search for: median

Hi All, I want to extract elements of elements in a list. Here's an example of what I mean: If I create a list: x <- as.list(100) for(loop in c(1:100)) { x[[loop]] <- summary(runif(100)) } > head(x) [[1]] Min. 1st Qu. Median Mean 3rd Qu. Max. 0.02271 0.25260 0.58130 0.52120 0.77270 0.99670 [[2]] Min. 1st Qu. Median Mean 3rd Qu. Max. 0.006796 0.259700 0.528100 0.515500 0.781900 0.993100 [[3]] Min. 1st Qu. Median Mean 3rd Qu. Max. 0.00927 0.22800 0.40780 0.46410 0.69460 0.98780 I want t...

Hello everyone, I am using *mlogit* to analyse my choice experiment data. I have *3 alternatives* for each individual and for each individual I have *9 questions*. I have a response from *516 individuals*. So it is a panel of 9*516 observations. I have arranged the data in long format (it contains 100 columns indicating different variables and identifiers). In mlogit I tried the following

...ng if...else and I hope you can shed some light on it. I am using R version 2.2.0.1. I have the following data frame: > head(dat2f) year tot_km3y [1,] 1964 0.1876854 [2,] 1965 0.1835116 [3,] 1966 0.1915012 [4,] 1967 0.1869758 [5,] 1968 0.2249865 [6,] 1969 0.1916011 I need to pick out the median year, and since there are an even number of data, I cannot use 'median' directly since it gives me a non-existent year/discharge. I found a way to get around that with the following: md <- dat2f[, list(med_year = max(year[which(abs(tot_km3y - median(tot_km3y)) == min(abs(tot_km3y - me...

Dear Sirs, I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, Chhattisgarh, India. While taking classes, I found the *by() *function producing following error when I use FUN=mean or median and some other functions, however, FUN=summary works. Given below is the output of the example I used on a built-in dataset "mtcars", along with error message reproduced herewith: > by(data=mtcars, INDICES=list(mtcars$am), FUN=mean) : 0 [1] NA ----------------------------------------...

Hi, I have a fairly simple linear regression using the lm function. There are about 100 variables and 30,000 rows of data. It runs fine and produces a decent looking R2 value. I'm interested in performing a stepwise variable selection to see if things can be cleaned up a bit. Calling the step function returns ONE iteration (all the variables) and then stops. No errors are reported.

Esteemed R UseRs, Regarding specifying arguments in functions and calling functions within functions: ## beginning ## ## some data ex <- rnorm(10) ex[5] <- NA ## example function Scale <- function(x, method=c("mean", "median")) { scl <- method scl(x) } ## both return NA Scale(ex, method=median) median(ex, na.rm=FALSE) ## both return the median Scale(ex, method="median") median(ex, na.rm=TRUE) ## 1. Why does the use of apostrophes have this effect when calling ## a fucntion within a function?...

Dear R Help List, I have a vector of n and a vector of n-1 and I want to use boot() to bootstrap the ratio of their respective medians. I want to eventually use boot.ci() to generate confidence intervals. Because the vectors are not equal in length, I tried a few things, but have yet to be successful. Try 1: > x <- runif(20) > > y <- c(runif(19), NA) > > median(x) [1] 0.522284 > > median(y[1:19])...

Hello, I am trying to calculate the median of numbers across each row for the data shown below , with the condition that if the number is negative, that it should be ignored and the median should be taken of only the positive numbers. For eg: data is in Column A,B,C. Column D and E demonstrates what I want to get as answer A B C...

...t;> Martin Maechler <maechler at stat.math.ethz.ch> >>>>> on Mon, 27 Feb 2017 10:42:19 +0100 writes: >>>>> Rob J Hyndman <Rob.Hyndman at monash.edu> >>>>> on Wed, 15 Feb 2017 21:48:56 +1100 writes: >> The generic stats::median method is defined as median <- >> function (x, na.rm = FALSE) {UseMethod("median")} >> I suggest that this should become median <- function (x, >> na.rm = FALSE, ...) {UseMethod("median")} >> This would allow additional S3 methods...

