Re: [R] NLME questions -- interpretation of results

From: <ctu_at_bigred.unl.edu>
Date: Wed, 02 Jul 2008 23:44:25 -0500

Hi Jenny, (I use the data you provide in the previous e-mail) For the 1st question, let me assume you only want to compare loc: A vs. B So you could specified your code like this: fmAB <- nlme(Y ~ SSlogis(X, Asym, R0, lrc),data = LAST,

             random = Asym ~1,
             fixed = Asym+R0+lrc ~ loc,
             start=c(0.97,0,
                     1.14,0,
                    -0.18,0))

> summary(fmAB)

Nonlinear mixed-effects model fit by maximum likelihood

   Model: Y ~ SSlogis(X, Asym, R0, lrc)
  Data: LAST

         AIC BIC logLik
   -31.02303 -19.70017 24.51152

Random effects:
  Formula: Asym ~ 1 | loc

         Asym.(Intercept)
StdDev:     1.549779e-06

  Formula: Asym ~ 1 | dir %in% loc
         Asym.(Intercept)   Residual
StdDev:     6.124237e-08 0.09426086

Fixed effects: Asym + R0 + lrc ~ loc
                       Value Std.Error DF   t-value p-value
Asym.(Intercept)  0.9477404 0.1037337 14  9.136280  0.0000
Asym.locB         0.0982456 0.4175971 14  0.235264  0.8174
R0.(Intercept)    1.1289387 0.0652189 14 17.310003  0.0000
R0.locB           0.1390656 0.3946717 14  0.352358  0.7298
lrc.(Intercept)  -0.2110057 0.0656513 14 -3.214036  0.0062
lrc.locB         -0.0820484 0.6826382 14 -0.120193  0.9060

Then you know Asym, R0, and lrc of loc B are not significant. Moreover, you can test the joint fixed effect by anova(fmAB)(Pinherio and Bate, 2000 Book, p 374)

for the 2nd question, How to get the fitted value for particular level? Based on this example, let me assume you want to get the fitted value of A/N.

then you could write a small code like this:
> FV<-data.frame(F.V=fitted(fmAB), group=summary(fmAB)$groups$dir)
> A.N<-FV[is.element(FV$group, c("A/N")),]

          F.V group
9 0.4209011 A/N
10 0.2726129 A/N

hope this is helpful~

Chunhao

Quoting Jenny Sun <jenny.sun.sun_at_gmail.com>:

