long
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Tue, Apr 19, 2016 at 5:29 PM, Michael Artz <michaelea...@gmail.com>
> wrote:
> > Again, IQR returns two both a .25 and a .75 value and it
; Cheers,
> > Bert
> > Bert Gunter
> >
> > "The trouble with having an open mind is that people keep coming along
> > and sticking things into it."
> > -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
> >
> >
> &g
wdunlap tibco.com
>
> On Tue, Apr 19, 2016 at 12:18 PM, Michael Artz <michaelea...@gmail.com>
> wrote:
>
>> Oh thanks for that clarification Bert! Hope you enjoyed your coffee! I
>> ended up just using the transform argument in the ddply function. It
&g
a <- data.frame(groupColumn=rep(1:5,1:5), col1=2^(0:14))
> >>> myIqr <- function(x) {
> >>> paste(round(quantile(x,0.25),0),round(quantile(x,0.75),0),sep="-")
> >>> }
> >>> ddply(data, ~groupColumn, summarise, col1_myIqr=myIqr(col1),
&
ot;)
> is not a function, it is an expression. ddplyr wants functions.
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Tue, Apr 19, 2016 at 7:56 AM, Michael Artz <michaelea...@gmail.com>
> wrote:
>
>> That didn't work Jim!
&g
mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Tue, Apr 19, 2016 at 7:56 AM, Michael Artz <michaelea...@gmail.com>
> wrote:
> > That didn't wor
> .col3_Range=iqr(datat$tenure)
>
> Jim
>
>
>
> On Tue, Apr 19, 2016 at 11:15 AM, Michael Artz <michaelea...@gmail.com>
> wrote:
> > Hi,
> > I am trying to show an interquartile range while grouping values using
> > the function ddply(). So my funct
Hi,
I am trying to show an interquartile range while grouping values using
the function ddply(). So my function call now is like
groupedAll <- ddply(data
,~groupColumn
,summarise
,col1_mean=mean(col1)
,col2_mode=Mode(col2)
olf <- predict(fittedModel,
> newdata=newdata)
>
>
> Bill Dunlap
> TIBCO Software
> wdunlap tibco.com
>
> On Fri, Apr 15, 2016 at 3:09 PM, Michael Artz <michaelea...@gmail.com>
> wrote:
>
>> I need the output to have groups and the probability any
t; decision rules "clusters" and probability attached to them. The examples
>> are sort of similar. You just provided links to general info about trees.
>>
>>
>>
>> Sent from my Verizon, Samsung Galaxy smartphone
>>
>>
>> ---- Original m
gers.com/a-brief-tour-of-the-trees-and-forests/
You can get the same kind of information from random forests, but it's
less straightforward. If you want a clear set of rules as in your golf
example, then you need rpart or similar.
Sarah
On Wed, Apr 13, 2016 at 6:02 PM, Michael Artz <michaelea.
t;
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, Apr 13, 2016 at 2:11 PM, Michael Artz <michaelea...
Also that being said, just because random forest are not the same thing as
decision trees does not mean that you can't get decision rules from random
forest.
On Wed, Apr 13, 2016 at 4:11 PM, Michael Artz <michaelea...@gmail.com>
wrote:
> Ok is there a way to do it with decision tree
eb resources, too.
>
> Cheers,
> Bert
>
>
> Bert Gunter
>
> "The trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed,
Hi I'm trying to get the top decision rules from a decision tree.
Eventually I will like to do this with R and Random Forrest. There has to
be a way to output the decsion rules of each leaf node in an easily
readable way. I am looking at the randomforrest and rpart packages and I
dont see
Hi I am having a problem with plot () and ggplot (). When I call one of
these functions, the plotting area starts to look as though it is working,
but nothijg ever is visible. Unless it was a dendrogram. Woth the bar
chart, the plotting area just had an x and y axis and nothing else. I tried
a
Hi,
I already have a dissimilarity matrix and I am submitting the results to
the elbow.obj method to get an optimal number of clusters. Am I reading
the below output correctly that I should have 17 clusters?
code:
top150 <- sampleset[1:150,]
{cluster1 <- daisy(top150
,
I don't get it, I thought the double index was to indicate and individual
element within a column(vector)?
I will stop using data.frame, thanks a lot!
