Greetings,
Can anyone suggest me if we can vectorize the following problem
effectively?
I have two datasets, one dataset is portfolio of stocks returns on a
historical basis and another dataset consist of a bunch of factors (again on
a historical basis). I intend to compute a rolling n-day sensitivitiesfor
each stock for each factor, so the output will be a data frame with
<ticker><dt><sensitivities>.
How would you go onto vector this situation effectively?
I end up with a psuedo code like this:
# For each date
For curr dt in all dates
# Get Universe of stocks as of that date
Get Universe for curr date
# Calculate Sensitivity for each factor between n days back
dt to curr date
sensitivity
sapply(univ{ticker},CalcSensitivity,n_days_back_dt,dt)
Next date
I would highly appreciate if the above logic could be improved (if at
all) by a more effective solution since I do get into such situations on a
regular basis.
Thanks in advance
Cheers
Manoj
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for each
education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ) {
l <- mydata$education[k]
mydata$md[k] <- median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end of
this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
_______________________________________________________
Instale o discador agora! http://br.acesso.yahoo.com/
Here I go again with ave(): mydata$md <- ave(mydata$income, mydata$education, FUN=median, na.rm=TRUE) IMHO it's one of the most under-rated helper functions in R. Andy> From: Dimitri Joe > > Hi there, > > I have a data frame (mydata) with 1 numeric variable (income) > and 1 factor (education). I want a new column in this data > with the median income for each education level. A obviously > inneficient way to do this is > > for ( k in 1: nrow(mydata) ) { > l <- mydata$education[k] > mydata$md[k] <- median(mydata$income[mydata$education==l],na.rm=T) > } > > Since mydata has nearly 30.000 rows, this will be done not > untill the end of this month. I thus need some help for > vectorizing this, please. > > Thanks, > > Dimitri > > [[alternative HTML version deleted]] > > > > > > > _______________________________________________________ > > Instale o discador agora! http://br.acesso.yahoo.com/ > > ______________________________________________ > R-help at stat.math.ethz.ch mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide! > http://www.R-project.org/posting-guide.html > > >
You can use tapply() to compute the medians, as in
meds <- tapply(mydata$inc,INDEX=mydata$ed,FUN=median)
then create a new column with the medians as
medianEd <- meds[mydata$ed]
Reid Huntsinger
-----Original Message-----
From: r-help-bounces at stat.math.ethz.ch
[mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Dimitri Joe
Sent: Friday, June 17, 2005 1:01 PM
To: R-Help
Subject: [R] vectorization
Hi there,
I have a data frame (mydata) with 1 numeric variable (income) and 1 factor
(education). I want a new column in this data with the median income for
each education level. A obviously inneficient way to do this is
for ( k in 1: nrow(mydata) ) {
l <- mydata$education[k]
mydata$md[k] <- median(mydata$income[mydata$education==l],na.rm=T)
}
Since mydata has nearly 30.000 rows, this will be done not untill the end of
this month. I thus need some help for vectorizing this, please.
Thanks,
Dimitri
[[alternative HTML version deleted]]
_______________________________________________________
Instale o discador agora! http://br.acesso.yahoo.com/
______________________________________________
R-help at stat.math.ethz.ch mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide!
http://www.R-project.org/posting-guide.html
These two lines worked for me: rst <- tapply(mydata$income, mydata$education, median) mydata$md <- rst[mydata$education] Here's my cheesy example:> mydata <- data.frame(income = round(rnorm(30000, 55000, 10000)),+ education = letters[rbinom(30000, 4, 1/2)+1])> rst <- tapply(mydata$income, mydata$education, median) > mydata$md <- rst[mydata$education] > head(mydata)income education md 1 66223 e 55094.5 2 56830 c 54966.0 3 58035 b 54937.5 4 74045 a 55213.5 5 61327 b 54937.5 6 64150 b 54937.5 Is this what you wanted? Kevin -----Original Message----- From: r-help-bounces at stat.math.ethz.ch [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Dimitri Joe Sent: Friday, June 17, 2005 10:01 AM To: R-Help Subject: [R] vectorization Hi there, I have a data frame (mydata) with 1 numeric variable (income) and 1 factor (education). I want a new column in this data with the median income for each education level. A obviously inneficient way to do this is for ( k in 1: nrow(mydata) ) { l <- mydata$education[k] mydata$md[k] <- median(mydata$income[mydata$education==l],na.rm=T) } Since mydata has nearly 30.000 rows, this will be done not untill the end of this month. I thus need some help for vectorizing this, please. Thanks, Dimitri [[alternative HTML version deleted]] _______________________________________________________ Instale o discador agora! http://br.acesso.yahoo.com/ ______________________________________________ R-help at stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
Hi,> -----Original Message----- > From: r-help-bounces at stat.math.ethz.ch > [mailto:r-help-bounces at stat.math.ethz.ch] On Behalf Of Dimitri Joe > Sent: Friday, June 17, 2005 7:01 PM > To: R-Help > Subject: [R] vectorization > > Hi there, > > I have a data frame (mydata) with 1 numeric variable (income) > and 1 factor (education). I want a new column in this data > with the median income for each education level. A obviously > inneficient way to do this is >I guess the attached code (incl. simulating your data structure) is not the most efficient way to do this, but at least (I hope so!) it does what you wanted it to do: ####################### Beginning of Example Code income <- runif(100) education <- as.factor(sample(c("high", "middle", "low"), size=length(income), replace=TRUE)) mydata <- data.frame(inc=income, edu=education) mymedians <- tapply(X=mydata$inc, INDEX=mydata$edu, FUN=median) mydata$medians <- ifelse(mydata$edu=="high", mymedians["high"], 0) mydata$medians <- ifelse(mydata$edu=="middle", mymedians["middle"], mydata$medians) mydata$medians <- ifelse(mydata$edu=="low", mymedians["low"], mydata$medians) head(mydata) mymedians ####################### End of Example Code Maybe one can increase the speed, but I think it is sufficient for your case of 30,000 cases as you can see from the timing on my desktop computer here (WinXP Pro SP2, P4, 3GHz, 512MB RAM):> time.check <- function(){+ income <- runif(30000) + education <- as.factor(sample(c("high", "middle", "low"), size=length(income), replace=TRUE)) + mydata <- data.frame(inc=income, edu=education) + + mymedians <- tapply(X=mydata$inc, INDEX=mydata$edu, FUN=median) + + mydata$medians <- ifelse(mydata$edu=="high", mymedians["high"], 0) + mydata$medians <- ifelse(mydata$edu=="middle", mymedians["middle"], mydata$medians) + mydata$medians <- ifelse(mydata$edu=="low", mymedians["low"], mydata$medians) + return(NULL) + }> system.time(time.check())[1] 0.36 0.02 0.38 NA NA> > version_ platform i386-pc-mingw32 arch i386 os mingw32 system i386, mingw32 status beta major 2 minor 1.0 year 2005 month 04 day 04 language R Best, Roland +++++ This mail has been sent through the MPI for Demographic Rese...{{dropped}}