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I have a data frame called "pose": DESCRIPTION QUANITY CLOSING.PRICE 1 WHEAT May/10 1 467.75 2 WHEAT May/10 2 467.75 3 WHEAT May/10 1 467.75 4 WHEAT May/10 1 467.75 5 COTTON NO.2 May/10 1 78.13 6 COTTON NO.2 May/10 3 78.13 7 COTTON NO.2 May/10 1 78.13

Dear HelpeRs, I have some data: "ice" <- structure(c(0.386, 0.374, 0.393, 0.425, 0.406, 0.344, 0.327, 0.288, 0.269, 0.256, 0.286, 0.298, 0.329, 0.318, 0.381, 0.381, 0.47, 0.443, 0.386, 0.342, 0.319, 0.307, 0.284, 0.326, 0.309, 0.359, 0.376, 0.416, 0.437, 0.548, 41, 56, 63, 68, 69, 65, 61, 47, 32, 24, 28, 26, 32, 40, 55, 63, 72, 72, 67, 60, 44, 40, 32, 27, 28, 33,

Dear All: I have some questions about the output of coxph. Below is the input and output: ---------------------------------------- > coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = + ovarian, x = TRUE) Call: coxph(formula = Surv(futime, fustat) ~ age + rx + ecog.ps, data = ovarian, x = TRUE) coef exp(coef) se(coef) z p age 0.147 1.158

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

I have a list of data frames: > str(data) List of 4 $ :'data.frame': 700773 obs. of 3 variables: ..$ V1: chr [1:700773] "200130446465779" "200070050127778" "200030633708779" "200010587002779" ... ..$ V2: int [1:700773] 0 0 0 0 0 0 0 0 0 0 ... ..$ V3: num [1:700773] 1 1 1 1 1 ... $ :'data.frame': 700773 obs. of 3 variables: ..$

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

Hi, I believe I have significantly improved [dpq]wilcox functions by implementing Harding's algorithm: Harding, E.F. (1984): An Efficient, Minimal-storage Procedure for Calculating the Mann-Whitney U, Generalized U and Similar Distributions, App. Statist., 33, 1-6 Results on my computer show (against R-2.9.1): > system.time( dwilcox( 800, 800, 80) ) user system elapsed 0.240

System: R 2.3.1 on Windows XP machine. I am building a logistic regression model for a sample of 100 cases in dataframe "d", in which there are 3 binary covariates: x1, x2 and x3. ---------------- > summary(d) y x1 x2 x3 0:54 0:50 0:64 0:78 1:46 1:50 1:36 1:22 > fit <- glm(y ~ x1 + x2 + x3, data=d, family=binomial(link=logit)) >

The optim documentation states (second from last sentence of Details Section) that "Any names given to par will be copied to the vectors passed to \code{fn} and \code{gr}." This does not seem to be the case when the method argument is set to "Brent". Consider finding an optimum with the "Brent" method and a fn argument that does not rely on a named par argument,

hi everyone! i want to check the autocorrelation function for a univariate time series (streamflow) in a data frame as below: < DF <- read.table("D:/file path....") < DF year jan feb mar apr ...... dec 1966 0.504 0.406 0.740 0.241 0.429 1967 0.683 0.529 0.780 0.443 0.503 . . . . what i first tried is: acf (DF, plot = TRUE)

Consider the following output [R2.2.0; Windows XP] > set.seed(160706) > X <- matrix(rnorm(40),nrow=10,ncol=4) > Xpc <- princomp(X,cor=FALSE) > summary(Xpc,loadings=TRUE, cutoff=0) Importance of components: Comp.1 Comp.2 Comp.3 Comp.4 Standard deviation 1.2268300 0.9690865 0.7918504 0.55295970 Proportion of Variance 0.4456907 0.2780929

Try to ping from client to server with its hostname. Sounds like dns problem. ping server Then try to ping its ip address. Then try to add server address to host file. Ex 192.168.8.30 server.example.com[1] server Best M On May 26, 2016 12:02, "Nico Speelman" <nico at speelmanrobben.nl[2]> wrote: Hello, I've been trying to add a new server to my Samba 4 Active directory, but

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,] 

Dear List, With regard to the question I previously raised, here is the result I obtained right now, brglm() does help, but there are two situations: 1) Classifiers with extremely high AIC (over 200), no perfect separation, coefficients converge. in this case, using brglm() does help! It stabilize the AIC, and the classification power is better. Code and output: (need to install package:

Hello I am using the step function in order to do backward selection for a linear model of 52 variables with the following commands: object<-lm(vars[,1] ~ (vars[,2:(ncol(predictors)+1)]-1)) BackS<-step(object,direction="backward") but it isn't dropping any if the variables in the model, but there are lots of not significant variables as you can see here >

# R for Windows will not send your bug report automatically. # Please copy the bug report (after finishing it) to # your favorite email program and send it to # # r-bugs at r-project.org # ###################################################### This report is joint from Richard Heiberger <rmh at temple.edu> and Burt Holland <bholland at temple.edu>. Burt Holland is the coauthor

Hi all, I'm seeing this failure on my PPC G4 box running TOT with llvm-gcc 4.2. Is anyone else seeing this? I'm sure it's related to the byval stuff that's recently gone into LLVM. I'm attaching the output of this command: $ llvm-gcc -emit-llvm -O3 -S -o - -emit-llvm /Users/wendling/llvm/ llvm.src/test/CFrontend/2008-01-25-ByValReadNone.c As you can see in it, there

Dear list colleagues, I'm trying to come up with a test question for undergraduates to illustrate comparison of means from a complex survey design. The data for the example looks roughly like this: mytest<-data.frame(harper=rnorm(500, mean=60, sd=1), party=sample(c("BQ", "NDP", "Conservative", "Liberal", "None", NA), size=500,

Full_Name: Adrian Custer Version: 0.90.0 OS: Linux on Thinkpad (pentium) and desktop (K6) Submission from: (NULL) (128.32.251.234) When I create a boxplot, and then try to overlay a lowess fit or just the points, the points do not appear in the highest level and the lowess curve does not reach the highest level. However, if I add one to each of the models, the problem is solved. I tried this

Hi there, I am doing multiple comparisons for data that is not normally distributed. For this purpose I tried both functions kruskal{agricolae} and kruskalmc{pgirmess}. It confuses me that these functions do not yield the same results although they are doing the same thing, don't they? Can anyone tell my why this happens and which function I can trust? kruskalmc() tells me that there are no