[R] SOM library - where do I find it
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 -- View this message in context: http://old.nabble.com/SOM-library---where-do-I-find-it-tp26415633p26415633.html Sent from the R help mailing list archive at Nabble.com. __ 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.
Re: [R] Logistic and Linear Regression Libraries
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 give me a values of 3.38? > > -- View this message in context: http://old.nabble.com/Logistic-and-Linear-Regression-Libraries-tp26140248p26141558.html Sent from the R help mailing list archive at Nabble.com. __ 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.
Re: [R] Logistic and Linear Regression Libraries
Hi Bill,
Thanks for you comments. You may be right in that my ability to use the
software may be the problem. I was using lm to fit a model with 'target'
values of 0 or 1. I then discovered there was a lrm model as well, so just
replaced lm with lrm and expected it to be fine. Then I found that the lrm
model was predicting values greater than 1 for logistic regression - what am
I doing wrong?
Initially I was just looking at AUC, which are similar for both models,
although different enough for me to be concerned. My data is highly
correlated so I want to use a Hessian based algorithm rather then a
non-hessian based algorithm, which is why I asked the initial question as to
what other logistic regression models existed in R.
Anyway, do you know why the lrm predict give me a values of 3.38?
model_lr <- lm(as.formula(paste(mytarget, " ~ . ")) , data=df_train)
model_lrA <- lrm(as.formula(paste(mytarget, " ~ . ")) , data=df_train)
scores_lr_test <- predict(model_lr, df_test)
scores_lr_train <- predict(model_lr, df_train)
scores_lrA_test <- predict(model_lrA, df_test)
scores_lrA_train <- predict(model_lrA, df_train)
print("scores")
print(scores_lr_train[1])
print("scoresA")
print(scores_lrA_train[1])
print(colAUC(scores_lr_train,trainY))
print(colAUC(scores_lrA_train,trainY))
print(colAUC(scores_lr_test,testY))
print(colAUC(scores_lrA_test,testY))
[1] "scores"
1
0.9887154
[1] "scoresA"
1
3.389009
[,1]
0 vs. 1 0.9448262
[,1]
0 vs. 1 0.9487878
[,1]
0 vs. 1 0.9346953
[,1]
0 vs. 1 0.9357858
1 of 1[1] ""
Bill.Venables wrote:
>
> glm is not, and never was. part of the MASS package. It's in the stats
> package.
>
> Have you sorted out why there is a "big difference" between the results
> you get using glm and lrm?
>
> Are you confident it is due to the algorithms used and not your ability to
> use the software?
>
> To be helpful, if 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.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 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 difference. Is there a list anywhere detailing the
> options available which details the specific algorithms used?
>
> Thanks in advance,
>
> Phil
>
>
>
> --
> View this message in context:
> http://old.nabble.com/Logistic-and-Linear-Regression-Libraries-tp26140248p26140248.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> 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.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
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> and provide commented, minimal, self-contained, reproducible code.
>
>
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[R] Logistic and Linear Regression Libraries
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 difference. Is there a list anywhere detailing the options available which details the specific algorithms used? Thanks in advance, Phil -- View this message in context: http://old.nabble.com/Logistic-and-Linear-Regression-Libraries-tp26140248p26140248.html Sent from the R help mailing list archive at Nabble.com. __ 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.
Re: [R] plotting labels (not values) on xy plot
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 = labels(x))
>>
>> gives me something like c("text1","text2"..) plotted for each point
>>
>> text(x,y, labels = names(x))
>>
>> gives nothing
>>
>> print(x) gives me
>>
>> [,1]
>> text10.0
>> text20.0
>> text3 -0.029027359
>> text4 -0.088602806
>>
>>
>> so how do 'text1' written rather than the value? I know there is a way
>> but I
>> am just guessing at the moment.
>>
>> Any pointers greatly appreciated.
>>
>>
>>
>>
>>
>>
> Hi Phil,
> It looks like you have a 4x1 matrix. My first guess would be:
>
> text(x,y,rownames(x))
>
> Jim
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide
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[R] plotting labels (not values) on xy plot
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, labels = names(x))
gives nothing
print(x) gives me
[,1]
text10.0
text20.0
text3 -0.029027359
text4 -0.088602806
so how do 'text1' written rather than the value? I know there is a way but I
am just guessing at the moment.
Any pointers greatly appreciated.
--
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Re: [R] modifying model coefficients
Thanks for the tip. I actually had to learn a bit of matrix multiplication and ended up calculating the RMSE this way. scores_lr <- t(coeffs_alldata) %*% t(df_train) rmseTrain <- (mean(((scores_lr)- trainY)^2))^0.5 full script and results here if you are interested. http://ausdm09.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 > function? > > my_model <- lm(x ~ . , data=my_data) > my_scores <- predict(my_model, my_data) > > my_coeffs <- coef(my_model) > > ## here we manipulate my_coeffs > ## and then want to set the my_model > ## coefficients to the new values so we > ## predict using the new values > > my_model.coefs <- my_coeffs ?? how is this done? > > ?? so that this will work with the new coefficients > my_scores_new <- predict(my_model, my_data) > > Any code snippets would be appreciated very much. > > -- View this message in context: http://www.nabble.com/modifying-model-coefficients-tp25952879p25964671.html Sent from the R help mailing list archive at Nabble.com. __ 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.
[R] modifying model coefficients
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_model, my_data) my_coeffs <- coef(my_model) ## here we manipulate my_coeffs ## and then want to set the my_model ## coefficients to the new values so we ## predict using the new values my_model.coefs <- my_coeffs ?? how is this done? ?? so that this will work with the new coefficients my_scores_new <- predict(my_model, my_data) Any code snippets would be appreciated very much. -- View this message in context: http://www.nabble.com/modifying-model-coefficients-tp25952879p25952879.html Sent from the R help mailing list archive at Nabble.com. __ 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.
