similar to: hist single block plot issue

Dear friends, The user coordiante system in the R graphics is easy to understand as it simply corresponds to the range of values on the axes of the plot. However, sometimes, we want to kown the coordinates of a region in other system, e.g., the normal coordinate system whose origin is the the upper-left cornner of the drawing area which is used to set the hot regions in the HTML image maps.

Dear All, I distributed the HTML style one year ago ( http://www.biosino.org/pages/newhtm/r/schtml and http://www.biosino.org/pages/newhtm/r/tchtml). Now I polished it, and rewrote it with Latex. You can download it from the URL: http://www.biosino.org/R/R-doc/ ( http://www.biosino.org/R/R-doc/files/R-intro_cn.pdf). Wish it will be useful for Chinese R users. Any

Dear R help I have been trying to conduct a chi square test on two groups of variables to test whether there is any relationship between the two sets of variables chisq.test(oxygen, train) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128, p-value = 1 > chisq.test(oxygen) Pearson's Chi-squared test data: oxygen X-squared = 26.6576, df = 128,

Hi, I would like to know the similarity between variables, but I don't exactly how begin and as from what dataframe or matrix! I have a matrix where in row I have 'Good', 'Medium','Bad' and in columns I have my Criterions ! What function and package should I use? Thanks a lot Sabine --------------------------------- ils, photos et vidéos ! [[alternative

Dear All, I am not familar with R. I want to use PCA (principal components analysis) and MDS (multidimensional scaling). Can someone tell me which R package I should use for PCA and MDS? I appreciate your help in advance. Ray

Hi all I wish to draw an histogram... with dates but the following append, i don't know where is the problem, help(hist.Date) works and i don't see any usefull information on what i'm doing wrong... > hist.Date(dt_cycles) Error: couldn't find function "hist.Date" > hist.date(dt_cycles) Error: couldn't find function "hist.date" > cycles [1] 7 1

Can someone help me with a misunderstanding I'm having with hist? I expected, from the example below, that the number of bins would always be 10 and the length of the counts array the same. According to the help section 'breaks' can be a integer indicating the number of bins. From the example below, the number of bins (length of the counts array) varies. Am I wrong in expecting the

Full_Name: Alex Deckmyn Version: 2.4.0 OS: linux Submission from: (NULL) (193.190.63.62) specifying the "right" option in hist results in a warning when plot=F. The option is taken into account correctly, but a warning is issued anyway. When plot=T there is no warning. > hist(c(1,1.5),breaks=0:4)$counts [1] 1 1 0 0 > hist(c(1,1.5),breaks=0:4,right=T)$counts [1] 1 1 0 0 >

Dear developers, the current implementation of hist.default() calculates 'density' (and 'intensities') as dens <- counts/(n*h) where h has been calculated before as h <- diff(fuzzybreaks) which results in 'fuzzy' values for the density, see e.g. > tmp <- hist(1:10,breaks=c(-2.5,2.5,7.5,12.5),plot=FALSE) > print(tmp$density,digits=15) [1]

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

Hi, This is a very basic question, but I would like to undestand hist(). I thought that the hist( , freq=FALSE) should provide the relative frequencies (probabilities), and so they should sum 1, however: set.seed(2) ah <- hist(rnorm(100), freq=F) sum(ah$intensities) [1] 2 set.seed(2) bh <- hist(rlnorm(100), freq=F) sum(bh$intensities) [1] 0.4999996 I'm getting similar figures with

Hi there, # Consider the following example: A = 19 B = 20 A< B A==B hist(c(1:15,B,50),breaks=c(0,15,A,50),plot=F)$count hist(c(1:15,A,50),breaks=c(0,15,B,50),plot=F)$count # I was expecting the same results with the following values of A and B: A = 19.6019203953960 B = 19.6019204365543 A< B A==B hist(c(1:15,B,50),breaks=c(0,15,A,50),plot=F)$count# wrong

Dear all, I search the mail list about this topic and learn that no simple way is available to get "lsmeans" in R as in SAS. Dr.John Fox and Dr.Frank E Harrell have given very useful information about "lsmeans" topic. Dr. Frank E Harrell suggests not to think about lsmeans, just to think about what predicted values wanted and to use the predict

On Sun, 11 Dec 2005 clausen at econ.upenn.edu wrote: > Full_Name: Andrew Clausen > Version: 2.1.0 > OS: Debian GNU/Linux > Submission from: (NULL) (71.242.192.73) > > > Hi, > > When I type > > hist(x, freq=F) > > I get a density function, as I expect. However, if I type > > plot(hist(x, freq=F)) > > then I get the same output as if I had

Hi all, A friend send me a question on why does this: x<-rpois(100,1) sum( hist(x)$density ) Gives out "2" I tried this: sum( hist(x, freq =T)$density ) It didn't help. Then he came back with the following insight: # with breaks b<-c(0,0.9,1:8) sum(hist(x,breaks=b)$density) # Much more then 2 # but if we add weights according to the interval length

The following code is giving me problems. I want to export densities of a distribution to a csv file. At the bottom of the code I use the hist function to generate the densities. But hist is returning values greater than 1. I don't understand, why. Any help you can supply is greatly appreciated. # Set word path dir<-"~/Research/MR Distribution Analysis/" setwd(dir)

Full_Name: Stephan Schlueter Version: 1.9.0 OS: Submission from: (NULL) (217.184.109.24) During my studies, I found a fault in the hist()-function: If you have a vector x with values around zero and also bigger than 10,000,000 , there will be a shift of -max(x)/10,000,000 in the hist-datas. See my example: x<-runif(10000) hist(x,breaks=c(seq(-3,3,0.1)),prob=TRUE) #everything ok, but

Hi! How to find out which plot function is used when i call plot(hist.default(1:10,plot=F)) and all works fine ? The reason why I would like to know it is that after loading some self written R functions > plot(hist.default(1:10,plot=F)) Error in xy.coords(x, y, xlabel, ylabel, log) : x and y lengths differ > traceback() 5: stop("x and y lengths differ") 4: xy.coords(x, y,

hist() is designed so that the total area sums to 1. You should build you desired behavior using a barchart. ? David Sent from my iPhone > On Aug 5, 2023, at 11:50 PM, Ott Toomet <otoomet at gmail.com> wrote: > > ?Sorry if this topic has been discussed earlier. > > Currently, hist(..., log="y") fails with > >> hist(rexp(1000, 1), log="y")

Dear listers, A quick question about breaks in hist(). The histogram is highly screwed to the right, say, the range of the vector is [0, 2], but 95% of the value is squeezed in the interval (0.01, 0.2). My question is : how to set the breaks then make the histogram look even? Thanks in advance, Leaf