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normalised data file extract values to do calculation

Hello,

I have a normalized file referred to in my programme as "geneSummaries". For each gene across the samples I want to get the median. I am using a For loop to do this.

Normalised the .CEL files which point to geneSummaries

exprs(geneSummaries)-> num
for(i in num){
    geneSummaries[-1, ]
    geneSummaries[ ,i]
    Will be doing the calculation here....
}

So my Question is to do with pulling the data out of the file to do the calculation. I want it drop the 1st column as this is just the name of the sample. See below a sample example of it, when I opened it.

7945460  7.471390  7.256158  7.287770  7.545794 
7945462  7.609366  7.528324  7.324294  7.310791
7945475  5.375443  5.749566  5.519073  5.806861

So to do that I used geneSummaries[-1, ] but it doesn't work

And to pull information out of the ith column I used geneSummaries[ ,i] so can later do calculation, which also doesn't work.

Could someone suggest a idea on how to do this, please.

normalized extract matrix r
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updated by rmnc2007 • rmnc2007 • written by cara78 • cara78

1 answer

I'm assuming that geneSummaries is an eSet. If you just want the median, then it'd be a lot simpler to just:

medians <- apply(exprs(geneSummaries), 1, median)

I'm guessing that the rows are actually then probes/genes rather than the columns. The exprs() accessor doesn't return the row or column names, those are simply displayed when a matrix is shown on screen.

• 3 views • link
written by Devon Ryan Devon Ryan

Hello,

I have a another question please. I am doing this survival analysis and the plan is :

Normalised data

Determine number genes

For every ith gene on the array

Find median of ith gene across samples

Make gene expression 1 or 0 depending on whether or not raw expression value is above or below median (diGene)

gene_survival <- coxph(Surv(survival time, status)~diGene)

My code I have so far :

cel_files <- dir(data_directory, full.names = T, pattern = ".CEL")
print(cel_files)

norm_data <- just.gcrma(cel_files)

exprs(norm_data)-> num

# for loop
for(i in num){                       
medians <- apply(exprs(norm_data), 1, median)

    if(medians > norm_data){
       diGene <- 1;
    } else {
       diGene <- 0;
    }
print(diGene);
}

My question is to do with the comparison if(medians > norm_data) I need to compare to the raw expression but there is an error

Error in medians > norm_data :
  comparison (6) is possible only for atomic and list types"

Am I wrong in thinking that norm_data is the raw data (UN-normalised) or those raw expression mean before Quality control and probeset filtration done?

Thanks

• 0 views • link
updated by Ram Ram • written by cara78 • cara78

Firstly, remove the for(i in num) loop. That'll iterate over every value in the matrix (not to mention that medians <- apply(exprs(geneSummaries), 1, median) will produce the matrix of row medians in one line), which isn't what you want (and anyway, it isn't used).

Secondly, medians > geneSummaries doesn't make sense since medians is a vector of values and geneSummaries is an eSet object (thus leading to that error).

Finally, regressing on whether a genes expression in a subject is above/below the mean is a bad idea. Firstly, you'll be performing enough regressions that your power will be atrocious. Secondly, even if a gene happened to correlate itself, this isn't exactly meaningful/useful if the variance of the gene is itself really small.

I would recommend reassessing your approach before spending more time on it.

• 0 views • link
written by Devon Ryan Devon Ryan

Ok thanks will have another look at it.

• 0 views • link
written by cara78 • cara78

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