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
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On Jan 16, 2010, at 7:09 PM, che wrote:> > hello every one, > > How to function more than one loop in R?for (i in 1:3) { for (j in 1:2) { for (k in letters[1:4]) { print(paste(i , j, k) ) } } } I'm afraid I develop a strong suspicion that a problem is homework when I am told it should be solved with a particular control structure and I then see no code. -- David.> 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-tp1015851p1015851.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.David Winsemius, MD Heritage Laboratories West Hartford, CT
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.
Really thanks very much, with your help i was able to write a prober code to
count the aminoacids in all the cleaved and noncleaved and then to display
the results in a matrix with 8 column, i used only two loops instead of
three. The code is working but i still have warning telling me that:
?In if (x$Label[i] == nc) { ... :
the condition has length > 1 and only the first element will be used)?
So please can you help me with this warning what is the reason of it as i
can?t understand exactly what does it mean?
Here is the code that i am using, and the data file is attached:
x<-read.table("C:/Uni/502/CA2/hiv.dat", header=TRUE)
num<-nrow(x)
AA<-c('A','C','D','E','F','G','H','I','K','L','M','N','P','Q','R','S','T','V','W','Y')
nc<-x$Label[61:308]
c<-x$Label[nc]
noncleaved<-function(x)
{
y<-matrix(0,20,8)
colnames(y)<-c("N4","N3","N2","N1","C1","C2","C3","C4")
for(i in 1:num){
if (x$Label[i]==nc)
{
for(j in 1:8){
res<-which(AA==substr(x$Peptide[i],j,j))
y[res, j]=y[res, j]+1
}
}
}
return (y/274*100)
}
cleaved<-function(x)
{
y<-matrix(0,20,8)
colnames(y)<-c("N4","N3","N2","N1","C1","C2","C3","C4")
for(i in 1:num){
if (x$Label[i]==c)
{
for(j in 1:8){
res<-which(AA==substr(x$Peptide[i],j,j))
y[res, j]=y[res, j]+1
}
}
}
return (y/113*100)
}
http://n4.nabble.com/file/n1288922/hiv.dat hiv.dat
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here you are the whole code, and the data is attached:> x<-read.table("C:/Uni/502/CA2/hiv.dat", header=TRUE) > num<-nrow(x) > AA<-c('A','C','D','E','F','G','H','I','K','L','M','N','P','Q','R','S','T','V','W','Y') > nc<-x$Label[61:308] > c<-x$Label[nc] > noncleaved<-function(x)+ { + y<-matrix(0,20,8) + colnames(y)<-c("N4","N3","N2","N1","C1","C2","C3","C4") + for(i in 1:num){ + if (x$Label[i] %in% nc) + { + for(j in 1:8){ + res<-which(AA==substr(x$Peptide[i],j,j)) + y[res, j]=y[res, j]+1 + } + } + } + return (y/274*100) + }> > cleaved<-function(x)+ { + y<-matrix(0,20,8) + colnames(y)<-c("N4","N3","N2","N1","C1","C2","C3","C4") + for(i in 1:num){ + if (x$Label[i] %in% nc) + { + for(j in 1:8){ + res<-which(AA==substr(x$Peptide[i],j,j)) + y[res, j]=y[res, j]+1 + } + } + } + return (y/113*100) + }> > hi<-function(x)+ { + height<-rep(0,8) + for (j in 1:8){ + height[j]<-sum(round(cleaved(x)[,j])) + max.height<-max(height) + } + plot(c(0,10*8),c(0,max.height+20),col="white") + }> recta<-function(x)+ { + colmap<-c("#FFFFFF", "#FFFFCC", "#FFFF99", "#FFFF66", "#FFFF33", + "#FFFF00", "#FFCCFF", "#FFCCCC", "#FFCC99", "#FFCC66", "#FFCC33", + "#FFCC00", "#FF99FF", "#FF99CC", "#FF9999", "#FF9966", "#FF9933", + "#FF9900", "#FF33FF", "#FF33CC") + for (j in 1:8){ + xx<-j*10 + for(k in 1:20){ + yy<-round(cleaved(x)[k,j]) + rect(xx,yy,xx+10,yy+round(cleaved(x)[k,j]),col=colmap[k]) + } + } + } -- View this message in context: http://n4.nabble.com/More-than-on-loop-tp1015851p1290019.html Sent from the R help mailing list archive at Nabble.com.
hopefully it is here, two files, one of them is .dat and the others is .txt, just in case. http://n4.nabble.com/file/n1290026/hiv.dat hiv.dat http://n4.nabble.com/file/n1290026/hiv.txt hiv.txt -- View this message in context: http://n4.nabble.com/More-than-on-loop-tp1015851p1290026.html Sent from the R help mailing list archive at Nabble.com.
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 > > > -- > View this message in context: http://n4.nabble.com/More-than-on-loop-tp1015851p1457645.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. >-- Jim Holtman Cincinnati, OH +1 513 646 9390 What is the problem that you are trying to solve?