search for: 0.605

Displaying 20 results from an estimated 29 matches for "0.605".

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2004 Nov 06
3
how to read this matrix into R
the following the the lower.tri matrix in a file named luxry.car and i want to read it in R as a lower.tri matrix.how can i do? i have try to use help.search("read"),but no result what i want. 1.000 0.591 1.000 0.356 0.350 1.000
2005 Apr 05
1
extracting Proportion Var and Cumulative Var values from factanal
Hi R users, I need some help in the followings: I'm doing factor analysis and I need to extract the loading values and the Proportion Var and Cumulative Var values one by one. Here is what I am doing: > fact <- factanal(na.omit(gnome_freq_r2),factors=5); > fact$loadings Loadings: Factor1 Factor2 Factor3 Factor4 Factor5 b1freqr2 0.246 0.486 0.145
2009 Jan 31
1
thurston case 5
Hi, I hope some one can help. I need to compute Thurston's case 5 on a large set of data. I have gotten as far as computing the proportional preference matrix but the next math is beyond me. Here us my matrix 0.500 0.472 0.486 0.587 0.366 0.483 0.496 0.434 0.528 0.500 0.708 0.578 0.633 0.554 0.395 0.620 0.514 0.292 0.500 0.370 0.557 0.580 0.615 0.329 0.413 0.422 0.630 0.500 0.783 0.641 0.731
2011 Jan 20
1
Inverse Prediction with splines
Hello, I have fit a simple spline model to the following data. Data x y 0 1.298 2 0.605 3 0.507 4 0.399 5 0.281 6 0.203 7 0.150 8 0.101 Model Sp.1=lm(y~bs(x,df=4)) Now I wish to inverse predict the x for y=.75, say. Optimize works fine for a polynomial but I can figure out how to get the spline model into the function argument. Can anyone help me out. Thanks!! Jeff Jeff Morris Sanofi
2007 Oct 17
3
how to repeat the results of a generated probabilities
hello, I want to simulate 200 times the mean of a joint probability (y1) and 200 times the mean of another joint distribution (y2), that is I'm expecting to get 200 means of y1 and 200 means of y2. y1 and y2 are probabilities that I calculate from the marginal prob. (z1 and z2 respectively) multiple by the conditional prob. (x1 and x2 respectively), which I generaterd from the binomial
2009 Sep 22
1
odd (erroneous?) results from gls
A couple weeks ago I posted a message on this topic to r-help, the response was that this seemed like odd behavior, and that I ought to post it to one of the developer lists. I posted to r-sig-mixed-models, but didn't get any response. So, with good intentions, I decided to try posting once more, but to this more general list. The goal is (1) FYI, to make you aware of this issue, in case it
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,
2012 Jul 02
1
How to get prediction for a variable in WinBUGS?
Dear all,I am a new user of WinBUGS and need your help. After running the following code, I got parameters of beta0 through beta4 (stats, density), but I don't know how to get the prediction of the last value of h, the variable I set to NA and want to model it using the following code.Does anyone can given me a hint? Any advice would be greatly appreciated.Best
2007 Jan 06
2
Bootstrapping Confidence Intervals for Medians
I apologize for this post. I am new to R (two days) and I have tried and tried to calculated confidence intervals for medians. Can someone help me? Here is my data: institution1 0.21 0.16 0.32 0.69 1.15 0.9 0.87 0.87 0.73 The first four observations compose group 1 and observations 5 through 9 compose group 2. I would like to create a bootstrapped 90% confidence interval on the difference of
2011 Apr 05
6
simple save question
Hi, When I run the survfit function, I want to get the restricted mean value and the standard error also. I found out using the "print" function to do so, as shown below, print(km.fit,print.rmean=TRUE) Call: survfit(formula = Surv(diff, status) ~ 1, type = "kaplan-meier") records n.max n.start events *rmean *se(rmean) median 200.000
2006 Mar 20
0
Estimating Daily Survival
R Users, I was wondering if someone might point me in the right direction. I am using a Cox model (survival package) to evaluate survival of pen-reared birds (time to event data collected daily) and I have been trying to determine how I can estimate a 'daily' survival rate and std error from the results of a Cox model? Using survfit (see input data below) I computed the predicted
2012 Apr 08
0
Need help interpreting output from rcorrp.cens with Cox regression
Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with the "old" comorbidity measure, Charlson's Comorbidity Index. I have nearly 10,000
2012 Aug 17
0
REPOST: Need help interpreting output from rcorrp.cens with Cox regression
I am reposting my message from April 8th because I never received a response to the original post: Dear R-listers, I am an MD and clinical epidemiologist developing a measure of comorbidity severity for patients with liver disease. Having developed my comorbidity score as the linear predictor from a Cox regression model I want to compare the discriminative ability of my comorbidity measure with
1999 Mar 10
1
lty=2
On Wed, Mar 10, 1999 at 03:14:08PM +0000, Simon Bond wrote: > Dear all, > > I'm using R 63.2 on windows NT, when I use > > > lines(x,y, lty=2) > > it produces a dashed line between the first pair of points and then reverts > back to a solid line. It produces different colours perfectly ok, but it's > not really a solution when the plot needs to be printed
2009 Jun 27
1
Regression; how to get t-values for all parameters estimates
Dear all, Even after a couple of hours looking at old messages I still haven't found a solution for my problem. I'm trying to fit an additive linear regression model with 2 effects, both fixed, to some dataset. The function contrasts(effectA) <- contr.sum can gaurantee that the coefficients per parameter sum to one, and the function dummy.coef provices the estimates of all
2011 Nov 28
2
problem in reading file
Hi, I have a file that looks like this : one,0 two,0.591,0 three,0.356,0.350,0 four,-0.098,0.072,0.380,0 five,0.573,0.408,0.382,0.062,0 six,0.156,0.232,0.517,0.424,0.303,0 seven,0.400,0.414,0.611,0.320,0.401,0.479,0 eight,0.282,0.375,0.512,0.346,0.308,0.463,0.605,0 nine,0.519,0.484,0.467,0.167,0.455,0.311,0.574,0.557,0 I want to create a data matrix out of it, so I tried this :
2001 Nov 05
1
Problem to transfer Splus functions
Hello I would like to transfer some Splus functions in R. But I have a problem first about this assignation in Splus : xnom <- deparse(substitute(x)) I am a bad programmer : I don't understand the R help How to modify these functions ? Thank you very much for your help Here are the four functions and a data test
2010 Sep 16
0
problems trying to reproduce structural equation model using the sem package
Hello, I've been unsuccessfully trying to reproduce a sem from Grace et al. (2010) published in Ecological Monographs: http://www.esajournals.org/doi/pdf/10.1890/09-0464.1 The model in question is presented in Figure 8, page 81. The errors that I've been getting are: 1. Using a correlation matrix: res.grace <- sem(grace.model, S = grace, N = 190) Warning message: In sem.default(ram
2006 Jun 23
1
looping through a data frame
Hi- I am having trouble with the syntax of looping through the rows and columns of a data frame. I have a table with 17 observations for 84 lines at n=5-10 per line. So the table is ~700x17. I want to pull out the median and stdev for each line and put it in a dataframe with rowname = linename. So I have tried the following.... #read in the table input.table <- read.table(file =
2006 Nov 23
2
random effect question and glm
consider p as random effect with 5 levels, what is difference between these two models? > p5.random.p <- lmer(Y ~p+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1)) > p5.random.p1 <- lmer(Y ~1+(1|p),data=p5,family=binomial,control=list(usePQL=FALSE,msV=1)) in addtion, I try these two models, it seems they are same. what is the difference between these two model. Is