[R] concatenating factor from list
Hi, I've run into a ridiculous problem I can't find any solutions for in the archives or help pages: data(barley) cutYield - with(barley, by(yield, variety, cut, breaks = c(0, 30, 60, 90))) As in this example, I'm using 'by' to return a factor for each level of another factor. The problem is that 'by' returns a list of the factors, and I need all these factors concatenated. No problem, I said: unlist(cutYield) which returns an 'integer' class object, so it's no longer a factor. The same happens with: do.call(c, cutYield) I could recreate the factor from the integer level codes returned above, but this is not good because this means redoing the job already done by the function called in 'by', making the code more prone to errors. Thanks in advance for any pointers, -- Sebastian __ R-help@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
Re: [R] concatenating factor from list
Is this ok or is it what you are trying to avoid: factor(unlist(lapply(cutYield, as.character))) On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote: Hi, I've run into a ridiculous problem I can't find any solutions for in the archives or help pages: data(barley) cutYield - with(barley, by(yield, variety, cut, breaks = c(0, 30, 60, 90))) As in this example, I'm using 'by' to return a factor for each level of another factor. The problem is that 'by' returns a list of the factors, and I need all these factors concatenated. No problem, I said: unlist(cutYield) which returns an 'integer' class object, so it's no longer a factor. The same happens with: do.call(c, cutYield) I could recreate the factor from the integer level codes returned above, but this is not good because this means redoing the job already done by the function called in 'by', making the code more prone to errors. Thanks in advance for any pointers, -- Sebastian __ R-help@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 __ R-help@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
Re: [R] concatenating factor from list
Gabor Grothendieck [EMAIL PROTECTED] wrote: Is this ok or is it what you are trying to avoid: factor(unlist(lapply(cutYield, as.character))) Thank you Gabor. The problem with that is what if some levels do not appear in any member of cutYield? In that case, the factor created above would contain fewer levels than those present in every member of cutYield. Cheers, -- Sebastian P. Luque __ R-help@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
Re: [R] concatenating factor from list
Since all components of cutYield have the same levels, one could do this to ensure that all levels are represented: factor(unlist(lapply(cutYield, as.character)), levels = levels(cutYield[[1]])) On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote: Gabor Grothendieck [EMAIL PROTECTED] wrote: Is this ok or is it what you are trying to avoid: factor(unlist(lapply(cutYield, as.character))) Thank you Gabor. The problem with that is what if some levels do not appear in any member of cutYield? In that case, the factor created above would contain fewer levels than those present in every member of cutYield. Cheers, -- Sebastian P. Luque __ R-help@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 __ R-help@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
Re: [R] concatenating factor from list
Sebastian Luque [EMAIL PROTECTED] wrote: Gabor Grothendieck [EMAIL PROTECTED] wrote: Is this ok or is it what you are trying to avoid: factor(unlist(lapply(cutYield, as.character))) Thank you Gabor. The problem with that is what if some levels do not appear in any member of cutYield? This addition to your code takes care of that, although it's a bit expensive: factor(unlist(lapply(cutYield, as.character)), levels = unique(unlist(lapply(test, levels Thanks! -- Sebastian P. Luque __ R-help@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
Re: [R] concatenating factor from list
On 3/15/06, Sebastian Luque [EMAIL PROTECTED] wrote: Sebastian Luque [EMAIL PROTECTED] wrote: Gabor Grothendieck [EMAIL PROTECTED] wrote: Is this ok or is it what you are trying to avoid: factor(unlist(lapply(cutYield, as.character))) Thank you Gabor. The problem with that is what if some levels do not appear in any member of cutYield? This addition to your code takes care of that, although it's a bit expensive: factor(unlist(lapply(cutYield, as.character)), levels = unique(unlist(lapply(test, levels In thinking about this a bit more here is another solution which does not disassemble and reassemble the factors and so may be more along the lines you were looking for: do.call(rbind, lapply(cutYield, function(x) data.frame(x = x)))$x __ R-help@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
