That makes a lot of sense. Thank you, Rui!
On Mon, Jun 24, 2013 at 3:41 PM, Rui Barradas <ruipbarra...@sapo.pt> wrote:
> Hello,
>
> From the help page for ?adf.test: "The p-values are interpolated from
> Table 4.2, p. 103 of Banerjee et al. (1993)."
>
> I believe it's a problem with your data. Putting a print statement in the
> code for adf.test() gave me the following:
>
>
> Call:
> lm(formula = yt ~ xt1 + 1 + tt + yt1)
>
> Residuals:
> ALL 3 residuals are 0: no residual degrees of freedom!
>
> Coefficients: (1 not defined because of singularities)
> Estimate Std. Error t value Pr(>|t|)
> (Intercept) -24.319 NA NA NA
> xt1 -3.115 NA NA NA
> tt 18.087 NA NA NA
> yt1 NA NA NA NA
>
> Residual standard error: NaN on 0 degrees of freedom
> Multiple R-squared: 1, Adjusted R-squared: NaN
> F-statistic: NaN on 2 and 0 DF, p-value: NA
>
>
>
> And the test statistic becomes NaN (Not a Number). It's computed as
> coefficients[2, 1]/coefficients[2, 2] or -3.115/NA.
> So when it tries to interpolate the p-value, the variable 'interpol' is
> equal to NaN and the if test fails.
>
> With a standard error like NA, maybe you don't have enough data points to
> run the tests. (Only 3 residuals, all zero, like seen above.)
>
>
> Rui Barradas
>
> Em 24-06-2013 20:12, Olga Musayev escreveu:
>
>> Rui-- thanks so much for the help!
>>
>> I'm getting this error though, which is leaving me stumped:
>>
>> test<-lapply(ids, function(i) {
>> if(!any(is.na <http://is.na>(df[df$ID==i,3])**))
>>
>> {adf.test(df[df$ID==i, 3])} else {NA} })
>>
>> Error in if (interpol == min(tablep)) warning("p-value smaller than
>> printed p-value") else warning("p-value greater than printed p-value") :
>> missing value where TRUE/FALSE needed
>>
>> Any idea what this could mean?
>>
>>
>>
>> On Sun, Jun 23, 2013 at 4:54 PM, Rui Barradas <ruipbarra...@sapo.pt
>> <mailto:ruipbarra...@sapo.pt>> wrote:
>>
>> Hello,
>>
>> Sorry, I forgot to Cc the list.
>>
>> Rui Barradas
>>
>> Em 23-06-2013 21:44, Rui Barradas escreveu:
>>
>> Hello,
>>
>> See if the following does what you want.
>>
>> lapply(seq_len(obsv), function(i) adf.test(df[df$ID == i, 3]))
>>
>>
>> Hope this helps,
>>
>> Rui Barradas
>>
>> Em 23-06-2013 19:12, Olga Musayev escreveu:
>>
>> Short question: Is it possible to use statistical tests,
>> like the
>> Augmented
>> Dickey-Fuller test, in functions with for-loops? If not, are
>> there any
>> alternative ways to scale measures?
>>
>> Detailed explanation: I am working with time-series, and I
>> want to flag
>> curves that are not stationary and which display pulses,
>> trends, or level
>> shifts.
>>
>> df
>>
>>
>> DATE ID VALUE2012-03-06 1 5.672012-03-07 1
>> 3.452012-03-08 1 4.562012-03-09 1 20.302012-03-10
>> 1
>> 5.102012-03-06 2 5.672012-03-07 2 3.452012-03-08
>> 2
>> 4.562012-03-09 2 5.282012-03-10 2 5.102012-03-06
>> 3
>> 5.672012-03-07 3 7.802012-03-08 3 8.792012-03-09
>> 3
>> 9.432012-03-10 3 10.99
>>
>> You can see, object 2 is stationary, but 3 exhibits a
>> trend and 1 has a
>> pulse at 3/09.
>>
>> What I want, in pseudo-code:
>>
>> flag<- list()
>> for (i in 1:length(obsv)) {
>> if adf.test(i) FAIL {
>> append(flag, i)
>> }}
>>
>> What I have so far:
>>
>> library(tseries)
>> adf.test(df[which(df$ID==1), 3])
>>
>> Augmented Dickey-Fuller Test
>>
>> data: dataDickey-Fuller = 11.1451, Lag order = 16, p-value
>> = 0.01null
>> hypothesis: non-stationary
>>
>> adf.test(df[which(df$ID==2), 3])
>>
>> Augmented Dickey-Fuller Test
>>
>> data: dataDickey-Fuller = 11.1451, Lag order = 16, p-value
>> = 0.99
>> alternative hypothesis: stationary
>>
>> adf.test(df[which(df$ID==3), 3])Augmented Dickey-Fuller
>> Test
>>
>>
>> data: dataDickey-Fuller = 11.1451, Lag order = 16, p-value
>> = 0.04null
>> hypothesis: non-stationary
>>
>> How can I use this output in a for-loop? Thank you in
>> advance!
>>
>> [[alternative HTML version deleted]]
>>
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