I'm having a very frustrating problem, trying to find the inverse distance
squared weighted interpolants of some weather data.
I have a data frame of weights, which sum to 1. I have attached the weights
data. I also have a data frame of temperatures at 48 grid points, which I
have also attached.
Now, all I need to do is multiply all of the rows of the temperature data
frame by the weights (element-wise), and sum across the columns.
However, when I try to use the most obvious approach,
temp3880W <- weight3880*temp[,(3:50)]
temp3880W <- rowsum(temp3880W)
I get the wrong result:
head(temp3880W)
1 2 3 4 5 6
-0.4904454 -1.2728543 -1.5360133 -0.2687030 62.3048012 6.2610305
I've only been successful by using a for loop which is far too slow:
temp3880 <- rep(0,length(temp$Year))
for (i in 1:length(temp$Year)) {
wmul <- weight3880*as.vector(temp[i,(3:50)])
temp3880[i] <- sum(wmul)
}
This gives the result
head(temp3880)
[1] -6.936374 -9.617799 -7.227260 1.135293 8.973817 13.632454
Can anyone point out to me what is going wrong here? I've tried the first
approach with smaller data frames and vectors and it seems to work fine, so
I must be making a mistake somewhere...
Thank you!
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http://r.789695.n4.nabble.com/file/n4632406/temp3880.csv temp3880.csv http://r.789695.n4.nabble.com/file/n4632406/weight3880.csv weight3880.csv Here are the files I promised to upload. -- View this message in context: http://r.789695.n4.nabble.com/rowSums-problem-tp4632405p4632406.html Sent from the R help mailing list archive at Nabble.com.
Hello, The files you've uploaded are the weights file and the results file, not the original temp.csv. So this is untested but it seems you have a standard matrix multiply problem. temp3880W <- temp[, 3:50] %*% weight3880 Hope this helps, Rui Barradas Em 05-06-2012 15:48, alonis10 escreveu:> I'm having a very frustrating problem, trying to find the inverse distance > squared weighted interpolants of some weather data. > > I have a data frame of weights, which sum to 1. I have attached the weights > data. I also have a data frame of temperatures at 48 grid points, which I > have also attached. > > Now, all I need to do is multiply all of the rows of the temperature data > frame by the weights (element-wise), and sum across the columns. > > However, when I try to use the most obvious approach, > > temp3880W<- weight3880*temp[,(3:50)] > temp3880W<- rowsum(temp3880W) > > > I get the wrong result: > > > head(temp3880W) > 1 2 3 4 5 6 > -0.4904454 -1.2728543 -1.5360133 -0.2687030 62.3048012 6.2610305 > > > > I've only been successful by using a for loop which is far too slow: > > temp3880<- rep(0,length(temp$Year)) > > for (i in 1:length(temp$Year)) { > wmul<- weight3880*as.vector(temp[i,(3:50)]) > temp3880[i]<- sum(wmul) > } > > > This gives the result > > head(temp3880) > [1] -6.936374 -9.617799 -7.227260 1.135293 8.973817 13.632454 > > > > Can anyone point out to me what is going wrong here? I've tried the first > approach with smaller data frames and vectors and it seems to work fine, so > I must be making a mistake somewhere... > > Thank you! > > > > > > -- > View this message in context: http://r.789695.n4.nabble.com/rowSums-problem-tp4632405.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.
I am having a problem visualizing what you are doing here. What I see is a temp file of 760 elements. From my point of view and reading in your data it would have a have dim(760, 1) but your code seems to suggest that it is not a single vector. Again I must be missing something completely as I would have thought that the output would have been a vector. Could exporting the csv files have somehow mixed up the data structure that you are working with? You might want to provide the two data files using the dput() command. John Kane Kingston ON Canada> -----Original Message----- > From: vashchyshyn.i at gmail.com > Sent: Tue, 5 Jun 2012 07:48:51 -0700 (PDT) > To: r-help at r-project.org > Subject: [R] rowSums problem > > I'm having a very frustrating problem, trying to find the inverse > distance > squared weighted interpolants of some weather data. > > I have a data frame of weights, which sum to 1. I have attached the > weights > data. I also have a data frame of temperatures at 48 grid points, which I > have also attached. > > Now, all I need to do is multiply all of the rows of the temperature data > frame by the weights (element-wise), and sum across the columns. > > However, when I try to use the most obvious approach, > > temp3880W <- weight3880*temp[,(3:50)] > temp3880W <- rowsum(temp3880W) > > > I get the wrong result: > > > head(temp3880W) > 1 2 3 4 5 6 > -0.4904454 -1.2728543 -1.5360133 -0.2687030 62.3048012 6.2610305 > > > > I've only been successful by using a for loop which is far too slow: > > temp3880 <- rep(0,length(temp$Year)) > > for (i in 1:length(temp$Year)) { > wmul <- weight3880*as.vector(temp[i,(3:50)]) > temp3880[i] <- sum(wmul) > } > > > This gives the result > > head(temp3880) > [1] -6.936374 -9.617799 -7.227260 1.135293 8.973817 13.632454 > > > > Can anyone point out to me what is going wrong here? I've tried the first > approach with smaller data frames and vectors and it seems to work fine, > so > I must be making a mistake somewhere... > > Thank you! > > > > > > -- > View this message in context: > http://r.789695.n4.nabble.com/rowSums-problem-tp4632405.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.____________________________________________________________ FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop!