It's a bug resulting from the new svyquantile() implementation. It's fixed in
the development version, which you can get from r-forge here:
https://r-forge.r-project.org/R/?group_id=1788
-thomas
Thomas Lumley
Professor of Biostatistics
From: Anthony
No, you can't (at the moment), though it shouldn't be too hard to extend.
I can't run your example, though. I get:
Error in eval(expr, envir, enclos) : object 'M' not found
-thomas
Thomas Lumley
Professor of Biostatistics
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From: Chris
number (number 4 in my exemple)
?
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rank tests under complex
sampling. Biometrika, 100, 831-842.
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to compute probabilities, which are then used to compute weights.
The code works in terms of probabilities because that's fairly
standard in textbooks. It makes it easier for me to get the formulas
right.
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on
the database backend part of the process.
You might try MonetDB and its R interface -- it is fast for
aggregation operations, and either the current version or the upcoming
version has dplyr support.
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=L96ludyhFBsC
(look for single in the whole book to find it).
Or set options(survey.lonely.psu) to one of the other values. But merging
strata is probably better.
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the variables in your dataset is to use the database-backed designs and put
the data in something like SQLite.
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many people would have to understand how they
are scoped.
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\\2\n,
fake
)
)
That is, match three word/period sequences, match a word, match a period,
and output the first two things.
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.
This was a recent (well, 2007) change in behaviour. Previously the function
did some tricks to make either approach work, which could be described as
'clever' or 'too clever by half'.
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anesthetictransfusionunits 0.1964 0.0214 9.17 0.0001
From: Thomas Lumley tlum...@uw.edu
Date: Tuesday, February 25, 2014 at 3:09 PM
To: Nathan Leon Pace, MD, MStat n.l.p...@utah.edu
Cc: r help list r-help@r-project.org
Subject: Re: [R] SEs rms cph vs survey svycoxph
On Tue, Feb
as sampling
weights.
Actually, lm() treats them as precision weights, not sampling weights, but
that's still the explanation.
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with survfit.coxph for independent
sampling when the model fits well, but are larger when the model fits
poorly.
That is, the note is for the survival curve rather than the coefficients.
It's still surprising that there's a big difference, but I think we need
more information.
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this accelerated-failure parametrisation of the Weibull with
survreg() in the survival package.
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(including quantiles) for
unequal-probability samples.
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, and effectively
constant average time to query (inverse-Ackerman amortized complexity).
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for a month or so).
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to use sampling weights in the past,
but according to post I found online from Thomas Lumley in mid-2012, R is
currently not equipped to be able to do this.
His post is here:
http://r.789695.n4.nabble.com/sampling-weights-for-multilevel-models-tp4632947p4632955.html
Does anyone know
, schrieb Thomas Lumley:
On Fri, May 3, 2013 at 2:27 AM, Sebastian Weirich
sebastian.weir...@iqb.hu-berlin.de wrote:
Hello,
I want to specify a linear regression model in which the metric outcome
is predicted by two factors and their interaction. glm() computes effects
for each factor level
, and what actually
happened, as the posting guidelines request.
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information at http://www.biostat.washington.edu/suminst/sisg/schedule
-thomas
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' is not lognormal. It is a model with normal
errors and log link, ie.
y ~ N(mu, sigma^2)
log(mu) = x \beta
-thomas
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reproduce the
reported error.
-thomas
-pd
On Apr 12, 2013, at 04:51 , Thomas Lumley wrote:
I don't get an error message (after I correct the missing line break
after
the comment
b- sapply(a, Cfun, upper=1)
b
[1] 1.583458e-54 7.768026e-50 2.317562e-45 4.206260e-41
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. It shouldn't be too hard to get it to extend that to
the last censoring time, but the reason it isn't too hard is that the curve
and standard error estimates don't change after the last failure time, so
it won't be particularly useful.
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sample by sampling from it with probability inversely proportional to the
original sampling probability.
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:
capacity 10^3 m^3
(with ^ denoting superscript (i.e. each '3' as superscript).
What did you try?
Anyhow, this works
text(1,1,expression(capacity~10^3~m^3))
-thomas
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that there cannot be any ties in the differences. If there are
ties, the function uses a Normal approximation.
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in all three cases.
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at differences in
medians. A permutation test or a bootstrap confidence interval is probably
the best way to do this.
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.
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whether there's some strange non-convexity going on or
whether the variable is just being put into the calculations backwards.
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to the sum square total in both the
Y scale and Z scale. Y is a normal distribution and Z is log normally
distributed. Where is the error?
Also, is there a way to calculate the weighted sum square?
-thomas
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at all if yyrandom[1] is more of a problem.
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or other
order statistic and thus write down an integral that gives the exact
variance. Better still, it's a polynomial, so you could evaluate the
integral exactly.
-thomas
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the branch cut goes on
the phase coordinate.
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in the
fourth decimal place, the matrix stops being singular.
-thomas
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] into the formula with
survdiff(eval(bquote(Surv(survival.m, survival) ~ .(names[i]))), data=svsv)
Even after you fix that, there's another problem, which is that your
code doesn't actually use the result from survdiff() in any way.