[R] populating an array
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
}
}
Warning messages:
1: In my_array[i][j] = 5 :
number of items to replace is not a multiple of replacement length
2: In my_array[i][j] = 5 :
number of items to replace is not a multiple of replacement length
my_array
[,1] [,2]
[1,]50
[2,]50
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Re: [R] field index given name.
Thanks - would never have guessed that. I eventually got the following to do
what I want...
> colprob <- array(dim=NCOL(iris))
> for(i in 1:NCOL(iris)){
+ colprob[i]=
+ ifelse(names(iris)[i] == 'Species',1,0.5)
+ }
> colprob
[1] 0.5 0.5 0.5 0.5 1.0
Schalk Heunis-2 wrote:
>
> 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 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$species] = 1 #this doesn't work
>> > colprob
>> [1] 0.5 0.5 0.5 0.5 0.5
>>
>>
>>
>>
>>
>> --
>> View this message in context:
>> http://www.nabble.com/field-index-given-name.-tp25851216p25851216.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
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>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide
>> http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
>>
>
> [[alternative HTML version deleted]]
>
> __
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>
>
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[R] field index given name.
Hi,
How do I access the index number of a field given I only know the field
name?
eg - I want to set the probability of the field 'species' higher than the
other fields to use in sampling.
> colprob <- array(dim=NCOL(iris))
> for(i in 1:NCOL(iris)){colprob[i]=0.5}
> colprob[iris$species] = 1 #this doesn't work
> colprob
[1] 0.5 0.5 0.5 0.5 0.5
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Re: [R] field names as function parameters
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 "$",
>
> d = data.frame(a=1:3)
> mytarget = "a"
> d[[mytarget]]
>
>
> HTH,
>
> baptiste
>
> 2009/10/11 tdm :
>>
>> Hi,
>>
>> I am passing a data frame and field name to a function. I've figured out
>> how
>> I can create the formula based on the passed in field name, but I'm
>> struggling to create a vector based in that field.
>>
>> for example if I hard code with the actual field name
>>
>> Y = df$Target, everything works fine.
>>
>> but if I use the passed in parameter name, it doesn't give me what I
>> want,
>>
>> Y = df$mytarget
>>
>>
>> Here is the function,
>>
>> # trying to pass field name to a function
>> logistictest <- function(df,mytarget)
>> {
>>
>> #library for AUC calculation
>> library(caTools)
>>
>> #build logistic model
>> mytarget <- deparse(substitute(mytarget))
>> myformula <- paste(mytarget," ~ .")
>> myformula <- deparse(substitute(myformula))
>> logistic_reg <- glm(myformula , data=df, family=binomial(link="logit"))
>> print("model build OK")
>>
>> #score up
>> scores <- predict(logistic_reg, type="response", df)
>> print("model scored OK")
>>
>> #calc AUC
>> Y = df$mytarget
>>
>> auc <- colAUC(scores,Y)
>> print("auc calculated OK")
>>
>> }
>>
>> logistictest(df=trainset,mytarget=Target)
>>
>>
>> [1] "model build OK"
>> [1] "model scored OK"
>> Error in as.vector(x, mode) : invalid 'mode' argument
>>
>> --
>> View this message in context:
>> http://www.nabble.com/field-names-as-function-parameters-tp25838606p25838606.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> __
>> 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.
>>
>
> __
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[R] field names as function parameters
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$Target, everything works fine.
but if I use the passed in parameter name, it doesn't give me what I want,
Y = df$mytarget
Here is the function,
# trying to pass field name to a function
logistictest <- function(df,mytarget)
{
#library for AUC calculation
library(caTools)
#build logistic model
mytarget <- deparse(substitute(mytarget))
myformula <- paste(mytarget," ~ .")
myformula <- deparse(substitute(myformula))
logistic_reg <- glm(myformula , data=df, family=binomial(link="logit"))
print("model build OK")
#score up
scores <- predict(logistic_reg, type="response", df)
print("model scored OK")
#calc AUC
Y = df$mytarget
auc <- colAUC(scores,Y)
print("auc calculated OK")
}
logistictest(df=trainset,mytarget=Target)
[1] "model build OK"
[1] "model scored OK"
Error in as.vector(x, mode) : invalid 'mode' argument
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[R] passing field name parameter to function
Hi,
I am passing a data frame and field name to a function. I've figured out how
I can create the formula based on the passed in field name, but I'm
struggling to create a vector based in that field.
for example if I hard code with the actual field name
Y = df$Target, everything works fine.
but if I use the passed in parameter name, it doesn't give me what I want,
Y = df$mytarget
Here is the function,
# trying to pass field name to a function
logistictest <- function(df,mytarget)
{
#library for AUC calculation
library(caTools)
#build logistic model
mytarget <- deparse(substitute(mytarget))
myformula <- paste(mytarget," ~ .")
myformula <- deparse(substitute(myformula))
logistic_reg <- glm(myformula , data=df, family=binomial(link="logit"))
print("model build OK")
#score up
scores <- predict(logistic_reg, type="response", df)
print("model scored OK")
#calc AUC
Y = df$mytarget
auc <- colAUC(scores,Y)
print("auc calculated OK")
}
logistictest(df=trainset,mytarget=Target)
[1] "model build OK"
[1] "model scored OK"
Error in as.vector(x, mode) : invalid 'mode' argument
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