ss wrote:
Dear Erik,

Thanks! The 'data' is matrix but all(apply(data[,3:85], 2, class) == "numeric")
is false.

 > class(data)
[1] "matrix"
 > a<- rowMeans(data[,3:85],na.rm = TRUE)
Error in rowMeans(data[, 3:85], na.rm = TRUE) : 'x' must be numeric
 > all(apply(data[,3:85], 2, class) == "numeric")
[1] FALSE
 >

What else should I do?

try str(data) , do you have a character matrix? That would explain the error message.

I would consider storing your data in a data.frame, as you appear not to have homogeneous types, and then your analysis should be trivial.

Erik


I appreciate!

Allen



On Thu, Jun 12, 2008 at 4:55 PM, Erik Iverson <[EMAIL PROTECTED] <mailto:[EMAIL PROTECTED]>> wrote:

    Hello -


    ss wrote:

        Hi all,

        I have a matrix called 'data', which looks like:

            data[1:4,1:4]

             Probe_ID       Gene_Symbol  M1601           M1602
        1 A_23_P105862    13CDNA73            -1.6            0.16
        2  A_23_P76435      15E1.2            0.18            0.59
        3 A_24_P402115      15E1.2            1.63           -0.62
        4 A_32_P227764      15E1.2           -0.76           -0.42

            dim(data)

        [1] 23963    85


    Do you really have a matrix, or a data.frame?

    Try

     > class(data)



        What I want to do is to make a new matrix called 'data2', which
        would be
        transformed
        by subtracting the mean of each row from matrix 'data'. There
        are some 'NA's
        in the
        matrix and I do want to keep it.


    See ?scale



        I tried to take 'mean's from each row first by using:

        a<- rowMeans(data[,3:85],na.rm = FALSE)

        but I got:

            a<- rowMeans(data[,3:85],na.rm = FALSE)

        Error in rowMeans(data[, 3:85], na.rm = FALSE) : 'x' must be numeric

        Can anybody suggest me how to get around this?


    Figure out what you are giving the rowMeans function.
    If you really have a matrix, then

    all(apply(data[,3:85], 2, class) == "numeric") should be TRUE.




        Thank you very much!

        Allen

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