similar to: levelplot/heatmap question

Does anyone know if there is a way to run multivariate linear regression in R? I tried using the lm function (e.g., lm(dv1, dv2~iv1+iv2+iv3), but got error messages. Is my syntax wrong, or do I need a particular package? Thanks, Jeff-- ________________________________________________________ Jeffrey J. Lusk, Ph.D. Postdoctoral Research Associate Department of Forestry &

# I've tried to make this easy to paste into R, though it's probably so simple you won't need to. # I have some data (there are many more variables, but this is a reasonable approximation of it) # here's a fabricated data frame that is similar in form to mine: my.df <- data.frame(replicate(10, round(rnorm(100, mean=3.5, sd=1)))) var.list <- c("dv1",

Dear all, I try to conduct a SEM / two stage least squares regression with the following equations: First: X ~ IV1 + IV2 * Y Second: Y ~ a + b X therein, IV1 and IV2 are the two instruments I would like to use. the structure I would like to maintain as the model is derived from economic theory. My problem here is that I have trouble solving the equations to get the reduced form so I can run

Dear all,  I hope this is the right place to ask this question. I am reviewing a research where the analyst(s) are using a linear regression model. The dependent variable (DV) is a continuous measure. The independent variables (IVs) are a mixture of linear and categorical variables. The author investigates whether performance (DV - continuous linear) is a function of age (continuous IV1 -

I have data from a design in which items are completely nested within subjects. Subject is the only second-level factor, but I have multiple first-level factors (IVs). Say there are 2 such independent variables that I am interested in. What is the proper syntax to fit a mixed-effects model with a by-subject random intercept, and by-subject random slopes for both the 2 IVs? I can

Hi all, I have data from an experiment with 3 independent variables, 2 are within and 1 is between. In addition to the dependent variable, I have a covariate that is a single measure per subject. Below I provide an example generated data set and my approach to implementing the ANCOVA. However the output confuses me; why does the covariate only appear in the first strata? Presumably it should

Dear R-Users, I have a trivial problem, but extensive googling and ??'ing did not solve it: I want to obtain the sums of squares from a summary.aov() object for later use. Example: > DV <- rnorm(100) > IV1 <- as.factor(rep(c("male", "female"), times = 50)) > IV2 <- as.factor(rep(c("young", "old"), times = 50)) > >

Hi, I would like to simulate data with a binary outcome and a set of predictors that are correlated. I want to be able to fix the number of event (Y=1) vs. non-event (Y=0). Thus, I fix this and then simulate the predictors. I have 2 questions: 1. When the predictors are continuous, I can use mvrnorm(). However, if I have continuous, ordinal and binary predictors, I'm not sure how to simulate

Hello everyone, I have tried to look for this everywhere and so far have no luck. I have a plotMeans(DV, IV1, IV2, error.bars="se") graph that plots my data (DV-continuous, IVs are factors, IV1 - two levels, IV2-four levels). I am trying to increase a scale of my y-axis (to be consistent with my other graphs), but unfortunately nothing works with "plotMeans" function, which

Hello, In order to experiment with OpenCA, I have built an RPM for redhat 7.2 of a recent OpenSSL snapshot (the binary rpm on the OpenCA was built with the different target directories and libraries. Unfortunately these recent OpenSSL snapshots seems to break all OpenSSH tarballs and RPMs that I have been able to find. None of them seem to compile successfully, even the snapshots at

Greetings, Apologies for such a simple question, but I have been trying to figure this out for a few days now (I'm quite new to R), have read manuals, list posts, etc., and not finding precisely what I need. I am accustomed to working in Stata, but I am currently working through the multilevel package right now. In Stata, I would include the statement "if model1 == 1" at the end

Hi there, as I'm not sure to understand the coloring levelplot uses, I'm looking for another easy way to create a heatmap like this: library(lattice) mat <- matrix(seq(1,5, length.out = 12), nrow = 3) mat[1,2] <- 3.5 my.at <- seq(0.5,5.5, length.out = 6) my.col.regions <- rainbow(5) graph <- levelplot(t(mat[nrow(mat):1, ] ), at = my.at, col.regions = my.col.regions)

Hi, How do I anchor z=0 to the white color in a levelplot so that the color changes from cyan to magenta precisely as z changes from negative to positive? Also is it easy to change color scheme, say to blue/red as it's more dramatic? Is there a better function for showing heatmap with a color bar? Thanks in advance for any help, I've played with image, heatmap and levelplot a little and

On Wed, Sep 2, 2015 at 5:36 AM, James Molloy <james at jamesmolloy.co.uk> wrote: > Hi, > > Coremark really isn't a good enough test - have you run the LLVM test suite > with this patch, and what were the performance differences? For the test suite single source benches, the 235 tests improved performance, 2 regressed and 705 were unchanged. That seems very optimistic.

Dear R users, I am trying to print heatmaps in a loop (with a pause). Idea is to visualize changing correlations over time and for testing I wrote this simple (reproducible) code below. My problem is that levelplot() does not produce any output when I run the code (though heatmap does). Ideally I would like to use levelplot() as it produces a neat index on the side indicating the color and the

Hi there, I'd like to create a heatmap from my matrix with a) a defined color range (lets say from yellow to red) b) using striking colors above and below a certain threshold (above = "green", below = "blue") Example matrix (there should be a few outliers generated...) + simple levelplot without outliers marked: library(lattice) my.mat <- matrix(rnorm(800), nrow =

Hi all, I have been struggling with ANOVAs on R. I am new to R, so I created a simple data frame, and I do some analyses on R just to learn R and then check them on SPSS to make sure that I am doing fine. Here is the problem that I've run into: when we use the aov function, it uses SS Type I as default (on SPSS it is Type III). Then I used the Anova function under cars package using the

On Mon, Aug 31, 2015 at 5:52 PM, Jake VanAdrighem <jvanadrighem at gmail.com> wrote: > Do you have some specific performance measurements? Averaging 4 runs of 10000 iterations each of Coremark on my X86_64 desktop showed: -O2 performance: +2.9% faster with the L.E.V. pass -Os size: 1.5% smaller with the L.E.V. pass In the case of Coremark, the benefit comes mainly from the matrix

Hi all, I just try to get familiar with levelplot() for generating heatmaps. I have x, y, z-data stored in a file: e.g.: "file.csv": 0 0 0.1 0 1 0.5 0 2 0.4 1 0 0.3 1 1 0.4 1 2 0.6 ... I can use scan() for generating the matrix for R: inp <- scan("file.csv", list(0, 0, 0)) How can I feed inp into levelplot()? Thanks a lot for your help Richard -- Richard M?ller . Am

While rototilling packet.c, I did some looking at cipher_encrypt in cipher.c. It ends up that for SSH_CIPHER_NONE in cipher_encrypt, it uses memcpy. However, it also appears that dest and src can be equal in cipher_encrypt. On most sane libc implementations, memcpy == memmove. However, ANSI C makes no such guarantee, and some implementations out there are bound to try to optimize memcpy