R version 2.9.2 (2009-08-24) - for windows
> library(SOM)
Error in library(SOM) : there is no package called 'SOM'
Where can I get the SOM library from?
Thanks in advance
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OK, I think I've figured it out, the predict of lrm didn't seem to pass it
through the logistic function. If I do this then the value is similar to
that of lm. Is this by design? Why would it be so?
1 / (1 + Exp(-1 * 3.38)) = 0.967
tdm wrote:
>
>
> Anyway, do you know
f disappointing, I think the answer to your question is
> "no". You will need to seek out the algorithms from the published
> information on them individually.
>
> W.
>
> From: r-help-boun...@r-project.org [r-help-boun...@r-project.
Hi all,
I'm trying to discover the options available to me for logistic and linear
regression. I'm doing some tests on a dataset and want to see how different
flavours of the algorithms cope.
So far for logistic regression I've tried glm(MASS) and lrm (Design) and
found there is a big differenc
Thanks Jim - that worked.
Jim Lemon-2 wrote:
>
> On 10/20/2009 01:10 PM, tdm wrote:
>> I have 2 vectors, x and y and have done an xy plot.
>>
>> I want to plot the label (name?) of the vector on the plot rather than
>> the
>> value.
>>
>> tex
I have 2 vectors, x and y and have done an xy plot.
I want to plot the label (name?) of the vector on the plot rather than the
value.
text(x,y, labels = x)
gives me the value of x.
text(x,y, labels = labels(x))
gives me something like c("text1","text2"..) plotted for each point
text(x,y, la
usdm09.freeforums.org/post34.html#p34
tdm wrote:
>
> I have build a model but want to then manipulate the coefficients in some
> way.
>
> I can extract the coefficients and do the changes I need, but how do I
> then put these new coefficients back in the model so I can use the predict
&g
I have build a model but want to then manipulate the coefficients in some
way.
I can extract the coefficients and do the changes I need, but how do I then
put these new coefficients back in the model so I can use the predict
function?
my_model <- lm(x ~ . , data=my_data)
my_scores <- predict(my
Hi,
Can someone please give me a pointer as to how I can set values of an array?
Why does the code below not work?
my_array <- array(dim=c(2,2))
my_array[][] = 0
my_array
[,1] [,2]
[1,]00
[2,]00
for(i in seq(1,2,by=1)){
for(j in seq(1,2,by=1)){
my_array[i][j] = 5
}
}
te:
>
> Hi Phil
> Try the following
>> which(names(iris)=='Species')
> [1] 5
>
> HTH
> Schalk Heunis
>
> On Mon, Oct 12, 2009 at 8:53 AM, tdm wrote:
>
>>
>> Hi,
>>
>> How do I access the index number of a field given I only kn
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
> colprob <- array(dim=NCOL(iris))
> for(i in 1:NCOL(iris)){colprob[i]=0.5}
> colprob[iris$species] = 1 #t
Thanks, just the clue I needed, worked a treat.
baptiste auguie-5 wrote:
>
> Hi,
>
> I think this is a case where you should use the ?"[[" extraction
> operator rather than "$",
>
> d = data.frame(a=1:3)
> mytarget = "a"
> d[[
Hi,
I am passing a data frame and field name to a function. I've figured out how
I can create the formula based on the passed in field name, but I'm
struggling to create a vector based in that field.
for example if I hard code with the actual field name
Y = df$Target, everything works fine.
bu
Hi,
I am passing a data frame and field name to a function. I've figured out how
I can create the formula based on the passed in field name, but I'm
struggling to create a vector based in that field.
for example if I hard code with the actual field name
Y = df$Target, everything works fine.
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