practical to loop over 2million rows?
This is a classic example from my tag line: Tell me what you want to do, not how you want to do it. For example you provided no information as to what the objects were. I hope that 'stratID' is at least of length one greater than 'x' based on your loops. Also on the last iteration you are trying to access an element outside of x (x[length(x) + 1]). The first part is easy for setting 'y' indx <- !is.na(x) y[indx] <- x[indx] For the second part you can do something like: indx <- head(stratID, -1) == tail(stratID, -1) # get the comparison but since you did not provide any data, the rest is left to the reader.
On Wed, Oct 10, 2012 at 5:16 PM, David Winsemius <dwinsemius at comcast.net> wrote:
On Oct 10, 2012, at 1:31 PM, Jay Rice wrote:
New to R and having issues with loops. I am aware that I should use vectorization whenever possible and use the apply functions, however, sometimes a loop seems necessary. I have a data set of 2 million rows and have tried run a couple of loops of varying complexity to test efficiency. If I do a very simple loop such as add every item in a column I get an answer quickly. If I use a nested ifelse statement in a loop it takes me 13 minutes to get an answer on just 50,000 rows. I am aware of a few methods to speed up loops. Preallocating memory space and compute as much outside of the loop as possible (or use create functions and just loop over the function) but it seems that even with these speed ups I might have too much data to run loops. Here is the loop I ran that took 13 minutes. I realize I can accomplish the same goal using vectorization (and in fact did so).
You should describe what you want to do and you should learn to use the vectorized capabilities of R and leave the for-loops for process that really need them
y<-numeric(length(x)) for(i in 1:length(x)) ifelse(!is.na(x[i]), y[i]<-x[i],
Instead : y[!is.na(x)] <- x[!is.na(x)] # No loop.
ifelse(strataID[i+1]==strataID[i], y<-x[i+1], y<-x[i-1]))
When you index outside the range of the length of x you get NA as a result. Furthermore you are setting y to be only a single element. So I think 'y' will be a single NA at the end of all this.
strataID <- sample(1:2, 10, repl=TRUE) strataID
[1] 1 1 2 2 1 2 2 2 2 1
for(i in 1:length(x)) {ifelse(strataID[i+1]==strataID[i], y<-x[i+1], y<-x[i-1])}
y
[1] NA There is no implicit indexing of the LHS of an assignment operation. How long is strataID? And why not do this inside a dataframe?
Presumably, complicated loops would be more intensive than the nested if statement above. If I write more efficient loops time will come down but I wonder if I will ever be able to write efficient enough code to perform a complicated loop over 2 million rows in a reasonable time. Is it useless for me to try to do any complicated loops on 2 million rows, or if I get much better at programming in R will it be manageable even for complicated situations?
You will gain efficiency when you learn vectorization. And when you learn to test your code for correct behavior.
Jay
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______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD Alameda, CA, USA
______________________________________________ R-help at r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it.