If I understand the problem correctly, then I’d suggest this:
ndays <- nrow(Wb30)
for (iday in 2:ndays) {
Wb30$Water_Balance[iday] <- Wb30$Water_Balance[iday-1] +
Wb30$Rainfall[iday] - Wb30$Evaporation[iday]
Wb30$Water_Balance[iday] <- min(Wb30$Water_Balance[iday], 100)
Wb30$Water_Balance[iday] <- max(Wb30$Water_Balance[iday], 0)
}
It may be possible without a loop, but I think a loop is easier to
understand in the current circumstances. And is likely to be fast enough
that it does not matter.
-
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 10/23/14, 5:33 AM, "Frederic Ntirenganya" <ntfr...@gmail.com> wrote:
>Dear All,
>
>I want to calculate water balance using the following formula:
>Water balance today = Water balance yesterday + Rainfall - Evaporation
>
>This is a sample of data I am using:
>
>head(Wb30)
> May Rainfall Evaporation Water_Balance
>1 7 0 5 0
>2 8 10 5 NA
>3 9 0 5 NA
>4 10 0 5 NA
>5 11 2 5 NA
>6 12 23 5 NA
>
>The following is the loop am trying to use but it is not working well.
>
>#Water balance today = Water balance yesterday + Rainfall - Evaporation
>#If Water balance today < 0 then Water balance today = 0
>#If Water balance today > 100 then Water balance today = 100
>wb=c()
>for (w in 1:length(Wb30$Water_Balance)){
> if(w<0 & w>100){w<-0 & w<-100}
> #print (w);
> for (i in 1:length(Wb30$Rainfall)){
> for (j in 1:length(Wb30$Evaporation)){
> }
> }
> wb<-Wb30$Water_Balance[w] + Wb30$Rainfall[i+1]-Wb30$Evaporation[j+1]
>}
>wb
>
>Any suggest of what I am missing for it to work correctly is welcome.
>
>
>Regards,
>
>Frederic Ntirenganya
>Maseno University,
>African Maths Initiative,
>Kenya.
>Mobile:(+254)718492836
>Email: fr...@aims.ac.za
>https://sites.google.com/a/aims.ac.za/fredo/
>
> [[alternative HTML version deleted]]
>
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>and provide commented, minimal, self-contained, reproducible code.
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