R help - Feb 2007 - Function to assign quantiles?

Hi.

If I call the quantiles function as follows:

qvec = quantiles(dvals,probs=seq(0,1,0.1))

the results will return a vector something like the following example:

      0%    10%    20%    30%     40%     50%    60%    70%      80%     90% 
     100%
    56.0   137.3   238.4   317.9   495.8   568.5   807.4  1207.7  1713.0  
2951.1  8703.0

Now I want to assign the deciles, 1 - 10, to each observation, so, in the 
above example, if dvals[322] = 256, I want to assign qvals[322] = 3, and if 
dvals[7216] = 1083, I want qvals[7216] = 7, etc.  I would think there would 
be a function, or some very quick code to do that, but I couldn't find it.

Here's what I have now.  It works, but I figure there must be a better way:

asdc <- function(q){max(1,which(qvec<q))}
qvals = apply(matrix(dvals,nrow=1),2,asdc)


Any suggestions?  Thanks!


--  TMK  --
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On Thu, 2007-02-15 at 17:50 -0500, Talbot Katz wrote:
> Hi.
> 
> If I call the quantiles function as follows:
> 
> qvec = quantiles(dvals,probs=seq(0,1,0.1))
> 
> the results will return a vector something like the following example:
> 
>       0%    10%    20%    30%     40%     50%    60%    70%      80%    
90%
>      100%
>     56.0   137.3   238.4   317.9   495.8   568.5   807.4  1207.7  1713.0  
> 2951.1  8703.0
> 
> Now I want to assign the deciles, 1 - 10, to each observation, so, in the 
> above example, if dvals[322] = 256, I want to assign qvals[322] = 3, and if
> dvals[7216] = 1083, I want qvals[7216] = 7, etc.  I would think there would
> be a function, or some very quick code to do that, but I couldn't find
it.
> 
> Here's what I have now.  It works, but I figure there must be a better
way:
> 
> asdc <- function(q){max(1,which(qvec<q))}
> qvals = apply(matrix(dvals,nrow=1),2,asdc)
> 
> 
> Any suggestions?  Thanks!
Take a look at ?cut and use the output of quantile() to define the
'breaks' argument:

x <- rnorm(100)
> cut(x, breaks = quantile(x, probs = seq(0, 1, 0.1)), 
      include.lowest = TRUE, labels = 1:10)
  [1] 2  1  9  10 9  3  1  7  5  2  4  1  10 8  4  10 9  9  5  3  5  8 
 [23] 6  9  7  1  10 1  10 9  3  3  9  3  5  5  6  4  8  10 6  2  2  8 
 [45] 6  3  6  9  1  4  7  10 8  7  5  3  1  10 2  1  7  8  8  2  8  8 
 [67] 3  10 4  6  5  1  4  6  4  4  5  9  9  7  2  8  2  5  1  3  7  6 
 [89] 4  6  4  6  2  7  2  10 7  3  5  7 
Levels: 1 2 3 4 5 6 7 8 9 10


HTH,

Marc Schwartz