Displaying 20 results from an estimated 54 matches for "0.146".
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0.14
2004 Jun 23
4
CRIS port of klibc
klibc now runs on the CRIS archtitecture. The patch below is against
0.146. Most of the changes are trivial but the new files in libgcc
requires some comments.
__negdi2.c: The CRIS port fallbacks to C-code for negdi2. The code in
__negdi2.c is copied from libgcc2.c but with modified typenames.
I would really appreciate if someone could check if it seams sane.
crisarith.c: CRIS has no built-in
2009 Nov 03
2
about the cox result
Hi all:
I finished cox analysis like this:
fit_cox<-coxph(Surv(dat$Time, dat$death) ~ dat$CD4 + strata(dat$gender),data=dat);
> fit_cox
Call:
coxph(formula = Surv(data_ori$Time, data_ori$death) ~ data_ori$drug +
strata(data_ori$gender), data = data_ori)
coef exp(coef) se(coef) z p
data_ori$drugddI 0.216 1.24 0.146 1.47 0.14
Likelihood ratio test=2.17 on
2009 Dec 08
4
Split comma separated list
Hi all,
I'm a beginner user of R. I am stuck at what I thought was a very obvious
problem, but surprisingly, I havent found any solution on the forum or
online till now.
My problem is simple. I have a file which has entries like the following:
#ID Value1 List_of_values
ID1 0.342 0.01,1.2,0,0.323,0.67
ID2 0.010 0.987,0.056,1.3,1.5,0.4
2010 Feb 17
2
extract the data that match
Hi r-users,
I would like to extract the data that match. Attached is my data:
I'm interested in matchind the value in column 'intg' with value in column 'rand_no'
> cbind(z=z,intg=dd,rand_no = rr)
z intg rand_no
[1,] 0.00 0.000 0.001
[2,] 0.01 0.000 0.002
[3,] 0.02 0.000 0.002
[4,] 0.03 0.000 0.003
[5,] 0.04 0.000 0.003
[6,]
2011 Dec 07
1
MIXED MODEL WITH REPEATED MEASURES
I am trying to specify a mixed model for my research, but I can't quite get
it to work. I've spent several weeks looking thru various online sources to
no avail. I can't find an example of someone trying to do precisely what I'm
trying to do. I'm hoping some smart member of this mailing list may be able
to help.
First off, full disclosure: (1) I'm an engineer by trade, so
2006 Jul 06
2
KPSS test
Hi,
Am I interpreting the results properly? Are my conclusions correct?
> KPSS.test(df)
---- ----
KPSS test
---- ----
Null hypotheses: Level stationarity and stationarity around a linear trend.
Alternative hypothesis: Unit root.
----
Statistic for the null hypothesis of
level stationarity: 1.089
Critical values:
0.10 0.05 0.025 0.01
0.347 0.463
2006 Jul 06
1
Access values in kpssstat-class
Hi,
How can I access the Values stored in kpssstat-class given by KPSS.test function and store it in a variable.
For example:
>x <- rnorm(1000)
>test <- KPSS.test(ts(x))
>test
---- ----
KPSS test
---- ----
Null hypotheses: Level stationarity and stationarity around a linear trend.
Alternative hypothesis: Unit root.
----
Statistic for the null
2008 May 08
2
acf function
Dear all,
I have an annual time-series of population numbers and I would like to
estimate the auto-correlation. Can I use acf() function and judge
whether auto-correlation is significant by the plots? The acf array, eg:
Autocorrelations of series 'x$log.s.r', by lag
0 1 2 3 4 5 6 7 8 9
10 11 12
1.000 0.031 -0.171
2006 May 14
1
Suggestion for system.time()
Hi, people. A tiny suggestion for the system.time function.
Could the returned vector have names? These could be like:
c("User", "System", "Elapsed", "Sub.User", "Sub.System")
That would then produce self-documenting output.
--
Fran?ois Pinard http://pinard.progiciels-bpi.ca
2009 Jan 04
1
R/octave/matlab etc.
