What answer do you want for the following data?
x <- c(2,2,3,4,4,4,4,5,5,5,3,1,1,0,0,0,1,1,1)
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Tue, Dec 22, 2015 at 11:34 PM, Makram Belhaj Fraj <
belhajfraj.mak...@gmail.com> wrote:
> Dear colleagues
> i need your generous help to solve the following problem
>
> I have a soil moisture time series qWC1 (61 values)
> > qWC1
> 75.33336 75.20617 75.20617 74.95275 74.95275 74.70059 74.70059 74.70059
> 74.57498 74.44968 74.32469 74.07563 85.57237 90.40123 90.73760 90.73760
> 90.73760 90.73760 90.90648 91.07582 91.24564 90.90648 86.82135 80.69793
> 79.30393 78.62058 78.21484 77.81226 77.67876 77.41279 77.28032 76.88495
> 76.75383 76.75383 76.49260 76.36249 76.23270 76.23270 76.10325 75.97412
> 75.84532 75.71685 75.71685 75.71685 75.71685 75.46087 75.46087 75.46087
> 75.33336 75.20617 75.20617 75.20617 75.20617 75.20617 75.20617 75.07930
> 75.07930 75.07930 74.95275 74.95275 74.95275
>
> I want to measure consecutive increases corresponding to irrigation and
> consecutive decreases corresponding to recharge
> I wrote the following code and it does not calculate for each increment in
> i?
> also note that I choose to not use diff command in time series because I
> want also that "plateaux" corresponding to a minimum of 2 equal
> consecutive values are accounted as positive differences=irrigations so
> when x[i+1]==x[i] the difference y might be equal to the previous value xi
>
> following the code i wrote
>
> x<-ts(qWC1,start=1, end=61, frequency=1)
> x[1]
> plot(x, type="h", col = "green")
> y<-rep(0,61)
> for (i in 1:61) {
> if (x[i+1] > x[i]){
> y[i]==x[i+1]-x[i]
> } else if (x[i+1]==x[i]){
> y[i]=x[i+2]-x[i]
> } else {
> y[i]==x[i+1]-x[i]
> }
>
> }
> plot(y, type="h", col = "blueviolet")
>
> Many thank
> Makram
>
> [[alternative HTML version deleted]]
>
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