R help - Apr 2008 - If statements for vectors

Dear Sirs,

I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the
'standard' R-package.

I am trying to select a list of genes which all have expression values below
a certain threshold.
I have done this by creating a vector which has 0s where the expression is
greater than the threshold and 1s where it is less than or equal to it.
Multiplying this vector by the expression values produces a list of 0s and
expression values below the threshold value.

However, now I need to remove the 0s.  I thought that this would be
relatively trivial but it appears it isn't!!!


The dimension of the list (with the 0s and values) is 506994.  So I wrote
the following:

for(i in 1:506994) {
if(exp2[i] > 0) {
exp3 <- append(1,exp2[i])
}
return(exp3)
}

where exp2 is the vector of 0s and threshold values.
However I have since discovered that 'if' does not work on vectors.  The
suggestions I have seen on this forum include 'ifelse' which I don't
believe
to be relevant in this situation and 'sapply' which again I don't
think is
relevant.

Can anyone therefore tell me how I can produce a new vector from an old one
by removing the 0 entries?


Thank you for your help,

Laura
University of Warwick

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On Wed, Apr 9, 2008 at 4:07 PM, Laura Bonnett <L.J.Bonnett at
warwick.ac.uk> wrote:
> Dear Sirs,
>
>  I am using both the Bioconductor adds on (Affy, AffyPLM,...) and the
>  'standard' R-package.
>
>  I am trying to select a list of genes which all have expression values
below
>  a certain threshold.
>  I have done this by creating a vector which has 0s where the expression is
>  greater than the threshold and 1s where it is less than or equal to it.
>  Multiplying this vector by the expression values produces a list of 0s and
>  expression values below the threshold value.
>
>  However, now I need to remove the 0s.  I thought that this would be
>  relatively trivial but it appears it isn't!!!
>
Without a working example from you, I have no way to test this
proposal.  But if I were you, I would get the index values of the
right cases in one step, and then use that to choose the ones I want.

theGoodOnes <- which(exp2 >= 0)
exp3 <- exp2[theGoodOnes]


This can be crammed into one line, but I'd do it in two just to make
sure it is correct, at least the first time.

My example:
> x <- rnorm(100)
> which(x >= 0)
 [1]  9 11 12 13 14 16 19 20 21 22 24 25 28 30 32 34 35 36 38 40 41 42 43 44 46
[26] 47 49 50 53 54 57 59 61 63 64 66 68 69 70 72 75 78 80 81 82 83 88 90 97
98
> theGoodOnes <- which(x>=0)
> newX <- x[theGoodOnes]
> newX
 [1] 0.89285908 0.51753998 1.18485887 2.10003705 0.54535841 1.28313738
 [7] 1.34092172 0.76064356 0.02201121 0.80808363 0.04578730 0.23045983
[13] 1.04306626 0.12694184 0.89706863 0.86302992 1.53471660 0.51192410
[19] 1.00366834 1.76923470 0.49508470 1.27888454 0.76706729 1.46340483
[25] 1.69315538 0.50537603 0.18422329 0.72968629 0.45490526 2.18208183
[31] 0.71926926 0.68915832 1.49076770 0.48763971 0.39273110 0.80709549
[37] 0.22099019 0.38103757 0.14626929 0.63933750 1.26643194 3.33091910
[43] 2.50341609 2.05286611 0.31986095 0.64548972 0.34712937 0.04075135
[49] 0.07206342 0.20325505


-- 
Paul E. Johnson
Professor, Political Science
1541 Lilac Lane, Room 504
University of Kansas