hey Jim.
brilliant, very short and productive, wish that i can have such skill in the
future, i will try to learn about all functions that you used.
thanks very much for helping me, i really appreciate it.
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hello,
i appreciate your help, your help, comments, and suggestion really are so
helpful to develop not only my R skills, but also my programming language
sense. as i said, i am not breaking any academic rules, and this is a
softwar i have to develop to deal with my project after two months along
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
One quick comment about looking at the graphs you provided, why aren't
all 8 columns the same height given that each column should have the
same number of amino acids in them. FOr the cleaved case is it 114
and even after normalizing, the column sums should be the same -- 100.
Are the graphs real
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
It is not entirely clear what you are trying to do. Can you explain
what the matrix that you are creating out of 'cleaved' represents?
"Tell me what you want to do; not how you want to do it". It is hard
to follow code when you have not explained what it is doing. THere
appear to be all kinds o
yes, but the outcome graphs are almost the same, that mean it does not
calculated in a cumulative way , if you apply the following code, then run
hi(x), and then recta(x), you will see how the shape are similar to the
frequency of Amino Acid in the matrix. i am looking for a code that can do
thi
It sounds like you want to use 'barplot' like below given that it
appears that the value in x.c would be the matrix you want to graph:
x.c <- cleaved(x)
barplot(x.c, col=c("#FF", "#CC", "#99", "#66", "#33",
"#00", "#FFCCFF", "#FF", "#FFCC99", "#FFCC66", "#FFCC33",
70% yes, the problem is i am trying to produce a graph similar to the one in
attachments in this message, which represents the frequency of each letter
"aminoacid" in the cleaved function and the noncleaved function. some thing
else i added to the attachments is the pattern which seemingly working
Does this do what you want? You were not initializing the plot before
calling 'rect'. also it appears that you only have to call cleaved(x)
once to get the matrix and then use it in the loop -- more efficient
that way.
x<-read.table("C:/hiv.txt", header=TRUE)
num<-nrow(x)
AA<-c('A','C','D','E','
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
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Sent f
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(0,2
can i ask your help again, please excuse my questions:
It is working perfectly now, i still have the last part which i tried a lot
with but still i can’t translate it properly for the computer through R. I
need to draw rectangular based on the frequency of each residue, actually i
found the patter
Notice that 'nc' is multivalued (nc<-x$Label[61:308]). If you want to
check if x$Label[i] is one of the values in 'nc', then use %in%:
if (x$Label[i] %in% nc)
The error happens because 'if' can only have a single conditional
expressions and your original 'if' statement would have resulted i
Really thanks very much, with your help i was able to write a prober code to
count the aminoacids in all the cleaved and noncleaved and then to display
the results in a matrix with 8 column, i used only two loops instead of
three. The code is working but i still have warning telling me that:
“In
I found the data you sent with the help of David Winsemius. Are you just
trying to count the number of amino acids in each of the groups? Is it
something like this:
> # split by the Label
> x <- split(amino, amino$Label)
> # now count the number of aminos in each group
> lapply(x, function(.lab)
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 wrote:
>
> Thank you very much for your help,
>
> you have been ex
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
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 homewor
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 “no
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