> Thank you for your reply Chunhao!
>
> I attached only part of the test data and that is why you might not
> be able to get convergence. Sorry.
>
> I have a couple more questions:
>
> For the second question you answered, how to specify the correct
> length of starting values. I tried using the length of levels in
> each of the parameters in the start list but found:
>
>> fm1 <- nlme(DIFN ~ SSlogis(SVA, Asym, R0, lrc),data = LAST,fixed =
>> Asym + R0 + lrc ~ dir %in% loc,random = Asym ~ 1,start =list(Asym =
>> c(1,1,1,1), R0 = c(1,1,1,1), lrc = c(-5,-2,-2,-2)))
> Error in nlme.formula(DIFN ~ SSlogis(SVA, Asym, R0, lrc), data = LAST, :
> start must have a component called "fixed"
>
> I've got two loc levels (A,B) with four group levels(N,E,S,W); How
> I am gonna define the list and the component called"fixed"?
>
> My another question is about the fitted value of the model. If I
> want to calculate adjusted R square, I have to get fitted(fm1).
> WHich has values like this;
>
> >fitted(fm1)[1:40]
> AB/N AB/N AB/E AB/S AB/W AB/W
> AB/W AB/W AB/W AB/W
> 0.6541876 0.7421748 0.8408251 0.5879220 0.4889387 0.6129576
> 0.5097593 0.6195679 0.5152567 0.5680860
> AB/W AB/W AB/W AB/W AB/W AB/W
> AB/N AB/N AB/N AB/E
> 0.4724423 0.8128148 0.7674529 0.7106698 0.6553155 0.6074771
> 0.5036201 0.5464105 0.6062978 0.6878438
> AB/N AB/N AB/N AB/S AB/S AB/S
> AB/S AB/S AB/S AB/S
> 0.7792725 0.8411961 0.7942503 0.7354845 0.5895700 0.6781973
> 0.6286886 0.5212052 0.8864748 0.8370021
> AB/S AB/S AB/N AB/N AB/N AB/N
> AB/E AB/E AB/E AB/E
> 0.7750731 0.7147024 0.6625288 0.5492599 0.5959280 0.6612426
> 0.7501786 0.8498928 0.6274681 0.7118615
>
> My question is how to get the fitted values for specified group
> levels (eg. values for AB/E)?
>
> Thank all very much!
>
> Jenny
>
>
>> Hi Jenny,
>> I try your code but I did not get in converge in fm3 (see the below).
>> For the first question, you could use fm1 to interpret the result
>> without bothering fm2 and fm3. It means that R0 and lrc can be treated
>> as pure fixed effects (Pinherir and Bates, 2000 Book).
>>
>> For the second question, your want to know "is AB/E different from the AB/S"
>>
>> The simplest way is to change your fixed statement:
>> fixed = Asym+R0+lrc ~ dir %in% loc
>> and specify the correct length of starting values.
>>
>> If I am wrong please correct me~
>>
>> Hope this helpful.
>>
>> Chunhao Tu
>>
>>> test<-read.table(file="C:\\Documents and
>>> Settings\\ado_cabgfaculty\\Desktop\\sun.txt", header=T)
>>> LAST<-groupedData(Y~X|loc/dir, data=test)
>>>
>>> fm1 <- nlme(Y ~ SSlogis(X, Asym, R0, lrc),data = LAST,
>> + random = Asym ~1,
>> + fixed = Asym+R0+lrc ~ 1,
>> + start=c(Asym = 0.97, R0 = 1.14, lrc = -0.18))
>>> fm2 <- update(fm1, random = pdDiag(Asym + R0 ~ 1))
>>> fm3 <- update(fm2, random = pdDiag(Asym+R0+lrc~ 1))
>> Error in nlme.formula(model = Y ~ SSlogis(X, Asym, R0, lrc), data = LAST, :
>> Step halving factor reduced below minimum in PNLS step
>>
>>
>>
>>
>> Quoting Jenny Sun <jenny.sun.sun_at_gmail.com>:
>>
>>> My special thanks to Chunhao Tu for the suggestions about testing
>>> significance of two locations.
>>>
>>> I used logistic models to describe relationships between Y and X at
>>> two locations (A & B). And within each location, I have four groups
