Thanks Jim. It works. I need to specify row.names as missing and then
get rid of the first column to make it as a matrix.
x<-read.table(file="x.txt",header=TRUE,row.names=NULL,na.strings = "NA")
x <- as.matrix(x[,-1])
y<-read.table(file="y.txt",header=TRUE,row.names=NULL,na.strings = "NA")
y <- as.matrix(y[,-1])
z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
dim(z) <- dim(x)
z
Best,
Allen
On Nov 11, 2007 11:02 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> Here is the way to read the data and convert it. Your data was a
> dataframe with the first column being the id:
>
> > x <- read.table(textConnection("id b1 b2 b3
> + a1 2 4 6
> + a2 1 2 NA
> + a3 4 6 NA"), header=TRUE)
> > y <- read.table(textConnection("id b1 b2 b3
> + a1 NA 4 4
> + a2 2 2 NA
> + a3 1 2 2"), header=TRUE)
> > # look at what x & y are:
> > str(x)
> 'data.frame': 3 obs. of 4 variables:
> $ id: Factor w/ 3 levels "a1","a2","a3": 1 2 3
> $ b1: int 2 1 4
> $ b2: int 4 2 6
> $ b3: int 6 NA NA
> > str(y)
> 'data.frame': 3 obs. of 4 variables:
> $ id: Factor w/ 3 levels "a1","a2","a3": 1 2 3
> $ b1: int NA 2 1
> $ b2: int 4 2 2
> $ b3: int 4 NA 2
> > # to convert to matrix, get rid of first column
> > x <- as.matrix(x[,-1])
> > y <- as.matrix(y[,-1])
> > z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> > dim(z) <- dim(x)
> > z
> [,1] [,2] [,3]
> [1,] 2.0 4 5
> [2,] 1.5 2 NaN
> [3,] 2.5 4 2
> > is.na(z) <- is.nan(z)
> > z
> [,1] [,2] [,3]
> [1,] 2.0 4 5
> [2,] 1.5 2 NA
> [3,] 2.5 4 2
> >
> >
>
>
>
> On Nov 11, 2007 10:47 PM, affy snp <[EMAIL PROTECTED]> wrote:
> > Hi,Jim. I created two txt files as:
> >
> > x.txt
> >
> > id b1 b2 b3
> > a1 2 4 6
> > a2 1 2 NA
> > a3 4 6 NA
> >
> > y.txt
> > id b1 b2 b3
> > a1 NA 4 4
> > a2 2 2 NA
> > a3 1 2 2
> >
> >
> > I tried it one more time but got different z:
> >
> > > x<-read.table(file="x.txt",header=TRUE,row.names=1,na.strings = "NA")
> > Warning message:
> > In read.table(file = "x.txt", header = TRUE, row.names = 1, na.strings =
> > "NA") :
> > incomplete final line found by readTableHeader on 'x.txt'
> > > x
> > b1 b2 b3
> > a1 2 4 6
> > a2 1 2 NA
> > a3 4 6 NA
> > > y<-read.table(file="y.txt",header=TRUE,row.names=1,na.strings = "NA")
> > Warning message:
> > In read.table(file = "y.txt", header = TRUE, row.names = 1, na.strings =
> > "NA") :
> > incomplete final line found by readTableHeader on 'y.txt'
> > > y
> > b1 b2 b3
> > a1 NA 4 4
> > a2 2 2 NA
> > a3 1 2 2
> > > z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> > > dim(z) <- dim(x)
> > Error in dim(z) <- dim(x) :
> > dims [product 9] do not match the length of object [3]
> > > z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> > > z
> > b1 b2 b3
> > 2.000000 3.333333 4.000000
> > >
> >
> >
> > Allen
> >
> > On Nov 11, 2007 10:41 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> > > What did your text files look like? It would appear that there was
> > > not a line feed on the last line of the file. Also what does 'str' of
> > > x and y show? It appears that one is a data frame and one is a
> > > matrix. That might be causing some of the problems.
