I am trying to use the Dunn test for a comparison but I am getting an error: "Error in Psort[1, i] : incorrect number of dimensions"
the data I am trying to use is this sort of idea (but sample size is bigger):
Frequency Height
1 10
2 11
1 9
1 8
2 15
1 9
2 11
2 13
the code I used was
dunnTest(Height ~ Frequency,
data=Data,
method="bh")
is my problem that my frequency is only split into two groups? cause for another factor my frequency had three groups and it worked fine. If this is the problem, is there another test I can do that will perform a similar/the same function?
Thanks!
The Dunn test is equivalent to the Wilcox test (wilcox.test) if you adjust values of input parameters (disable the exact calculation of p value, disable the continuity correction, more here). For your data, one obtains:
> wilcox.test(df$Frequency, df$Height, correct = FALSE, exact = FALSE)
Wilcoxon rank sum test
data: df$Frequency and df$Height
W = 0, p-value = 0.0006346
alternative hypothesis: true location shift is not equal to 0
I think you are using the dunnTest function from the FSA package. This function fails for two groups.
Data
df <- read.table(text="Frequency Height
1 10
2 11
1 9
1 8
2 15
1 9
2 11
2 13", header=TRUE)
Related
I'm new to clmm and run into the following problem:
I want to obtain the optimal sample size for my study with R using powerSim and powerCurve. Because my data is ordinal, I'm using a clmm. Study participants (VPN) should evaluate three sentence types (SH1,SM1, and SP1) on a 5 point likert scale (evaluation.likert). I need to account for my participants as a random factor while the sentence types and the evaluation are my fixed factors.
Here's a glimpse of my data (count of VPN goes up to 40 for each of the parameters, I just shortened it here):
VPN parameter evaluation.likert
1 1 SH1 2
2 2 SH1 4
3 3 SH1 5
4 4 SH1 3
...
5 1 SM1 4
6 2 SM1 2
7 3 SM1 2
8 4 SM1 5
...
9 1 SP1 1
10 2 SP1 1
11 3 SP1 3
12 4 SP1 5
...
Now, with some help I created the following model:
clmm(likert~parameter+(1|VPN), data=dfdata)
With this model, I'm doing the simulation:
ps1 <- powerSim(power, test=fixed("likert:parameter", "anova"), nsim=40)
Warning:
In observedPowerWarning(sim) :
This appears to be an "observed power" calculation
print(ps1)
Power for predictor 'likert:parameter', (95% confidence interval):
0.00% ( 0.00, 8.81)
Test: Type-I F-test
Based on 40 simulations, (0 warnings, 40 errors)
alpha = 0.05, nrow = NA
Time elapsed: 0 h 0 m 0 s
nb: result might be an observed power calculation
In the above example, I tried it with 40 participants but I already also ran a simulation with 2000000 participants to check if I just need a huge amount of people. The results were the same though: 0.0%.
lastResult()$errors tells me that I'm using a method which is not applicable for clmm:
not applicable method for'simulate' on object of class "clmm"
But besides the anova I'm doing here, I've also already tried z, t, f, chisq, lr, sa, kr, pb. (And instead of test=fixed, I've also already tried test=compare, test=fcompare, test=rcompare, and even test=random())
So I guess there must be something wrong with my model? Or are really none of these methods applicaple for clmms?
Many thanks in advance, your help is already very much appreciated!
I'm trying to make an adjusted survival curve based on a weighted cox regression performed on a case cohort data set in R, but unfortunately, I can't make it work. I was therefore hoping that some of you may be able to figure it out why it isn't working.
In order to illustrate the problem, I have used (and adjusted a bit) the example from the "Package 'survival'" document, which means im working with:
data("nwtco")
subcoh <- nwtco$in.subcohort
selccoh <- with(nwtco, rel==1|subcoh==1)
ccoh.data <- nwtco[selccoh,]
ccoh.data$subcohort <- subcoh[selccoh]
ccoh.data$age <- ccoh.data$age/12 # Age in years
fit.ccSP <- cch(Surv(edrel, rel) ~ stage + histol + age,
data =ccoh.data,subcoh = ~subcohort, id=~seqno, cohort.size=4028, method="LinYing")
The data set is looking like this:
seqno instit histol stage study rel edrel age in.subcohort subcohort
4 4 2 1 4 3 0 6200 2.333333 TRUE TRUE
7 7 1 1 4 3 1 324 3.750000 FALSE FALSE
11 11 1 2 2 3 0 5570 2.000000 TRUE TRUE
14 14 1 1 2 3 0 5942 1.583333 TRUE TRUE
17 17 1 1 2 3 1 960 7.166667 FALSE FALSE
22 22 1 1 2 3 1 93 2.666667 FALSE FALSE
Then, I'm trying to illustrate the effect of stage in an adjusted survival curve, using the ggadjustedcurves-function from the survminer package:
library(suvminer)
ggadjustedcurves(fit.ccSP, variable = ccoh.data$stage, data = ccoh.data)
#Error in survexp(as.formula(paste("~", variable)), data = ndata, ratetable = fit) :
# Invalid rate table
But unfortunately, this is not working. Can anyone figure out why? And can this somehow be fixed or done in another way?
