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11885383
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2012-08-09T14:12:29.547
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2012-08-09T16:30:43.007
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2012-08-09T16:30:43.007
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342362
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It should be pretty quick if you go down to level of the math of the linear regression. If X is the independent variable and Y is the dependent variable. The coefficients are given by <code>Beta = inv(t(X) %*% X) %*% (t(X) %*% Y)</code> I'm a little confused about which variable you want to be the dependent and which one is the independent but hopefully solving a similar problem below will help you as well. In the example below I take 1000 variables instead of the original 5 and do not introduce any NA's. <pre><code>require(xts) data <- matrix(sample(1:10000, 1500000, replace=T), 1500, 1000, byrow = TRUE) # Random data data <- xts(data, order.by = as.Date(1:1500, origin = "2000-01-01")) NR <- nrow(data) # number of observations NC <- ncol(data) # number of factors obs <- 30 # required number of observations for rolling regression analysis </code></pre> Now we can calculate the coefficients using Joshua's TTR package. <pre><code>library(TTR) loop.begin.time <- Sys.time() in.dep.var <- data[,1] xx <- TTR::runSum(in.dep.var*in.dep.var, obs) coeffs <- do.call(cbind, lapply(data, function(z) { xy <- TTR::runSum(z * in.dep.var, obs) xy/xx })) loop.end.time <- Sys.time() print(loop.end.time - loop.begin.time) # prints the loop runtime </code></pre> Time difference of 3.934461 secs <pre><code>res.array = array(NA, dim=c(NC, NR, obs)) for(z in seq(obs)) { res.array[,,z] = coredata(data - lag.xts(coeffs, z-1) * as.numeric(in.dep.var)) } res.sd <- apply(res.array, c(1,2), function(z) z / sd(z)) </code></pre> If I haven't made any errors in the indexing <code>res.sd</code> should give you the standardized residuals. Please feel free to fix this solution to correct any bugs. 
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1. This table or related slice is empty.
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1. PORolling regression over multiple columns
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1. USSameer
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1. USSameer
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2. VO
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1. CO+1 for the direct approach.
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