search for: freqs

...ain about this question as being unrealistic, consider the possibility that you generated the frequency lists of two corpora (here, a and b) that are so large that you cannot combine them into one (a.and.b<-c(a, b)) and generate a frequency list of that combined vector (table(a.and.b)) ... joint.freqs a b d e f g i j 3 1 3 1 5 5 1 1 joint.freqs<-vector(length=length(sort(unique(c(names(freq.list.a), names(freq.list.b)))))) # You generate an empty vector joint.freqs (i) that is as long as there are different types in both a and b (but note that, as requested, this information is not taken fro...

Greetings R Community, I am working with the ifelse function and it is returning something unexpected. In the code the line with the MODE1 assignment the output is a vector [1] 4 5 6 but when I put the MODE1 object into the ifelse function [R}'s output for MODE1 is the first number from the string (4). Why is this? Given the supplied vector of x I would assume both the MODE1 and ifelse()

Dear list, I have a problem with freq from prettyR. Please have a look at my syntax with a litte example: library(prettyR) #Version 1 test.df<-data.frame(q1=sample(1:4,8,TRUE), gender=sample(c("f","m"),8,TRUE)) test.df freq(test.df) #No error message #Version 2 test.df<-data.frame(gender=sample(c("f","m"),8,TRUE), q1=sample(1:4,8,TRUE)) test.df

#Hello, #I have three data frames, X,Y and Z with two columns each and different numbers of rows. # creation of data frame X X.alleles <- c(1,5,6,7,8) X.Freq <- c(0.35, 0.15, 0.05 , 0.10, 0.35) Loc1 <- cbind( X.alleles,X.Freq) X <- data.frame(Loc1) #creation of data frame Y Y.alleles <- c(1,4,6,8) Y.Freq <- c(0.35, 0.35, 0.10, 0.20 )

In http://tolstoy.newcastle.edu.au/R/e2/help/06/10/3064.html a method was given for converting a frequency table to an expanded data frame representing each observation as a set of factors. A slightly modified version was later included in the NCStats package, only on http://rforge.net/ (and it has too many dependencies to be useful). I've tried to make it more general, allowing an input

Hi all, I have a basic question concerning graphs in R. I?m using the par() function and I?m working with multiple figures by row (mfrow) but my the hight of my figures become compressed. I have 4 rows and 2 columns (because I want to plot 8 histograms (freq = FALSE ) on it. I know I can adapt my margins with for example ?oma? and ?mai? but I don?t know how to choose the size of the figure? I

I have written the following function to calculate the weighted mean difference for univariate data (see http://www.xycoon.com/gini_mean_difference.htm for a related formula). Unsurprisingly, the function is slow (compared to sd or mad) for long vectors. I wonder if there's a way to make the function faster, short of creating an external C function. Thanks very much for your advice. gmd

I write a function to get the frequency and prop of a variable. freq<-function(x,digits=3) {naa<-is.na(x) nas<-sum(naa) if (any(naa)) x<-x[!naa] n<-length(x) ta<-table(x) prop<-prop.table(ta)*100 res<-rbind(ta,prop) rownames(res)<-c("Freq","Prop") cat("Missing value(s) are",nas,".\n") cat("Valid case(s)

Dear useRs, I have written a very simple function to compute some probabilities on words (function is below). The function includes a which() statement applied to a vector of characters (word.split): sapply (word.split, function(x) which(letters==x)). This statement worked as expected when used outside the global function : > word <- "hello" > (word.split <-

Hi, Is there any documentation on enabling hypervisor support for both C and P-state control? On xen-unstable and linux-2.6.18-xen.hg, if I enable cpuidle=1 on the xen command line and then run xenpm, I will get output for C-states (shown below) but it complains that "Xen cpufreq is not enabled!" cpu id : 0 total C-states : 2 idle time(ms) : 73264 C0

Hi, i'd like to plot some data that I have with the value on the x axis and freq on the y axis. So, I need to calculate the freq a value is seen within my data vector for example, say i have a vector of data data=c(1,1,1,4,5,5,6) I want values<-c(1,4,5,6) freq<-c(3,1,2,1) in order to enable me to plot this. Sorry, i'm new to R. What is standard

In R0.61, In hist(), should the line rel.freqs <- counts/(sum(x) * diff(breaks)) computing the relative frequencies or height of the rectangles in a histogram not be rel.freqs <- counts/(sum(counts) * diff(breaks)) instead, or do I misunderstand something? Thanks, Philippe -- --------------------------------------------------------...

...how to make density histograms) It doesn't seem easy to make density on the y-axis with the current hist(). The freq argument only lets you choose counts (TRUE) or relative frequency (FALSE). I would like to suggest that freq (or some renamed version of the argument) take on the values counts freqs densities So that you can easily do a density histogram. The strange thing to me is that if you do h<-hist(...) you get h$intensities, which is the densities. So hist calculates the densities, but doesn't let you plot them? May I suggest you call it $densities? I don't know why you cal...

Hi Everybody I need to create a lot of frequency tables with frequencies and percentages (and cumilative freq and % as well) for a report. freq() in prettyR give more or less what I need. I am trying to export the result of freq() to html but the html doesn't look look the console output. See the following example library(prettyR) library(Hmisc) x <- matrix(sample(1:3, 12,

Dear all, I have done further modifications on the histogram functions that I reported earlier this year, and I hope this can be of use and perhaps included in the distribution. I have been using this stuff a couple of months myself, and while it is nothing sophisticated, it has it's applications. :-) I did a few small modifications today to make it a bit more compact. I have modified the

Dear Rich, I think that it's generally a bad idea to give statistical (as opposed to simply technical) advice by email without knowing the context of the research. I think that you'd do well to seek help from a statistician, and not just do what I suggest below. Interpolating the data only makes sense if there's no random component to the response (mag in your data). Otherwise, it

I am trying to perform goodness of fit test using R. I am using this website http://wiener.math.csi.cuny.edu/Statistics/R/simpleR/stat013.html for help. However, I am unable to carry out the test successfully. My code follows. It is taken from the website just mentioned. freq=c(22,21,22,27,22,36) # frequencies obtained after rolling the dice 150 times. prob=c(1,1,1,1,1,1)/6 # specify expected

Hi, #working directory data1 #changed name data to data1.? Added some files in each of sub directories a1, a2, etc. ?indx1<- indx[indx!=""] lapply(indx1,function(x) list.files(x)) #[[1]] #[1] "a1.txt"??????? "mmmmm11kk.txt" #[[2]] #[1] "a2.txt"??????? "mmmmm11kk.txt" #[[3]] #[1] "a3.txt"??????? "mmmmm11kk.txt" #[[4]] #[1]

Hello All, Am learning to create tables with n(%) formatting using R. Below is a working example. I think this is not bad but wondered if there are better ways of doing it. Although it can be quite humbling, seeing good code helps me assess my progress as an R programmer. Ultimately want to have code that I can turn into a function. Will then use the output produced to make tables using

Dear libvirt experts, I can not instantiate even a simple machine when using the 'vbox' hypervisor: s14% virt-install --connect=vbox:///session --virt-type vbox --name vtest --memory 500 ERROR cannot obtain CPU freq: No such file or directory s14% virsh -c vbox:///session błąd: cannot obtain CPU freq: No such file or directory (1) How to fix this error? The VirtualBox driver seems