search for: sturges

Dear All, According to the Sturges rule, the number of classes of a histogram is the closest integer to 1 + logb(n,base=2) where n is the number of observations. The function hist(), by default, uses the Sturges rule. However, the code x <- 1:200 hist(x) produces a histogram with 10 classes and not 9 classes as determined by...

...of cells for the histogram, * a character string naming an algorithm to compute the number of cells (see Details), * a function to compute the number of cells. In the last three cases the number is a suggestion only. B) The default for 'breaks' is '"Sturges"': see 'nclass.Sturges'. If I look at the code for nclass.Sturges() I see function (x) ceiling(log2(length(x)) + 1) and, for length(X) = 10000, this gives 15. This is not related to any of the numbers of breaks I actually got, in any way obvious to me. So: Question 1: hist()...

...the my functions ('er' is a mention to EasieR, because I'm trying to do a package for myself and the my students): #2. Tables from data.frames #2.1---er.table.df.br (User define breaks and right)------------ er.table.df.br <- function(df, breaks = c('Sturges', 'Scott', 'FD'), right = FALSE) { if (is.data.frame(df) != 'TRUE') stop('need "data.frame" data') dim_df <- dim(df) tmpList <- list() for (i in 1:dim_df[2]) { x <- as.matrix(df[ ,i])...

...# Make final table names(res) <- c('Class limits', 'fi', 'fr', 'fr(%)', 'fac', 'fac(%)') return(res) } # # Common function # tb.make.table.II <- function (x, k, breaks=c('Sturges', 'Scott', 'FD'), right=FALSE) { x <- na.omit(x) # User defines only x and/or 'breaks' # (x, {k,}[breaks, right]) if (missing(k)) { brk <- match.arg(breaks) switch(brk, Sturges = k <- nclass.Stu...

Buenas. Sé que en R hay multitud de datasets y me haría falta alguno que trataran de variables relacionadas con salud, sobre todo para aprender más acerca de cómo realizar una regresión logística binomial o multinomial. Gracias..

Hi Group, I've a vector of 1000 numeric values for which I want to draw a histogram. I've read this vector into R with no variable name.I mean only the 1000 values, which makes V1 the name of the variable by default?? Then I tried > hist(V1, breaks = "Sturges", +      freq = NULL, probability = !freq, +      include.lowest = TRUE, right = TRUE, +      density = NULL, angle = 45, col = NULL, border = NULL, + main = paste("Histogram of" , V1name), + V1lim = range(breaks), ylim = NULL, + V1lab = V1name, ylab, + axes = TRUE, plot = TRUE, labe...

...m a newbie in R and could not find help on this problem. I am trying to plot an histogram with probabilities in the y axis. This is the code I am using: #TLC uniform n=30 mi=1; mx=6 nrep=1000 xbar=rep(0,nrep) for (i in 1:nrep) {xbar[i]=mean(runif(n,min=mi,max=mx))} hist(xbar,prob=TRUE,breaks="Sturges",xlim=c(1,6),main=paste("n =",n), xlab="Média", ylab="Probabilidade") curve(dnorm(x,mean=mean(xbar),sd=sd(xbar)),add=TRUE,lwd=2,col="red") The problem is that I am getting greater than 1 probabilities in the Y axis? Is there a way to correct this? Many...

Hi all: As to hist,the help file says:" R's default with equi-spaced breaks (also the default) is to plot the counts in the cells defined by breaks." I wanna know how the break is calculated by R? In other words: break = (max - min)/(number of group) but how the "number of group" is calculated by R? Thanks!

Scott's SpanDSP FAQ mentions that the senders' TSID (20 bytes sent by sending fax machine) is made available, but doesn't mention where or how. Is there a variable or something that's set? I didn't see it in http://www.voip-info.org/tiki-index.php?page=Asterisk+variables Thanks, Woody

...ncy by the size of the intervals being used. So I could divide the series by this interval length, and then plot the relative frequency. The problem is determining the interval length that will be used in advance. I'm supposed to manually pick the # of intervals between 20 and 40 (not use sturges). However, when I try different values for the nbreaks parameters, incrementing from 20 - 40, the histogram only changes at a couple of points on that interval. Anyone have any suggestions as to how I should create this plot? Or an explanation as to why my nbreaks parameter does not seem to f...

...y default?? No it defaults to nothing. Type ls() to see this. You must explicitly name the variable: V1 <- whatever Be careful, you likely are reading it in as a data.frame. Try class(V12) or str(V1) to see what data structure you have. Then I tried > > > hist(V1, breaks = "Sturges", > +????? freq = NULL, probability = !freq, > +????? include.lowest = TRUE, right = TRUE, > +????? density = NULL, angle = 45, col = NULL, border > = NULL, > + main = paste("Histogram of" , V1name), > + V1lim = range(breaks), ylim = NULL, > + V1lab = V1name, yla...