On Thu, Apr 7, 2016 at 9:29 PM, David Winsemius <dwinsem...@comcast.net>
wrote:
>
> > On Apr 7, 2016, at 6:46 PM, Michael
== 'blue'
>
> Hadley
>
> On Thu, Apr 7, 2016 at 6:52 AM, David Barron <dnbar...@gmail.com> wrote:
> > ifelse is vectorised, so just use that without the loop.
> >
> > colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
> >
> > David
>
== 'blue' + 0
>
> Hope this helps,
>
> Rui Barradas
>
>
> Citando David Barron <dnbar...@gmail.com>:
>
> ifelse is vectorised, so just use that without the loop.
>
> colordata$response <- ifelse(colordata$color == 'blue', 1, 0)
>
> David
>
> On
data.frame.$columnToAdd["CurrentColumnName" == "ConditionMet"] <- 1
Can someone please explain to me why the above command gives all NAs to
columnToAdd? I thought this was possible in R to do logical expression in
the index of a data frame
[[alternative HTML version deleted]]
use you are giving a
> vector of numbers (the answer f gave you) to the second argument of lapply
> instead of a function.
> --
> Sent from my phone. Please excuse my brevity.
>
> On April 7, 2016 7:31:18 AM PDT, Michael Artz <michaelea...@gmail.com>
> wrote:
>>
>> I
response calculation. The result is a data
> frame (list of columns) with no column names, so I give the new columns
> names based on the old column names. You could choose different names, e.g.
>
> names(responses) <- paste0( "response", 1:2 )
>
> but you have to be caref
gt;
> On 7 April 2016 at 12:41, Michael Artz <michaelea...@gmail.com> wrote:
>
>> Hi I'm not sure how to ask this, but its a very easy question to answer
>> for
>> an R person.
>>
>> What is an easy way to check for a column value and then assigne a new
&g
Hi I'm not sure how to ask this, but its a very easy question to answer for
an R person.
What is an easy way to check for a column value and then assigne a new
column a value based on that old column value?
For example, Im doing
colordata <- data.frame(id = c(1,2,3,4,5), color = c("blue",
Maybe it's not the article itself for sale. Sometimes a company will
charge a fee to have access to its knowledge base. Not because it owns all
of the content, but because the articles, publications, etc have been
tracked down and centralized. This is also the whole idea behind paying a
company
Thank you everyone I got it!
I needed to install munsell was all. I was giving a typo when I tried to
install munsell
On Mon, Mar 28, 2016 at 12:01 PM, Michael Artz <michaelea...@gmail.com>
wrote:
> Thanks. SessionInfo() did not show it.
>
> This is the error when I tr
‘ggplot2’
On Mon, Mar 28, 2016 at 11:57 AM, Jeff Newmiller <jdnew...@dcn.davis.ca.us>
wrote:
> Post plain text only please.
>
> Are you sure it loaded? Verify with sessionInfo()...
> --
> Sent from my phone. Please excuse my brevity.
>
> On March 28, 2016 9:21:56 AM PDT,
Hi,
I am getting the error,
Error: could not find function "createDataPartition"
when I do the code
dataFrame_data <- createDataPartition(data$colA, p=.7, list=FALSE)
even though I have run already
install.packages("caret", dependencies = c("Depends", "Imports",
"Suggests"))
and
016, at 1:27 PM, Michael Artz <michaelea...@gmail.com>
> wrote:
> >
> > Hi,
> > I am trying to use the summary from the glm function as a data source.
> I
> > am using the call sink() then
> > summary(logisticRegModel)$coefficients then sink().
>
> S
Hi,
I am trying to use the summary from the glm function as a data source. I
am using the call sink() then
summary(logisticRegModel)$coefficients then sink(). The independent
variables are categorical and thus there is always a baseline value for
every category that is omitted from the glm
.2497
tenure -0.0702813 0.0077113 -9.114 < 2e-16
***
TotalCharges 0.0004276 0.874 4.892 9.97e-07
***
On Thu, Mar 10, 2016 at 4:05 PM, David Winsemius <dwinsem...@comcast.net>
wrote:
>
> > On Mar 10, 2016, at 8:08 AM, Michael Artz &
HI all,
I have the following error -
> resultVector <- predict(logitregressmodel, dataset1, type='response')
Warning message:
In predict.lm(object, newdata, se.fit, scale = 1, type = ifelse(type == :
prediction from a rank-deficient fit may be misleading
I have seen on internet that there
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