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(vv)))
api00api99
105.7489 112.8504
## delta method for standard error of square root of variance
sqrt(vcov(vv)[api00,api00]/(4*coef(vv)[api00,api00]))
[1] 6.555219
-thomas
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unit, right?
and I am estimating an average among voting units?
You want a ratio estimator
svyratio(~candidate1, ~totalVotes, design)
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the function.
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. the simple
model just with an intercept and the variable GNI is not shown in the plot,
why?
You asked for the two best models of each size, so you get the two
best models of each size.
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information on the
individual level we are confident to be able to correct the sample weights
for that.
That sounds plausible
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Why have you asked this question three times?
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--+---
Male | 47.96.585346.8149.11
Female | 52.04.585350.8953.19
On Sun, Sep 23, 2012 at 4:30 PM, Thomas Lumley tlum...@uw.edu wrote:
On Sat, Sep 22, 2012 at 2:51 AM, Anthony Damico ajdam...@gmail.com
they aren't]
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Harrell comes to mind) like it because they believe stochastic
ordering is a reasonable assumption in the problems they work in, not
because they think you can do non-parametric testing in its absence.
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the shift is in the same direction across the whole
distribution.
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for all errors.
tryCatch() is also quieter.
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parametrisation. In a model with only an intercept, that would
be exp(intercept).
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(df)
b - df$str == 12
This saves more time than I expected, about 100ms per evaluation on my
computer.
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firm=, firms, and date=, begindts, and date =, enddts)
lapply(queries, sqlQuery, channel=my.database.connection)
will return a list of data frames, one for each set of values.
-thomas
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numbers from such a distribution?
Not directly, as far as I know, but you can easily simulate X|Xc by
transforming uniform random numbers using the inverse CDF, and Y|X=x
is univariate Normal with mean linear in x and variance independent of
x.
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#---
test.statistic dfp.value
1 18.82551 5 0.00207139
(At least I got internal consistency. I see you copied Thomas Lumley, which
is a good idea. I'll be happy to get corrected on any point. I'm adding the
maintainer of epiR to the recipients.)
Yes, that will give internal
won't be appropriate for
the survey design.
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-subset(alldata, !is.na(LBXTC))
design1 - svydesign(id=~SDMVPSU,
strata=~SDMVSTRA,nest=TRUE,weights=~WTMEC2YR,data=somedata)
svyby(~HI_CHOL,~race+RIAGENDR,design=subset(postStratify(design1,~race+RIAGENDR+agecat,racegenderage),RIDAGEYR=20),svymean,na.rm=TRUE)
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The function step() uses AIC. As far as I know, no-one has yet
constructed valid analogues of AIC,BIC,CIC, ... under complex sampling
(Alastair Scott and I are looking into it), so p-values are the only
option, making the process even less useful.
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()
Since the problem seems to be data-dependent, and happens with fairly
high frequency, you might want to use trace() to stick some sort of
data summary in before the call to rqbr, to see if anything obvious is
wrong with the data being passed in before things go wrong.
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();
REAL(a)[0]=rgamma(5000,1);
PutRNGstate();
UNPROTECT(1);
return (a);
}
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the ANSI standard actually *required* a diagnostic for the
incompatible return types.
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I'd be prepared to
implement it anyway.
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. The example on the
withReplicates() help page shows how to do this for quantile
regression, and it should be similar.
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more than observations with low weight. Your
code does the opposite.
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formatted the code in the call to
eval(), and the good formatting is how R would print it without any
source constraints, so this is probably something to do with the
keep.source= argument source().
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(), design=mydesign)
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,
you need some other software.
I do have longer-term plans to add multilevel modelling capabilities
to the survey package, but it's harder than it may appear.
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with the package vcd assocstats but
without considering the survey package.
You can use svytable() to generate an estimated population table and
then feed that to assocstats().
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, by setting nbest fairly large
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2 3 4 ...
..$ : NULL
As the help says, the default is predictions of the linear predictor.
To get predictions of the probability, use type=response
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=
M3_11_AUTOPERCEPCIONSALUDGENERAL)
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I may be overthinking this and the answer is much simpler)
You don't need to create two new variables; you just need a year variable
svychisq(~MyVar+Year, BothYears, statistic=”Chisq”, na.rm= TRUE)
tests whether MyVar is independent of Year.
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structures of the package: if you
want to see the weights, use the weights() function.
The components of $postStrata are used in standard error computations.
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Professor of Biostatistics
University of Auckland
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has been evaluated.
-thomas
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Thomas Lumley
Professor of Biostatistics
University of Auckland
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PLEASE do read the posting guide http://www.R-project.org/posting
the same model. Like SAS, Stata by default assumes
that random effects are independent of each other, so your Stata model
has correlation between the random effects forced to zero. The R
model estimates the correlation, and finds it to be far from zero
(-0.69).
-thomas
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Thomas Lumley
Professor
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
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Thomas Lumley
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