Classifying values by interval
On 08/31/2011 06:00 PM, Ted Harding wrote:
Greetings All! As is often the case on this list, the answer may well be under my nose but I can't see it! I am looking for a "smart" way to do the following. Say I have a vector of values, X. I set up bins" for X, say with breaks at B = c(b1,b2,...,b11) covering the range of X, i.e. bins numbered 1:10. The value x is in bin i if B[i]< x<= B[i+1] What I seek is a vector, of the same length as X, which for each x in X gives the number of the bin that x is in. Clearly this can be done in an "unsmart" way by looping through all of X along with something like which( (B[1:10]< X[j])& (X[j]<= B[2:11]) ) However, I feel that this naturally occurring task must have received a smarter solution! The hist() function already does this implicitly, since it has to decide which bin a value in X should be counted in. But it apparently then discards this information, since there is nothing relevant in the return values from hist(). So is there a "smart" function somewhere for this? The motivation here is that I have multivariate data, (X,Y,Z,...) and I wish to study how it behaves in each different bin for X. So the "bin index", ixB aY, derived for X can be applied to select corresponding subsets of the other variables. Rather than doing it the clumsy way each time, e.g. according to Y[(B[i]< X)& (X<= B[j+1])] I would like to have the bin index permanently available -- for example it allows easy logical combinations of bins, such as Y[(ixB==j1) | (ixB==j2)], or Y[(ixB %in% ixB0)].
Hi Ted, Are you looking for something like this? x<-sample(1:10,20,TRUE) x [1] 5 10 10 9 1 1 1 7 2 1 2 1 1 1 9 7 8 5 6 8 binx<-cut(x,breaks=0:10) as.numeric(binx) [1] 5 10 10 9 1 1 1 7 2 1 2 1 1 1 9 7 8 5 6 8 As binx is a factor, coercing it to numeric should return the bin number for each value. Jim