search for: equidist

I am running ARMA model on time series data. Data consists of price changes in S&P 500 during the day, recorded at 4 predefined time stamps( 8:30, 10:30,10:15, 3:12, hence non equidistant) My question is, do I need to worry about the data being non equidistant? If so, how do I modify my series to account for it? Or is it safe to assume data as being equidistant,as it may not have a strong influence on the final forecast? Thanks! -Ritz -- View this message in context: htt...

...pointwise confidence interval at x=0.5 contains the true value at 0.5 # i.e. what is the so called "coverage rate"? # ========================================================================================= pos = which(x==0.5) sum(abs(estlp[pos,] - m(x[pos])) <= 1.96*selp[pos,]) # equidistant x outputs 946 # non-equidistant x outputs 938 sum(abs(estss[pos,] - m(x[pos])) <= 1.96*sess[pos,]) # equidistant x outputs 895 # non-equidistant x outputs 936 This basically tells...

...r to what i get with plot(x,y,type='o'). The problem is that i have to many data-points for that method, the symbols will just "melt together", as in plot(seq(1,500), seq(1,500), type='o') - just one big line. What i want is to have a configurable number of symbols spread equidistantly over the whole graph, regardless of the number or position of data-points. Does anyone have an idea how to achieve this? Thanks, Timo

This might not possible in R but I thought I would give it shot. I am have to set up a 40 x 40 cm grid of 181 points equidistant from each other. Is there any way to produce a graph with R that can do this for me? Actual sizes are unimportant as long it is to scale. Thanks -- View this message in context: http://www.nabble.com/Grid-building-in-R-tp18371874p18371874.html Sent from the R help mailing list archive at Nabble...

...s: hist(islands,breaks=foo$breaks,freq=T) ##... this lead to the warning message: Warning message: the AREAS in the plot are wrong -- rather use `freq=FALSE'! in: plot.histogram(r, freq = freq, col = col, border = border, angle = ##The reason for this seems to be, that the breaks are NOT ##equidistant (despite foo$equidist being TRUE!): > foo$breaks [1] -0.0018 2000.0018 4000.0018 6000.0018 8000.0018 10000.0018 [7] 12000.0018 14000.0018 16000.0018 18000.0018 ##Correcting this (by changing the first element of foo$breaks): corr.breaks <- c(+0.0018,2000.0018,4000.0018,6000.0018,8...

...t a wireframe of a function which is defined on the unit square using the lattice library (for trellis-like plots). The plot contains z-values of about 100 (only in the neighborhood of zero) although most of the z-values are in the range of -let's say- 0 to 10. If I evaluate this function on an equidistant grid of 25 points on the unit square the plot quality is not very good, but I can see the rough shape of the plot (i.e. I can see the that the points in the range of 0 to 10 have different heights). If I increase the grid points to 100 (so the function is evaluated on an equidistant grid with 10...

Hello R Users, How to make a variable equidistance with time i.e. how to interpolate a variable if it is not sampled at equal time interval. Many thanks, Regards, Yogesh [[alternative HTML version deleted]]

Hi, I build a graph taking into account the times: 1, 2,4,6,8,10,12,15,18,21,24,28,32 and 48. Be that the scale of the X axis does not look right. It seems equidistant. (graph attached) What changes have I to do in the following commands so that the scale be correct? interaction.plot(Tempo, Trat, Valor, ylim=c(0, 2), las=1, lty=c(1,2,3,4), lwd=3, bty='l', col=c('red','blue','magenta','gre...

...o something like this? (The example does not work but I don't know why...) data <- rnorm(100) # as example, of course this is not skewed... h <- hist(data, plot=FALSE) mh <- 5 hh <- list(h$breaks[0:(mh+1)], h$counts[0:mh],h$intensities[0:mh],h$density[0:mh],h$mids[0:mh],h$xname,h$equidist) names(hh) <- names(h) plot(hh) Antje

Hi all, I am trying to find a way to rename R objects with names pulled from a vector of names. For example, I have a data frame, my.data.frame, and a list of names, my.names. My.names is simply the column names of my.data.frame. I want save the histogram with the column name as the name of the object. for (i in 1:ncol(my.data.frame) { tmp<-hist(my.data.frame[,i])

...c(if (include.lowest) -diddle else diddle, rep(diddle, length(breaks) - 1)) else c(rep(-diddle, length(breaks) - 1), if (include.lowest) diddle else -diddle) breaks <- breaks + fuzz ... The addition of the fuzz variable to the breaks causes hist to think of no equidistant breaks (even if they are) and therefore a density plot is generated by default. Removing these lines the behaviour is like the one in version 1.2.1 and earlier. Is there an alternative doing this with the new version? And what about the border color parameter in barplot? Many thanks in advan...