>>> (N,E,S,W)representing directions. So the test data can be arranged as:
>>>
>>> Y X dir loc
>>> 0.6295 0.8667596 S A
>>> 0.7890 0.7324820 S A
>>> 0.4735 0.9688875 S A
>>> 0.7805 1.1125239 S A
>>> 0.8640 0.9506174 E A
>>> 0.9445 0.6582157 E A
>>> 0.8455 0.5558860 E A
>>> 0.9380 0.3304870 E A
>>> 0.4010 1.1763090 N A
>>> 0.2585 1.3202890 N A
>>> 0.3750 1.1763090 E A
>>> 0.3855 1.3202890 E A
>>> 0.3020 1.1763090 S A
>>> 0.2300 1.3202890 S A
>>> 0.3155 1.1763090 W A
>>> 0.8890 0.6915861 W B
>>> 0.9185 0.6149019 W B
>>> 0.9275 0.5289258 W B
>>> 0.8365 0.9507088 S B
>>> 0.7720 0.8842165 N B
>>> 0.8615 0.8245123 N B
>>> 0.9170 0.7559687 W B
>>> 0.9590 0.6772720 W B
>>> 0.9900 0.5872023 W B
>>> 0.9940 0.4849064 W B
>>> 0.7500 0.9560776 W B
>>>
>>>
>>> The data is grouped using:
>>>
>>>> LAST<-groupedData(Y~X|loc/dir, data=test)
>>>
>>> I then used logistic models to define the relationship between Y and
>>> X, and got fm1, fm2, and fm3 as follows:
>>>
>>> --------------------------
>>>> fm1 <- nlme(DIFN ~ SSlogis(SVA, Asym, R0, lrc),data = LAST,fixed =
>>>> Asym + R0 + lrc ~ 1,random = Asym ~ 1,start =c(Asym = 1, R0 = 1,
>>>> lrc = -5))
>>>> fm2 <- update(fm1, random = pdDiag(Asym + R0 ~ 1))
>>>> fm3 <- update(fm2, random = pdDiag(Asym + R0 + lrc ~ 1))
>>>> anova(fm1,fm2,fm3)
>>> ------------------------------------------------------------
>>> ANOVA showed:
>>>
>>>> anova(fm1,fm2,fm3)
>>> Model df AIC BIC logLik Test L.Ratio p-value
>>> fm1 1 7 -1809.913 -1774.304 910.9564
>>> fm2 2 9 -1805.774 -1758.295 910.8871 1 vs 2 0.1386696 0.9999
>>> fm3 3 12 -1801.822 -1742.473 910.9109 2 vs 3 0.0475543 0.9666
>>>
>>> ** question: do the results show that fm1 could represent the
>>> results of fm2 and fm3?
>>>
>>>> coef(fm1)
>>> Asym R0 lrc
>>> AB/E 0.9148927 1.389432 -0.3009858
>>> AB/N 0.8775250 1.389432 -0.3009858
>>> AB/S 0.9247592 1.389432 -0.3009858
>>> AB/W 0.8479180 1.389432 -0.3009858
>>> BC/E 0.8791908 1.389432 -0.3009858
>>> BC/N 0.8414229 1.389432 -0.3009858
>>> BC/S 0.9169323 1.389432 -0.3009858
>>> BC/W 0.8817838 1.389432 -0.3009858
>>>
>>> ** question: how could I know if any of the models is significantly
>>> different from the other ones? (eg. AB/E is different from the AB/S)?
>>>
>>>> summary(fm1)
>>> Nonlinear mixed-effects model fit by maximum likelihood
>>> Model: DIFN ~ SSlogis(SVA, Asym, R0, lrc)
>>> Data: LAST
>>> AIC BIC logLik
>>> -1809.913 -1774.304 910.9564
>>>
>>> Random effects:
>>> Formula: Asym ~ 1 | loc
>>> Asym
>>> StdDev: 2.303402e-05
>>>
>>> Formula: Asym ~ 1 | dir %in% loc
>>> Asym Residual
>>> StdDev: 0.03208693 0.1741559
>>>
>>> Fixed effects: Asym + R0 + lrc ~ 1
>>> Value Std.Error DF t-value p-value
>>> Asym 0.8855531 0.015375906 2783 57.59355 0
>>> R0 1.3894322 0.009418047 2783 147.52869 0
>>> lrc -0.3009858 0.012833066 2783 -23.45393 0
>>> Correlation:
>>> Asym R0
>>> R0 -0.440
>>> lrc -0.452 0.150
>>>
>>> Standardized Within-Group Residuals:
>>> Min Q1 Med Q3 Max
>>> -4.1326757 -0.6117037 0.1082112 0.6575250 3.3297270
>>>
>>> Number of Observations: 2793
>>> Number of Groups:
>>> loc dir %in% loc
>>> 2 8
>>>
>>>
>>> I have marked all the codes and questions(**). Any answers and
>>> suggestions are appreciated.
>>>
>>> Have a good day!
>>>
>>> Jenny
>>>
>>>
>
>



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