search for: cumprod

...r more rows than columns. I started with apply and after some thought realized that a loop of columns might be faster and it was substantially faster (see below). Because the likelihood function is called many times I?d like to speed it up even more if possible. Below is an example showing the cumprod.matrix and prod.matrix looping functions that I wrote and some timing comparisons to the use of apply for different column and row dimensions. At this point I?m better off with looping but I?d like to hear of any further suggestions. Thanks ?jeff > prod.matrix=function(x) + { + y=x[,1] + f...

Full_Name: J. Sisk Version: 1.6.1 OS: Linux (RedHat 8) Submission from: (NULL) (67.119.41.66) Suppose you make a data-frame like so: xxx <- data.frame(a=10,b=20,c=30,d=40) Then cumprod(xxx[1,]) returns > cumprod(xxx[1,]) a b c d 1 10 20 30 40 The documentation for cumprod says that it should work on "numerical objects", and this is a data-frame, but it seems like it ought to "do the right thing" in this case. If this is redundant, my apologies.

...feel free to have a laugh but would appreciate if someone can give me a hint. Otherwise I guess I'll have to give RCpp a try..... ## Bench mark the recursion functions loopRec <- function(x, alpha){ n <- length(x) y <- double(n) for(i in 1:n){ y[i] <- sum(cumprod(rep(alpha, i)) * rev(x[1:i])) } y } loopRec(c(1, 2, 3), 0.5) ## This is a crazy solution, but worth giving it a try. charRec <- function(x, alpha){ n <- length(x) exp.mat <- matrix(rep(x, each = n), nc = n, byrow = TRUE) up.mat <- matrix(eval(parse(text = past...

...0.9, 1.1, 0.9, 1.1) I need to come up with function to get the following output o[1] = a[1] o[2] = a[2] o[3] = a[3] o[4] = o[3]*[f3] # Because a[3] is NA o[5] = o[4]*[f4] # Because a[4] is NA; This looks like recursive calculations; If the rest of the elements we NA, I would use a * c(rep(1, 3), cumprod(f[3:9])), but that's not the case o[6] = a[6] # Not NA anymore o[7] = a[7] o[8] = o[7]*f[7] # Again a[8] is NA o[9] = o[8]*f[8] o[10] = a[10] # Not NA Even though my explanation may seems complex, in reality the requirement is pretty simple and in Excel is achieved with a very short formula....

I have a dataframe of numeric values with 30 ?rows? and 7 ?columns?. For each column, beginning at ?row? 12 and down to ?row? 30, I wish to calculate the ?rolling 12 row product?. I.e., within each column, I wish to multiply all the values in row 1:12, 2:13, 19:30. I wish to save the results as a new dataframe, which will have 19 rows and 7 columns.

Is there anything like cumsum and cumprod but which allows you to apply an arbitrary function instead of sum and product? In other words, I want a function cumfunc(x, f) that returns a vector, so that for all n up to the length of x cumapply(x,f)[n] = f(x[1:n]) This would give cumsum and cumprod as special cases when f=sum or f=prod...

...63344 -0.0066314769 0.0075063818 [6] -0.0033548024 0.0015647863 0.0119815982 -0.0021430336 0.0044617167 [11] 0.0053447708 -0.0005590323 0.0063195781 0.0073059640 -0.0181872678 [16] -0.0098094568 0.0013679040 -0.0028490887 -0.0131129191 0.0126610358 z1 <- ifelse(is.na(y1), 10000, 10000*cumprod(1+y1)) z1 [1] 9931.369 9830.957 9664.162 9600.074 9672.136 9639.688 9654.772 9770.451 [9] 9749.513 9793.012 9845.354 9839.850 9902.034 9974.378 9792.971 9696.907 [17] 9710.172 9682.506 9555.541 9676.524 y <-c(NA, rnorm(19,0, .013)) y [1] NA 0.0056258152 -0.0117690116 0.01639...

...s()) ## DEFINE VARIABLES mu=0.05 ; sigma=0.20 ; dt=.25 ; T=50 ; sims=1e5 timesteps = T/dt ## MAKE PHI AND DS phi = matrix(rnorm(timesteps*sims), nrow=sims, ncol=timesteps) ds = mu*dt + sigma * sqrt(dt) * phi ## USE APPLY TO CALCULATE ROWWISE CUMULATIVE PRODUCT system.time(y1 <- apply(1+ds, 1, cumprod)) ## UNTRANSFORM Y1, BECAUSE ROW APPLY FLIPS THE MATRIX y1=t(y1) ## USE FOR LOOP TO CALCULATE ROWWISE CUMULATIVE PRODUCT y2=matrix(NA,nrow(ds),ncol(ds)) system.time( for (i in 1:nrow(ds)){ y2[i,]<-cumprod(1+ds[i,]) } ) ## COMPARE RESULTS TO MAKE SURE THEY DID THE SAME THING str...

