Ah OK, I didn't get your question then.
a dist-object is actually a vector of numbers with a couple of attributes.
You can't just cut out values like that. The hclust function needs a perfect
distance matrix to use the calculations.
shortcut is easy : just do f <- f/2*max(f), and all values are b
I can't run your code.
Please, just give me whatever comes on your screen when you run:
dput(q)
On Fri, May 28, 2010 at 10:57 PM, Ayesha Khan
wrote:
> I assume my matrix should look something like this?..
>
> >round(distance, 4)
>P00A P00B M02A M02B P04A P04B M06A M06B P0
I assume my matrix should look something like this?..
>round(distance, 4)
P00A P00B M02A M02B P04A P04B M06A M06B P08A
P08B M10A
P00B 0.9678
M02A 1.0054 1.0349
M02B 1.0258 1.0052 1.2106
P04A 1.0247 0.9928 1.0145 0.9260
P04B 0.9898 0.9769 0.9875 0.9855 0.6075
M06A 1.0159 0.
v <- dput(x,"sampledata.txt")
dim(v)
q <- v[1:10,1:10]
f =as.matrix(dist(t(q)))
distB=NULL
for(k in 1:(nrow(f)-1)) for( m in (k+1):ncol(f)) {
if(f[k,m] <2) distB=rbind(distB,c(k,m,f[k,m]))
}
#now distB looks like this
> distB
[,1] [,2] [,3]
[1,]12 1.6275568
[2,]13 0
Yes Joris. I did try that and it does produce the results. I am now
wondering why I wanted a matrix like structure in the first place. However,
I do want 'f' to contain values less than 2 only. but when i try to get rid
of values greater than 2 by doing N <- (f[f<2], f strcuture disrupts and
hclust
errr, forget about the output of dput(q), but keep it in mind for next time.
f = dist(t(q))
hclust(f,method="single")
it's as simple as that.
Cheers
Joris
On Fri, May 28, 2010 at 10:39 PM, Ayesha Khan
wrote:
> v <- dput(x,"sampledata.txt")
> dim(v)
> q <- v[1:10,1:10]
> f =as.matrix(dist(t(q)))
Hi Ayesha,
I wish to help you, but without a simple self contained example that shows
your issue, I will not be able to help.
Try using the ?dput command to create some simple data, and let us see what
you are doing.
Best,
Tal
Contact
Details:---
Thanks Tal & Joris!
I created my distance matrix distA by using the dist() function in R
manipulating my output in order to get a matrix.
distA =as.matrix(dist(t(x2))) # x2 being my original dataset
as according to the documentaion on dist()
For the default method, a "dist" object, or a matrix (of
As Tal said.
Next to that, I read that column1 (and column2?) are supposed to be seen as
factors, not as numerical variables. Did you take that into account somehow?
It's easy to reproduce the error code :
> n <- NULL
> if(n<2)print("This is OK")
Error in if (n < 2) print("This is OK") : argument
Hi Ayesha,
hclust is a way to go (much better then trying to invent the wheel here).
Please add what you used to create:
distA
And create a sample data set to show us what you did, using
dput
Best,
Tal
Contact
Details:---
Con
i have a matrix with the following dimensions
136 3
and it looks something like
[,1] [,2] [,3]
[1,] 402 675 1.802758
[2,] 402 696 1.938902
[3,] 402 699 1.994253
[4,] 402 945 1.898619
[5,] 424 470 1.812857
[6,] 424 905 1.816345
[7,] 470 905 1.871252
[8,
Hi all.
I have alrge microarray dat set that i would like to analyze using
hierarchical clustering. The problem is when i use the command below,
> hc<- hclust(dist(array), "ave")
i get get this feedback...
Error in as.vector(x, mode) :
cannot coerce type 'closure' to vector of type 'any'
Can som
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