Re: [R] counting Na/not NA by groups by column

2010-06-10 Thread Henrique Dallazuanna
Try this also: rowsum(`mode<-`(!is.na(m[,-(1:2)]), 'numeric'), m[,2]) On Wed, Jun 9, 2010 at 10:03 PM, steven mosher wrote: > # create a matrix with some random NAs in it > > m<-matrix(NA,nrow=15,ncol=14) > > m[,3:14]<-52 > > m[13,9]<-NA > > m[4:7,8]<-NA > > m[1:2,5]<-NA > > m[,2]<-rep(1800:1804

Re: [R] counting Na/not NA by groups by column

2010-06-09 Thread steven mosher
thats beautiful > apply(m[, 3:14], 2, + function(x) tapply(x, m[,2], function(x) sum(!is.na(x [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] 1800332332333 3 3 3 1801332332333 3

Re: [R] counting Na/not NA by groups by column

2010-06-09 Thread Erik Iverson
Hello, steven mosher wrote: # create a matrix with some random NAs in it m<-matrix(NA,nrow=15,ncol=14) m[,3:14]<-52 m[13,9]<-NA m[4:7,8]<-NA m[1:2,5]<-NA m[,2]<-rep(1800:1804, by=3) y<-order(m[,2]) m<-m[y,] m[,1]<-rep(1:3,by=5) # what we want is a result that looks like this 1800 3 3

[R] counting Na/not NA by groups by column

2010-06-09 Thread steven mosher
# create a matrix with some random NAs in it > m<-matrix(NA,nrow=15,ncol=14) > m[,3:14]<-52 > m[13,9]<-NA > m[4:7,8]<-NA > m[1:2,5]<-NA > m[,2]<-rep(1800:1804, by=3) > y<-order(m[,2]) > m<-m[y,] > m[,1]<-rep(1:3,by=5) > m [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,