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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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
.
From: r-help-boun...@r-project.org [r-help-boun...@r-project.org] On
Behalf Of tdm [ph...@philbrierley.com]
Sent: 31 October 2009 16:53
To: r-help@r-project.org
Subject: [R] Logistic and Linear Regression Libraries
Hi all,
I'm trying to discover the options
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 why the lrm predict
/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
function?
my_model - lm(x
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,
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.
text(x,y, labels = x)
gives me the value of x.
text(x,y, labels
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 -
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
}
}
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$species] = 1 #this
the following
which(names(iris)=='Species')
[1] 5
HTH
Schalk Heunis
On Mon, Oct 12, 2009 at 8:53 AM, tdm ph...@philbrierley.com wrote:
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than
Hi,
I am passing a data frame and field name to a function. I've figured out how
I can create the formula based on the passed in field name, but I'm
struggling to create a vector based in that field.
for example if I hard code with the actual field name
Y = df$Target, everything works fine.
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 $,
d = data.frame(a=1:3)
mytarget = a
d[[mytarget]]
HTH,
baptiste
2009/10/11 tdm ph...@philbrierley.com:
Hi
Hi,
I am passing a data frame and field name to a function. I've figured out how
I can create the formula based on the passed in field name, but I'm
struggling to create a vector based in that field.
for example if I hard code with the actual field name
Y = df$Target, everything works fine.
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