Using dates in R for Theil-Sen

Question

I am trying to use dates as my X variable in a Theil-Sen slope estimation and I am having difficulty using the R package zyp

library(zyp)

myDates <- as.Date(c("2009-11-24","2009-12-03","2010-01-19","2010-02-18","2010-03-04"))
myData <- c(10.17,10.91,11.72,12.5,13.43)
a <- c("a","b","c","d","e")
df = data.frame(a,myData,myDates)
zyp.sen(myDates~myData, df)

And then I get the following error:

    Error in `+.Date`(y - slope * x, intercept) : 
      binary + is not defined for "Date" objects

I am also trying to use zyp.trend.dataframe(df, 1) and I get essentially nothing:

  a lbound trend trendp ubound tau sig nruns autocor valid_frac linear intercept
  1 a     NA    NA     NA     NA  NA  NA    NA      NA         NA     NA        NA
  2 b     NA    NA     NA     NA  NA  NA    NA      NA         NA     NA        NA
  3 c     NA    NA     NA     NA  NA  NA    NA      NA         NA     NA        NA
  4 d     NA    NA     NA     NA  NA  NA    NA      NA         NA     NA        NA
  5 e     NA    NA     NA     NA  NA  NA    NA      NA         NA     NA        NA

I am assuming that I am accessing zyp incorrectly. Does anyone know what I am doing wrong ?

TIA

never used the zyp package, but I'd look at the documentation for zyp.sen. The error says it doesn't like date objects, so check what types are valid for the method. — DMT, Oct 03 '14 at 21:49
Just looked at the docs, not very descriptive but I'd pass the dates in as integers, then convert back after if you want to keep the dates. http://cran.r-project.org/web/packages/zyp/zyp.pdf — DMT, Oct 03 '14 at 21:50
Hmm. From the zyp documentation I interpret "zyp.trend.dataframe: trend the Sen’s slope (trend) per unit time" as time (e.g. date) do you think that is wrong ? — user918967, Oct 03 '14 at 21:52
That's the output, which could be a date, but I don't know. You've got to get to the output though, so you've got to change something to fix your input. Since it complains about myDates being a date object, you'll have to try some other type. Since the example in the documentation uses integers, that's why I suggested it. It is unfortunate they don't specify what type of input is expected — DMT, Oct 03 '14 at 22:00

score 1 · Accepted Answer · edited Oct 04 '14 at 03:43

1

Converting Dates to their underlying integer values should succeed:

myDates <- as.integer( 
as.Date(c("2009-11-24","2009-12-03","2010-01-19","2010-02-18","2010-03-04")))
myData <- c(10.17,10.91,11.72,12.5,13.43)
a <- c("a","b","c","d","e")
df = data.frame(a,myData,myDates)
 zyp.sen(myDates~myData, df)

Call:
NULL

Coefficients:
Intercept     myData  
 14204.66      36.12

edited Oct 04 '14 at 03:43

user918967

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answered Oct 03 '14 at 22:51

IRTFM

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Turns out we had a similar idea. I changed it from dates to days between samples with the following code: myDateDay <- c(0,myDates[2]-myDates[1],myDates[3]-myDates[1],myDates[4]-myDates[1],myDates[5]-myDates[1]) – user918967 Oct 04 '14 at 02:24

Using dates in R for Theil-Sen

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