It's not clear whether your numbers are tab or space-separated, I will
assume space-separated. My lowtech (and not R) solution would be to
dump the output into a text file (call it data.in), then run a sed
command to first replace two initial spaces from each line, then
replace initial spaces with
What would be the sophisticated R method for reading the data shown below
into a list? The data is output from a numerical model. Pasting the
second block of example R commands (at the end of the message) results in a
failure ("Error in scan...line 2 did not have 6 elements"). I no doubt
could c
Given these two kinds of data set, and data set2 was obtained through the
weighted Euclidean formula. Can we estimate the weight parameter for each
variable in R based on the steepest descent method ? Thanks very much.
data set 1:
9 164 78 0 0 32.8 0.148 45
4 134 72 0 0 23.8 0.277 60
5 166 72 19
Thanks for the reply.
> I suppose you can set up a contrast matrix that would make the intercept
> equal to the overall mean, but the definition would depend on the group sizes.
Any suggestion on how to set up one?
Erlis
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R-help@r-project.org mailin
On Thu, 10 Nov 2016, danilo.car...@uniparthenope.it wrote:
Thank you for your hints, now the goodness of fit test provides me good
results, but surprisingly for me the three-component model turns out to be
worse than the two-component one (indeed, I focused on the three-component
mixture becau
> On 10 Nov 2016, at 16:30 , erlis ruli wrote:
>
> Here comes the questions. Is it possible to modify Helmert contrasts in order
> to use weighted means instead of means of means?
I think not. Not in general. I suppose you can set up a contrast matrix that
would make the intercept equal to t
Hello,
Just read the help page ?paste, in particular the argument collapse.
x <- LETTERS[1:9]
paste(x, collapse = "")
Hope this helps,
Rui Barradas
Em 10-11-2016 17:14, Ferri Leberl escreveu:
Dear All,
If I have a vector V consisting of 9 strings — how can I paste them into a
single string
Dear All,
If I have a vector V consisting of 9 strings — how can I paste them into a
single string without programming a loop?
Thank you in advance!
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Given these two kinds of data set, and data set2 was obtained through the
weighted Euclidean formula. Can we estimate the weight parameter for each
variable in R based on the steepest descent method ? Thanks very much.
data set 1:
9 164 78 0 0 32.8 0.148 45
4 134 72 0 0 23.8 0.277 60
5 166 72 19
Hi all!
Suppose that we have got a response y and an unbalanced treatment x with three
levels or groups.
The treatment is unbalanced by design. Indeed, the first group has 3
replications and the other two have two replications each.
For instance, in R the data might look like this:
y = c(66.
I think you answered your own question. For loops are not a boogeyman... poor
memory management is.
Algorithms that are sensitive to evaluation sequence are often not very
re-usable, and certainly not parallelizable. If you have a specific algorithm
in mind, there may be some advice we can give
You are mistaken. apply() is *not* vectorized. It is a disguised loop.
For true vectorization at the C level, the answer must be no, as the
whole point is to treat the argument as a whole object and hide the
iterative details.
However, as you indicated, you can always manually randomize the
index
nBuyMat <- data.frame(matrix(rnorm(28), 7, 4))
nBuyMat
nBuy <- nrow(nBuyMat)
sample(1:nBuy, nBuy, replace=FALSE)
sample(1:nBuy)
sample(nBuy)
?sample
apply(nBuyMat[sample(1:nBuy,nBuy, replace=FALSE),], 1, function(x) sum(x))
apply(nBuyMat[sample(nBuy),], 1, function(x) sum(x))
The defaults for sa
Is there a way to use vectorization where the elements are evaluated in a
random order?
For instance, if the code is to be run on each row in a matrix of length nBuy
the following will do the job
for (b in sample(1:nBuy,nBuy, replace=FALSE)){
}
but
apply(nBuyMat, 1, function(x))
will be run
Thank you for your hints, now the goodness of fit test provides me
good results, but surprisingly for me the three-component model turns
out to be worse than the two-component one (indeed, I focused on the
three-component mixture because the two-component one exhibits a low
p-value).
In a
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