I'd echo a lot of what has been said about this by the folk who have
been making R work so well. One of the main difficulties is that the
environment of computations affects relative performance. e.g., what
settings did a distro package builder choose. I note that my 3 GHz Dual
Core machine running Ubuntu 8.04 gets
octave 3.0.0
octave:6> tic; a = a + 1; toc
Elapsed time is 0.120027
2003 Nov 03
1
svm in e1071 package: polynomial vs linear kernel
I am trying to understand what is the difference between linear and
polynomial kernel:
linear: u'*v
polynomial: (gamma*u'*v + coef0)^degree
It would seem that polynomial kernel with gamma = 1; coef0 = 0 and degree
= 1
should be identical to linear kernel, however it gives me significantly
different results for very simple
data set, with linear kernel
2009 Feb 23
1
why results from regression tree (rpart) are totally inconsistent with ordinary regression
Hi,
In my analysis of impacts of insecticide-treated bednets on malaria, I
look at the relationship between malaria incidence and mosquito
behaviors. The condensed data set is copied here. Ordinary regression
(lm) shows that Incidence was negatively related to Mortality. This
makes sense because the latter reflected the strength of killing
mosquitoes by insecticide-treated nets. Since the
2006 Jan 30
1
weights argument in the lmer function in lme4
I suspect the weights argument is not having any effect.
Package: Matrix
Version: 0.995-2
Date: 2006-01-19
Beginning with this:
Browse[1]> resp.lmer <- lmer(SensSSC ~ Block + Season + (1 | Plot) + (1 | Ma) + (1 | Pa) +
+ (1 | MaPa), weights = SensSSC.N, data = xx)
I group the output into a table with my ran.eff function and get this:
2010 Jun 26
1
predict newdata question
Hi:
I am using a subset of the below dataset to predict PRED_SUIT for
the whole dataset but I am having trouble with 'newdata'. The model
was created with 153 records and want to predict for 208 records.
wolf2 <- structure(list(gridcell = c(367L, 444L, 533L, 587L, 598L, 609L,
620L, 629L, 641L, 651L, 662L, 674L, 684L, 695L, 738L, 748L, 804L,
805L, 872L, 919L, 929L, 938L, 950L, 958L,
2004 Dec 02
3
Dominant factors in aov?
Hi all,
I'm using R 2.0.1. for Windows to analyze the influence of following factors
on response Y:
A (four levels)
B (three levels)
C (two levels)
D (29 levels) with
E (four replicates)
The dataset looks like this:
A B C D E Y
0 1 1 1 1 491.9
0 1 1 1 2 618.7
0 1 1 1 3 448.2
0 1 1 1 4 632.9
250 1 1 1 1 92.4
250 1 1 1 2 117
250 1 1 1 3 35.5
250 1 1 1 4 102.7
500 1 1 1 1 47
500 1 1 1 2 57.4
2009 Feb 26
1
using predict method with an offset
Hi,
I have run into another problem using offsets, this time with
the predict function, where there seems to be a contradiction
again between the behavior and the help page.
On the man page for predict.lm, it says
Offsets specified by offset in the fit by lm will not be included in
predictions, whereas those specified by an offset term in the formula
will be.
While it indicates nothings about
2010 Dec 21
2
please Help me on a repeated measures anova
I currently work on a draft of an aquatic bioassessment. The conditions
tested are the following: ER river water T dechlorinated water control 0.5 +
0.5mg / L of malate T + 1 dechlorinated water control + 1g / L of malate T
ED dechlorinated water control SED + ER + river water sediment SED ED +
sediment + water dechlorinated. It is the result of AChE in muscle (fillet
of fish). The production of
2010 Feb 04
2
help needed using t.test with factors
I am trying to use t.test on the following data:
date type INTERVAL nCASES MTF SDF MTO SDO
nFST MF nOBS MO MB BIASCV BIASEV ME MAE
RMSE CRCF
2001-06-15 avn GE1.00 4385 0.246 0.300 1.502
0.556 1367 1.373 4385 1.502 1.471 0.285 0.164
-1.256 1.266 1.399 0.056
2001-06-15 avn
2006 Jun 14
2
lmer binomial model overestimating data?
Hi folks,
Warning: I don't know if the result I am getting makes sense, so this
may be a statistics question.
The fitted values from my binomial lmer mixed model seem to
consistently overestimate the cell means, and I don't know why. I
assume I am doing something stupid.
Below I include code, and a binary image of the data is available at
this link:
2012 May 26
2
Assessing interaction effects in GLMMs
Dear R gurus
I am running a GLMM that looks at whether chimpanzees spend time in shade
more than sun (response variable 'y': used cbind() on counts in the sun and
shade) based on the time of day (Time) and the availability of shade
(Tertile). I've included some random factors too which are the chimpanzee
in question (Individual) and where they are in a given area (Zone). There
are