On 13-12-02 7:49 PM, Bill wrote:
It seems so inefficient. I mean the whole first vector will be evaluated. Then if the second if is run the whole vector will be evaluated again. Then if the next if is run the whole vector will be evaluted again. And so on. And this could be only to test the first element (if it is false for each if statement). Then this would be repeated again and again. Is that really the way it works? Or am I not thinking clearly?
Read the manual. Duncan Murdoch
On Mon, Dec 2, 2013 at 4:48 PM, Duncan Murdoch <murdoch.dun...@gmail.com <mailto:murdoch.dun...@gmail.com>> wrote: On 13-12-02 7:33 PM, Bill wrote: ifelse ((day_of_week == "Monday"),1, ifelse ((day_of_week == "Tuesday"),2, ifelse ((day_of_week == "Wednesday"),3, ifelse ((day_of_week == "Thursday"),4, ifelse ((day_of_week == "Friday"),5, ifelse ((day_of_week == "Saturday"),6,7))))))) In code like the above, day_of_week is a vector and so day_of_week == "Monday" will result in a boolean vector. Suppose day_of_week is Monday, Thursday, Friday, Tuesday. So day_of_week == "Monday" will be True,False,False,False. I think that ifelse will test the first element and it will generate a 1. At this point it will not have run day_of_week == "Tuesday" yet. Then it will test the second element of day_of_week and it will be false and this will cause it to evaluate day_of_week == "Tuesday". My question would be, does the evaluation of day_of_week == "Tuesday" result in the generation of an entire boolean vector (which would be in this case False,False,False,True) or does the ifelse "manage the indexing" so that it only tests the second element of the original vector (which is Thursday) and for that matter does it therefore not even bother to generate the first boolean vector I mentioned above (True,False,False,False) but rather just checks the first element? Not sure if I have explained this well but if you understand I would appreciate a reply. See the help for the function. If any element of the test is true, the full first vector will be evaluated. If any element is false, the second one will be evaluated. There are no shortcuts of the kind you describe. Duncan Murdoch
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