similar to: request

Displaying 20 results from an estimated 10000 matches similar to: "request"

2008 Sep 11
3
periodicity validation
There is a series of data contains time in fixed step and energy varying with time, how to test its periodicity?In R, it seems there is no direct tools since I have search the R manual with periodic and I have not found any related topic. Thanks a lot
2008 Oct 20
2
calculating mean for samples
Hi everyone, > does any one knows how can I calculate mean for different samples > i.e. I have a data like this: > > s1 s2 s3 s4 > 1 0 0 0 1 > 2 1 0 1 0 > 3 0 0 0 0 > 4 0 0 0 0 > 5 0 1 0 1 > 6 1 0 0 0 > 7 0 0 0 0 > 8 0 0 0 0 > 9 0 0 0 0 > 10 0 0 0 1 > > I need to make 5 different sample with 5
2008 Nov 04
2
Zoo seems to be running slow in R 2.8.0 windows
R version 2.8.0 (2008-10-20) i386-pc-mingw32 locale: LC_COLLATE=English_United States.1252;LC_CTYPE=English_United States.1252;LC_MONETARY=English_United States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] StreamMetabolism_0.01 chron_2.3-24 zoo_1.5-4 loaded
2008 Sep 20
1
fitting a hyperbole
I have got a data set that is Gross Primary Productivity ~ Total Suspended Solids it is a hyperbola just like: plot(1/c(1:1000)) how do I model this relationship so that I can get all of the neat things that lm gives residuals etc. etc. so that I can see if my eyeball model stands up. Thanks for any help, pointers, or good things to read. -- Stephen Sefick Research Scientist Southeastern
2008 Aug 28
1
abline of an lm fit not correct
mac osx 10.5.4 R 2.7.1 I have fit a model d<-lm(y~x) with an R^2 of 0.963 but when I issue the command abline(d) the line is below where it ought to be. Looks like the right slope, but not the right intercept. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large
2008 Oct 23
1
Reversing xlim qplot
I would like to be able to reverse the xlim on qplot this is the code that I am using qplot(a[,"River.Mile"], a[,26] ,ylab=colnames(a)[26], xlab="RiverMile", xlim=rev(c(60, 216)))+geom_smooth()+scale_x_continuous(breaks=c(215,202,198,190,185,179,148,119,61),
2008 Oct 29
1
Macro stuff to work on up through august 2007
Title says it all remember cast() with sum as the aggregation function -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being
2008 Nov 09
1
maptools sunrise sunset function
##This is a function that I am trying to write to calculate sunrise and sunset and works "mostly", but returns nonsensical values. What am I #missing? Thanks in advance. ###remember to include maptools as dependence### library(maptools) sunrise.set <- function(lat, long, date, timezone="UTC", num.days=1){ #this needs to be long lat# lat.long <- matrix(c(long, lat),
2008 Oct 22
1
plotting matrix
a <- c(1:26) b <- rnorm(25) e <- rnorm(25) f <- rnorm(25) g <- data.frame(b,e, a,f) I would like to plot a agianst all possibilities and then shoot it out to a pdf one graph per page. I think it would be okay to have this as a lattice plot or a ggplot with many graphs per page. I can figure all of that out I think, but I need something like r <- as.matrix(g) plot(.~a, data=r)
2008 Aug 29
2
non-parametric Anova and tukeyHSD
I have insect data from twelve sites and like most environmental data it is non-normal mostly. I would like to preform an anova and a means seperation like tukey's HSD in a nonparametric sense (on some sort of central tendency measure - median?). I am searching around at this time on the internet. Any suggestions, books, etc. would be greatly appreciated. -- Stephen Sefick Research
2008 Oct 20
5
Staging area for data before read into R
I am wondering if there is a better alternative than Excel for data storage that does not require database knowledge (I will eventually have to learn this, but it is not on my immediate todo list). I need something that is not limited to 256 columns... I don't need any of the built in functions in excel just a spreadsheet like program with cells that hold data in a data.frame format for a
2008 Sep 15
4
getting data into correct format for summarizing ... reshape, aggregate, or...
