[R] R Data

Fowler, Mark M@rk@Fow|er @end|ng |rom d|o-mpo@gc@c@
Thu Feb 14 14:44:07 CET 2019


I am not sure I would use the word ‘accounted’, more like discounted (tossed out).

From: Spencer Brackett <spbrackett20 using saintjosephhs.com>
Sent: February 14, 2019 9:21 AM
To: Fowler, Mark <Mark.Fowler using dfo-mpo.gc.ca>
Cc: R-help <r-help using r-project.org>; Sarah Goslee <sarah.goslee using gmail.com>; Caitlin Gibbons <bioprogrammer using gmail.com>; Jeff Newmiller <jdnewmil using dcn.davis.ca.us>
Subject: Re: R Data

Mr. Fowler,

Thank you! This information is most helpful. So from my understanding, I can use the regression coefficients shown (via the coding I originally sent, to generate a continuous distribution with what is essentially a line of best fit? The data added here had some 30,000 variables (it is genomic data from TCGA), does this mean that any none NA data is being accounted for in said distribution?

Best,

Spencer Brackett



On Thursday, February 14, 2019, Fowler, Mark <Mark.Fowler using dfo-mpo.gc.ca<mailto:Mark.Fowler using dfo-mpo.gc.ca>> wrote:
Hi Spencer,

The an1 syntax is adding regression coefficients (or NAs where a regression could not be done) to the downloaded and processed data, which ends up a matrix. The cbind function adds the regression coefficients to the last column of the matrix (i.e. bind the columns of the inputs in the order given). Simple example below. Not actually any need for the separate cbind commands, could have just used an1=cbind(an,p,t). The cbind function expects all the columns to be of the same length, hence the use of the tryCatch function to capture NA's for failed regression attempts, ensuring that p and t correspond row by row with the matrix.

 x=seq(1,5)
 y=seq(6,10)
 z=seq(1,5)
xyz=cbind(x,y,z)
xyz
   x  y z
[1,] 1  6 1
[2,] 2  7 2
[3,] 3  8 3
[4,] 4  9 4
[5,] 5 10 5
dangs=rep(NA,5)
xyzd=cbind(xyz,dangs)
xyzd
     x  y z dangs
[1,] 1  6 1    NA
[2,] 2  7 2    NA
[3,] 3  8 3    NA
[4,] 4  9 4    NA
[5,] 5 10 5    NA

-----Original Message-----
From: R-help <r-help-bounces using r-project.org<mailto:r-help-bounces using r-project.org>> On Behalf Of Spencer Brackett
Sent: February 14, 2019 12:32 AM
To: R-help <r-help using r-project.org<mailto:r-help using r-project.org>>; Sarah Goslee <sarah.goslee using gmail.com<mailto:sarah.goslee using gmail.com>>; Caitlin Gibbons <bioprogrammer using gmail.com<mailto:bioprogrammer using gmail.com>>; Jeff Newmiller <jdnewmil using dcn.davis.ca.us<mailto:jdnewmil using dcn.davis.ca.us>>
Subject: [R] R Data

Hello everyone,

The following is a portion of coding that a colleague sent. Given my lack of experience in R, I am not quite sure what the significance of the following arguments. Could anyone help me translate? For context, I am aware of the downloading portion of the script... library(data.table) etc., but am not familiar with the portion pertaining to an1 .

library(data.table)
anno = as.data.frame(fread(file =
"/rsrch1/bcb/kchen_group/v_mohanty/data/TCGA/450K/mapper.txt", sep ="\t", header = T)) meth = read.table(file = "/rsrch1/bcb/kchen_group/v_mohanty/data/TCGA/27K/GBM.txt", sep  ="\t", header = T, row.names = 1) meth = as.matrix(meth) """ the loop just formats the methylation column names to match format"""
colnames(meth) = sapply(colnames(meth), function(i){
  c1 = strsplit(i,split = '.', fixed = T)[[1]]
  c1[4] = paste(strsplit(c1[4],split = "",fixed = T)[[1]][1:2],collapse =
"")
  paste(c1,collapse = ".")
})
exp = read.table(file =
"/rsrch1/bcb/kchen_group/v_mohanty/data/TCGA/RNAseq/GBM.txt", sep = "\t", header = T, row.names = 1) exp = as.matrix(exp) c = intersect(colnames(exp),colnames(meth))
exp = exp[,c]
meth = meth[,c]
m = apply(meth, 1, function(i){
  log2(i/(1-i))
})
m = t(as.matrix(m))
an = anno[anno$probe %in% rownames(m),]
an = an[an$gene %in% rownames(exp),]
an = an[an$location %in% c("TSS200","TSS1500"),]

p = apply(an,1,function(i){
  tryCatch(summary(lm(exp[as.character(i[2]),] ~ m[as.character(i[1]),]))$coefficient[2,4], error= function(e)NA)
})
t = apply(an,1,function(i){
  tryCatch(summary(lm(exp[as.character(i[2]),] ~ m[as.character(i[1]),]))$coefficient[2,3], error= function(e)NA)
})
an1 =cbind(an,p)
an1 = cbind(an1,t)
an1$q = p.adjust(as.numeric(an1$p))
summary(lm(exp["MAOB",] ~ m["cg00121904",]$coefficient[2,c(3:4)]
###############################################

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