search for: fivenum

Dear useRs, how can I construct a box&whisker plot based on the vector "fivenum"? The challenge I face is as follow: I have a table such as x | fivenum --------------- ... | (.....) ... | (.....) and so forth.... For each observation x I have generated a vector containing the fivenum estimates. The first challenge is to group my fivenum vectors into groups...

I've just started using R and am still a neophyte, but I found the following curious result. I'm using the current version of R (2.5.1 (2007-06-27) ). Why are the results for the third quartile different in the output from the summary and fivenum commands? For the following data set 457 514 530 530 538 560 687 745 745 778 786 790 792 821 821 822 822 828 845 850 886 886 886 913 1050 1050 1065 1065 1065 1065 1090 1130 Summary yi...

Full_Name: Richard Huggins Version: 1.7.1 OS: windows 2000 Submission from: (NULL) (131.172.4.44) > x<-rnorm(100,2,1) > mean(x) [1] 1.73299 > summary(fivenum(x)) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7320 2.3840 3.7910 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. -0.3655 1.1070 1.7430 1.7330 2.3830 3.7910 > y<-log(abs(x)) > summary(fivenum(y)) Min. 1st Qu. Median Mean 3rd...

Full_Name: David Leong Version: 2.7.2 OS: Windows XP Submission from: (NULL) (12.187.86.2) It appears for the fivenum function, there is a strait average between two ranked samples. This should be a linear interpolation between ranks. For the following data set #data CaCO3<-c(130.8,129.9,131.5,131.2,129.5,132.7,131.5,127.8,133.7,132.2,134.8,131.7,133.9,129.8,131.4,128.8,132.7,132.8,131.4,131.3) fivenum report...

I'm having trouble understanding the output from as.windrose(). For one thing, data on a boundary between sectors seem to be left out of the counts. I assume that explains the missing point in the output below (angle 45). Shouldn't one side of each sector interval be open, to include values such as my 45 in the example? Also, why does the angle 180 in my input apparently not result in

...I've just started using R and am still a neophyte, but I found the >> following curious result. I'm using the current version of R >> (2.5.1 (2007-06-27) ). >> >> Why are the results for the third quartile different in the output >> from the summary and fivenum commands? For the following data set >> >> 457 514 530 530 538 560 687 745 >> 745 778 786 790 792 821 821 822 822 >> 828 845 850 886 886 886 913 1050 >> 1050 1065...

Dear all i use R only a few days and don't understand the difference between fivenum(x) und summary(x). > x [1] 20.77 22.56 22.71 22.99 26.39 27.08 27.32 27.33 27.57 27.81 28.69 29.36 [13] 30.25 31.89 32.88 33.23 33.28 33.40 33.52 33.83 33.95 34.82 > fivenum(x) [1] 20.770 27.080 29.025 33.280 34.820 > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 20.77...

When all else fails, read the help page... ?fivenum says to look at ?boxplot.stats, and the "Details" section of ?boxplot.stats has, well, details. Tukey had reasons to call those hinges rather than quartiles. Andy > -----Original Message----- > From: Knut M. Wittkowski [mailto:kmw at rockefeller.edu] > Sent: Thursday, May 29,...

...a clue will be sufficient to get this show on the road. I am trying to graph tide prediction data in various ways. I haven't been able to answer this question after pouring over TFM. I read the datafile theight <- read.table("tides",header=TRUE) I do get a summary() or fivenum() summary(theight) fivenum(theight) But when I run stem(theight), I receive CONSISTENTLY the following message, no matter when I have done: > stem(theight) Error in stem(theight) : stem: x must be numeric The file looks like this: Height 0.225970 0.141066 0...

I'm using the image.plot() function (fields package), but I want to enlarge the characters of the legend (as they are too small to be read in a combined figure), but there is no way I can find a command to do this. I can enlarge the legend bar (with legend.witdh), axis character size (cex.axis) or the total legend size (legend.shrink), but not the character size of the legend characters

Okay, so I have a homework projecr for R, and we had to input the following link as some sort of data: nb10 <- read.table("http://www.adjoint-functors.net/su/web/314/R/NB10"). Afterwards, we have to use fivenum(nb10) to find max, min, quantiles, and sd, but I'm okay with this. The next question is where I'm stuck. The question is as follows; Compare the median (use the quantiles or median(nb10$V1)) to the mean. I found the mean to be 408.8 and the median to be 404. Can you please explain to me...