I would like to reformat this data frame into something that I can produce some descriptive statistics. I have been playing around with the reshape package and maybe this is not the best way to proceed. I would like to use RiverMile and constituent as the grouping variables to get the summary statistics: 198a 198b mean mean sd sd ... ... etc. for all of these. I have tried
2008 Oct 27
3
Problem in installation of "Rgraphviz" package
Dear R users,   I am not so used to this R software. I have to use the package " Rgraphviz" but found some problem in the installation process. I download this package and store in R library but i am not getting this package in R installation list. I made review in google search net and use the following command:   ### source("http://bioconductor.org/biocLite.R")
2008 Sep 25
1
Splitting row names up and then adding up the columns associated with criteria from the parts of the site coding (help)
d <- c("upwd1201", "upwd0502", "upwd0702", "upwd1002", "upwd1102", "upwd0203", "upwd0503", "upwd0803", "upwd0104", "upwd0704", "upwd0804", "upwd1204", "upwd0805", "upwd1005", "upwd0106", "dnwd1201", "dnwd0502",
2008 Sep 24
2
lattice xyplot symbols instead of colors and legend matching plot symbols or colors
I would like to use the data below where the plots are close to what I want. Instead of color I would like to use different symbols, and have the symbols in the legend match the graphs. I am also going to add a regression line to these I know about the type="r" (which is fine for these particular graphs) argument, but it fits the subsets instead of the entire data set-- should I use a
2008 Nov 01
2
Hidden line algorithms and a different kind of waterfall
This is not the same as the recent thread on a waterfall graph. I'm thinking about the rolling FFT display used in acoustics and other spectrum analysis tasks. Here's an example of a very fancy 3-D waterfall display: http://www.ultimaserial.com/UltimaWaterfall.html I was just wondering if there are any simple hidden-line tools in R that I could use to draw simple waterfall displays.
2008 Sep 06
1
plot a list
i have a list of 6 each containing a dataframe of 96 observations as a zoo object. Is there a way to plot these in one frame par(mfrow=c(3,2)) this is what I tried lapply(d, FUN=plot) I can provide data, list is large. thanks -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so
2008 Oct 06
1
easier way to do this without a loop? (successive euclidean distances between points)
a <- c(1:10) b <- c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11) d <- c(21:30) z <- data.frame(a,b,d) library(fields) results <- c() for(i in 1:(length(rownames(z))-1)){ results[i] <- rdist(z[i,], z[(i+1),]) } results.1 <- data.frame(results) f <- rownames(z) r <- f[-1] rownames(results.1) <- r colnames(results.1) <- f[1] this does what I want it to do - is
2008 Oct 19
1
zoo in ggplot2
library(zoo) d<-(structure(c(1.39981554315924, 0.89196314359498, 0.407816250252697, 0.823496839063978, 1.14429021220358, 1.23971035967413, 0.960868900583432, 0.927685306209829, 1.22072345292821, 0.249842897450642, 1.00879641624694, 0.925372139878243, 0.317259909172362, 0.382677149697482), index = structure(c(11808, 11869, 11961, 11992, 12084, 12173, 12265, 12418, 12600, 12631, 12753, 12996,
2008 Nov 10
1
how to vectorize a for loop
#Is there a way to vectorize the for loop #maybe a fancy indexing trick? #thanks d=0.5 L=20 x=seq(20, by=1, length.out=20) reflecting <- function(pre, d, L){ r=L-1 x=rep(0, L) for(j in 2:r){ x[j]=((1-(2*d))*pre[j])+(d*pre[(j+1)])+(d*pre[(j-1)]) } x[1]=((1-d)*pre[1])+(d*pre[2]) x[L]=((1-d)*pre[L])+(d*pre[(L-1)]) y=x } f = reflecting(x, d, L) -- Stephen Sefick Research Scientist