similar to: Please Help pairwise.t.test!!

Dear R users, Something is strange in summary and IQR. Suppose, I have a data set and I would like to find the Q1, Q2, Q3 and IQR. x<-c(2,4,11,12,13,15,31,31,37,47) > summary(x) Min. 1st Qu. Median Mean 3rd Qu. Max. 2.00 11.25 14.00 20.30 31.00 47.00 > IQR(x) [1] 19.75 However, I test the same data set in SAS "proc univariate", and SAS shows that

Hi R users, I try to build a function to compute odds ratio and relative risk however something wrong. I stuck for many hours but I really don't know how to solve it. Would someone please give me a hint? > OR.RR<-function(x){ + x <- as.matrix(any(dim(x)==2)) + OR<-(x[1,1]*x[2,2])/(x[1,2]*x[2,1]) + RR<-(x[1,1]/(sum(x[1,])))/(x[2,1]/(sum(x[2,]))) +

Dear R users, I have a question how to use 2 "ifelse" to sort my data. Such as from 11 to 20 assign to A; 6 to 10 assign to B, and the rest of them assign to C a<-1:20 tt<-ifelse(a>10, "A",no=ifelse( 5< a <=10, "B", "C")) Many Thanks Chunhao -- View this message in context:

Hi R users, I google the website and I found that there are three ways to perform repeated measure ANOVA: aov, lme and lmer. http://www.mail-archive.com/r-help at stat.math.ethz.ch/msg58502.html But the questions are which one is good to use and how to do post-hoc test? I use the example that is provided in the above link and I try > tt<-aov(p.pa~group*time+Error(subject/time),data=P.PA)

Hi R users, I have a question. How can I use for loop to do pair comparisons. For example, > x<-c(1,2,3) > result<-matrix(data=NA, nrow=choose(3,2), ncol=1) > for(i in 1: length(x)) + { result[i,]<-ifelse(x[i] > x[i+1], yes="Big", no="Small") + result} > result [,1] [1,] "Small" [2,] "Small" [3,] NA

Hi R users, Is is possible for me to use the try function with boot? I would to do the bootstraping with a nonlinear model(it works well when R < 1000). But it does not work very well (when R is large) thus I try to use "try" to resolve. I put the try function in two cases: case1: put the try in front of the boot > c1.try<-try(boot(c1data, statistic = c1.fun,

Hi R users Is there anything wrong in "is" function? (R 2.7.2) I believe that everyone will agree that "7" is an integer, right? but why R shows 7 is not an integer > is.integer(7) [1] FALSE > is(7,"integer") [1] FALSE > is(as.integer(7), "integer") [1] TRUE Thank you very much in advance Chunhao

Hi R users Is there anything wrong in "is" function? (R 2.7.2) I believe that everyone will agree that "7" is an integer, right? but why R shows 7 is not an integer > is.integer(7) [1] FALSE > is(7,"integer") [1] FALSE > is(as.integer(7), "integer") [1] TRUE Thank you very much in advance Chunhao

Hi R users, I have one simple question but I really don't know why I can't get it work. The data was adopted from Efron's An introduction to the bootstrap book. > nlaw LSAT GPA [1,] 576 3.39 [2,] 635 3.30 [3,] 558 2.81 [4,] 578 3.03 [5,] 666 3.44 [6,] 580 3.07 [7,] 555 3.00 [8,] 661 3.43 [9,] 651 3.36 [10,] 605 3.13 [11,] 653 3.12 [12,] 575 2.74

Dear R-help, I have a question about pairwise.t.test and adjustment for multiple comparisons for paired data points. I have the following data: n=c("x", "x", "x", "x", "x", "x", "x", "x", "x", "x", "y", "y", "y", "y", "y", "y",

First thanks for Jinsong's suggestions I would like to do a bootstrap in a nonlinear model. But it fails to converge in most of time. (it did converge if I just use nls without boot). Thus, I use "try" function to resolve my problem. This following code is from Jinsong's suggestion. h1a.nls<-nls(density~nmf(time, alpha, delta, psi, tau, gamma),data=h1a,

# I'm doing a pairwise.t.test on a large dataset and need the output in a data frame so I can work further with it, e.g. so I can export it to a spreadsheet. Is there any way to coerce the results to an exportable format? # For example, if I do: test <- pairwise.t.test(numbers, factors, p.adj="bonferroni") # and then write.table(test, file="output.csv",

Hi there, I am trying to do multiple pairwise Wilcoxon signed rank tests in a manner similar to: a <- c(runif(1000, min=1,max=50), rnorm(1000, 50), rnorm(1000, 49.9, 0.5), rgeom(1000, 0.5)) b <- c(rep("group_a", 1000), rep("group_b", 1000), rep("group_c", 1000), rep("group_d", 1000)) pairwise.wilcox.test(a, b, alternative="two.sided",

Good afternoon Running R2.10.0 on Windows I have a data frame that includes (among much else) a factor (In_2006) and a continuous variable (math_3_4). I would like to find the 2 cases for In_2006 = 0 that are closest to each case where In_2006 = 1. My data looks like In_2006 math_3_4 0 55.1 1 51.6 1 18.1 1 26.6 1 14.1

Proposal That a new mailing list be established that pertains exclusively to R documentation. The purpose of the list would be to discuss weak sections of the documentation and establish fixes for those weak spots. Pro If it works, there would be better documentation. It would be an excellent opportunity for newish and/or less technical people to contribute to R. In some respects such people

I am familiar with pairwise t-tests, corrections for multiple testing, etc. however I have a problem whose answer I have not found after extensive R-help archive and Google searching. What I have done in the past: I have N items which are measured, exposed to a condition, and then measured again. I wish to know if the condition changes the items so I can perform a t-test. Better yet I can

Hello, this is probably more a statistical question than an R-specific problem, but I'll risk it. I've fitted a Cox Proportional hazard model with one factor Treatment (seven levels) as a predictor variable. The general Null hypothesis (all groups show the same survival behaviour) is clearly rejected. Now, is there any (statistically sensible) way of doing pairwise comparisons and/or

Hello Running R2.9.2 on Windows XP I am puzzled by the performance of LME in situations where there are missing data. As I understand it, one of the strengths of this sort of model is how well it deals with missing data, yet lme requires nonmissing data. Thus, m1.mod1 <- lme(fixed = math_1 ~ I(year-2007.5)*TFC_, data = long, random =

Hi, We could use qqplot to see how two distributions are different from each other. To show better how they are different (departs from the straight line), how is it possible to plot the straight line that goes through them? I am looking for some thing like qqline for qqnorm. I thought of abline but how to determine the slope and intercept? Best wishes, Carol

Good afternoon Using R 2.9.2 on a machine running Windows XP I have a longitudinal data set, with data on schools and their test scores over a four year period. I have centered year, and run the following m1.mod1 <- lme(fixed = math_1 ~ I(year-2007.5)*TFC_, data = long, random = ~I(year-2007.5)|schoolnum, na.action = "na.omit") where