similar to: Output results to a single postscript document

Displaying 20 results from an estimated 1000 matches similar to: "Output results to a single postscript document"

2009 Feb 03
1
How to show variables used in lm function call?
Hello R users, I am new to R and am wondering if anyone can help me out with the following issue: I wrote a function to build ts models using different inputs, but when R displays the call for a model, I cannot tell which variables it is using because it shows the arguments instead of the real variables passed to the function. (e.g Call: lm(formula = dyn(dep ~ lag(dep, -1) + indep)) --->
2013 Mar 21
4
easy way of paste
Hello, Is there a better way to use paste such as: a = paste(colnames(list.indep)[1],colnames(list.indep)[2],colnames(list.indep)[3],colnames(list.indep)[4],colnames(list.indep)[5],sep="+") > a [1] "aa+dummy1+dummy2+bb+cc" I tried a = paste(colnames(list.indep)[1:5],sep="+") > a [1] "aa" "dummy1" "dummy2"
2008 May 22
1
How to account for autoregressive terms?
Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates <- as.Date(data.df[,1]) selection<-which(dates>=as.Date("1986-1-1") & dates<=as.Date("2007-12-31")) dep <- ts(data.df[selection,c("dep")]) indep.ret1
2012 Sep 29
1
Unexpected behavior with weights in binomial glm()
Hi useRs, I'm experiencing something quite weird with glm() and weights, and maybe someone can explain what I'm doing wrong. I have a dataset where each row represents a single case, and I run glm(...,family="binomial") and get my coefficients. However, some of my cases have the exact same values for predictor variables, so I should be able to aggregate up my data frame and
2010 Feb 16
3
converting character vector "hh:mm" to chron or strptime 24 clock time vectors
Hi All, I am attempting to work with some data from loggers. I have read in a .csv exported from MS Access that already has my dates and times (in 24 clock format), (with StringsAsFactors=FALSE). > head(tdata) LogData date time 1 77.16 2008/04/24 02:00 2 61.78 2008/04/24 04:00 3 75.44 2008/04/24 06:00 4 89.43 2008/04/24
2010 Nov 03
4
Drawing circles on a chart
Dear Group, I have the following data matrix which is a timeseries. > dput(tData) structure(list(A = c(0.2, 0.13, 0.05, 0.1, 0.02, 0.18, 0.09, 0.06, 0.13), B = c(0.15, 0.06, 0.09, 0.02, 0.03, 0.12, 0.01, 0.15, 0.06), C = c(-0.1, 0, -0.07, -0.06, -0.05, -0.05, -0.06, -0.08, -0.07), D = c(-0.15, -0.05, -0.1, -0.03, -0.13, -0.04, -0.1, -0.04, -0.15), E = c(-0.17, -0.16, -0.08, -0.07, -0.09,
2012 Jan 15
1
Need help interpreting the logit regression function
Hello R community, I have a question about the logistic regression function. Specifically, when the predictor variable has not just 0's and 1's, but also fractional values (between zero and one). I get a warning when I use the "glm(formula = ... , family = binomial(link = "logit"))" which says: "In eval(expr, envir, enclos) : non-integer #successes in a binomial
2002 Mar 15
1
calibration/inverse regression?
I wonder if anyone out there has written a routine to solve the simple linear calibration problem? - fit regression of y vs x - estimate the value x0 (with 95% CI) that gives y0 Thanks for any help. Bill -.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.- r-help mailing list -- Read http://www.ci.tuwien.ac.at/~hornik/R/R-FAQ.html Send "info",
2010 Mar 16
3
function arguments: name of an object vs. call producing the object?
