search for: 0.356

the following the the lower.tri matrix in a file named luxry.car and i want to read it in R as a lower.tri matrix.how can i do? i have try to use help.search("read"),but no result what i want. 1.000 0.591 1.000 0.356 0.350 1.000

Hello, I am trying to filter a data set like below so that the peaks in the Phase value are more obvious and can be identified by a peak finding function following the useful advise of Carl Witthoft. I have written the following for(i in length(data$Phase)){ newphase=if(abs(data$Phase[i+1]-data$Phase[i])>6){ data$Phase[i+1] }else{data$Phase[i] } } I get the following error which I have not

I've just found the lavaan package, and I really appreciate it, as it seems to succeed with models that were failing in sem::sem. I need some clarification, however, in the output, and I was hoping the list could help me. I'll go with the standard example from the help documentation, as my problem is much larger but no more complicated than that. My question is, why is there one latent

On Wed, Dec 04, 2019 at 01:45:54AM +0200, Nir Soffer wrote: > On Mon, Dec 2, 2019 at 12:28 PM Richard W.M. Jones <rjones@redhat.com> wrote: > > > > > > I have pushed some parts of these patches in order to reduce the delta > > between your patches and upstream. However still some problems with > > the series: > > > > Patch 1: Same problem with

Dear R community members, I think I am asking a very simple question, but I really looked up in the faqs and manuals and found nothing helpful. I am trying to plot a data frame with the following structure (this is just a small extract): glo conc odor line series X1 X2 X3 X4 X5 X6 X7 X8 X9 X10 X11 X12 X13 1 0 AIR LN1 UP -0.488

I have pushed some parts of these patches in order to reduce the delta between your patches and upstream. However still some problems with the series: Patch 1: Same problem with scale as discussed before. Patch 2: At least the documentation needs to be updated since it no longer matches what is printed. The idea of collecting the time taken in each operation is good on its own, so I pushed

Hello, I've been looking through ?phantom and ?expression and this forum for examples of how I might be able to manipulate some of the names that appear on the y-axis of the barplot below. For example, the "gw" in "ECgw" would appear as a subscript...or "qr" would be the theta symbol followed by subscript "r". My attempts haven't even come close

Hi Everyone, I am uncertain that I am writing the contrast statements correctly. Basically, I'm unsure when to use a -1 and a 1 when writing the contrasts. Specifically I am interested in comparing the slopes between different temperature regimes. Temperature is therefore a factor. Time and percent are numerical. Using the gmodels package I made the following model:

Dear all, I am attempting to explain patterns of arthropod family richness (count data) using a regression model. It seems to be able to do a pretty good job as an explanatory model (i.e. demonstrating relationships between dependent and independent variables), but it has systematic problems as a predictive model: It is biased high at low observed values of family richness and biased low at

We are two french students and we have a problem concerning an exercize. We don't know how to resolve it. It would be fantastic if someone can help us. Thanks. Description: This study examined how the metabolic cost of locomotion varied with speed, stride frequency and body mass. Cost was determined by measuring oxygen consumption (?vo2?), analyzing the oxygen content in air inhaled and

Dear R-ophiles, I've found something very odd when I apply convolve to ever larger vectors. Here is an example below with vectors ranging from 2^11 to 2^17. There is a funny bump up at 2^12. Then it gets very slow at 2^16. > for( i in 11:20 )print( system.time(convolve(1:2^i,1:2^i,type="o"))) user system elapsed 0.002 0.000 0.002 user system elapsed 0.373

I was wondering if somebody could explain why I get different results here: >treats[,2]<-as.factor(treats[,2]) >treats[,5]<-as.factor(treats[,5]) >treats[,4]<-as.factor(treats[,4]) #there are 'c' on more days than I have 'h2o2', where treats[,4] is the day. I only want 'c' that correspond to the same days that I have a 'h2o2' also.

Dear R list members, I've turned up a strange discrepancy between results obtained from the lme function in the nlme package in R and results obtained with lme in S-PLUS. I'm using version 3.1-24 of nlme in R 1.4.1 under Windows 2000, and both S-PLUS 2000 and 6.0, again under Windows 2000. I've noticed discrepancies in a couple of instances. Here's one, using data from Bryk

On Mon, Dec 2, 2019 at 12:28 PM Richard W.M. Jones <rjones@redhat.com> wrote: > > > I have pushed some parts of these patches in order to reduce the delta > between your patches and upstream. However still some problems with > the series: > > Patch 1: Same problem with scale as discussed before. I addressed this here:

Good evening R-users! I have the following problem: i want to get a weighted crosstable along with the adjusted standardized residuals test Example: a<-c(1,1,1,1,1,2,2,2,3,3) b<-c(10,10,23,24,33,45,46,70,21,66) c<-c(3,3,2,3,4,1,1,1,3,3) d<-c("a", "b","b","c","a","a","a", "b",

Hi, Couldn't is.unsorted() bail out immediately here (after comparing the first 2 elements): > x <- 20000000:1 > system.time(is.unsorted(x), gcFirst=TRUE) user system elapsed 0.084 0.040 0.124 > x <- 200000000:1 > system.time(is.unsorted(x), gcFirst=TRUE) user system elapsed 0.772 0.440 1.214 Thanks! H.

flush is usually called once but it can take significant time, so we need to include it. Here is an example run with this change: $ ./nbdkit --foreground \ --unix /tmp/nbd.sock \ --exportname '' \ --filter stats \ file file=/var/tmp/dst.img \ statsfile=/dev/stderr \ --run 'qemu-img convert -p -n -f raw -O raw -T none /var/tmp/fedora-30.img

Dear All, I'm not sure of the interpretation of interactions with contrasts. Can anyone help? I do an ANCOVA, dryweight is covariate, block and treatment are factors, c4 the response variable. model<-aov(log(c4+1)~dryweight+treatment+block+treatment:block) summary(model); Df Sum Sq Mean Sq F value Pr(>F) dryweight 1 3.947 3.947 6.6268 0.01076 *

Hi, I have a file that looks like this : one,0 two,0.591,0 three,0.356,0.350,0 four,-0.098,0.072,0.380,0 five,0.573,0.408,0.382,0.062,0 six,0.156,0.232,0.517,0.424,0.303,0 seven,0.400,0.414,0.611,0.320,0.401,0.479,0 eight,0.282,0.375,0.512,0.346,0.308,0.463,0.605,0 nine,0.519,0.484,0.467,0.167,0.455,0.311,0.574,0.557,0 I want to create a data matrix out of it, so I tried this :

I am connecting from a PC to a Linux system running CentOS release 5.5 (Final) and it is extremely slow to render plots to the X11 device. This is not R's fault but I wonder if anyone can offer guidance so I can help the system administrators address the problem. I can connect to the Linux server using a NoMachine NX client for Windows or using X-Win32. I also have access to R running