The generic stats::median method is defined as median <- function (x, na.rm = FALSE) {UseMethod("median")} I suggest that this should become median <- function (x, na.rm = FALSE, ...) {UseMethod("median")} This would allow additional S3 methods to be developed with additional arguments....

Hi, 1.) The "median" function does not work well. Please refer to the data below (same data is attached as txt-delimited). This is what I try to do in R: median ( dataf [2:9] ) I get warning: "needs numeric data" 2.) BUT if apply the median to a single vector: median ( dataf [,2]]...

Hi is anybody aware of a function to calculate a the median for specified columns in a dataframe or matrix - so analogous to rowMeans? Thanks Max --

...4 PM, Akhilesh Singh < akhileshsingh.igkv at gmail.com> wrote: > Dear Sirs, > > I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, > Chhattisgarh, India. > > While taking classes, I found the *by() *function producing following error > when I use FUN=mean or median and some other functions, however, > FUN=summary works. > > Given below is the output of the example I used on a built-in dataset > "mtcars", along with error message reproduced herewith: > > > by(data=mtcars, INDICES=list(mtcars$am), FUN=mean) > : 0 > [1] NA &g...

...Method("normalize", "MatrixLike") >message("str(norm.meth)") >str(norm.meth) >message("show(norm.meth at .Data)") >show(norm.meth at .Data) Last show() displays this: function (object, ...) { .local <- function (object, method = c("median", "vs", "tukey"), calc.medians = TRUE, sweep.cols = calc.medians, recalc.after.sweep = sweep.cols, ...) { .do_normalize(object, method = match.arg(method), calc.medians = calc.medians, sweep.cols = sweep.cols,...

Hello, I get > summary(E) level nodes ave_nodes time Min. : 1 Min. : 1.00 Min. : 10.71 Min. : 0.0000 1st Qu.: 237414 1st Qu.: 2.00 1st Qu.: 19.70 1st Qu.: 0.0100 Median : 749229 Median : 3.00 Median : 27.01 Median : 0.0100 Mean : 767902 Mean : 49.85 Mean : 98.89 Mean : 0.2296 3rd Qu.:1111288 3rd Qu.: 7.00 3rd Qu.: 80.38 3rd Qu.: 0.0300 Max. :2097152 Max. :70632.00 Max. :4451.67 Max. :338.9200 ave_time...

Hi, I tried looking through google search on whether there's a way to computer the median for each row of a nxn matrix and return the medians for each row for further computation. And also if the number of columns in the matrix are even, how could I specify which median to use? Thank you very much! -- Edward Chen [[alternative HTML version deleted]]

...ail.com> wrote: > >> Dear Sirs, >> >> I am Professor at Indira Gandhi Krishi Vishwavidyalaya, Raipur, >> Chhattisgarh, India. >> >> While taking classes, I found the *by() *function producing following >> error >> >> when I use FUN=mean or median and some other functions, however, >> FUN=summary works. >> >> Given below is the output of the example I used on a built-in dataset >> "mtcars", along with error message reproduced herewith: >> >> > by(data=mtcars, INDICES=list(mtcars$am), FUN=mean)...

...flect the information for use with corresponding fields in GUI (in a different package), to provide default values, argname as key for UI label lookups, etc. So I want something much more like the formals of the implementation: { "object", "method": c("median", "vs", "tukey"), "calc.medians": TRUE, "sweep.cols": calc.medians, "recalc.after.sweep": sweep.cols, "?" } not those of the generic: { "object", "?" } From: Michael Lawre...

Hello. I need calculate the median of several column of a data.frame, in a new column of this data frame, but the median operator only calculate from a vector. I have made a functionc that calculate the median but it is very slow. Are there any method in any package to calculate this? Best regards. Jose Sierra. A...

...that I can line up the text when I stretch it wide. i.e.: This is typical R returns: > summary(a) V1 V2 V3 V4 F:625 Min. :20020103 EMC : 34 Min. : 9300300 L:944 1st Qu.:20020530 BRW : 27 1st Qu.: 9323865 N:615 Median :20020807 Q : 23 Median : 9355230 Mean :20020761 DSS : 22 Mean : 9404488 3rd Qu.:20021015 LU : 20 3rd Qu.: 9410738 Max. :20021231 SLR : 20 Max. :13191609 (Other):2038...