similar to: Retrieve results from chisq.test programmatically

Hello, I received the following warning when running chi-square; n Is there a way to catch the 'error' code of 'warning' after run chisq.test(x)? n What does this error mean? Thank you for your help. [[alternative HTML version deleted]]

Dear R-Helpers In R 2.5.1, the command library(Rgraphviz) fails on my Windows (XP SP2) system with error popup "The procedure entry point Rf_allocString could not be located in the dynamic link library R.dll". Thanks in advance for any suggestion in solving the error. My D. Coyne Imagination is more important than knowledge... (Albert Einstein) mcoyne@boninc.com

Hi All, I'm new with R; this is a basic question. I was given a matrix I of (nrow, ncol), I would like to create another matrix A with some data in the matrix I, say [1,4] (row 1, column 4) to [271,19000] (row 271, column 19000). How do I do this? Please help. Thank you very much. --mc [[alternative HTML version deleted]]

Hi, everyone I've been trying to make a scatter plot of POSIXct format data to see the difference between two start time. Like X: [1] "1995-05-26 19:00:00" "1995-05-27 01:00:00" "1995-05-27 07:00:00" [4] "1995-06-29 01:00:00" "1995-06-29 07:00:00" "1995-06-30 13:00:00" [7] "1995-06-30 19:00:00" "1995-07-01

I don't know if this is the correct forum to ask the following question; however, when I search the aroma.affymetrix discussion group, it suggested that I should posted the question to r-help. Here it goes. I followed the instructions on aroma.affymetrix trying to install the packages; following are the steps: > install.packages(c("R.oo", "R.utils"),

R-newbie question I have 2 matrices (a) P1 has only one column of 32K rows (b) PC has 2 column {P, C} of 3200 rows Every values in P1 matches with a value in PC[,p] (column p). I would like to use Which to search for all value in P1 that matchex PC[,p] and get the PC[,c]. However because P1 and PC does not have the same number of rows, I got lots of 'NA'. Thanks for your

Hi, I place a t.test in a loop and would like to continue to process the loop even when t.test encounter error. How do I do that? For example, in one iteration, the data is completely constant and t.test gives error, the entire program terminates. I would like to write the information out to a file, and the loop should continue. Thanks My D. Coyne [[alternative HTML version

Hi, everyone I got some problems when trying to subscript the value of vector and list, by using calculated indices. Here is the vector I am generated lon<-rep(0,886);lat<-rep(0,691) for (i in 1:886){ lon[i]<-112+0.05*(i-1) } for (i in 691:1){ lat[i]<--44.5+0.05*(691-i) } For a given location of lon(xp) and lat(yp), I would like to calculate the position of them,

Hello, Newbie question and hope you can help . I have two vector V1 and V2, where length(V2) = length of (V1) * 2; length(V1) ~ 16,000. For each member in V1, I need to compare 2 element of V2 for equality i.e. for (I in 1:length (V1)) { if ( v2[i] == v1[i] & v2[i+1]==v1[i] ){ statement_1 statement_2 . } } This for-loop is too slow

Thanks for the advice. Going to look into this more this evening and hopefully get everything working. Thanks again! On Thu, Nov 5, 2015 at 11:56 AM, Dirk Eddelbuettel <edd at debian.org> wrote: > > On 5 November 2015 at 11:44, Michael Coyne wrote: > | What do you have installed as far as libcurl goes? Would you mind > | listing your packages and greping for libcurl entries.

I'm ordering "The R Book" and hope to learn a lot more about R. In a meantime, I have a matrix of digits and I would like to look for a column that contains zeroes and remove the entire column from the matrix. I can write a piece of R code to do that (and it works); however, I think the R-experts have more tricks to do such a 'simple' thing. Many thanks in advance.

Hi, In the chisq.test(), if the expected frequency for some categories is <5, there will be a warning message which says Warning message: Chi-squared approximation may be incorrect in: chisq.test(x, p = probs) I am wondering whether there are some methods to get rid of this mistake... Seems the ?chisq.test() doesn''t provide more options to solve this problem. Or, the only choice is

> On 28 Dec 2017, at 13:08 , Kurt Hornik <Kurt.Hornik at wu.ac.at> wrote: > >>>>>> Jan Motl writes: > >> The chisq.test on line 57 contains following code: >> STATISTIC <- sum(sort((x - E)^2/E, decreasing = TRUE)) > > The preceding 2 lines seem relevant: > > ## Sorting before summing may look strange, but seems to be >

Hello everybody, I'm seeing some strange behavior on R 1.8.1 on Intel/Linux compiled with gcc 3.2.2. The p-value calculated from the chisq.test function is incorrect for some input values: > chisq.test(matrix(c(0, 1, 1, 12555), 2, 2), simulate.p.value=TRUE) Pearson's Chi-squared test with simulated p-value (based on 2000 replicates) data: matrix(c(0, 1, 1,

I am getting "Chi-squared approximation may be incorrect in: chisq.test(x)" with the data bleow. Frequency distribution of number of male offspring in families of size 5. Number of Male Offspring N 0 518 1 2245 2 4621 3 4753 4 2476 5

Full_Name: Reginaldo Constantino Version: 2.8.0 OS: Ubuntu Hardy (32 bit, kernel 2.6.24) Submission from: (NULL) (189.61.88.2) For many tables, chisq.test with simulate.p.value=TRUE gives a p value that is obviously incorrect and inversely proportional to the number of replicates: > data(HairEyeColor) > x <- margin.table(HairEyeColor, c(1, 2)) >

Dear list! I would like to calculate "chisq.test" on simple data set with 70 observations, but the output is ''Warning message:'' Warning message: In chisq.test(tabele) : Chi-squared approximation may be incorrect Here is an example:         tabele <- matrix(c(11, 3, 3, 18, 3, 6, 5, 21), ncol = 4, byrow = TRUE)         dimnames(tabela) <- list(        

I'm using R1.7.0 runing with Win XP. Thanks, ...Tao ???????????????????????????????????????????????????????? >x [,1] [,2] [1,] 149 151 [2,] 1 8 >t(x) [,1] [,2] [1,] 149 1 [2,] 151 8 >chisq.test(x, simulate.p.value=T, B=100000) Pearson's Chi-squared test with simulated p-value (based on 1e+05 replicates) data: x X-squared = 5.2001, df =

for example, an expression such as chisq(df=1,ncp=0) ? thanks -- View this message in context: http://www.nabble.com/how-to-calculate-chisq-value-in-R-tp15338943p15338943.html Sent from the R help mailing list archive at Nabble.com.

Where does the value 2.2e-16 come from in p-values for chisq tests such as those reported below? > Anova(cm.mod2) Analysis of Deviance Table (Type II tests) Response: Freq LR Chisq Df Pr(>Chisq) B 11026.2 1 < 2.2e-16 *** W 7037.5 1 < 2.2e-16 *** Age 886.6 8 < 2.2e-16 *** B:W 3025.2 1 < 2.2e-16 *** B:Age 1130.4 8 < 2.2e-16 *** W:Age 332.9 8 < 2.2e-16 *** --- Signif.