...@ -120,7 +123,9 @@ for(i in seq_len(NROW(y1)-1)) polygon(c(x1, rev(x1)), c(y1[i+1,], rev(y1[i,])), col = col[i], border = border) - axis(1) + if (is(x, "POSIXct")) { + axis.POSIXct(1, x) + } else axis(1) equidist <- any(diff(y1[,1L]) < tol.ylab) if(equidist) -- Seb

...[7] 1.4492754 3.6231884 0.0000000 0.0000000 0.7246377 $density [1] 7.2463754 2.8985507 10.8695652 13.7681159 5.7971014 3.6231884 [7] 1.4492754 3.6231884 0.0000000 0.0000000 0.7246377 $mids [1] 0.15 0.17 0.19 0.21 0.23 0.25 0.27 0.29 0.31 0.33 0.35 $xname [1] "data" $equidist [1] TRUE attr(,"class") [1] "histogram"

Dear R-Listers, I have spatial data on an equidistant rectangular grid, similar to topographic data. I know that there are quite a few R-packages or base functions that provide nice iso-contours plot, but I don't want a plot, just the smoothed isocontour line of ONE level (e.g. 10 mm). Data sets are large, so it would be preferable if the avai...

...esentation of frequencies, the 'counts' component of the result; if 'FALSE', probability densities, component 'density', are plotted (so that the histogram has a total area of one). Defaults to 'TRUE' _iff_ 'breaks' are equidistant (and 'probability' is not specified). I therefore expect that the following command: > h <- hist(StockReturns, freq=FALSE) where StockReturns has the following data in it: > sourcedata$StockReturns [1] -0.006983 0.111565 0.053782 0.027966 0.068956 0.165424 -0...

...; measurements (i.e. 4 times an hour) of a catchment ........ > Maybe time series are easier, but in > > ts(data = NA, start = X,... > > X should be a number or a vector. how does this coresponds to a > data and hour (e.g. april 26,2002, 15:00:00)? If your observations are equidistant, e.g. you've got 24 hourly measurements per day, you could do something like this for the above example: R> rain <- ts(rain, start = c(26, 15), freq = 24) R> flow <- ts(flow, start = c(26, 15), freq = 96) ........... ========================================================...

...it=lm(y ~ 1 + x) pred=predict(fit, interval="prediction") matplot(x,pred,type="l",ylab="y") points(x,y) I would have expected the lower and upper prediction intervals to converge as x increases (and the noise decreases), but they seem to remain virtually equidistant. Can anyone explain (a) the behavior that I see, and (b) how to obtain curves that do bound the noise? Thanks, Ronnen. -- Ronnen Levinson, Ph.D. \/ RML27 at cornell.edu scientist || http://ronnen.com Lawrence Berkeley National Lab /\ fax...

...freq=F) > h <- hist(x, freq=F) > summary(h) # Length Class Mode #breaks 11 -none- numeric #counts 10 -none- numeric #intensities 10 -none- numeric #density 10 -none- numeric #mids 10 -none- numeric #xname 1 -none- character #equidist 1 -none- logical # The help file says that <h$density> holds the values # plotted in the probability histogram. If that's the # case, I'd expect that the sum of h$density for a histogram where freq=F would equal 1.0 ... However: > sum(h$density) #returns the value :...

Hi, I have a large amount of data that I would like to create a histogram of and plot and do things with in R. It is pretty much impossible to read the data into R, so I have written a program to bin the data and now have a list of counts in each bin. Is it possible to somehow import this into R and use hist(), so I can, for instance, plot the probability density? I have looked at the help page

In an experimental setup we obtain z-data samples at equidistant grid points. The surf.gls (Kriging) algorithm produces an error under this circumstance when performing the Choleski decomposition. A workaround is to dither the grid coordinates using (x <- rnorm(length(x)) ; y<- rnowm(length(y))). Question: Is this an expected behaviour of the surf...