Hello,
Thanks to everyone for the multiple answers. Josh, thanks for the
function. My data 12 datasets have over 500,000 rows so your answer
greatly appreciated.
Cheers,
Daisy
On Thu, Jan 27, 2011 at 9:10 PM, Joshua Wiley wrote:
> Hi,
>
> This problem seemed deceptively simple to me. After cha
Hi,
This problem seemed deceptively simple to me. After chasing a
considerable number of dead ends, I came up with fg(). It lacks the
elegance of Dennis' solution, but (particularly for large datasets),
it is substantially faster. I still feel like I'm missing something,
but
##
Hi:
Try this:
f <- function(x) ifelse(sum(x > 0) == 1L, names(which(x > 0)), NA)
g <- function(x) ifelse(sum(x > 0) == 2L, names(which(x == 0L)), NA)
> apply(df1[, 3:5], 1, f)
[1] "ANN" "CTA" "GLM" NA"ANN" NANANANA"CTA"
> apply(df1[, 3:5], 1, g)
[1] NANANANAN
Here's a solution., maybe not the most elegant but works.
df.r = df1[, c(3:5)]; # restricted data
nNonZero = apply(df.r!=0, 1, sum);
one = nNonZero==1;
oneZero = nNonZero==2;
whichOne = apply(df.r[one, ]!=0, 1, which);
whichZero = apply(df.r[oneZero, ]==0, 1, which);
colNames = colnames(df.r);
Hello,
I am not sure where to begin with this problem or what to search for
in r-help. I just don't know what to call this.
If I have 5 columns, the first 2 are the x,y, locations and the last
three are variables about those locations.
x<-seq(1860,1950,by=10)
y<-seq(-290,-200,by=10)
ANN<-c(3,0,0
5 matches
Mail list logo
