I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) +
a3(b1+b2+b4) + a4(b1+b2+b3)
or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i
I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i
* b_i
would appreciate some help.
Thank you.
--
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Have you read the Introduction to R tutorial? You seem unaware of even the most basic stuff, especially vectorization and subscripting. Also, this smells like homework, and we don't do homework here. -- Bert On Thu, Oct 18, 2012 at 12:33 PM, djbanana <karl79264219 at gmail.com> wrote:> I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + > a3(b1+b2+b4) + a4(b1+b2+b3) > > or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i > > I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i > * b_i > > would appreciate some help. > > Thank you. > > > > -- > View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.-- Bert Gunter Genentech Nonclinical Biostatistics Internal Contact Info: Phone: 467-7374 Website: http://pharmadevelopment.roche.com/index/pdb/pdb-functional-groups/pdb-biostatistics/pdb-ncb-home.htm
I am aware of the most basic stuff, especially vectorization and subscripting. I only gave a simple example with length 4. I need to do that for vectors of length 190. Are there any in-built commands? Or should I write the loops myself? Thank you. -- View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678p4646705.html Sent from the R help mailing list archive at Nabble.com.
Assuming that you actually mean
a1(b2+b3+b4) + a2(b1+b3+b4) + a3(b1+b2+b4) + a4(b1+b2+b3)
^ ^ ^
this might give you what you want:
x <- data.frame( a = sample( 1:10, 4 ), b = sample( 11:20, 4 ) )
x
a b
1 1 16
2 7 15
3 8 19
4 4 13
for( i in 1 : length( x$a ) )
x$c[i] <- x$a[i] * ( sum( x$b ) - x$b[i] )
x
a b c
1 1 16 47
2 7 15 336
3 8 19 352
4 4 13 200
Rgds,
Rainer
On Thursday 18 October 2012 12:33:39 djbanana wrote:> I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) +
> a3(b1+b2+b4) + a4(b1+b2+b3)
>
> or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i
>
> I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i
> * b_i
>
> would appreciate some help.
>
> Thank you.
>
>
>
> --
> View this message in context:
http://r.789695.n4.nabble.com/summation-coding-tp4646678.html
> Sent from the R help mailing list archive at Nabble.com.
>
> ______________________________________________
> R-help at r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
Hi, I think I solved it myself by writing loops. What I meant is: are there in-built functions in R that calculate the following: a1(b2+...+b190) + a2(b1+b3+...+b190) + ... I managed to solve it, quite similar to what you just emailed. Thanks anyway! -- View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678p4646712.html Sent from the R help mailing list archive at Nabble.com.
You asked for a loop, you got one... Vectorized is easier and faster: x$c <- x$a * ( sum( x$b ) - x$b ) Rgds, Rainer On Friday 19 October 2012 09:38:35 you wrote:> Hi, > > I think I solved it myself by writing loops. > > What I meant is: are there in-built functions in R that calculate the > following: > > a1(b2+...+b190) + a2(b1+b3+...+b190) + ... > > I managed to solve it, quite similar to what you just emailed. > > Thanks! > > On 19 October 2012 09:20, Rainer Schuermann <rainer.schuermann at gmx.net>wrote: > > > Is it possible that you mean > > a1(b2+b3+b4) + a2(b1+b3+b4) + a3(b1+b2+b4) + a4(b1+b2+b3) > > ^ ^ ^ > > > > > > On Thursday 18 October 2012 12:33:39 djbanana wrote: > > > I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + > > > a3(b1+b2+b4) + a4(b1+b2+b3) > > > > > > or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i > > > > > > I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} > > a_i > > > * b_i > > > > > > would appreciate some help. > > > > > > Thank you. > > > > > > > > > > > > -- > > > View this message in context: > > http://r.789695.n4.nabble.com/summation-coding-tp4646678.html > > > Sent from the R help mailing list archive at Nabble.com. > > > > > > ______________________________________________ > > > R-help at r-project.org mailing list > > > https://stat.ethz.ch/mailman/listinfo/r-help > > > PLEASE do read the posting guide > > http://www.R-project.org/posting-guide.html > > > and provide commented, minimal, self-contained, reproducible code. > >
On 18-10-2012, at 21:33, djbanana wrote:> I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + > a3(b1+b2+b4) + a4(b1+b2+b3) > > or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i > > I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i > * b_i >This is partly TeX notation not summation notation, whatever that may be. And these "sums" make no sense: where is index j used in the first? where is it used in the second? Solutions starting from you explicit formula, from slowest to fastest sum( outer(a,b,"*") ) - sum(a*b) sum(sapply(1:length(a),function(k) a[k]*sum(b[-k]))) sum(convolve(a,b)) - sum(a*b) # sum(convolve(a,rev(b), type="o")) - sum(a*b) Berend> would appreciate some help. > > Thank you. > > > > -- > View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678.html > Sent from the R help mailing list archive at Nabble.com. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.
On 19-10-2012, at 10:38, Berend Hasselman wrote:> > On 18-10-2012, at 21:33, djbanana wrote: > >> I would like to code the following in R: a1(b1+b2+b3) + a2(b1+b3+b4) + >> a3(b1+b2+b4) + a4(b1+b2+b3) >> >> or in summation notation: sum_{i=1, j\neq i}^{4} a_i * b_i >> >> I realise this is the same as: sum_{i=1, j=1}^{4} a_i * b_i - sum_{i=j} a_i >> * b_i >> > > This is partly TeX notation not summation notation, whatever that may be. > And these "sums" make no sense: where is index j used in the first? where is it used in the second? > > Solutions starting from you explicit formula, from slowest to fastest > > sum( outer(a,b,"*") ) - sum(a*b) > sum(sapply(1:length(a),function(k) a[k]*sum(b[-k]))) > sum(convolve(a,b)) - sum(a*b) # sum(convolve(a,rev(b), type="o")) - sum(a*b) >And from Nabble another solution (not posted to the R-help list (yet?) ) which turns out to be the quickest (obviously) sum(a*(sum(b)-b)) Berend> Berend > > >> would appreciate some help. >> >> Thank you. >> >> >> >> -- >> View this message in context: http://r.789695.n4.nabble.com/summation-coding-tp4646678.html >> Sent from the R help mailing list archive at Nabble.com. >> >> ______________________________________________ >> R-help at r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > ______________________________________________ > R-help at r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code.