search for: tabulators

Hi, To clarify the default behavior that Boris is referencing below, note the definition of the 'bin' argument to the tabulate() function: bin: a numeric vector ***(of positive integers)***, or a factor. Long vectors are supported. I added the asterisks for emphasis. This is also noted in the examples used for the function in ?tabulate at the bottom of the help page. The second

Peter, I thought this one was noted and fixed, but I could be wrong. R : Copyright 1999, The R Development Core Team Version 0.63.3 (March 6, 1999) .... [Previously saved workspace restored] > tabulate(1:10, 5) Process R:1 segmentation fault at Sat Apr 3 17:48:34 1999 -- (The following contact details become official on 1 May 1999, but the email

|> 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

Hi, I'd like to ask you a question again. It is basically about data frames, NAs and tabulate function. I have this data frame. I already used this in one of the previous questions of mine. It intentionally looks this simple, my real 'df' dataframe is much bigger actually and again, I am not willing to annoy anyone with huge databases... So, my database: id

Hi all, I have a question about apply function. Is that possible to pass some non-default arguments in the function we want to apply ? For example: if "mat" is a matrix and I want to use the "tabulate" function on its row. The command apply(mat,1,tabulate) works but I have problem with this one apply(mat, 1, tabulate(nbins=4)). Any clue ? Thanks, Eric -- Eric

tabulate has .C("R_tabulate", as.integer(bin), as.integer(length(bin)), as.integer(nbins), ans = integer(nbins), PACKAGE="base")$ans The implementation of R_tabulate has if(x[i] != R_NaInt && x[i] > 0 && x[i] <= *nbin) and so copes with (silently drops) NA. Perhaps the .C could have NAOK=TRUE? This is useful in apply'ing tabulate to

I am having hard time tabulating data in a dataframe, and getting a single "table" for an answer. I am trying to tabulate all "counts" for given "status" on a given date. I have a data frame such as: delta_ts status count 1 2008-02-27 CLOSED 3 2 2008-02-27 NEW 56 3 2008-02-27 RESOLVED 5 4 2008-02-21 ASSIGNED 1 5

I asked the question in Rmetrics subforum, but for some reason, almost two weeks later, it keeps saying I haven't been approved to post yet there. I'll try again here on the open forum. I wanted to have a script that tabulates results by trade. I.e. instead of tabulating each day as a trade event, you enter on one date, exit another, and simply tabulate metrics based on one trade period.

Dear all, it looks like that tabulate() does not check for the bounds of the input. Reproducible example: > b <- 1:2 > tabulate(b[1:100]) [1] 1 1 > R.version _ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status Patched major 2 minor 11.1 year 2010 month 09 day

Hi all, Please, is there any way of controlling factors in row/columns when using ftable/xtabs? As far as I can see, the last cross-clasifing variable in the formula will appear in columns. The previous ones, in rows. For instance, is it possible to make tension and replicate appear in columns? ftable(xtabs(breaks ~ wool + tension + replicate, data = warpbreaks)) After some years using SAS

Dear R-Developers, please find attached a set of drafted functions for flat file import and export, partially extending existing functions, partially completely written as new code. I thought you might be interested in those functions and the accompanying utilities for padding and trimming. Main features are - supports several formats, i.e. fixed width and CSV (with one exception) - supports

Hi all, I have a data set x1 x2 x3 1 2 1 1 2 3 2 1 2 1 2 1 3 1 1 I want to tabulate in the following way. 1 2 3 x1 3 2 1 x2 2 3 0 x3 3 1 1 It is just like frequency distribution Any help is highly appreciated [[alternative HTML version deleted]]

Hello, I wonder if someone could send me suggestions on how to solve the following problem: I have a vector of an arbitrary size (ex. data<-c(10,10,11,10,12,11,10,12,11,11,10,11)) and use the table function, which gives the following result 10 11 12 5 5 2 that''s fine, but what I would like to do now is: construct new classes based on the number of classes from table, 10

R-helpers: I have a data frame containing 4 factor variables (let's say A,B,C, and D) and 1 numerical variable (N). I would like to produce a cross-tabulated data frame in which A,B,C are individual columns, each factor of D is its own column, and the field is calculated as a given function of N (I would like to have two output data frames, one with the mean(N) and one with the

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

Hi, I have a huge number of categorical variables, say at least 10000, and I put them into a matrix, each column is one variable. The question is: how can I make all of the pairwise cross tabulation tables efficiently? The straightforward solution is to use for-loops by looping two indexes on the table() function, but it was just too slow. Is there a more efficient way to do that? Any guidance

Hi R, This is a cross tabulation question. Suppose that, > d=read.table("clipboard",header=F) > d V1 V2 V3 A One Apple A One Cake A One Cake B One Apple B One Apple B One Apple > table(d$V2,d$V3) Apple Cake One 4 2 But, I don't want the count to be like the above. Here, it is counting the

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

I have the following relatively simple problem. Say we have three factors, and we want to create a cross-tabulation against each of the other two: x <- factor(rbinom(5, 1, 1/2)) y <- factor(rbinom(5, 1, 1/2)) z <- factor(rbinom(5, 1, 1/2)) table(x,y) table(x,z) This looks like: y x 0 1 0 2 0 1 1 2 z x 0 1 0 1 1 1 2 1 I would like to get (surely this will

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 > }