similar to: lmer output

Hello, My name is Rodrigo, I am using R program and I have a trouble. I am trying to do a dendrogram with genetics information. Let me explain... The Similarity Matrix was already did, and with this matrix I want to construct a dendrogram. So, the distance is done. I need to transform this matrix (that I have) in a dendrogram, I woud be very grateful if someone could help me. PS: I am sending

Dear R users, how can I extrapolate values listed in the summary of an lm model but not directly available between object values such as the the standard errors of the calculated parameters? for example I got a model: mod <- lm(Crd ~ 1 + Week, data=data) and its summary: > summary(mod) Call: lm(formula = Crd ~ 1 + Week, data = data, model = TRUE, y = TRUE) Residuals: Min

Dear Group, I am using the ecdf function as follows: cawa.cdp <- ecdf(cawaocc$LEFF80) summary(cawa.cdp) Empirical CDF: 223 unique values with summary Min. 1st Qu. Median Mean 3rd Qu. Max. 0.07918 1.35700 1.68600 1.61000 1.91200 2.70000 I can see by the summary that the y-value for the 3rd quartile is 1.912. How can I obtain the x-value for a specified y-value (e.g., 0.8)?

R 2.12 windows 7 I am summary data that I would like to make into a 2x2 table representing counts positive vs. negative counts: 28/289 20/276 My table should look something like the following: group1 group2 Positive 28 20 Negative 289 276 How can a (1) create the 2x2 table (2) run a chi square test on the table? I have tried the following code, but I

Hello R-community, I am trying to populate a column (doy) in a large dataset with the first column number that exceeds the value in another column (thold) using the 'apply' function. Sample data: pt D1 D17 D33 D49 D65 D81 D97 D113 D129 D145 D161 D177 D193 D209 D225 D241 D257 1 39177 0 0 0 0 0 0 0 0 0.4336 0.4754 0.5340667 0.5927334 0.6514 0.6966

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,

I have a fairly large set of data with the following attributes: >str(raw.data) `data.frame': 1429 obs. of 16 variables: $ TStamp :`POSIXlt', format: chr "2001-11-25 02:00:00" "2001-11-25 01:55:00" "2001-11-25 01:50:00" "2001-11-25 01:45:00" ... $ iPDT.AHU14.14: num 0.0122 0.0125 0.0120 0.0120 0.0122 ... $ iPDT.AHU14.15: num 0.0121

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On Mon, Feb 24, 2003 at 05:26:32PM -0800, H. Peter Anvin wrote: > Greg has taken on the job of integrating klibc with the kernel, but > please post klibc bug reports to the klibc mailing list at > <klibc@zytor.com>. To avoid annoying warning from gcc I had to check for compatibility with -falign-* like we do in arch/i386/Makefile. check_usergcc located in top-level makefile to

Hello, This is an urgent request. I want to convert a list of 3 elements into a matrix and I am not sure how to do it. The list looks like this: List of 3 $ : num [1:15364, 1] 0.133 0.622 0.588 1.024 0.583 ... ..- attr(*, "dimnames")=List of 2 .. ..$ : chr [1:15364] "6420681" "3610072" "2260458" "60689" ... .. ..$ : NULL $ :

Dear all, Last autumn there was some discussion on the list of the warning Warning message: fitted probabilities numerically 0 or 1 occurred in: (if (is.empty.model(mt)) glm.fit.null else glm.fit)(x = X, y = Y, when fitting binomial GLMs with many 0 and few 1. Parts of replies: "You should be able to tell which coefficients are infinite -- the coefficients and their standard errors will

I am trying to use R lm() with quantitative and qualitative predictors, but am getting different results than those that I get in SAS. In the R ANOVA table documentation I see that "Type-II tests corresponds to the tests produced by SAS for analysis-of-variance models, where all of the predictors are factors, but not more generally (i.e., when there are quantitative predictors)." Is

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

Hello, I'm using VAR models in R in order to obtain impulse responses of stock market shock on US economy. I have series of quarterly changes in real gdp, S&P 500 and quarterly level of unemployment for 1985 - 2012 period. My series are stationary. So I did all the steps below. However I don't understand what do irf function results mean. These are the cumulative orthogonal responses

I have a file, each column of which is a separate year, and each row of each column is mean precipitation for that month. Looks like this (except it goes back to 1964). month X2000 X2001 X2002 X2003 X2004 X2005 X2006 X2007 X2008 X2009 1 1.600 1.010 4.320 2.110 0.925 3.275 3.460 0.675 1.315 2.920 2 2.960 3.905 3.230 2.380 2.720 1.880 2.430 1.380

Dear colleagues, How can I do to "relevel" the intercept? I need that the treatment "Db" be the intercept, and have p-values for the comparisons with the others treatments. I used the function "relevel" but it did not work out to have what I want. Thanks for your help, Xavier T1 <- read.table(file.choose(), h=T) > head(T1) treatment replicate Time

Having spent the last few weeks trying to decipher R, I feel I may finally be getting somewhere, but i'M still in need of some advice and all my tutors seem to be on holiday! Basically a bit of background, I have data collected on a population of Lizards which includes age,sex, and body condition. I collected data myself this year and I have data previously collected from 1999, 2002 and

Hi list, I?ve spend a LOT of time reading and trying many versions, from Fink to Brew and also compiling source. OSX 10.11.3 NUT: v2.7.4 Driver: nutdrv_atcl_usb I have the same problem that has been reported, that Mac OSX grabs the USB device so NUT can?t use it. 0.001924 Starting UPS: UDups 0.001956 1 remaining attempts 0.001966 exec: /sw/lib/nut/nutdrv_atcl_usb -a UDups -u root

Hello friends and fellow R users, I have successfully tabulated and entered my survival data into R and have generated survival curves. But I would like to be able to determine what the survival rates are now at one month, three months, six months and one year. I have a data set, via.wall, which I have entered into R, and which generates the following Surv object: Surv(Days,Status==1) [1]

Target: FreeBSD 6.2-STABLE on i386 autoconf says: configure:2122: checking build system type configure:2140: result: i386-unknown-freebsd6.2 [...] configure:2721: gcc -v >&5 Using built-in specs. Configured with: FreeBSD/i386 system compiler Thread model: posix gcc version 3.4.6 [FreeBSD] 20060305 [...] objdir != srcdir, for both llvm and gcc. Release build. llvm-gcc 4.2 from source.