similar to: Getting all possible contingency tables

Hello again, let say we have following data: Dat1 <- structure(list(factor.sample.LETTERS.1.3...6..replace...T.. = structure(c(1L, 3L, 2L, 1L, 3L, 3L), .Label = c("A", "B", "C"), class = "factor"), factor.sample.letters.1.2...6..replace...T.. = structure(c(2L, 2L, 1L, 1L, 2L, 1L), .Label = c("a", "b"), class =

Dear all, I was encountering with a typical Matching problem and was wondering whether R can help me to solve it directly. Let say, I have 2 vectors of equal length: vector1 <- LETTERS[1:6] vector2 <- letters[1:6] Now I need to match these 2 vectors with all possible ways like: (A,B,C,D,E) & (a,b,c,d,e) is 1 match. Another match can be (A,B,C,D,E) & (b,a,c,d,e), however there

I am trying to create contingency tables (to evaluate prior to performing Pearson's Chi-Squared test for independence). I would like to see column and row totals as well as expected and observed values and cell counts. I tried to use the package "contingency. tables" but get the following warning: package ‘contingency.tables’ is not available (for R versions 2.15.2) Is there

Dear all Seems that puzzles always come in packs. I was asked to help with some statistics in blood analysis. (You can not refuse your wife's asks :-). She has contingency table for values IgVH mutation and ZAP expression. I can do chi-square test (in R) and get a results, and with some literature I can try explain them. However she found an article in which they use SPSS and use

I'd like to carry out a Monte Carlo simulation test where given data is a contingency table. I think this is something to do with using rmultinonom(), but I'm not sure how to code this, to simulate contingency tables. Could anyone please help with how to use R to simulate contingency tables like this? -- View this message in context:

Dear all, I know this problem was discussed many times in forum, however unfortunately I could not find any way out for my own problem. Here I am having Memory allocation problem while generating a lot of random number. Here is my description: > rnorm(50000*6000) Error: cannot allocate vector of size 2.2 Gb In addition: Warning messages: 1: In rnorm(50000 * 6000) : Reached total allocation

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,

@Eric - with this approach I am getting below error : Error in FUN(X[[i]], ...) : unused argument (list()) On Sun, Mar 4, 2018 at 10:18 PM, Eric Berger <ericjberger at gmail.com> wrote: > Hi Christofer, > You cannot assign to list(...). You can do the following > > myList <- list(...)[!names(list(...)) %in% 'mc.cores'] > > HTH, > Eric > > On Sun, Mar

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Hi again, I am struggling to extract the number part from below string : "\"cm_ffm\":\"563.77\"" Basically, I need to extract 563.77 from above. The underlying number can be a whole number, and there could be comma separator as well. So far I tried below : > library(stringr) > str_extract("\"cm_ffm\":\"563.77\"",

Dear all, when I start R, I want that the console window should be in the Maximized size automatically. Can somebody help me how to achieve that? Thanks and regards,

Hello again, I want to remove "$" sign and replace with nothing in my text. Therefore I used following code: > gsub("$|,", "", "$232,685.35436") [1] "$232685.35436" However I could not remove '$' sign. Can somebody help me why is it so? Thanks and regards

Dear all. Let say I have a group of codes which will be used in many places in my overall R-code files. These group of codes will be used within a for-loop (with a big length, like 10000 times) and also many other places outside of that for loop. As this group of codes are being used in many places, I thought to put them within a user-defined function. Here my question is, is there any speed

Hi Joshua, thanks for your prompt reply. However as I said, sum() function I used here just for demonstrating the problem, I have other custom function to implement, not necessarily sum() I am looking for a generic solution for above problem. Any better idea? Thanks, On Fri, Aug 11, 2017 at 12:04 AM, Joshua Ulrich <josh.m.ulrich at gmail.com> wrote: > Use a `width` of integer index

That's fine. The issue is how you called Lapply_me(). What did you pass as the argument to FUN? And if you did not pass anything that how is FUN declared? You have not shown that in your email. On Sun, Mar 4, 2018 at 7:11 PM, Christofer Bogaso < bogaso.christofer at gmail.com> wrote: > My modified function looks below : > > Lapply_me = function(X = X, FUN = FUN, Apply_MC =

Hi friends, there is methods() function to see the all available methods for a particular function, for example: > head(methods("print")) [1] "print.acf" "print.anova" "print.aov" "print.aovlist" "print.ar" "print.Arima" In this list, there are some functions which are asterisked like print.acf().

I happened to see these: > round(.5, 0) [1] 0 > round(1.5, 0) [1] 2 > round(2.5, 0) [1] 2 > round(3.5, 0) [1] 4 > round(4.5, 0) [1] 4 What is the rule here? Should not round(.5, 0) = 1, round(2.5, 0) = 3 etc? Thanks and regards,