similar to: How to hold a value(Mean sq) with a string

Hi all: I have always used SPSS to estimate weekly covariance based on a linear regression model but have to hard code the model Std. Error and the Mean-Square and then execute one week a the time. I was wondering if someone could give me an idea on how to estimate weekly(WK) covariance using the summary and anova of "dfr"(lineal model below). I have to do this for 52

Hello all! I have a problem to extract values greater that for example 1820. I try this code: x[x[,1]>1820,]->x1 Please help me! Thank you! The data structure is: structure(c(2.576, 1.728, 3.434, 2.187, 1.928, 1.886, 1.2425, 1.23, 1.075, 1.1785, 1.186, 1.165, 1.732, 1.517, 1.4095, 1.074, 1.618, 1.677, 1.845, 1.594, 1.6655, 1.1605, 1.425, 1.099, 1.007, 1.1795, 1.3855, 1.4065, 1.138, 1.514,

Hi I am looking for a couple of pointers using glm (family = binary). 1. I want to add all the products of my predictive features as additional features (and I have 23 of them). Is there some easy way to add them? 2. I want to drop each feature in turn and get the most significant, then drop two and get the next most significant, etc. Is there some function that allows me to do this?

Dear all, Does any one knows why should I get the following error message, when trying to do a simple data.frame?? DataF<-data.frame(Subject,BiomR,Spp,Capas,Litter,Herbs,LitterD,MaxCanH,DDifS p,DSSp,Slope, CanDens,NearestSp) Erro em data.frame(Subject, BiomR, Spp, Capas, Litter, Herbs, LitterD, : arguments imply differing number of rows: 202, 0 The data I am using

Dear all, I'm using the mgcv library by Simon Wood to fit gam models with interactions and I have been reading (and running) the "factor 'by' variable example" given on the gam.models help page (see below, output from the two first models b, and b1). The example explains that both b and b1 fits are similar: "note that the preceding fit (here b) is the same as

Hello, I tried to do a 'sem' analysis for data of how blueberry consumption by birds is influenced by a pollution gradient, using distance and vegetation structural and composition variables, but I got the following error message: Error in sem.default(ram = ram, S = S, N = N, param.names = pars, var.names = vars, : S must be a square triangular or symmetric matrix This may be very

Dear R-help, I am having trouble with your scatterplot3d program. For help with this problem I was directed to your address by Martin Maechler at " r-core-bounces at r-project.org." I'm also sending a CC to " r-core-owner at r-project.org" as I'm not yet certain of the proper address to use for this. I have R version 2.8.1 and have downloaded 'scatterplot3d.'

Dear forum members, How can I force a negative slope in a linear regression even though the slope might be positive? I will need it for the purpose of determining the trend due reasons other than biological because the biological (genetic) trend is not positive for these data. Thanks. Julia Example of the data: [1] 1.254 1.235 1.261 0.952 1.202 1.152 0.801 0.424 0.330 0.251 0.229

Hello, I'm a beginner with R and it's the first time I'm using the R-help list... I hope I'm in the right place, if not: Sorry!! I need to do non linear regressions on a data set which columns are: "river.name" "Portata" "PTG.P" "PO4.P" "NT.N" "NH4.N" "NO3.N" "BOD5" "SiO2"

Hi, I need a help in a survival analysis using survreg function with weibull distribution from survival package. Look the data sample: ########## Start of script dados <- structure(list(TFD = c(20L, 34L, 1L, 2L, 3L, 3L, 50L, 26L, 1L, 50L, 21L, 3L, 13L, 11L, 22L, 50L, 50L, 1L, 50L, 9L, 50L, 1L, 13L, 50L, 50L, 1L, 6L, 50L, 50L, 50L, 36L, 3L, 46L, 10L, 50L, 1L, 18L, 3L, 36L, 37L, 50L, 7L, 1L,

Dear friends, After running the lm() model, we can get summary resluts like the following: Coefficients: Estimate Std. Error t value Pr(>|t|) x1 0.11562 0.10994 1.052 0.2957 x2 -0.13879 0.09674 -1.435 0.1548 x3 0.01051 0.09862 0.107 0.9153 x4 0.14183 0.08471 1.674 0.0975 . x5 0.18995 0.10482 1.812 0.0732 . x6 0.24832 0.10059 2.469 0.0154 * x7

Dear all, The following is a example that I run and hope to get a linear model. However, I find the lm() can not give correct coefficients for the linear model. I hope it's just my own mistake. Please help. TIA. Regards, Jinsong > x [1] 3.760216 3.997288 3.208872 3.985417 3.265704 3.497505 2.923540 3.193937 [9] 3.102787 3.419574 3.169374 2.928510 3.153821 3.100385 3.768770 3.610583

Dear R-helpers, I'm stuck with a little problem that surely has an easy solution but I can't think of a way to solve it. I'd really appreciate any help you can offer me! I'll provide a small example. Given a dataframe data.txt that looks like this: ID freq Var Var_mean Ratio_mean Var_median Ratio_median Var_sum Ratio_min Var_max Ratio_max Var_min

Hi! I use simtest fonction of multcomp package to compile a Dunnett's test. I have 10 treatments and one control group, so i create a matrix with: m<-matrix(0,10,11) m[1,1]<--1 m[1,2]<-1 m[2,1]<--1 m[2,3]<-1 m[3,1]<--1 m[3,4]<-1 m[4,1]<--1 m[4,5]<-1 m[5,1]<--1 m[5,6]<-1 m[6,1]<--1 m[6,7]<-1 m[7,1]<--1 m[7,8]<-1 m[8,1]<--1 m[8,9]<-1

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)) >

I sent this to r-devel the other day but didn't get any takers. This may not be a bug but rather an inconsistency. I'm not sure if this is intentional. summary.lm stores weighted residuals whereas I think most users will want print.summary.lm to summarize unweighted ones as if saying summary(resid(fit)). > set.seed(1) > dat <- data.frame(y = rnorm(15), x = rnorm(15), w = 1:15)

Hi Everyone, I'm continuing to run into trouble with polyfit. I'm using the fitting function of the form; fit <- lm(y ~ poly(x,degree,raw=TRUE)) and I have found that in some cases a polynomial of certain degree can't be fit, the coefficient won't be calculated, because of a singularity. If I use orthogonal polynomials I can fit a polynomial of any degree, but I don't get

i cannot see caller ID of the call originated from outside zap channel. i hv configured both zapata.conf and extensions.conf. i m right now in india i think asterisk only supports Bellcore enable caller ID. so is it the same bug of BT caller ID problem in UK? or it is the bug of my asterisk configuration? i hv enabled callerID from my TELCO. -------------- next part -------------- An HTML

Another issue to consider is T1 framing. If your application is putting bits onto the T1 at the rate of 1.544 Mbit/s then the T1 would need to be unframed. I don't believe this is an option in zaptel! If however, it is putting bits on at a rate of 1.536 Mbit/s and adding 8000 bit/s for framing then you may be able use the suggestion below. Don Pobanz On Thursday, April 03, 2003 3:28 PM,

Hello, I work on large matrices and found something interesting. For multiplication of matrices, the order has a huge influence on computing time when one of them is a sparse matrix. In the below example, M is a full matrix and A is a sparse matrix in a regular matrix class. A %*% M takes much more time than M %*% A; moreover, t(t(M) %*% t(A)) is much faster than A %*% M with same result. I