.../><o:p></o:p> I would like to create a histogram from a data collumn consisting of 4 classes (0; 0.05;0.5;25;75). Due to the difference in scale the classes 0;0.05 and 0.5 are displayed within one combined bin by default with the code:Hist(x, scale="percent", breaks="Sturges"). How can I display them all, or as unique classes, leaving out the rest of the x_axes scale? There was no enlightenment in the help function and the command : breaks=5 did not do the trick.<o:p></o:p> Thanks in advance <o:p></o:p> Chris<o:p></o:p> Ver...

...misspelled (as 'Friedman') in src/library/graphics/man/hist.Rd. As a result, the help page currently implies that breaks = "Fried" is a valid argument to hist, but results in an error: > hist(rnorm(100), breaks = "Fried") Error in match.arg(tolower(breaks), c("sturges", "fd", "freedman-diaconis", : 'arg' should be one of sturges, fd, freedman-diaconis, scott > sessionInfo() R version 2.5.0 Under development (unstable) (2006-11-06 r39797) i686-pc-linux-gnu ...

...m, 5 ) ) ^ 2 ) I have read the help for ?cut, ?table, ?hist, and ?split, but am stumped for which one to use in this case--if any. How do you calculate c( mean(o[1:5]), mean(o[6:10]), ... ) for an arbitrary length vector using an appropriate number of bins (fixed at 5, or perhaps calculated using Sturges' formula)? I have also posted a more detailed version of this question on StackOverflow: http://stackoverflow.com/questions/3073365/root-mean-square-deviation-on-binned-gam-results-using-r Many thanks. Dave [[alternative HTML version deleted]]

Full_Name: Reinhold Koch Version: 1.6.1 OS: redhat 8.0 Submission from: (NULL) (131.152.84.111) I came across rather weird behavior of the breaks in hist: hist(1:3) gives the expected result, besides an unnecessary gap between 2nd and 3rd column hist(1:4) always merges up the first two columns, also if I resort to hist.default(1:4,breaks=1:4). hist.default(1:4, include.lowest=F) gives an

...quot;D:\\data 12.24.06\\AllMamushiCorrected5.8.10_7.12.10.csv", header=TRUE, strip.white=TRUE, na.strings="") attach(gb)   superhist2pdf <- function(x, filename = "super_histograms.pdf", dev = "pdf", title = "Superimposed Histograms", nbreaks ="Sturges") { junk = NULL grouping = NULL for(i in 1:length(x)) { junk = c(junk,x[[i]]) grouping <- c(grouping, rep(i,length(x[[i]]))) } grouping <- factor(grouping) n.gr <- length(table(grouping)) xr <- range(junk) histL <- tapply(junk, grouping, hist, breaks=nbreaks, plot = FALSE) maxC...

...s and cannot judge the importance yet.) I'd be very thankful for any hints! Berry PS: I recently read about barcharts in lattice, but by now I'm already used to my function. (And I learned a lot writing it a couple of years ago). # Function horiz.hist <- function(Data, breaks="Sturges", col="transparent", las=1, ylim=range(HBreaks), labelat=pretty(ylim), labels=labelat, border=par("fg"), ... ) {a <- hist(Data, plot=FALSE, breaks=breaks) HBreaks <- a$breaks HBreak1 <- a$breaks[1] hpos <<- function(Pos) (Pos-HBreak1)*(length(HBreaks...

i want to plot the histogram and the curve in the same graph.if i have a set of data ,i plot the histogram and also want to see what distribution it was.So i want to plot the curve to know what distribution it like. -- View this message in context: http://www.nabble.com/how-to-plot-the-histogram-and-the-curve-in--the-same-graph-tp20082506p20082506.html Sent from the R help mailing list archive at

Using R-2.12.1 and latticeExtra-0.6-14, I would like to understand why a rootogram displaying samples from the Poisson distribution looks like I expected it, whereas a rootogram using the normal distribution does not: library(latticeExtra) rootogram(~rpois(1000, lambda = 50), dfun = function(x) dpois(x, lambda = 50)) rootogram(~rnorm(1000), dfun = function(x) dnorm(x,mean(x),sd(x))) I

I have my peers registered to SER.asterisk seems to be sending mwi for the peers seen in the sip show peers CLI command. i have my ser server registered with asterisk as a type=friend and all clients register to ser.how do i get mwi to work for these clients registered to SER. Thank you, -AA