Hi all
I have a problem with the cpquery function in the bnlearn package.
I have constructed a hybrid network (using a mix of continuous and discrete
variables).
The network is named "fitted".
I am interested in predicting the probability of observing a value greater
that a particul
> On Jul 28, 2016, at 7:37 PM, Jun Shen wrote:
>
> Because in reality the NA may appear in one variable but not others. For
> example for ID=1, CL may be NA but not for others, For ID=2, V1 may be NA
> etc. To keep all the IDs and all the variables in one data frame, it's
> inevitable to see som
How did you try to find the answer before posting? Some possibilities might be
go ogling [1] or perusing CRAN to find [2]...
Note that HTML tends to mangle code... please follow the Posting Guide and send
plain text email to this list.
[1] http://bfy.tw/6y3o
[2] https://cran.r-project.org/web/v
Thank you very much Jim. It works beautifully.
Best regards,
On Fri, Jul 29, 2016 at 11:56 AM, Jim Lemon wrote:
> Hi Roslina,
> Try this:
>
> aggbalok_mth[aggbalok_mth$year %in% 2009:2014,]
>
> Jim
>
>
> On Fri, Jul 29, 2016 at 1:12 PM, roslinazairimah zakaria
> wrote:
> > Dear r-users,
>
Hi Roslina,
Try this:
aggbalok_mth[aggbalok_mth$year %in% 2009:2014,]
Jim
On Fri, Jul 29, 2016 at 1:12 PM, roslinazairimah zakaria
wrote:
> Dear r-users,
>
> I would like to extract year from 2009 to 2014 with the corresponding month
> and rain amount.
>
> I tried this:
> aggbalok_mth[aggbalo
Dear r-users,
I would like to extract year from 2009 to 2014 with the corresponding month
and rain amount.
I tried this:
aggbalok_mth[aggbalok_mth$year == 2009:2014, ] but some of the data is
missing.
> dput(aggbalok_mth[aggbalok_mth$year == 2009:2014, ] )
structure(list(month = c(1, 7, 2, 8, 3
Because in reality the NA may appear in one variable but not others. For
example for ID=1, CL may be NA but not for others, For ID=2, V1 may be NA
etc. To keep all the IDs and all the variables in one data frame, it's
inevitable to see some NA
On Thu, Jul 28, 2016 at 10:22 PM, Jeff Newmiller
wrot
Why not remove it yourself before passing it to those functions?
--
Sent from my phone. Please excuse my brevity.
On July 28, 2016 5:51:47 PM PDT, Jun Shen wrote:
>Dear list,
>
>I write a small function to calculate multiple stats on multiple
>variables
>and export in a format exactly the way I
The trouble is getting the figs to look right for different users who
happen to have different display settings. Nearly all my users will be on
MS Windows, and, for the parameter set you gave, the figs will look
different, depending on which "display size" parameter value is active and
the screen r
Hi Dan
This is one of the features/problems of proportional graphing – getting it
exactly right.
Anyway good graphs usually take time
I wonder if aspect in both base and lattice may be of use.
Also I have a default par settings for base graphics where I sometimes change
the mai comm
Dear list,
I write a small function to calculate multiple stats on multiple variables
and export in a format exactly the way I want. Everything seems fine until
NA appears in the data.
Here is my function:
do.stats <- function(data, stats.func, summary.var)
as.data.frame(signif(sappl
Hi Gang,
This is one way:
gangdat<-read.table(text="subject QMemotion yi
s1 75.1017 neutral -75.928276
s2 -47.3512 neutral -178.295990
s3 -68.9016 neutral -134.753906
s1 17.2099 negative -104.168312
s2 -53.1114 negative -182.373474
s3 -33.0322 negative -137.
Although others may be better able to decipher your messed up html
post than I, you are more likely to get a helpful response if you
**follow the posting guide** and post in plain text only, using
?dput() to input example data in a form that makes it easy for R
helpers to input your data, scrutiniz
Hi R users,
I would like to convert a string into a data frame by creating a separator
(ie pipe) between each potential fields (then using *read.table* function).
ie. Here is the dummy input data for illustration
(4 x 5)
Date Type Description
Thanks, Duncan. This is close to what I was looking for. But I'm not using
lattice. And the fontsize$text and fontsize$points are independent of
display settings in Windows (screen resolution and 'size of objects').
I need to make my graphs [esp. placement and size of text(), mtext()] look
good re
One thing to watch out for are there always two samples (one of each type)
for each subject? You had better sort by the emotion to make sure that
when you do the difference, it is always with the data in the same
order.Here is an example of some of these cases where they are ignored:
> libra
Hi Jim and Jeff,
Thanks for the quick help!
Sorry I didn't state the question clearly: I want the difference
between 'neutral' and 'negative' for each subject. And another person
offered a solution for it:
aggregate(cbind(QM, yi) ~ subject, data = mydata, FUN = diff)
On Thu, Jul 28, 2016 at 4:
What represents the difference when multiple values are present? sd?