> > >
> > >
> > > On Nov 11, 2007 10:30 PM, affy snp <[EMAIL PROTECTED]> wrote:
> > > > Hi Jim,
> > > >
> > > > Thanks a lot! I am wondering why I ended up getting the result as
> > > > follows:
> > > >
> > > > > x<-read.table(file="x.txt",header=TRUE,row.names=1,na.strings = "NA")
> > > > Warning message:
> > > > In read.table(file = "x.txt", header = TRUE, row.names = 1, na.strings
> > > > = "NA") :
> > > > incomplete final line found by readTableHeader on 'x.txt'
> > > > > x
> > > > b1 b2 b3
> > > > a1 2 4 6
> > > > a2 1 2 NA
> > > > a3 4 6 NA
> > > > > y<-as.matrix(read.table(file="y.txt",header=TRUE,row.names=1,na.strings
> > > > > = "NA"))
> > > > Warning message:
> > > > In read.table(file = "y.txt", header = TRUE, row.names = 1, na.strings
> > > > = "NA") :
> > > > incomplete final line found by readTableHeader on 'y.txt'
> > > > > y
> > > > b1 b2 b3
> > > > a1 NA 4 4
> > > > a2 2 2 NA
> > > > a3 1 2 2
> > > > > z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> > > > > z
> > > > b1 b2 b3 <NA> <NA> <NA> <NA> <NA>
> > > > 2.333333 3.500000 3.500000 2.750000 3.500000 4.000000 2.750000 4.000000
> > > > <NA>
> > > > 4.000000
> > > > > dim(z) <- dim(x)
> > > > > z
> > > > [,1] [,2] [,3]
> > > > [1,] 2.333333 2.75 2.75
> > > > [2,] 3.500000 3.50 4.00
> > > > [3,] 3.500000 4.00 4.00
> > > > > is.na(z) <- is.nan(z)
> > > > > z
> > > > [,1] [,2] [,3]
> > > > [1,] 2.333333 2.75 2.75
> > > > [2,] 3.500000 3.50 4.00
> > > > [3,] 3.500000 4.00 4.00
> > > > >
> > > >
> > > >
> > > > Allen
> > > >
> > > >
> > > > On Nov 11, 2007 5:27 PM, jim holtman <[EMAIL PROTECTED]> wrote:
> > > > > Here is one way of doing it:
> > > > >
> > > > > > x
> > > > > [,1] [,2] [,3]
> > > > > [1,] 2 4 6
> > > > > [2,] 1 2 NA
> > > > > [3,] 4 6 NA
> > > > > > y
> > > > > [,1] [,2] [,3]
> > > > > [1,] NA 4 4
> > > > > [2,] 2 2 NA
> > > > > [3,] 1 2 2
> > > > > > z <- mapply(function(a,b)mean(c(a,b), na.rm=TRUE), x, y)
> > > > > > dim(z) <- dim(x)
> > > > > > z
> > > > > [,1] [,2] [,3]
> > > > > [1,] 2.0 4 5
> > > > > [2,] 1.5 2 NaN
> > > > > [3,] 2.5 4 2
> > > > > > # to change it to NA
> > > > > > is.na(z) <- is.nan(z)
> > > > > > z
> > > > > [,1] [,2] [,3]
> > > > > [1,] 2.0 4 5
> > > > > [2,] 1.5 2 NA
> > > > > [3,] 2.5 4 2
> > > > >
> > > > > >
> > > > > >
> > > > >
> > > > >
> > > > > On Nov 11, 2007 4:52 PM, affy snp <[EMAIL PROTECTED]> wrote:
> > > > > > Dear list,
> > > > > >
> > > > > > I am new to R and very inexperienced. Sorry for the trouble.
> > > > > > I have two txt files and want to merge them by taking the average.
> > > > > > More specifically, for example, the txt file1, with row names and
> > > > > > column names,
> > > > > > consists of 238000 rows and 196 columns. Each column corresponds
> > > > > > to a sample. The data is mixed with numeric or NA. So what I plan to
> > > > > > do is:
> > > > > >
> > > > > > (1) Take the 1st column from txt file 1 and txt file 2, calculate
> > > > > > the average
> > > > > > if both numbers are numeric. If one is numeric and the other one is
> > > > > > NA or
> > > > > > the opposite, just use the numeric; If both are NA, then use NA, Do
> > > > > > all this
> > > > > > for all columns
> > > > > > (2) Create txt file 3 with the numbers from the above and add the
> > > > > > row names and
> > > > > > column names.
> > > > > >
> > > > > > So an illustrative example could be:
> > > > > >
> > > > > > txt file 1
> > > > > >
> > > > > > A B C
> > > > > > row1 2 4 6
> > > > > > row2 1 2 NA
> > > > > > row3 4 6 NA
> > > > > >
> > > > > > txt file 2
> > > > > >
> > > > > > A B C
> > > > > > row1 NA 4 4
> > > > > > row2 2 2 NA
> > > > > > row3 1 2 2
> > > > > >
> > > > > > then txt file 3 will be created as:
> > > > > >
> > > > > > A B C
> > > > > > row1 2 4 5
> > > > > > row2 1.5 2 NA
> > > > > > row3 2.5 4 2
> > > > > >
> > > > > > Any help will be appreciated.
> > > > > >
> > > > > > Thanks!
> > > > > > Allen
> > > > > >
> > > > > > ______________________________________________
> > > > > > R-help@r-project.org mailing list
> > > > > > https://stat.ethz.ch/mailman/listinfo/r-help
> > > > > > PLEASE do read the posting guide
> > > > > > http://www.R-project.org/posting-guide.html
> > > > > > and provide commented, minimal, self-contained, reproducible code.
> > > > > >
> > > > >
> > > > >
> > > > >
> > > > > --
> > > > > Jim Holtman
> > > > > Cincinnati, OH
> > > > > +1 513 646 9390
> > > > >
> > > > > What is the problem you are trying to solve?
> > > > >
> > > >
> > >
> > >
> > >
> > > --
> > >
> > > Jim Holtman
> > > Cincinnati, OH
> > > +1 513 646 9390
> > >
> > > What is the problem you are trying to solve?
> > >
> >
>
>
>
> --
>
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
>
> What is the problem you are trying to solve?
>
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