search for: summat

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)...

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 help mailing li...

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 suggesti...

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])}))})) ------------------------...

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 t...

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 kn...

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]...

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 y...

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 Darnell Duke University Marine Lab 135 Duke Marine...

...fthymiou 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 implementing Kahan summation here. Kahan summation is described in @article{ goldberg91what, author = "David Goldberg", title = "What Every Computer Scientist Should Know About Floating-Point Arithmetic"...

...39;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] 18 [[2]] [1] 168 $`Absolute value` [1] 1 4 6 7 What I would like is for the names of the result to be the same as the names of funlist (as is the case in components 1 and 3 above), but if some components of f...

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 adap...

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 > y2=sum(sapply(1:2,function(i){sum(sapply(1:i,function(j){(3^(sum(sapply(i:j,function(l){x[l]}))))}))})) > y2 [1] 819 But it didn't work, it gave another solution Any help or recommendations

...<- function(N,f,m, x, t ){ B <- c(1:(x+t)) C <- (sum(Probability(N, f, B, 1, x, t))) M = 2 if (M<=max(m)) { B <- c(1:(x+t)) append(C,sum(Probability(N, f, M, B, x, t)), after=M-1) M = M+1} else {return(C)} } -- View this message in context: http://r.789695.n4.nabble.com/Summation-resulting-in-a-list-tp3660514p3660514.html Sent from the R help mailing list archive at Nabble.com.

...r 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 following double summation: sum_{i=1}^k sum_{j=1}^{n_i}(x_{ij}-mean(x_i))*(y_{ij}-mean(y_i)) i is from 1 to k, indexing the ith subject id; and j is from 1 to n_i, indexing the jth observation for the ith subject. in the above expression, mean(x_i) is the mean of x values for the ith subject, mean(y_i) is the mean of y...

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

...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 (for some i), and d=0.01 the whole summation is turning out to be infinite. Is there a way to get round that in R? Can anyone suggest any computational trick to calculate S when c_i/d>710 (for some i)? Any suggestions would be much appreciated....

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 # ru...

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 ( Xi * Xj)) ) / n] Unfortunately this formula stays difficult in iterative programming, because I have to ke...