Dear people I've learnt that by using the "boxplot.stats" command in the "grDevices" library I can get the 5-number summaries of a boxplot, plus other important information, like the confidence interval around the median. I'm interested in knowing the actual formula to used in that package to calculate that confidence interval. Can someone help me with this? Cheers,

Why am I getting a wrong result for quartiles? here is my code: > cbiomass = c(910, 1058, 929, 1103, 1056, 1022, 1255, 1121, 1111, 1192, > 1074, 1415) > summary(cbiomass) > IQR(cbiomass) The result R gives me is: For the summary > Min. 1st Qu. Median Mean 3rd Qu. Max. 910 1048 1088 1104 1139 1415 For IQR > 91.25 ********* The true Q1 is 1039

Hi, This message is to let the R developers know that the pdfs for the first two volumes of the printed edition of R reference manual are available at, http://www.network-theory.co.uk/R/base/vol1.pdf http://www.network-theory.co.uk/R/base/vol2.pdf I've divided the manual into "core commands" of the base package (control flow, and related programming commands, etc) in Volume 1

...vel/tests' make[1]: *** [test-all-basics] Fehler 1 make[1]: Leaving directory `/home/suapm095/Download/R-devel/tests' make: *** [check] Fehler 2 At the end of the file "tests/Examples/base-Ex.Rout.fail" I found: > i39 <- sapply(3:9, seq) # list of vectors > sapply(i39, fivenum) [,1] [,2] [,3] [,4] [,5] [,6] [,7] [1,] 1.0 1.0 1 1.0 1.0 1.0 1 [2,] 1.5 1.5 2 2.0 2.5 2.5 3 [3,] 2.0 2.5 3 3.5 4.0 4.5 5 [4,] 2.5 3.5 4 5.0 5.5 6.5 7 [5,] 3.0 4.0 5 6.0 7.0 8.0 9 > > hist(replicate(100, mean(rexp(10)))) Fehler in...

...", 8), rep("blue", 4), rep("pink", 8), rep("blue", 4), rep("red", 8))) ################################################################################ # Histogram depicting Hinge Spread with boxplot layout(matrix(c(1, 2)), heights=c(2, 1)) fn <- fivenum(nnum) LH <- fn[2] UH <- fn[4] HS <- UH - LH SHS <- HS/10 BR <- seq(LH - SHS*30, UH + SHS*30, SHS) par(mar=c(0,2,3,2)) hist(nnum, breaks=BR, col=c(rep("red", 30), rep("blue", 10), rep("red", 30)), xlim=c(60,140), axes=FALSE, xlab="", y...

Hi all: If I have a dataframe of N columns.If I wanna get the min(or max,or mean...etc)of the whole dataframe,how to do it quickly? What I can do is only: min(data[,1:ncol(data)]) Any other suggestion? Thanks! best [[alternative HTML version deleted]]

Dear Sir, I have data, coming from tests, consisting of 300 values. Is there a way in R with which I can confirm this data to 68-95-99.8 rule or three-sigma rule? I need to look around percentile ranks and prediction intervals for this data. I, however, used SixSigma package and used ss.ci() function, which produced 95% confidence intervals. I still am not certain about percentile ranks

...ir2) air.glm<-glm(mi~pm10, family=poisson, data=air) #air.gam<-gam(mi~s(pm10), family=poisson) qres[q]<-as.numeric(coef(air.cl)) glmres[q]<-as.numeric(coef(air.glm)[2]) #gamres[q]<-as.numeric(coef(air.gam)[2]) } par(mfrow=c(2,1)) plot(density(qres)) plot(density(glmres)) fivenum(qres) fivenum(glmres) # Berserk script # this shows that even for very few cases, my intended way of doing TS C-C is quite time-consuming. # #case<-c(5,0,2,1) #airq<-c(10,2,3,3) # #nctrl<-0*case #for (i in 1:length(case)) #{ # nctrl[i]<-sum(case[-i]) #} # #airca<-re...

...{ # Basic data manipulation this <- ecg[subscripts, ] placebos <- subset(ecg, variable == this$variable[1] & trt == 'Placebo' & !is.na(resp)) medians <- ddply(placebos, .(time), summarize, median = median(resp), h1 = fivenum(resp)[2], h2 = fivenum(resp)[4] ) medians$x <- as.numeric(medians$time) # Structure, including reference lines panel.grid(-1, 0) panel.abline(v = c(2.5, 7.5), col = trellis.par.get('reference.line')$col) if(current.row() == 2...