Essentially, I'm looking for a way to graphically illustrate the effect of a continuous variable in a weighted cox regression performed on a case cohort data set, so I would, generally, also be interested in hearing if there are other alternatives than the adjusted survival curves?
Two reasons it is throwing errors.
The ggadjcurves function is not being given a coxph.object, which it's halp page indicated was the designed first object.
The specification of the variable argument is incorrect. The correct method of specifying a column is with a length-1 character vector that matches one of the names in the formula. You gave it a vector whose value was a vector of length 1154.
This code succeeds:
fit.ccSP <- coxph(Surv(edrel, rel) ~ stage + histol + age,
data =ccoh.data)
ggadjustedcurves(fit.ccSP, variable = 'stage', data = ccoh.data)
It might not answer your desires, but it does answer the "why-error" part of your question. You might want to review the methods used by Therneau, Cynthia S Crowson, and Elizabeth J Atkinson in their paper on adjusted curves:
https://cran.r-project.org/web/packages/survival/vignettes/adjcurve.pdf
I have some actual data that I am afraid is somewhat nasty.
It's essentially a Positive Negative Binomial distribution (without any zero counts). However, there are some outliers that seem to cause some bad calculations to occur (maybe underflow or NaNs?) The first 8 or so entries are reasonable, but I'm guessing the last few are causing some problems with the fitting.
Here's the data:
> df
counts t
1 1968 1
2 217 2
3 55 3
4 26 4
5 11 5
6 5 6
7 8 7
8 3 8
9 1 10
10 1 11
11 1 12
12 1 13
13 1 15
14 1 18
15 1 26
16 1 59
This command runs for a while and then spits out the error message
> vglm(counts ~ t, data=df, family = posnegbinomial)
Error in if (take.half.step) { : missing value where TRUE/FALSE needed
BUT, if I rerun this cutting off the outliers, I get a solution for posnegbinomial
> vglm(counts ~ t, data=df[1:9,], family = posnegbinomial)
Call:
vglm(formula = counts ~ t, family = posnegbinomial, data = df[1:9,])
Coefficients:
(Intercept):1 (Intercept):2 t
7.7487404 0.7983811 -0.9427189
Degrees of Freedom: 18 Total; 15 Residual
Log-likelihood: -36.21064
If I try the family pospoisson (Positive Poisson: no zero values), I get a similar error "argument is not interpretable as logical".
I do notice that there are a number of similar questions in Stackoverflow about missing values where TRUE/FALSE is needed, but with other R packages. This indicates to me that perhaps the package writers need to better anticipate calculations might fail.
I think your proximal problem is that the predicted means for the negative binomial for your extreme values are so close to zero that they are underflowing to zero, in a way that was not anticipated/protected against by the package authors. (One thing to realize about nonlinear optimization/fitting is that it is always possible to break a fitting method by giving it extreme data ...)
I couldn't get this to work in VGAM, but I'll offer a couple of other suggestions.
plot(log(counts)~t,data=dd)
And eyeballing the data to get an initial estimate of parameter values (at least for the mean model):
m0 <- lm(log(counts)~t,data=subset(dd,t<10))
I thought I might be able to get vglm() to work by setting starting values, but that didn't actually pan out, even when I have fairly good values from other platforms (see below).
glmmADMB
The glmmADMB package can handle positive NB, via family="truncnbinom":
library(glmmADMB)
m1 <- glmmadmb(counts~t, data=dd, family="truncnbinom")
(there are some warning messages ...)
bbmle::mle2()
This requires a little bit more work: it failed with the standard model, but works if I set a floor on the predicted mean ...
library(VGAM) ## for dposnegbin
library(bbmle)
m2 <- mle2(counts~dposnegbin(size=exp(logk),
munb=pmax(exp(logeta),1e-7)),
parameters=list(logeta~t),
data=dd,
start=list(logk=0,logeta=0))
Again warning messages.
Compare glmmADMB, mle2, simple truncated lm fit ...
cc <- cbind(coef(m2),
c(log(m1$alpha),coef(m1)),
c(NA,coef(m0)))
dimnames(cc) <- list(c("log_k","log_int","slope"),
c("mle2","glmmADMB","lm"))
## mle2 glmmADMB lm
## log_k 0.8094678 0.8094625 NA
## log_int 7.7670604 7.7670637 7.1747551
## slope -0.9491796 -0.9491778 -0.8328487
This is in principle also possible with glmmTMB, but it runs into the same kinds of problems as vglm() ...
