similar to: R-beta: spline problems(?)

Hallo, I am totally confused why Excel and R give me totally different regression results for the data below. If you know the solution, please enlighten me. In Excel I used LINEST() and Data>Data Analysis>Regression and both (fortunately) gave the same result. The coefficients were: /b0=1.16, b1=0.957, b2=0.024, R2=0.0027, adjusted R2=-0.017/ In R I used

I'm trying to write a function to find the means over factors of the responses in a mlm (something I would do easily in SAS with PROC SUMMARY). The not-working stub of a function to do what I want is below, and my problem is that I don't know how to call aggregate (or some other function) in the context of terms in a linear model extracted from a lm/mlm object. means.mlm <-

Hello everyone, I would like to test two regression slopes:do they differ significantly?The data and commands I've used so far: x<-8.5:32.5 #Vektor x y<-c( NA , NA , 5.67 , 6.53 , 6.83, 7.41 , 7.93 , 8.5 , 8.86, 9.46 , 9.82 , 10 ,10.35 , 10.7 ,11.03 ,11.37 ,11.61 ,11.84, 12.12, 12.39 ,12.67 ,12.96, 13.28 ,13.47, 13.65) #Vektor y (regression<-lm(y~x)) summary(regression)

Dear list, First my information: platform i386-pc-linux-gnu arch i386 os linux-gnu system i386, linux-gnu status major 2 minor 3.1 year 2006 month 06 day 01 svn rev 38247 language R version.string Version 2.3.1 (2006-06-01) Now my question: How is it possible that a command in an R script is not

On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > Ralf, et al., > > > > Attached is the latest version of my autovectorization patch. llvmdev > > has been CC'd (as had been suggested to me); this e-mail contains > > additional benchmark results. > > > > First, these are preliminary

On Sat, 2011-10-29 at 15:16 -0500, Hal Finkel wrote: > On Sat, 2011-10-29 at 14:02 -0500, Hal Finkel wrote: > > On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > > > Ralf, et al., > > > > > > Attached is the latest version of my autovectorization patch. llvmdev > > > has been CC'd (as had been suggested to me); this e-mail contains > >

Hello. I am trying to do a non-linear fit using the 'nls' command. The data that I'm using is as follows pH k 1 3.79 34.21 2 4.14 25.85 3 4.38 20.45 4 4.57 15.61 5 4.74 12.42 6 4.92 9.64 7 5.11 7.30 8 5.35 5.15 9 5.67 3.24 with a transformation of pH to H <- 10^-pH When using the nls command for a set of parameters - a, b and c, I receive two sets of errors: >

I have a data.frame named "df". The dput of df is at the bottom of this e-mail. What I'd like to do is replace the "n/a " values with NA. On Mac OSX, it works to do this: df[df == "n/a"] <- NA However, it does not work on Ubuntu. See below. Thanks in advance, Garrett > x <- df[27, 4] # complete data.frame dput is below > dput(x) "n/a?"

Dear R-helpers, I want to make a series of boxplots on several numeric univariates with two group variables (species and population, population nested in species, and with population as the X-axis). In order to get a proper order of the individual populations in X-axis, I need to assign a wanted order to the factor (population). I used the levels() function to do this assignment, but it seemed

Hello I?m need fitt growth curve with data length-age. I want to evaluate which is the function that best predicts my data, to do so I compare the Akaikes of different models. I'm now need to evaluate if changing the initial values changes the parameters and which do not allow to estimate the model. To do this I use the function nls(); and I randomize the initial values (real positive number).

Hi Folks, I have just started using dplyr and could use some help getting unstuck. It could well be that dplyr is not the package to be using, but let me just pose the question and seek your advice. Here is my basic data frame. head(h) subject ageGrp ear hearingGrp sex freq L2 Ldp Phidp NF SNR 1 HALAF032 A L A F 2 0 -23.54459 55.56005 -43.08282

On Sat, 2011-10-29 at 12:30 -0500, Hal Finkel wrote: > Ralf, et al., > > Attached is the latest version of my autovectorization patch. llvmdev > has been CC'd (as had been suggested to me); this e-mail contains > additional benchmark results. > > First, these are preliminary results because I did not do the things > necessary to make them real (explicitly quiet the

On 2016-05-25 19:13, Kelly Lesperance wrote: > Hdparm didn?t get far: > > [root at r1k1 ~] # hdparm -tT /dev/sda > > /dev/sda: > Timing cached reads: Alarm clock > [root at r1k1 ~] # Hi Kelly, Try running 'iostat -xdmc 1'. Look for a single drive that has substantially greater await than ~10msec. If all the drives except one are taking 6-8msec, but one is very

Ralf, et al., Attached is the latest version of my autovectorization patch. llvmdev has been CC'd (as had been suggested to me); this e-mail contains additional benchmark results. First, these are preliminary results because I did not do the things necessary to make them real (explicitly quiet the machine, bind the processes to one cpu, etc.). But they should be good enough for discussion.

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

On Mon, Dec 16, 2013 at 04:16:29PM -0800, Michael Dalton wrote: > Commit 2613af0ed18a ("virtio_net: migrate mergeable rx buffers to page frag > allocators") changed the mergeable receive buffer size from PAGE_SIZE to > MTU-size, introducing a single-stream regression for benchmarks with large > average packet size. There is no single optimal buffer size for all >

R 2.15.1 OS X and Windows Colleagues, I have a file that looks that this: TABLE NO. 1 PTID TIME AMT FORM PERIOD IPRED CWRES EVID CP PRED RES WRES 2.0010E+03 3.9375E-01 5.0000E+03 2.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 1.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 0.0000E+00 2.0010E+03 8.9583E-01

Dear all: I have been thinking to import below one data file (.txt)into R by read.table(..,skip=1, header=T). But How can I deal with the repeated rows of TABLE NO.1 and names of data variables in the middle of this data file. The similar block will be repeated 100 times, here only show 4 of them and within each block, data records also can vary, here only paste 4 rows for example. I appreciate

R 2.15.1 OS X Colleagues, I am reading a 1 GB file into R using read.table. The file consists of 100 tables, each of which is headed by two lines of characters. The first of these lines is: TABLE NO. 1 The second is a list of column headers. For example: TABLE NO. 1 COL1 COL2 COL3 COL4 COL5 COL6 COL7 COL8 COL9 COL10

Dear all: I have been thinking to import below one data file (.txt)into R by read.table(..,skip=1, header=T). But How can I deal with the repeated rows of TABLE NO.1 and names of data variables in the middle of this data file. The similar block will be repeated 100 times, here only show 4 of them and within each block, data records also can vary, here only paste 4 rows for example. I appreciate