[R] ifelse on a series of rows for multiple criteria

[David Winsemius]

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
> 		 	   		
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David Winsemius, MD
Heritage Laboratories
West Hartford, CT