...##### #  Function calculating a scenario of results : ############################################# ResultScenar <- function(FirstYear,OptimYear,Price,CumulRetentionCoef,Commission,IndividualExp,NbOfFutureSimYear,       IndivMeanLossAmount,CPIRate,LossInflaRate,Alpha) { CumulLossInflaRate <- cumprod(1+LossInflaRate) CumulCPIRate <- cumprod(1+CPIRate) y <- numeric(NbOfFutureSimYear) delta <- OptimYear - FirstYear if (OptimYear==FirstYear)  { y[1] <CumulRetentionCoef[1]*( (Price[1]*(1-Commission[delta +1])-IndivMeanLossAmount[delta +1])*CumulLossInflaRate[delta +1]                - I...

Hello, everyone I wonder if there is in R somewhere a function similar to cumsum(). The function calculates a statistic (say mean or standard deviation) buy adding consequtively one more data point. So, say I have a timeseries of 100 observations. I start by calculating mean of first 30 observations Then I add one observation and calculate mean of 31 observations Then I add one more observation

Hi all, I was wondering whether it is possible to use the lapply() function to alter the value of the input, something in the spirit of : a1<-runif(100) a2<-function(i){ a1[i]<-a1[i-1]*a1[i];a1[i] } a3<-lapply(2:100,a2) Something akin to a for() loop, but using the lapply() infrastructure. I haven't been able to get rapply() to do this. The reason is that the "real"

What is the R way of computing cumulative growth rates given a series of discrete values indexed . For instance, given a matrix of 20 observations for each of 5 series (zz), what is the most straight forward technique in R for computing cumulative growth (zzcum) ? It seems for the solution I'm after might be imbedding the following cum growth rate calc as a function into a function call

Hi! I noticed that cumsum behaves different than the other cumulative functions wrt. NaN values: > values <- c(1,2,NaN,1) > for ( f in c(cumsum, cumprod, cummin, cummax)) print(f(values)) [1] 1 3 NA NA [1] 1 2 NaN NaN [1] 1 1 NaN NaN [1] 1 2 NaN NaN The reason is that cumsum (in cum.c:33) contains an explicit check for ISNAN. Is that intentional? IMHO, ISNA would be better (because it would make the behavior consistent with the other...

...(d))))) (d is a vector containing, in effect, the numbers of levels of the factors in col1, col2, ... of the categorical variables. Therefore umd is a vector containing the numbers of possible combinations of factor levels in col1, col1&col2, ... ) But why not do it as umd <- as.integer(cumprod(d)) ?? [It can't be that this number could go out or range, since that would be equally true of exp(cumsum(log(d)))) ] With thanks, and best wishes to all, Ted. -------------------------------------------------------------------- E-Mail: (Ted Harding) <Ted.Harding at nessie.mcc.ac.uk>...

Hi all, I am looking for a function like cumprod() that works for matrix multiplication. In other words, I have matrices [M1, M2, ..., Mn], and I want to calculate [M1, M1%*%M2, M1%*%M2%*%M3, ..., M1%*%...%*%Mn] as quickly as possible. Right now I'm using a for() loop but it seems like there should be a faster way. Any help is appreciated!...

...there a function that replaces the following code? n=200 boot.x[1]=odhad+boot.res[1] #(boot.x[0]=1) for (j in 1:(n-1)) { boot.x[j+1]=odhad*boot.x[j]+boot.res[j+1] } This is nested in two other loops, and I am looking for some way to improve code performance I tried sapply and cumprod but no success. Thanks Jan -- View this message in context: http://r.789695.n4.nabble.com/Replacing-the-for-loop-for-time-series-buid-up-tp3094421p3094421.html Sent from the R help mailing list archive at Nabble.com.

Hello all, I want to define a vector like w[k+1]=w[k]*a/(b+k) for k=1,...,N-1 without use loop. Is it posible to do in R? Regards khazaei

I am trying to simulate the trajectory of the pension assets of one person. In C-like syntax, it looks like this: daily.wage.growth = 1.001 # deterministic contribution.rate = 0.08 # deterministic 8% Wage = 10 # initial Asset = 0 # initial for (10,000 days) { Asset += contribution.rate * Wage

Does anyone have any ideas how I could do a power calculation for a log rank test. I would like to know what the suggested sample sizes would be to pick a difference when the control to active are in a ratio of 80% to 20%. Thanks Dan -- ************************************************************** Daniel Brewer, Ph.D. Institute of Cancer Research Email: daniel.brewer at icr.ac.uk

All, I would like to find a simper method that I now have to find the product of all elements in a vector: #get product of vector elements: (1,2,3,4,5) vec.product <- exp(sum(log(c(1,2,3,4,5)))) I have not found a vector product function, if one has been written. Thanks, Gerard [[alternative HTML version deleted]]