[R] More than on loop??

[jim holtman]

Also this looks like homework, so I can not really reply with a
solution.  BTW, once you have the normalized matrix, barplot will
create your output without the complications of steps 8-13.  You will
have to use the data to put the text, but that again is relatively
easy with the data.

On Sat, Jan 30, 2010 at 3:38 PM, che <fadialnaji at live.com> wrote:
>
> Here is the the written instruction as i managed to get it from my professor,
> the graphs and data are attached:
>
> The graph below shows an example of the expected outcome of this course
> work. You may
> procude a better one. The graph for analysing the motifs of a set of
> peptides is designed
> this way
>
> • the graph is composed of columns of coloured rectangles
>
> • a column corresponding to a residue from “N4” to “C4”. Note that eight
> residues
> are denoted by “N4”, “N3”, “N2”, “N1”, “C1”, “C2”, “C3”, “C4”. “N4” means
> the
> 4th flanking residue of a cleavage site on the N-terminal side and “C3”
> means the 3rd
> flanking residue of a cleavage site on the C-terminal side. The cleavage
> occurs between
> “N1” and “C1”.
>
> • there are 20 rectangles in each column corresponding to 20 amino acids. A
> rectangular
> of an amino acid has a larger height if the corresponding amino acid has a
> larger
> frequency to occur at the residue, for instance, the rectangular of “S” in
> the first
> column for the cleaved peptides.
>
> • a letter of an amino acid is printed within a rectangular. Its font size
> depends on the
> frequency of the amino acid in a residue.
>
> In your package, you need to have the following functions
> 1. set a colour map using the following or your own design
> • colmap<-c("#FFFFFF", "#FFFFCC", "#FFFF99", "#FFFF66", "#FFFF33",
> "#FFFF00", "#FFCCFF", "#FFCCCC", "#FFCC99", "#FFCC66", "#FFCC33",
> "#FFCC00", "#FF99FF", "#FF99CC", "#FF9999", "#FF9966", "#FF9933",
> "#FF9900", "#FF33FF", "#FF33CC")
> 2. define a set of amino acids using string or other format if you want
> • amino.acid<-"ACDEFGHIKLMNPQRSTVWY"
>
> 3. read in the given peptide data (“hiv.dat”) using
> read.table(‘‘../data/hiv.dat’’,header=TRUE)
> • The data I sent to you should not be saved in the same directory where you
> save
> your R code!
> • The data is composed of two parts, cleaved (denoted by “cleaved”) and non
> cleaved (denoted by “noncleaved”). The first five lines of the data are
> shown
> below
> Peptide Label
> TQIMFETF cleaved
> GQVNYEEF cleaved
> KVFGRCEL noncleaved
> VFGRCELA 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]
>
> 4. detect the number of cleaved peptides and the number of non-cleaved
> peptides using
> • nrow(X)
>
> 5. define two matrices with initialised entries, one for positive peptides
> and one for neg-
> ative 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
> • name the columns of these two matrices using
> – c("N4","N3","N2","N1","C1","C2","C3","C4"),
>
> 6. 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
> }
> }
> }
>
> 7. make sure that each frequency matrix needs to be converted to a
> percentage, i.e. each
> entry in the matrix is divided by the number of cleaved or non-cleaved
> peptides and
> multiplied by 100. This converted frequency is named as the normalised
> frequency.
>
> 8. detect the maximum height of the normalised frequency each residue in
> cleaved or
> non-cleaved peptides using
> height<-rep(0,mer)
> for(j in 1:mer)
> height[j]<-sum(round(X.frequency[,j]))
> max.height<-max(height)
> • Note that the height of each column in a graph (see the graph on 3)
> corresponds
> to the summation of 20 frequencies of 20 amino acids for a residue.
>
> 9. draw a blank plot using the maximum height
> • plot(c(0,10*mer),c(0,max.height),col="white", • • •)
> • in this blank plot, you can add graphics as discussed below
>
> 10. determine the x coordinate, but it is recommended to use i*10 as the
> x-coordinate
> where i indexes the residues. The x-coordinate represents columns in the
> graph shown
> in 3. If there are 8 residues in peptides, there are 8 columns.
>
> 11. determine the y coordinate, which is cumulative (see next item below).
> The y-
> coordinate represents rows in the graph shown in 3. There are always 20 rows
> for
> 20 amino acids. Note that the rows cannot be aligned because the frequency
> of an
> amino acid in a residue varies.
> 12. draw a rectangular based on the frequency of each residue and each amino
> acid
> • rect(x,y,x+10,y+round(X.frequency[k,j]),col=colmap[k]), where k indi-
> cates an amino acid and j indicates a residue
> • after drawing this rectangular, the y-coordinate “y” should be increased
> by round(X.frequency[k,j])
> • after one column is drawn for one residue, the x-coordinate “x” should be
> in-
> creased by 10
> 13. plot a text at the corresponding position using
> • text((x+5),(y+round(X.frequency[k,j])/2),substr(amino.acid,k,k))
> 14. place two drawings in one plot using the par function
> http://n4.nabble.com/file/n1457645/cleaved.jpg cleaved.jpg
> http://n4.nabble.com/file/n1457645/noncleaved.jpg noncleaved.jpg
> http://n4.nabble.com/file/n1457645/hiv.dat hiv.dat
>
>
> --
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>
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>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

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