# [R] R Data

Spencer Brackett @pbr@ckett20 @end|ng |rom @@|ntjo@ephh@@com
Thu Feb 14 14:20:39 CET 2019

```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>
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> On Behalf Of Spencer Brackett
> Sent: February 14, 2019 12:32 AM
> To: 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: [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
> but am not familiar with the portion pertaining to an1 .
>
> library(data.table)
> "/rsrch1/bcb/kchen_group/v_mohanty/data/TCGA/450K/mapper.txt", sep ="\t",
> 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 = ".")
> })
> "/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)
> summary(lm(exp["MAOB",] ~ m["cg00121904",]\$coefficient[2,c(3:4)]
> ###############################################
>
>         [[alternative HTML version deleted]]
>
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