In a function, say foo.glm for glm objects I want to use the name of the object as a label for some output, but *only* if a glm object was passed as an argument, not a call to glm() producing that object. How can I distinguish these two cases? For example, I can use the following to get the name of the argument: foo.glm <- function(object) { oname <- as.character(sys.call())[2]
2012 May 25
1
Multiple rms summary plots in a single device
I would like to incorporate multiple summary plots from the rms package into a single device and to control the titles, and also to open a new device when I reach a specified number of plots. Currently I am only getting a single "plot(summary(" graph in the upper left- hand corner of each successive device. However, in the rms documention I see instances of a loop being used with
2002 Aug 12
1
question about cloud() in lattice package
Hi all, I have been previously been using scatterplot3d package to create some graphs but unfortunately it does not allow me to rotate the plot on all three axis. The cloud() function in the lattice package does allow me to do so. When I was using scatterplot3d I was using a script (Shown Below) to calculate the mean, quartiles and range limits for all three axis and I was representing that on the
2004 Aug 17
1
suggestion for ARMAacf()
hi, in 1.9.1, the return value from ARMAacf(pacf=TRUE) is not named by lags, contrary to ?ARMAacf. the simple fix is to move names(Acf) <- down after if(pacf), with an appropriate starting lag as pacf=TRUE appears to start at lag 1 (whereas pacf=FALSE starts at lag 0). for consistency, one could argue to append 1 for lag 0 for pacf=TRUE (or start pacf=F at lag 1). however, given the
2003 Sep 08
2
pacf lags
pacf in devel seems by default to return a different number of lags than 1.7.1 for $pacf. I don't see any mention of this in the NEWS file, or any change in the documentation, so I suspect it is and error, though it may be an undocumented improvement. (Newbie question: How is the simplest way to display a function like pacf.default that is not exported from a namespace?) Paul
2009 Jun 21
2
Help on qpcR package
I am using R on a Windows XP professional platform. The following code is part of a bigger one CODE press=function(y,x){ library(qpcR) models.press=numeric(0) cat("\n") dep=y print(dep) indep=log(x) print(indep) yfit=dep-PRESS(lm(dep~indep))[[2]] cat("\n yfit\n") print(yfit) yfit.orig=yfit presid=y-yfit.orig press=sum(presid^2)
2013 Apr 17
1
Bug in VGAM z value and coefficient ?
Dear, When i multiply the y of a regression by 10, I would expect that the coefficient would be multiply by 10 and the z value to stay constant. Here some reproducible code to support the case. *Ex 1* library(mvtnorm) library(VGAM) set.seed(1) x=rmvnorm(1000,sigma=matrix(c(1,0.75,0.75,1),2,2))
2009 May 20
1
stationarity tests
How can I make sure the residual signal, after subtracting the trend extracted through some technique, is actually trend-free ? I would greatly appreciate any suggestion about some Stationarity tests. I'd like to make sure I have got the difference between ACF and PACF right. In the following I am citing some definitions. I would appreciate your thoughts. ACF(k) estimates the correlation
2016 Apr 23
2
Data Frame Column Name Attribute
I am attempting to add a calculated column to a data frame. Basically, adding a column called "newcol2" which are the stock closing prices from 1 day to the next. The one little hang up is the name of the column. There seems to be an additional data column name included in the attributes (dimnames?). So when i run HEAD(DATAFRAMENAME) i get the column name = "Open". but
2003 Apr 02
2
pacf.mts
I am getting the following: *** Weave Errors *** Error in driver$runcode(drobj, chunk, chunkopts) : Error in eval(expr, envir, enclos) : couldn't find function "pacf.mts" *** Source Errors *** Error in eval(expr, envir, enclos) : couldn't find function "pacf.mts" make[1]: *** [checkVignettes] Error 1 I don't really understand the new namespace mechanism,
2005 Mar 30
2
Step error
Could anyone tell me what am I doing wrong? > pro<-function(indep,dep){ + d<-data.frame(indep) + form<-formula(lm(dep~.,data=d)) + forward<-step(lm(dep~X1,data=d),scope=form,trace=0,direction='f') + return(forward) + } > pro(m,q) Error in inherits(x, "data.frame") : Object "d" not found Where q is a vector with the dependent variable's
2005 Jan 07
3
Basic Linear Algebra
I don't normally have to go anywhere near this stuff , but it seems to me that this should be a straight-forward process in R. For the purposes of this enquiry I thought I would use something I can work out on my own. So I have my matrix and the right hand results from that matrix tdata <- matrix(c(0,1,0,-1,-1,2,0,0,-5,-6,0,0,3,-5,-6,1,-1,-1,0,0),byrow = T,ncol = 5) sumtd <-