[R] Classifying values by interval

[Jim Lemon]

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