R help - Feb 2010 - ifelse on a series of rows for multiple criteria

Dear all,

I am attempting to perform a calculation which counts the number of positive (or
negative) values based on the sample mean (on a per-row basis). If the mean
is>0 then only positive values should be counted, and if the mean is <0
then only negative values should be counted. In cases where the mean is equal to
zero, the value -99999 should be returned.

The following is an example of the data frame I'm working on (all values are
of class 'numeric').
> head(combdframe)
???????? V1?????? V2??????? V3???????? V4?????? V5??????? V6
1 -328.0999 3404.038? 791.7602? 211.23932 513.0479 -1178.079
2 -383.4249 3207.306? 808.7268? 141.20352 424.2388 -1164.402
3 -295.9050 2930.754? 918.1146?? 11.74804 464.2448 -1133.109
4 -326.8606 2703.638 1052.2824 -104.17344 246.2851 -1103.887
5 -296.7194 2663.987 1202.7648? -87.15331 255.1338 -1090.147
6 -227.1772 2619.096 1343.1529? -75.89626 381.6089 -1064.733


The mean of the first row is 571 and therefore a count of the positive values
should be returned for this row (=4). *If* the mean was -571, then a count of
the negative values would be returned (=2).
If the 7th row was composed of values 1.5, -1.5, 2.5, -2.5, 0? and 0 (i.e. the
mean = 0), then -99999 should be returned for this row.

I've attempted to construct this code as follows:

direction_func <- function(combdframe) {
??? ifelse(mean(i> 0), sum(i> 0), ifelse(mean(i < 0), sum(i < 0),
-99999))
??? }
??? 
for (i in nrow(combdframe)) {

direction <- apply(combdframe[i,],1, direction_func)

}


...but this, and varients on this, result in a bit of a mess!


Any guidance on how to perform this (whether it be a correction of the above or
a whole new approach) would be very much appreciated.

Many thanks,

Steve
 		 	   		  
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On Feb 5, 2010, at 10:48 AM, Steve Murray wrote:
>
> Dear all,
>
> I am attempting to perform a calculation which counts the number of  
> positive (or negative) values based on the sample mean (on a per-row  
> basis). If the mean is>0 then only positive values should be  
> counted, and if the mean is <0 then only negative values should be  
> counted. In cases where the mean is equal to zero, the value -99999  
> should be returned.
>
> The following is an example of the data frame I'm working on (all  
> values are of class 'numeric').
>
>> head(combdframe)
>          V1       V2        V3         V4       V5        V6
> 1 -328.0999 3404.038  791.7602  211.23932 513.0479 -1178.079
> 2 -383.4249 3207.306  808.7268  141.20352 424.2388 -1164.402
> 3 -295.9050 2930.754  918.1146   11.74804 464.2448 -1133.109
> 4 -326.8606 2703.638 1052.2824 -104.17344 246.2851 -1103.887
> 5 -296.7194 2663.987 1202.7648  -87.15331 255.1338 -1090.147
> 6 -227.1772 2619.096 1343.1529  -75.89626 381.6089 -1064.733
> The mean of the first row is 571 and therefore a count of the  
> positive values should be returned for this row (=4). *If* the mean  
> was -571, then a count of the negative values would be returned (=2).
> If the 7th row was composed of values 1.5, -1.5, 2.5, -2.5, 0  and 0  
> (i.e. the mean = 0), then -99999 should be returned for this row.
Try:
 > (rowSums(combframe > 0))*(rowMeans(combframe)>0) +
   (rowSums(combframe < 0))*(rowMeans(combframe) < 0) -
    99999*(rowSums(combframe) == 0)
      1      2      3      4      5      6      7
      4      4      4      3      3      3 -99999

The equality holds here but I worry about "==" when working with  
floating point numbers.
>
> I've attempted to construct this code as follows:
>
> direction_func <- function(combdframe) {
>     ifelse(mean(i> 0), sum(i> 0), ifelse(mean(i < 0), sum(i <
0),
> -99999))
>     }
>
> for (i in nrow(combdframe)) {
>
> direction <- apply(combdframe[i,],1, direction_func)
>
> }
>
>
> ...but this, and varients on this, result in a bit of a mess!
>
>
> Any guidance on how to perform this (whether it be a correction of  
> the above or a whole new approach) would be very much appreciated.
>
> Many thanks,
>
> Steve
> 		 	   		
> _________________________________________________________________
> We want to hear all your funny, exciting and crazy Hotmail stories.  
> Tell us now
>
> ______________________________________________
> 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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT

Hi Steve,

as far as i understood, you're trying to do this:

direction_func <- function(combdframe) {
     ifelse(mean(combdframe==0), -99999,
       sum((sign(mean(combdframe))*combdframe)>0))
}

direction<-apply(combdframe, 1, direction_func)
direction


cheers,
thomas

-- 
Thomas Liebig
Fraunhofer-Institut f?r Intelligente Analyse- und
Informationssysteme (IAIS)

Schloss Birlinghoven, D-53754 Sankt Augustin, Germany
Email:  thomas.liebig at iais.fraunhofer.de
Phone:  +49 2241 142050
Fax:    +49 2241 142072