Since there was no data, it is hard to propose a solution. I would estimate
that 'loops' are not required; the use of 'table' or one of the 'apply'
functions will probably provide the answer.
On Sat, Jan 16, 2010 at 8:04 PM, che <fadialn...@live.com> wrote:
>
> Thank you very much for your help,
>
> you have been excused to have a suspicion, but dont worry i am not
> cheating, it is not a home work, rather it is a pre-project task that i
> have
> to deal with in order to prepare to my project, and i cant understand this
> programming things alone, i tried my best but still i cant deal with it
> properly, i am studying master and PhD in bioinformatics, and i need to
> develop a good understanding of programming languages. still a beginner
> but
> i start to have some fears ... what ever you send me, i study it and know
> exactly how it works, and believe me that can help a lot to develop my
> skills. Moreover i am dealing with it in a very honesty way that does not
> break any academic regulations.
>
> thanks again i will try what you sent me ..
>
> Yours
>
> che wrote:
> >
> > hello every one,
> >
> > How to function more than one loop in R? I have the following problem to
> > be solved with the a method of three loops, can you help me please?
> >
> > The data is attached with this message.
> >
> > The data is composed of two parts, cleaved (denoted by cleaved) and non
> > cleaved (denoted by noncleaved).
> > to access to the ith peptide, you can use X$Peptide[i]
> > to access to the ith label, you can use X$Label[i]
> >
> > define a set of amino acids using string or other format if you want
> > amino.acid<-"ACDEFGHIKLMNPQRSTVWY"
> > define two matrices with initialised entries, one for cleaved peptides
> > and one for none-cleaved peptides
> > matrix(0,AA,mer),where AA is the number of amino acids, and mer is the
> > number
> > of residues detected from data using the nchar function
> > both matrices have the same size, the number of rows being equal to the
> > number
> > of amino acids and the number of columns being equal to the number of
> > residues
> > in peptides
> >
> >
> > use one three-loop structure to detect the frequency of amino acids in
> > cleaved peptides
> > and one three-loop structure to detect the frequency of amino acids in
> > non-cleaved
> > peptides. They should not be mixed in one three-loop structure. The best
> > way to
> > handle this is to use a function. The three-loop structure is exampled as
> > below
> > for(i in 1:num)#scanning data for all peptides, where num means the
> number
> > of peptides
> > {
> > for(j in 1:mer)#scanning all residues in a peptide
> > {
> > for(k in 1:AA)#scanning 20 amino acids
> > {
> > #actions
> > }
> > }
> > }
> > http://n4.nabble.com/file/n1015851/hiv.dat hiv.dat
> >
>
> --
> View this message in context:
> http://n4.nabble.com/More-than-on-loop-tp1015851p1015863.html
> Sent from the R help mailing list archive at Nabble.com.
>
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> PLEASE do read the posting guide
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>
--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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