In a case like this, if you can possibly work with matrices instead
of data frames, you might get significant speedup.
(More accurately, I have had situations where I obtained speed up by
working with matrices instead of dataframes.)
Even if you have to code character columns as numeric, it can be worth it.
Data frames have overhead that matrices do not. (Here's where
profiling might have given a clue) Granted, there has been recent
work in reducing the overhead associated with dataframes, but I think
it's worth a try. Carrying along extra columns and doing row
subsetting, rbinding, etc, means a lot more things happening in
memory.
So, for example, if all of your matching is based just on a few
columns, extract those columns, convert them to a matrix, do all the
matching, and then based on some sort of row index retrieve all of
the associated columns.
-Don
At 2:09 PM -0400 6/5/08, Daniel Folkinshteyn wrote:
Hi everyone!
I have a question about data processing efficiency.
My data are as follows: I have a data set on quarterly institutional
ownership of equities; some of them have had recent IPOs, some have
not (I have a binary flag set). The total dataset size is 700k+ rows.
My goal is this: For every quarter since issue for each IPO, I need
to find a "matched" firm in the same industry, and close in market
cap. So, e.g., for firm X, which had an IPO, i need to find a
matched non-issuing firm in quarter 1 since IPO, then a (possibly
different) non-issuing firm in quarter 2 since IPO, etc. Repeat for
each issuing firm (there are about 8300 of these).
Thus it seems to me that I need to be doing a lot of data selection
and subsetting, and looping (yikes!), but the result appears to be
highly inefficient and takes ages (well, many hours). What I am
doing, in pseudocode, is this:
1. for each quarter of data, getting out all the IPOs and all the
eligible non-issuing firms.
2. for each IPO in a quarter, grab all the non-issuers in the same
industry, sort them by size, and finally grab a matching firm
closest in size (the exact procedure is to grab the closest bigger
firm if one exists, and just the biggest available if all are
smaller)
3. assign the matched firm-observation the same "quarters since
issue" as the IPO being matched
4. rbind them all into the "matching" dataset.
The function I currently have is pasted below, for your reference.
Is there any way to make it produce the same result but much faster?
Specifically, I am guessing eliminating some loops would be very
good, but I don't see how, since I need to do some fancy footwork
for each IPO in each quarter to find the matching firm. I'll be
doing a few things similar to this, so it's somewhat important to up
the efficiency of this. Maybe some of you R-fu masters can clue me
in? :)
I would appreciate any help, tips, tricks, tweaks, you name it! :)
========== my function below ===========
fcn_create_nonissuing_match_by_quarterssinceissue = function(tfdata,
quarters_since_issue=40) {
result = matrix(nrow=0, ncol=ncol(tfdata)) # rbind for matrix is
cheaper, so typecast the result to matrix
colnames = names(tfdata)
quarterends = sort(unique(tfdata$DATE))
for (aquarter in quarterends) {
tfdata_quarter = tfdata[tfdata$DATE == aquarter, ]
tfdata_quarter_fitting_nonissuers = tfdata_quarter[
(tfdata_quarter$Quarters.Since.Latest.Issue > quarters_since_issue)
& (tfdata_quarter$IPO.Flag == 0), ]
tfdata_quarter_ipoissuers = tfdata_quarter[
tfdata_quarter$IPO.Flag == 1, ]
for (i in 1:nrow(tfdata_quarter_ipoissuers)) {
arow = tfdata_quarter_ipoissuers[i,]
industrypeers = tfdata_quarter_fitting_nonissuers[
tfdata_quarter_fitting_nonissuers$HSICIG == arow$HSICIG, ]
industrypeers = industrypeers[
order(industrypeers$Market.Cap.13f), ]
if ( nrow(industrypeers) > 0 ) {
if (
nrow(industrypeers[industrypeers$Market.Cap.13f >=
arow$Market.Cap.13f, ]) > 0 ) {
bestpeer =
industrypeers[industrypeers$Market.Cap.13f >= arow$Market.Cap.13f,
][1,]
}
else {
bestpeer = industrypeers[nrow(industrypeers),]
}
bestpeer$Quarters.Since.IPO.Issue =
arow$Quarters.Since.IPO.Issue
#tfdata_quarter$Match.Dummy.By.Quarter[tfdata_quarter$PERMNO ==
bestpeer$PERMNO] = 1
result = rbind(result, as.matrix(bestpeer))
}
}
#result = rbind(result, tfdata_quarter)
print (aquarter)
}
result = as.data.frame(result)
names(result) = colnames
return(result)
}
========= end of my function =============
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
--------------------------------------
Don MacQueen
Environmental Protection Department
Lawrence Livermore National Laboratory
Livermore, CA, USA
925-423-1062
______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
