search for: degreed

Very new to this but would appreciate a hand as this is driving me mad, I''m trying to update both degree and assigned from a single form, the degree section updates fine but assigned just retains the same values, any help much appreciated! my edit.rhtml file looks like: <h1>Editing degree</h1> <%= error_messages_for :degree %> <%= start_form_tag :action =>

Hi, I have plot a graphics by next way: par(mar=c(5,5,1,1), font.axis=3, ps=14, xaxs="i", yaxs="i", lab=c(18,18,7), font .lab=3); plot(l, b, type="n", xlim=c(0,360), ylim=c(-90,90), xlab="l, deg", ylab="b, deg"); points(l, b, cex=vel/60, col=1, pch=21); On this plot I have axis x (denoted by l) ranged in [0,360] with labeling step 20. But I

Full_Name: Murray H Smith Version: 1.7.1 OS: Windows2000 Submission from: (NULL) (202.36.29.1) This alleged bug is Windows specific and occurs when using Windows metafile plots. The problem does not occur in a Linux version. It does not occur in the pt rintout when a graphic is saved to a postscript file under Windows. The problem came to light when using plotmath to label a plot with

Hello help, I have changed around some graphing code and made it into a function. Previously they y label of the axis was inserted as text in its own layout box. text(1,1, expression(~degree~C),cex=1) This worked great and resulted in the symbol for degree. In the function, I have changed it so: text(1,1,paste(b_unit),cex=1) and b_unit<-expression(~degree~C) This now inserts ~degree~C

Full_Name: Russell Lenth Version: 2.6.2 OS: Windows XP Pro Submission from: (NULL) (128.255.132.36) The poly() function allows a higher-degree polynomial than it should, when raw=FALSE. For example, consider 5 distinct 'x' values, each repeated twice. we can fit a polynomial of degree 8: ===== R> x = rep(1:5, 2) R> y = rnorm(10) R> lm(y ~ poly(x, 8)) Call: lm(formula = y ~

Hi how can print x-axis labels in 45 degree in boxplot() (or plot in general)? I can use las=2 to print in 90 degree, but it looks ugly. Is there a simple option to do 45 degree easily? Thanks John

Hi All, I have location data set up as degrees: minutes: seconds (for example, 122:45:45) I need to get this data into decimal degree form (122.7625). I know that I need to use the formula Decimal degrees = Degrees + (Minutes/60) + (Seconds/3600) Does anyone know how I can get R to do this? The original data is all in one column (called 'lat') and the degrees, mins and secs are

Hello all. I was wondering if there is any way to adjust the denominator degrees of freedom in lme(). It seems to me that there is only one method that can be used. As has been pointed out previously on the list, the denominator degrees of freedom given by lme() do not match those given by SAS Proc Mixed or HLM5. Proc Mixed, for example, offers five different options for computing the

Hi all, I have two networks for the same group of the users. I want to compare individual's degree in different networks. But how could I get the degrere of the nodes according to its name? When I use degree(g1), I could only get a vector of the degree of each node. But when I turn to g2, I don't know whose degree it is so I can't match it with the degree(g2). Now I'm trying to

Hi, I am plotting degree distribution of a graph using the function, library(igraph) dd1 = degree.distribution(G) plot(dd1, xlab = "degree", ylab="frequency") I would like to plot the mean of the distribution as a vertical line in the attached plot. Please let me know how to do this. Thanks, Kumar http://r.789695.n4.nabble.com/file/n3334375/cdata3_dd.png cdata3_dd.png --

Hello. Douglas Bates has explained in a previous posting to R why he does not output residual degrees of freedom, F values and probabilities in the mixed model (lmer) function: because the usual degrees of freedom (obs - fixed df -1) are not exact and are really only upper bounds. I am interpreting what he said but I am not a professional statistician, so I might be getting this wrong... Does

In barplot for degree distribution x-axis is not seen. See the example below > g = barabasi.game(500, 0.4) > dd1 = degree.distribution(g) > plot(dd1, xlab="degree", ylab = "frequency") whereas barplot doesnot have any x-axis > barplot(dd1, xlab = "degree", ylab = "frequency") Please see the figures attached.

Hi all, I want to fit data called "metal" with a polynominal function as dP ~ a.0 + a.1 * U0 + a.2 * U0^2 + a.3 * U0^3 + a.4 * U0^4 The data set includes, the independant variable U0 and the dependant variable dP. I've seen that the combination of lm() and poly() can do that instead of using the nls() function. But I don't get how to interpret the results from the linear

Hi, I'm working with step.gam in gam package. I'm interested both in spline and lowess functions and when I define all the models that I'm interested in I get something like that: > gam.object.ALC<-gam(X143S~ALC,data=dane,family=binomial) >

Hello, I work with two packages sna and graph from CRAN resp. Bioconductor. Both packages have a function called "degree". Therefore one of the functions is masked by the other and which one gets called depends on the order of loading. The problem is that both package do not have a namespace, therefore calling the masked function with "package::degree" does not work.

Hi, Suppose I have the following lines at the end of a function: answer <- c(2, 1, 0, 4, 5) # In fact, answer will be generate in my # function print(answer) # Print the answer # Now, find the best fitted n degree polynomial print(paste("The best fit is with", which.min(answer) - 1, "-degree polynomial")) this will return:

Hello, I'm running mixed models in GAMM4 with 2 (non-nested) random intercepts and I want to include a spline term for one of my exposure variables. However, when I include a spline term, I always get reported degrees of freedom of less than 1, even when I know that my spline is using more than 1 degree of freedom. For example, here is the code for my model: >

Hello, I've just carried out my first good-looking model using glmmPQL, and the output makes perfect sense in terms of how it fits with our hypothesis and the graphical representation of the data. However, please could you clarify whether my degrees of freedom are appropriate? I had 106 subjects, each of them was observed about 9 times, creating 882 data points. The subjects were in 3

Goodmorning, This is a documentation related question about the B-spline function in R. In the help file it is stated that: "df degrees of freedom; one can specify df rather than knots; bs() then chooses df-degree-1 knots at suitable quantiles of x (which will ignore missing values)." So if one were to specify a spline with 6 degrees of freedom (and no intercept) then a basis

Dear all, I'm trying to plot the degree symbol by itself between two square brackets. I just want to have "K [?]". So far I got to: expression(K ~ group("[",1*degree,"]")) or expression(K ~ group("[",1^o,"]")) But it won't work without a number or letter. Any suggestions? Thanks, andre