[R] Computational problems in R

Steven McKinney smckinney at bccrc.ca
Sat Oct 25 03:50:11 CEST 2008


I suspect there's a deeper issue here.
sum(exp(yi)) when large yi occur is
problematic.  exp(yi) for yi>710 is
just a huge number, and summing additional
values only makes the overall sum larger as all
components of the summation are positive.
There's no way around that.

You could try this with Robin Hankins'
package "brobdingnag" which can handle
bunches of bizarrely large numbers.

What kind of process are you studying?
What kind of process generates values
such as exp(8/0.01) when other values
are much smaller?  Are these outliers
in an otherwise well-behaved
data set?  Perhaps then they need
to be set aside and investigated
separately, etc.


Steven McKinney

Statistician
Molecular Oncology and Breast Cancer Program
British Columbia Cancer Research Centre

email: smckinney +at+ bccrc +dot+ ca

tel: 604-675-8000 x7561

BCCRC
Molecular Oncology
675 West 10th Ave, Floor 4
Vancouver B.C. 
V5Z 1L3
Canada




-----Original Message-----
From: r-help-bounces at r-project.org on behalf of Duncan Murdoch
Sent: Fri 10/24/2008 4:04 PM
To: A.Noufaily
Cc: r-help at r-project.org
Subject: Re: [R] Computational problems in R
 
On 24/10/2008 12:42 PM, A.Noufaily wrote:
> Dear all,
> 
> I would be grateful if anyone can help me with the following:
> 
> My aim is to compute explicitely the sum S=A+B where A=sum(exp(c_i/d)),
> i=1,...,n;
> B, c_i, and d are real numbers with -Inf<B,c_i<+Inf; and d>0.
> The problem is that when c_i/d >710 (for some i) R is setting
> exp(c_i/d) to be equal to +Inf and hence the whole summation S.
> So in simple cases where for example c_i=8 (for some i), and d=0.01 the
> whole summation is turning out to be infinite.
> Is there a way to get round that in R?
> Can anyone suggest any computational trick to calculate S when c_i/d>710
> (for some i)?

Work on a log scale.  Use the identity that

A+B = A*(1 + B/A)
     = exp(log(A) + log(1 + B/A))

(where you chose A to be the biggest term in the sum).

Duncan Murdoch

> 
> Any suggestions would be much appreciated.
> 
> Regards,
> 
> Amy
> 
> 
> 
> 
> 
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