similar to: Multiple sets of proportion tests

Hi anonymous, ?prop.test states that it returns a list. And one of the element is 'p.value'. str() on the output of prop.test() reveals that too. So prop.test()$p.value or prop.test()["p.value"] should work. Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR

Thank you for clarifying this point but my main question was about how to modify my code to do the analysis correctly. The code I mentioned :- MyResults <- apply(Mydata, 2, function(x)prop.test(Mydata,c(200,100)) Results in this error : 'x' and 'n' must have the same length in the prop.test(x,n). How can I modify "x' or "n" arguments so the analysis

HI there I am running this model in negative binomial regression, using glm. I had no problems with running the model with a set of data, but now that i'm trying to run if for new one. I always have this same error when running the regression: > > #Run Regression > x=cbind(factor2ind(d$year),factor2ind(d$month_week)) > > out<- glm(cbind(influenza, n_sample) ~ x,

Hi, I am apparently not understanding some nuance about either the use of subset or more likely my ability to test for a numerical match using '='. Which is it? Thanks in advance. I've read a data file, reshaped it and then created MyResults by keeping only lines where the value column is greater than 0. So far so good. The data in MyResults looks good to me by eye. The

How can I get the results of lm() into a list so I can loop through the results? e.g. myResults[1] <- lm(...) myResults[2] <- lm(...) myResults[3] <- lm(...) ... myResults[15] <- lm(...) myResults[16] <- lm(...) so far every attempt I've tried doesn't work throwing a "number of items to replace is not a multiple of replacement length" error or simply not

Dear Paula, There are probably missing observations in your data set. Read the na.action part of the glm help file. na.exclude is most likely what you are looking for. Best regards, ir. Thierry Onkelinx Statisticus / Statistician Vlaamse Overheid / Government of Flanders INSTITUUT VOOR NATUUR- EN BOSONDERZOEK / RESEARCH INSTITUTE FOR NATURE AND FOREST Team Biometrie & Kwaliteitszorg /

Hi Duncan I tried Ecdf and/or wtd.quantile from Hmisc and it is working (probably). Ecdf(x, q=.5) Ecdf(x, weights=xw,col=2, add=T, q=.5) wtd.quantile(x) 0% 25% 50% 75% 100% 10 10 10 100 300 wtd.quantile(x, weights=xw, type="i/n") 0% 25% 50% 75% 100% 10.0000 138.8667 192.5778 246.2889 300.0000 But could you please be more specific in this? >

The loop : AllMAFs <- list() for (i in length(SeparatedGroupsofmealsCombs) { AllMAFs[[i]] <- apply( SeparatedGroupsofmealsCombs[[i]], 2, function(x)maf( tabulate( x+1) )) } gives these errors (I tried this many times and I'm sure I copied it entirely) :- Error in apply(SeparatedGroupsofmealsCombs[[i]], 2, function(x) maf(tabulate(x + : object 'i' not found > }

> On Nov 18, 2017, at 1:52 AM, Allaisone 1 <allaisone1 at hotmail.com> wrote: > > Although the loop seems to be formulated correctly I wonder why > it gives me these errors : > > -object 'i' not found > - unexpected '}' in "}" You probably did not copy the entire code offered. But we cannot know since you did not "show your code",

Although the loop seems to be formulated correctly I wonder why it gives me these errors : -object 'i' not found - unexpected '}' in "}" the desired output is expected to be very large as for each dataframe in the list of dataframes I expect to see maf value for each of the 600 columns! and this is only for for one dataframe in the list .. I have around 150-200

Thanks Boris , this was very helpful but I'm struggling with the last part. 1) I combined the first 2 columns :- library(tidyr) SingleMealsCode <-unite(MyData, MealsCombinations, c(MealA, MealB), remove=FALSE) SingleMealsCode <- SingleMealsCode[,-2] 2) I separated this dataframe into different dataframes based on "MealsCombination" column so R will recognize each meal

Always reply to the list. I am not a free, private consultant! "For example, if I have the values : 1 , 2 , 3 in each column, applying Tabulate () would calculate the frequency of 1 and 2 without 3" Huh?? > x <- sample(1:3,10,TRUE) > x [1] 1 3 1 1 1 3 2 3 2 1 > tabulate(x) [1] 5 2 3 Cheers, Bert Bert Gunter "The trouble with having an open mind is that people

Combine columns 1 and 2 into a column with a single ID like "33.55", "44.66" and use split() on these IDs to break up your dataset. Iterate over the list of data frames split() returns. B. > On Nov 17, 2017, at 12:59 PM, Allaisone 1 <allaisone1 at hotmail.com> wrote: > > > Hi all .., > > > I have a large dataset of around 600,000 rows and 600

On 18/11/2017 4:40 PM, Allaisone 1 wrote: > > The loop : > > > AllMAFs <- list() > > for (i in length(SeparatedGroupsofmealsCombs) { > AllMAFs[[i]] <- apply( SeparatedGroupsofmealsCombs[[i]], 2, function(x)maf( tabulate( x+1) )) > } > > > gives these errors (I tried this many times and I'm sure I copied it entirely) :- > > Error in

Thanks but a new error appeared with the loop : Error in x + 1 : non-numeric argument to binary operator I think this can be solved by converting columns (I,II,II,..600) into "numeric" instead of the current "int" type as shown below in the structure of "33_55" dataframe . $ 33_55:'data.frame': 256 obs. of 600 variables: ..$ MealsCombinations

Thank you for your effort Bert.., I knew what is the problem now, the values (1,2,3) were only an example. The values I have are 0 , 1, 2 . Tabulate () function seem to ignore calculating the frequency of 0 values and this is my exact problem as the frequency of 0 values should also be calculated for the maf to be calculated correctly. ________________________________ From: Bert Gunter

|> x <- sample(0:2, 10, replace = TRUE) |> x [1] 1 0 2 1 0 2 2 0 2 1 |> tabulate(x) [1] 3 4 |> table(x) x 0 1 2 3 3 4 B. > On Nov 10, 2017, at 4:32 AM, Allaisone 1 <allaisone1 at hotmail.com> wrote: > > > > Thank you for your effort Bert.., > > > I knew what is the problem now, the values (1,2,3) were only an example. The values I have are

I have a read a lot about the benefits of vectorization in R. I have a program that takes "almost forever" to run. A good way to see if I have learned something ... My problem can be summarized like this : I have a nonlinear function of several variables that I want to optimize over one letting the other describe a family of curves. In short, I wan't to optimize f(x,a,b) for several

Hi, What is cast telling me when it says the following? Aggregation requires fun.aggregate: length used as default What is 'length'? I've taken a small subset of data and wondered what EnTime vs ExTime might look like. cast is kind enough to give me a table but I don't understand the values in the table. They seem to sum up ro be the same as the total dimension of the data

Something like the following? AllMAFs <- list() for (i in length(SeparatedGroupsofmealsCombs) { AllMAFs[[i]] <- apply(SeparatedGroupsofmealsCombs[[i]], 2, function(x)maf(tabulate(x+1))) } (untested, of course) Also the solution is a bit generic since I don't know what the output of maf() looks like in your case, and I don't understand why you use tabulate because I would have