[R] need for help for solving operations in a vector

[Makram Belhaj Fraj]

Hi Petr,
I apologize It was my first time using r-help so I didn't know how to
replay to email to all or not,
I am replaying to all for this email,

Many thanks for the code, I am trying to use it,
please find attached the data file in csv,

following are the first lines of code to read the data and calculate qWC1

sdata<-read.csv("almondc_10augv.csv",head=TRUE,sep=",")

tint=sdata$scan #time intervall

mtime=sdata$mtime #measurement time

v1=sdata$vwc1 #value of moisture in percent

qWC1=200*v1 #conversion in mm to get the variable I am working on



On Tue, Dec 29, 2015 at 11:32 AM, PIKAL Petr <petr.pikal at precheza.cz> wrote:

> Hi
>
>
>
> You shall send your posts to the list, others can answer your questions
> and not only you can benefit from their answers too.
>
>
>
> As you did not post any data it is hard to say what are your issues. I
> believe that there are several values which are the same not only near the
> peak but also at the bottom part. If your data look like I remember and you
> want to keep all values near the peak value regardless they are slightly
> growing or falling, one approach can be to identify peak value, and select
> all values near the peak (the threshold is up to you).
>
>
>
> something like that
>
>
>
> sss<-smooth.spline(temp$theta, temp$int, nknots=length(temp$int)/2)
>
> peak <- which.max(predict(sss)$y)
>
> baseline <- min(predict(sss)$y)
>
> vyska <- predict(sss)$y[peak]
>
> # 50% threshold
>
> HM <- (vyska+baseline)/2
>
> plot(temp$theta, temp$int, col=(temp$int>HM)+1, pch=19)
>
> #10% threshold
>
> HM<-vyska-(vyska*.1)
>
> plot(temp$theta, temp$int, col=(temp$int>HM)+1, pch=19)
>
>
>
> > dput(temp)
>
> structure(list(theta = c(28.995, 29.005, 29.015, 29.025, 29.035,
>
> 29.045, 29.055, 29.065, 29.075, 29.085, 29.095, 29.105, 29.115,
>
> 29.125, 29.135, 29.145, 29.155, 29.165, 29.175, 29.185, 29.195,
>
> 29.205, 29.215, 29.225, 29.235, 29.245, 29.255, 29.265, 29.275,
>
> 29.285, 29.295, 29.305, 29.315, 29.325, 29.335, 29.345, 29.355,
>
> 29.365, 29.375, 29.385, 29.395, 29.405, 29.415, 29.425, 29.435,
>
> 29.445, 29.455, 29.465, 29.475, 29.485, 29.495, 29.505, 29.515,
>
> 29.525, 29.535, 29.545, 29.555, 29.565, 29.575, 29.585, 29.595,
>
> 29.605, 29.615, 29.625, 29.635, 29.645, 29.655, 29.665, 29.675,
>
> 29.685, 29.695, 29.705, 29.715, 29.725, 29.735, 29.745, 29.755,
>
> 29.765, 29.775, 29.785, 29.795, 29.805, 29.815, 29.825, 29.835,
>
> 29.845, 29.855, 29.865, 29.875, 29.885, 29.895, 29.905, 29.915,
>
> 29.925, 29.935, 29.945, 29.955, 29.965, 29.975, 29.985, 29.995
>
> ), int = c(329L, 330L, 318L, 287L, 315L, 344L, 333L, 324L, 334L,
>
> 366L, 339L, 374L, 375L, 335L, 415L, 371L, 413L, 382L, 408L, 406L,
>
> 407L, 440L, 475L, 465L, 516L, 510L, 490L, 550L, 663L, 647L, 628L,
>
> 721L, 789L, 814L, 890L, 923L, 1085L, 1102L, 1222L, 1356L, 1521L,
>
> 1729L, 1868L, 2120L, 2491L, 2656L, 3196L, 3599L, 4128L, 4536L,
>
> 5043L, 5310L, 5638L, 5792L, 5699L, 5374L, 4886L, 4473L, 4293L,
>
> 3757L, 3319L, 2934L, 2422L, 1998L, 1753L, 1397L, 1163L, 972L,
>
> 854L, 775L, 648L, 695L, 616L, 553L, 554L, 509L, 530L, 483L, 482L,
>
