R help - Apr 2001 - R Novice Question

Please forgive the relative lack of sophistication here -- I'm trying to 
learn R, remove the rust from my basic training in statistics, AND produce 
a useful result.

I have two vectors of equal length,  A and B, with a one-to-one 
correspondence between the elements (ie. A[i] corresponds to B[i]) which 
are provided by an SQL server upon which I can run any SQL statement.

I would like to partition A (like breaks in a histogram) into some number 
of equal-size partitions, and then calculate and plot the mean value of 
the B's that correspond to the A's in each partition (or, more
generally,
produce some descriptive statistics for each set of B's, and potentially 
try to draw some inferences about the B's in each partition).

I can imagine how to do this in Perl by writing a loop, but I suspect 
there may be a more natural way to do it in R, and would be grateful for 
any suggestions.

Thanks.

Regards,

Mike

___________________________________________________
Michael Nielsen
Synergy Plus Pty Ltd
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On Wed, 18 Apr 2001 MichaelNielsen at synergy.com.au wrote:
> Please forgive the relative lack of sophistication here -- I'm trying
to
> learn R, remove the rust from my basic training in statistics, AND produce
> a useful result.
>
> I have two vectors of equal length,  A and B, with a one-to-one
> correspondence between the elements (ie. A[i] corresponds to B[i]) which
> are provided by an SQL server upon which I can run any SQL statement.
>
> I would like to partition A (like breaks in a histogram) into some number
> of equal-size partitions, and then calculate and plot the mean value of
> the B's that correspond to the A's in each partition (or, more
generally,
> produce some descriptive statistics for each set of B's, and
potentially
> try to draw some inferences about the B's in each partition).
>
> I can imagine how to do this in Perl by writing a loop, but I suspect
> there may be a more natural way to do it in R, and would be grateful for
> any suggestions.
Use something like

tapply(B, cut(A, 5), mean)

cut() gives 5 (roughly) equal-sized pieces, and tapply partitions on the
values of the cut and applies the function.

For more complicated situations you might like to look at by().

-- 
Brian D. Ripley,                  ripley at stats.ox.ac.uk
Professor of Applied Statistics,  http://www.stats.ox.ac.uk/~ripley/
University of Oxford,             Tel:  +44 1865 272861 (self)
1 South Parks Road,                     +44 1865 272860 (secr)
Oxford OX1 3TG, UK                Fax:  +44 1865 272595

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