On Mon, Jun 13, 2011 at 2:11 PM, Patrizio Frederic <[email protected]> wrote: > On Mon, Jun 13, 2011 at 6:47 PM, Aparna <[email protected]> wrote: >> Hi Joshua >> >> While looking at the data, all the values seem to be in numeric. As i >> mentioned, >> the dataset is already in data.frame. >> >> As suggested, I used str(mydata) and got the following result: >> >> >> str(leu_cluster1) >> 'data.frame': 984 obs. of 100 variables: >> $ V2 : Factor w/ 986 levels "-0.00257361",..: 543 116 252 54 520 ... > > your data columns are not numeric but factors indeed. > you may try this one > > a <- as.character(rnorm(100)) # some numeric data > adf <- data.frame(matrix(a,10)) # which are misinterpreted as factors > adf > adf[,1] > class(adf[,1]) # check for the class of the first column > sapply(adf,function(x)class(x)) # check classes for all columns > > b <- sapply(adf,function(x)as.numeric(as.character(x))) #
But coercing to a character class first is not the recommended method. Also, I am leery about using sapply() with data frames, because it converts them to matrices, which can cause havoc, if you have different classes of data. You mentioned that as a first step, you had removed the names column from the data frame before trying to convert it to numeric. I would simply leave the names in, and then (supposing they are in column 101) leu_cluster1[, 1:100] <- lapply(leu_cluster1[, 1:100], function(x) as.numeric(levels(x))[x]) apply the conversion to numeric on only the necessary columns. This simplifies life because you are not making interim data sets. Using lapply() allows you to work with (potentially) different classes of data (although I realize in this particular case you are only dealing with one class). So long as you are assigning the results back into a data frame (as above), the methods for lapply will automatically conver the list back to a data frame. If you are concerned about this, just wrap the call in as.data.frame() leu_cluster1[, 1:100] <- as.data.frame(lapply( leu_cluster1[, 1:100], function(x) as.numeric(levels(x))[x])) Cheers, Josh > as.character: use levels literally, as.numeric: transforms in numbers > b # look at b > > class(b) # which is now a numeric matrix > > best regards > > PF > > -- > +----------------------------------------------------------------------- > | Patrizio Frederic, > | http://www.economia.unimore.it/frederic_patrizio/ > +----------------------------------------------------------------------- > > ______________________________________________ > [email protected] 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. > -- Joshua Wiley Ph.D. Student, Health Psychology University of California, Los Angeles http://www.joshuawiley.com/ ______________________________________________ [email protected] 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.
