search for: summated

Dear R experts I need to write a function that incorporates double summation, the problem being that the upper limit of the second summation is the index of the first summation, i.e: sum_{j=0}^{x} sum_{i=0}^{j} choose(i+j, i) where x variable or constant, doesn't matter. The following code obviously doesn't work: f=function(x) {j=0:x; i=0:j; sum( choose(i+j,i) ) } Can you help? Thanks

I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + a3(b1+b2+b4) + a4(b1+b2+b3) or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i * b_i would appreciate some help. Thank you. -- View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678.html Sent from the R

hello, I am new to R (convert from Stata) and I am wondering if there is an R command/function for generating a summated scale from, for example, four separate variables in a data set. In other words, I want to transform the variables x1, x2, x3, and x4 (which have high inter-item reliability) into one new summated variable x'. Can't seem to find it in the psy or psych package ... Thanks for any suggestion...

Dear R users... I made the R-code for this double summation computation http://www.nabble.com/file/p19213599/doublesum.jpg ------------------------------------------------- Here is my code.. sum(sapply(1:m, function(k){sum(sapply(1:m, function(j){x[k]*x[j]*dnorm((mu[j]+mu[k])/sqrt(sig[k]+sig[j]))/sqrt(sig[k]+sig[j])}))})) ------------------------------------------------- In fact, this is

Dear R users, I try to compute this summation, http://www.nabble.com/file/p25054272/dd.jpg where f(y|x) = Negative Binomial(y, mu=exp(x' beta), size=1/alp) http://www.nabble.com/file/p25054272/aa.jpg http://www.nabble.com/file/p25054272/cc.jpg In fact, I tried to use "do.call" function to compute each u(y,x) before the summation, but I got an error, "Error in X[i, ]

Hi all, I'm trying to model some data where the y is defined by y = summation[1 to 50] B1 * x + B2 * x^2 + B3 * x^3 Hopefully that reads clearly for email. Anyway, if it wasn't for the summation, I know I would do it like this lm(y ~ x + x2 + x3) Where x2 and x3 are x^2 and x^3. However, since each value of x is related to the previous values of x, I don't know how to do this.

Dear R users... I made the R-code for this triple summation computation http://www.nabble.com/file/p20517134/a.jpg ------------------------------------------------- Here is my code.. x=seq(.1,1,.1); l=10 y=seq(1,10); m=10 z=seq(.1,1,.1); n=10 sum(sapply(1:l, function(i) {sum(sapply(1:m, function(j) {sum(sapply(1:n, function(k){exp(x[i]*y[j]*z[k] )/gamma(y[j]+1)}))^(1.5) }))}))

Hi all, I have a very quick question on how to use the summation sign in R for the function. Here?s a basic example: the function is sum(i=1 to 5)log(1-xi^2) Id be grateful if someone knows how to do this without writing it out 5 times - I am looking sth along the lines of the following: computeR <- function(x) { return (-sum(log(1-x^2)) }^ thank you vm in advance! -- View this

I am relatively new to R, so this may be a much simpler question than it seems to me. I am trying to create a function that includes two summations, and can't figure out how. I am attaching the equation as a pdf file. This function will then be optimized over a chosen range of values, but right now I just need help with the function. Thank you. Zack Darnell Duke University -- Zack

Full_Name: Nick Efthymiou Version: R1.5.0 and above OS: Red Hat Linux Submission from: (NULL) (162.93.14.73) With the introduction of the functions rowSums(), colSums(), rowMeans() and colMeans() in R1.5.0, function "SEXP do_colsum(SEXP call, SEXP op, SEXP args, SEXP rho)" was added to perform the fast summations. We have an excellent opportunity to improve the accuracy by

Here's a simple example of the type of function I'm trying to write, where the first argument is a list of functions: myfun <- function(funlist, vec){ tmp <- lapply(funlist, function(x)do.call(x, args = list(vec))) names(tmp) <- names(funlist) tmp } > myfun(list("Summation" = sum, prod, "Absolute value" = abs), c(1, 4, 6, 7)) $Summation [1]

Hello, I have to compute a multiple summation (not an integration because the independent variables a are discrete) for all the values of a function of several variables f (x_1,...,x_n), that is sum ... sum f(x_1,...,x_n) x_1 x_n have you some suggestion? Is it possible? I know that for multiple integration there is the function adapt, but it has at most n=20. In my case n depends on the

Dear R Experts that's my original equation > x=c(2,4) > Y1=sum(sapply(1:2,function(i){sum(sapply(1:i,function(j){(3^(x[i]+x[j]))}))})) > y1 [1] 7371 I want to write the inside function (3^(x[i]+x[j])) in a more condensed form cause this will help me when the multiple summations are more than two I've tried the following form >

Probability <- function(N, f, m, b, x, t) { #N is the number of lymph nodes #f is the fraction of Dendritic cells (in the correct node) that have the antigen #m is the number of time steps #b is the starting position (somewhere in the node or somewhere in the gap between nodes. It is a number between 1 and (x+t)) #x is the number of time steps it takes to traverse the gap #t is the number

Dear helpers in this forum, This is a clarified version of my previous questions in this forum. I really need your generous help on this issue. > Suppose I have the following data set: > > id x y > 023 1 2 > 023 2 5 > 023 4 6 > 023 5 7 > 412 2 5 > 412 3 4 > 412 4 6 > 412 7 9 > 220 5 7 > 220 4 8 > 220 9 8 > ...... > Now I want to compute the

Hello! Maybe this is a trivial question as I'm still a new baby in R but I wish that u will help me. I want to calculate the following U= sum (t_j*v_j) where t_j is a vector and v_j is the matrix Thanks Dina

Dear all, I would be grateful if anyone can help me with the following: My aim is to compute explicitely the sum S=A+B where A=sum(exp(c_i/d)), i=1,...,n; B, c_i, and d are real numbers with -Inf<B,c_i<+Inf; and d>0. The problem is that when c_i/d >710 (for some i) R is setting exp(c_i/d) to be equal to +Inf and hence the whole summation S. So in simple cases where for example c_i=8

The algorithm does make a differece. You can use Kahan's summation algorithm (https://en.wikipedia.org/wiki/Kahan_summation_algorithm) to reduce the error compared to the naive summation algorithm. E.g., in R code: naiveSum <- function(x) { s <- 0.0 for(xi in x) s <- s + xi s } kahanSum <- function (x) { s <- 0.0 c <- 0.0 # running compensation for lost

Dear Sirs/Madam, I am a beginner to R, and I am currently working on a data matrix which looks like this: > head(oligo) ko:K00001 ko:K00003 ko:K00005 ko:K00008 ko:K00009 ko:K00010 ko:K00012 AAA 370 631 365 67 164 455 491 KAA 603 1208 170 157 68

Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ): H = sqrt [ (Si (Sj (Xi - Xj)?? ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj