Christoph Bier <[EMAIL PROTECTED]> writes:
Dave Cacela schrieb:
Christoph, I concur with the other respondents who questioned why someone would wish to calculate the median of a factor. However, with regard to your actual question, I suspect that median() is giving different answers because the two vectors are not both factors, i.e., that one of them is a character. Did you test that?
Yes, I did and find the same like Tony Plate:
is.factor(fbhint.spss1$V15.SPS) # = column 264[1] TRUEmode(fbhint.spss1$V15.SPS)[1] "numeric"is.factor(fbhint.spss1$V15.SP1) # = column 566[1] TRUEmode(fbhint.spss1$V15.SP1)[1] "numeric"
Using S, I have seen quirks in this regard that relate to import procedure and the value of the first element in the vector. In your case, the first elements differ in that one is NA while the other is "teils/teils".
It also occurs if the first element is the same. For example "wichtig" in columns 263 and 565.
I suspect the solution to the riddle was given earlier (I forget by
Yes, you're right. I just wanted to answer Dave's question.
Nevertheless I don't understand what I observed above, maybe because of a lack of knowledge/understanding of R. But I'm reading your and Brian Ripley's book ...
[...]
Best regards,
Christoph
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