I am having difficulties understanding how the multiclass.roc parameters should look like.
Here a snapshot of my data:
> head(testing.logist$cut.rank)
[1] 3 3 3 3 1 3
Levels: 1 2 3
> head(mnm.predict.test.probs)
1 2 3
9 1.013755e-04 3.713862e-02 0.96276001
10 1.904435e-11 3.153587e-02 0.96846413
12 6.445101e-23 1.119782e-11 1.00000000
13 1.238355e-04 2.882145e-02 0.97105472
22 9.027254e-01 7.259787e-07 0.09727389
26 1.365667e-01 4.034372e-01 0.45999610
>
I tried calling multiclass.roc with:
multiclass.roc(
response=testing.logist$cut.rank,
predictor=mnm.predict.test.probs,
formula=response~predictor
)
but naturally I get an error:
Error in roc.default(response, predictor, levels = X, percent = percent, :
Predictor must be numeric or ordered.
When it's a binary classification problem I know that 'predictor' should contain probabilities (one per observation). However, in my case, I have 3 classes, so my predictor is a list of rows that each have 3 columns (or a sublist of 3 values) correspond to the probability for each class.
Does anyone know how should my 'predictor' should look like rather than what it's currently look like ?
The pROC package is not really designed to handle this case where you get multiple predictions (as probabilities for each class). Typically you would assess your P(class = 1)
multiclass.roc(
response=testing.logist$cut.rank,
predictor=mnm.predict.test.probs[,1])
And then do it again with P(class = 2) and P(class = 3). Or better, determine the most likely class:
predicted.class <- apply(mnm.predict.test.probs, 1, which.max)
multiclass.roc(
response=testing.logist$cut.rank,
predictor=predicted.class)
Consider multiclass.roc as a toy that can sometimes be helpful but most likely won't really fit your needs.
This question is related to: Selecting Percentile curves using gamlss::lms in R
I can get centile curve from following data and code:
age = sample(5:15, 500, replace=T)
yvar = rnorm(500, age, 20)
mydata = data.frame(age, yvar)
head(mydata)
age yvar
1 12 13.12974
2 14 -18.97290
3 10 42.11045
4 12 27.89088
5 11 48.03861
6 5 24.68591
h = lms(yvar, age , data=mydata, n.cyc=30)
centiles(h,xvar=mydata$age, cent=c(90), points=FALSE)
How can I now get yvar on the curve for each of x value (5:15) which would represent 90th percentiles for data after smoothening?
I tried to read help pages and found fitted(h) and fv(h) to get fitted values for entire data. But how to get values for each age level at 90th centile curve level? Thanks for your help.
Edit: Following figure show what I need:
I tried following but it is correct since value are incorrect:
mydata$fitted = fitted(h)
aggregate(fitted~age, mydata, function(x) quantile(x,.9))
age fitted
1 5 6.459680
2 6 6.280579
3 7 6.290599
4 8 6.556999
5 9 7.048602
6 10 7.817276
7 11 8.931219
8 12 10.388048
9 13 12.138104
10 14 14.106250
11 15 16.125688
The values are very different from 90th quantile directly from data:
> aggregate(yvar~age, mydata, function(x) quantile(x,.9))
age yvar
1 5 39.22938
2 6 35.69294
3 7 25.40390
4 8 26.20388
5 9 29.07670
6 10 32.43151
7 11 24.96861
8 12 37.98292
9 13 28.28686
10 14 43.33678
11 15 44.46269
See if this makes sense. The 90th percentile of a normal distribution with mean and sd of 'smn' and 'ssd' is qnorm(.9, smn, ssd): So this seems to deliver (somewhat) sensible results, albeit not the full hack of centiles that I suggested:
plot(h$xvar, qnorm(.9, fitted(h), h$sigma.fv))
(Note the massive overplotting from only a few distinct xvars but 500 points. Ande you may want to set the ylim so that the full range can be appreciated.)
The caveat here is that you need to check the other parts of the model to see if it is really just an ordinary Normal model. In this case it seems to be:
> h$mu.formula
y ~ pb(x)
<environment: 0x10275cfb8>
> h$sigma.formula
~1
<environment: 0x10275cfb8>
> h$nu.formula
NULL
> h$tau.formula
NULL
So the model is just mean-estimate with a fixed-variance (the ~1) across the range of the xvar, and there are no complications from higher order parameters like a Box-Cox model. (And I'm unable to explain why this is not the same as the plotted centiles. For that you probably need to correspond with the package authors.)