I have been learning to use the R RecordLinkage package recently. On a very small example with linking 2 datasets, one with 8 rows and the other with 11, I get the results:
Linkage Data Set
8 records in data set 1
11 records in data set 2
8 record pairs
4 matches
4 non-matches
0 pairs with unknown status
Weight distribution:
[0.4,0.5] (0.5,0.6] (0.6,0.7] (0.7,0.8] (0.8,0.9] (0.9,1]
2 0 2 0 1 3
3 links detected
0 possible links detected
5 non-links detected
alpha error: 0.250000
beta error: 0.000000
accuracy: 0.875000
Classification table:
classification
true status N P L
FALSE 4 0 0
TRUE 1 0 3
What am failing to understand, is the relationship between the alpha error, beta error and accuracy with the Classification table. Where are the figures below coming from exactly, how are they calculated:
alpha error: 0.250000
beta error: 0.000000
accuracy: 0.875000
Any help greatly appreciated
Alpha and beta error are statistical measures, more commonly known as type I and type II error, respectively. In statistical terms, the alpha error is the probability of rejecting the null hypothesis given that it is true; the beta error is the probability of asserting the null hypothesis given that it is not true (compare, for example http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2996198/).
In the case of record linkage, the null hypthesis is that a record pair is a match, i.e. the two records represent the same entity. Thus, the alpha error is the probability of labelling a pair as non-match given that it is really a match (false negative). This error is calculated as: (number of matches classified as 'non-link') / (number of matches).[1] In the above example, there are 4 matches, of which 1 is not recognized, thus, the alpha error is 1 / 4 = 0.25.
Similarly, beta error is the probability of classifying a pair as match given that it is really a non-match (false positive). It is calculated as (number of non-matches classified as 'link') / (number of non-matches). In the above example, there is no false positive classification, so the beta error is 0. Let's assume a different classification table:
classification
true status N P L
FALSE 2 0 2
TRUE 1 0 3
In this case, there are 4 non-matches, of which 2 are falsely classified as links, so the beta error is 2 / 4 = 0.5.
Finally, accuracy is just the proportion of correct classifications among all pairs (see https://en.wikipedia.org/wiki/Evaluation_of_binary_classifiers#Single_metrics). In the classification table from the question, there are 7 correct classifications (4 non-matches, 3 matches), so accuracy is 7 / 8 = 0,875.
[1] I use '(non-)link' instead of '(non-)match' when I mean the outcome of the classification algorithm in contrast to the real status.
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
Using
clear
score group test
2 0 A
3 0 B
6 0 B
8 0 A
2 0 A
2 0 A
10 1 B
7 1 B
8 1 A
5 1 A
10 1 A
11 1 B
end
I want to scatter plot mean score by group for each test (same graph) with confidence intervals (the real data has thousands of observations). The resulting graph would have two sets of two dots. One set of dots for test==a (group==0 vs group==1) and one set of dots for test==b (group==0 vs group==1).
My current approach works but it is laborious. I compute all of the needed statistics using egen: the mean, number of observations, standard deviations...for each group by test. I then collapse the data and plot.
There has to be another way, no?
I assumed that Stata would be able to take as its input the score group and test variables and then compute and present this pretty standard graph.
After spending a lot of time on Google, I had to ask.
Although there are user-written programs, I lean towards statsby as a basic approach here. Discussion is accessible in this paper.
This example takes your data example (almost executable code). Some choices depend on the large confidence intervals implied. Note that if your version of Stata is not up-to-date, the syntax of ci will be different. (Just omit means.)
clear
input score group str1 test
2 0 A
3 0 B
6 0 B
8 0 A
2 0 A
2 0 A
10 1 B
7 1 B
8 1 A
5 1 A
10 1 A
11 1 B
end
save cj12 , replace
* test A
statsby mean=r(mean) ub=r(ub) lb=r(lb) N=r(N), by(group) clear : ///
ci means score if test == "A"
gen test = "A"
save cj12results, replace
* test B
use cj12
statsby mean=r(mean) ub=r(ub) lb=r(lb) N=r(N), by(group) clear : ///
ci means score if test == "B"
gen test = "B"
append using cj12results
* graph; show sample sizes too, but where to show them is empirical
set scheme s1color
gen where = -20
scatter mean group, ms(O) mcolor(blue) || ///
rcap ub lb group, lcolor(blue) ///
by(test, note("95% confidence intervals") legend(off)) ///
subtitle(, fcolor(ltblue*0.2)) ///
ytitle(score) xla(0 1) xsc(r(-0.25 1.25)) yla(-10(10)10, ang(h)) || ///
scatter where group, ms(none) mla(N) mlabpos(12) mlabsize(*1.5)
We can't compare your complete code or your graph, because you show neither.
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'm using the mclust() package for R to classify a univariate dataset, then assign classifications to new data using the Discriminate Analysis functionality. When trying to calculate the cross validation error rate using cv.MclustDA() I keep getting an error. Below is the code and error. The model object works fine, but something isn't right with doing the cross validation on that model object. Can anyone shed any light on what this error is? It's obviously failing based on the nfolds= argument, but changing the number doesn't help.
> DA_mclust_AmazData_3group=MclustDA(data=Combined_AmazData[,4], class=Combined_AmazData[,13])
> summary(DA_mclust_AmazData_3group)
------------------------------------------------
Gaussian finite mixture model for classification
------------------------------------------------
MclustDA model summary:
log.likelihood n df BIC
-Inf 29 18 -Inf
Classes n Model G
1 12 E 4
2 8 X 1
3 9 E 4
Training classification summary:
Predicted
Class 1 2 3
1 12 0 0
2 0 8 0
3 0 0 9
Training error = 0
>
> cv.MclustDA(DA_mclust_AmazData_3group)
cross-validating...
|
| 0%Error in data[-folds[[i]], , drop = FALSE] :
incorrect number of dimensions
The administrator for the Mclust package got in touch with me. This was a bug in the handling of the univariate case and will be fixed in the next version, which should be posted to CRAN soon.