> 406L, 451L, 422L, 403L, 393L, 396L, 348L, 367L, 428L, 345L, 384L,
>
> 330L, 342L, 312L, 313L, 323L, 328L, 340L, 322L, 330L, 305L, 311L
>
> )), .Names = c("theta", "int"), row.names = 100:200, class = "data.frame")
>
> >
>
>
>
> Cheers
>
> Petr
>
>
>
>
>
> *From:* Makram Belhaj Fraj [mailto:belhajfraj.makram at gmail.com]
> *Sent:* Tuesday, December 29, 2015 5:54 AM
> *To:* PIKAL Petr
> *Subject:* Re: [R] need for help for solving operations in a vector
>
>
>
> Hi Petr,
>
> I would like to thank you for your time
>
> i choose the following modification on the plotting so to keep only
> successive equals in red
>
> plot(qWC1, col=(c(0, diff(qWC1))==0 )+1)
>
> then, the red points I want to include in the irrigation period are the 3
> successive red points in the plateau (equal values close to the max)
>
> These points are very important as they are corresponding to saturation,
> we continue to irrigate with the same flow and values remains constant for
> a certain time, so I must add them to irrigation
>
>
>
> up to now I didn't succeed in adding this plateau values to
>  theirrigations using the diff(qWC1, lag=1)
>
> however, I wrote also some loops trying to catch all the irrigation and
> recharge separately and I still have some issues,
>
> following are the loops I used, with comments corresponding to the issues
>
> x<-qWC1
>
> length(x)
>
> irrig<-rep(1,61)
>
> for (i in 2:61) {
>
> if (x[i-1]<x[i]){
>
>     irrig[i]<-x[i]-x[i-1]
>
> }
>
> }
>
> rech<-rep(1,61)
>
> for (i in 2:61) {
>
> if (x[i-1]>x[i]){
>
>     rech[i]<-x[i-1]-x[i]
>
> }
>
> }
>
> plot(x, type = "l", col = "black", ylim = c(min(0), max(92)))
>
> lines(irrig, type = "l", col = "cornflowerblue", ylim = c(min(0),
> max(15))) lines(rech, type = "l", col = "brown", ylim = c(min(0), max(15)))
>
> temp<-irrig<1.000001 #logical command to identify low values into a
> temporary vector temp
>
> temp2<-as.numeric(irrig>1.000001) #logical command to identify high values
> with 1
>
> temp2
>
> temp3<-as.numeric(rech>1.000001)
>
> temp3
>
> irrig2<-irrig*temp2  #remove values inf 1 mm inirrig
>
> rech2<-rech*temp3    #same for rech
>
> plot(irrig2, type = "l", col = "cornflowerblue", ylim = c(min(0),
> max(15))) lines(rech2, type = "l", col = "brown", ylim = c(min(0), max(15)))
>
> Many thanks for your time and intellectual generosity
>
> Cheers
>
> Makram
>
>
>
> On Mon, Dec 28, 2015 at 12:15 PM, PIKAL Petr <petr.pikal at precheza.cz>
> wrote:
>
> Hi
>
> On top of answers you have got here is some plotting you need to answer
> yourself
>
> plot(qWC1, col=(c(0, diff(qWC1))>=0 )+1)
>
> Which from those red points you want to be included in irrigation period?
> All of them? Only part? Which part?
>
> Based on your figures you probably will not get 100% correct answer.
>
> Cheers
> Petr
>
>
>
> > -----Original Message-----
> > From: R-help [mailto:r-help-bounces at r-project.org] On Behalf Of Makram
> > Belhaj Fraj
> > Sent: Wednesday, December 23, 2015 8:35 AM
> > To: r-help at r-project.org; r-help-owner at r-project.org
> > Subject: [R] need for help for solving operations in a vector
> >
>
> >  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
> >
>
>
>
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