Can someone verify for me if the for loop below is really calculating the
nonzero min for each row of a matrix? I have a bug somewhere in the is
section of code. My first guess is how I am find the the nonzero min of each
row of my matrix. The overall idea is to make sure I am investing all of my
money, i.e. new.set is a set of indicator variables for each stock for a
particular portfolio, i.e. 0=did not buy, 1=bought. y are the stocks I could
still buy, assuming I have the money, data3[,5] are their cost, so for each
portfolio, i.e. the rows of new.set I have the option to purchase another
stock at a cost listed in the rows of variable remain. Obvisouly the
cheapest stock needs to have a cost>0 in order for me to be allowed to buy
it. My code is intended to weed out portfolios where I could have bought
another stock, by taking budget-portfolio cost - (cheapest available stock)
and subsetting new.set when this is negative, i.e. buying the cheapest
available stock puts me over budget. My problem is that my code is still
allowing examples like the following budget of 10, portfolio cost 6,
cheapest availabe stock 3 despite the diff variable being negative.
Any ideas?
y<-1-new.set[,6:26]
remain<-y*data3[,5]
minn<-matrix(0,nrow(remain),1)
for(q in 1:nrow(remain))
{
remainc<-remain[q,]
minn[q,]<-min(remainc[which(remainc>0)])
}
maxcost<-matrix(150,nrow(new.set),1)
diff<-maxcost[,1]-new.set[,5]-minn[,1]
new.set<-subset(new.set,diff<0)
--
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!!! Well, that strikes me as a fair bit of chutzpah. Some generous soul may well respond, but why not do your own debugging using R's debugging tools.It will serve you well in the long run to put in the effort now to learn them. Debugging is a major part of any programming. ?trace ?debug ?browser ... and the "error" settingt of options() are a place to start. If you check this list's archives you'll find other similar queries and helpful debugging tips. -- Bert On Sun, Jul 22, 2012 at 9:34 PM, wwreith <reith_william at bah.com> wrote:> Can someone verify for me if the for loop below is really calculating the > nonzero min for each row of a matrix? I have a bug somewhere in the is > section of code. My first guess is how I am find the the nonzero min of each > row of my matrix. The overall idea is to make sure I am investing all of my > money, i.e. new.set is a set of indicator variables for each stock for a > particular portfolio, i.e. 0=did not buy, 1=bought. y are the stocks I could > still buy, assuming I have the money, data3[,5] are their cost, so for each > portfolio, i.e. the rows of new.set I have the option to purchase another > stock at a cost listed in the rows of variable remain. Obvisouly the > cheapest stock needs to have a cost>0 in order for me to be allowed to buy > it. My code is intended to weed out portfolios where I could have bought > another stock, by taking budget-portfolio cost - (cheapest available stock) > and subsetting new.set when this is negative, i.e. buying the cheapest > available stock puts me over budget. My problem is that my code is still > allowing examples like the following budget of 10, portfolio cost 6, > cheapest availabe stock 3 despite the diff variable being negative. > > Any ideas? > > y<-1-new.set[,6:26] > remain<-y*data3[,5] > minn<-matrix(0,nrow(remain),1) > > for(q in 1:nrow(remain)) > { > remainc<-remain[q,] > minn[q,]<-min(remainc[which(remainc>0)]) > } > maxcost<-matrix(150,nrow(new.set),1) > diff<-maxcost[,1]-new.set[,5]-minn[,1] > new.set<-subset(new.set,diff<0) > > > > > -- > View this message in context: http://r.789695.n4.nabble.com/Bug-in-my-code-finding-nonzero-min-tp4637399.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > 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.-- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
Strictly, it's calculating the minimum of the positive values in each row. And you could do that with a bit less code using q[,1] <- apply(remain, 1, function(x) min(x[x>0]))> -----Original Message----- > From: r-help-bounces at r-project.org > [mailto:r-help-bounces at r-project.org] On Behalf Of wwreith > Sent: 23 July 2012 05:35 > To: r-help at r-project.org > Subject: [R] Bug in my code (finding nonzero min) > > Can someone verify for me if the for loop below is really > calculating the nonzero min for each row of a matrix? I have > a bug somewhere in the is section of code. My first guess is > how I am find the the nonzero min of each row of my matrix. > The overall idea is to make sure I am investing all of my > money, i.e. new.set is a set of indicator variables for each > stock for a particular portfolio, i.e. 0=did not buy, > 1=bought. y are the stocks I could still buy, assuming I have > the money, data3[,5] are their cost, so for each portfolio, > i.e. the rows of new.set I have the option to purchase > another stock at a cost listed in the rows of variable > remain. Obvisouly the cheapest stock needs to have a cost>0 > in order for me to be allowed to buy it. My code is intended > to weed out portfolios where I could have bought another > stock, by taking budget-portfolio cost - (cheapest available > stock) and subsetting new.set when this is negative, i.e. > buying the cheapest available stock puts me over budget. My > problem is that my code is still allowing examples like the > following budget of 10, portfolio cost 6, cheapest availabe > stock 3 despite the diff variable being negative. > > Any ideas? > > y<-1-new.set[,6:26] > remain<-y*data3[,5] > minn<-matrix(0,nrow(remain),1) > > for(q in 1:nrow(remain)) > { > remainc<-remain[q,] > minn[q,]<-min(remainc[which(remainc>0)]) > } > maxcost<-matrix(150,nrow(new.set),1) > diff<-maxcost[,1]-new.set[,5]-minn[,1] > new.set<-subset(new.set,diff<0) > > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/Bug-in-my-code-finding-nonzero-m > in-tp4637399.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > 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. >******************************************************************* This email and any attachments are confidential. Any use...{{dropped:8}}