--
Sent from my phone. Please excuse my brevity.
On July 28, 2016 1:40:16 PM PDT, Gang Chen wrote:
>With the following data in data.frame:
>
>subject QMemotion yi
> s1 75.1017 neutral -75.928276
> s2 -47.3512
Not sure what you mean by "nice way", but here is a dplyr solution:
> library(dplyr)
> mydata <- read.table(text = "subject QMemotion yi
+s1 75.1017 neutral -75.928276
+s2 -47.3512 neutral -178.295990
+s3 -68.9016 neutral -134.753906
+s1 17.2099 negative -10
Hello,
Just use ?sub.
x <- "35.84375_.100.71875"
y <- sub("_\\.", "_-", x)
Hope this helps,
Rui Barradas
Citando lily li :
> Thanks, but how to get the string like this:
> "35.84375_-100.71875" use the minus sign instead of dot.
>
> On Thu, Jul 28, 2016 at 2:38 PM, jim holtman wrote:
>> ju
add another step: (need to learn about regular expressions)
> a
[1] "X35.84375_.100.71875"
> a.new <- sub("^.", '', a)
> a.new
[1] "35.84375_.100.71875"
> sub("_.", "_-", a.new)
[1] "35.84375_-100.71875"
>
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me wh
Thanks, but how to get the string like this:
"35.84375_-100.71875" use the minus sign instead of dot.
On Thu, Jul 28, 2016 at 2:38 PM, jim holtman wrote:
> just strip off the first character:
>
> > a
> [1] "X35.84375_.100.71875"
> > a.new <- sub("^.", '', a)
> > a.new
> [1] "35.84375_.100.71875"
With the following data in data.frame:
subject QMemotion yi
s1 75.1017 neutral -75.928276
s2 -47.3512 neutral -178.295990
s3 -68.9016 neutral -134.753906
s1 17.2099 negative -104.168312
s2 -53.1114 negative -182.373474
s3 -33.0322 negative -137.420410
I can
just strip off the first character:
> a
[1] "X35.84375_.100.71875"
> a.new <- sub("^.", '', a)
> a.new
[1] "35.84375_.100.71875"
>
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Jul 28, 2016 at 3:51
Hi R users,
I have a string for example 'X35.84375_.100.71875', and I have another
dataframe df that I want to export with the transformed string name
'35.84375_-100.71875' with no extension. How to do this in R? Thanks for
your help.
a = 'X35.84375_.100.71875'
write.table(df, file='', row.names=
Basically using Reduce as an lapply in that example, but I think that was
caused by how people started talking about things in the first place =) But
the point is the accumulator can be anything as far as I can tell.
On Thu, Jul 28, 2016 at 12:14 PM, jeremiah rounds
wrote:
> Re:
> "What I'm tryi
Re:
"What I'm trying to
work out is how to have the accumulator in Reduce not be the same type as
the elements of the vector/list being reduced - ideally it could be an S3
instance, list, vector, or data frame."
Pretty sure that is not true. See code that follows. I would never solve
this task i
Hello R Experts,
I am using the Survey Package in R to do some initial descriptive stats for my
dissertation. With the outputs for both the svymean and the barplot, I would
like the value labels to be displayed for the variable-it would make the
descriptive statistics much easier to interpret.
Er, I failed to include the step to write the repaired data to a file...
fnamenobom <- "nobom.xml"
cat( paste( txt, collapse="\n" ), file=fnamenobom )
xmlfile <- xmlTreeParse( fnamenobom )
--
Sent from my phone. Please excuse my brevity.
On July 28, 2016 11:20:23 AM PDT, Jeff Newmiller
wrot
On Thu, Jul 28, 2016 at 2:46 PM, Arthur Rodrigues Stilben
wrote:
>
>
> Em 28-07-2016 15:45, Arthur Rodrigues Stilben escreveu:
>>
>> Sarah,
>>
>> First of all, thanks for reply.
>>
>> Second, It really works, but in fact I would to like to set individuals
>> universe groups for each fuzzy_variable
Thanks for reply, Bert.
Em 28-07-2016 15:31, Bert Gunter escreveu:
Below.
-- 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 Thu, Jul 28, 2016 at 11:11
Em 28-07-2016 15:45, Arthur Rodrigues Stilben escreveu:
Sarah,
First of all, thanks for reply.
Second, It really works, but in fact I would to like to set
individuals universe groups for each fuzzy_variable. Something like this:
> test1 = fuzzy_variable( a = fuzzy_trapezoid( corners = c( 0
Thanks very much Don! You alternative method does make it much faster.
However I think I made a mistake in my previous coding. Sorry.
Instead of checking whether each row sum is larger than 8, what I wanted is
to
check whether each column sum is lager than 8. See the code below.
So in this case,
Below.
-- 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 Thu, Jul 28, 2016 at 11:11 AM, Arthur Rodrigues Stilben
wrote:
> Sorry, I forgot to mention:
>
>
As Jeff suggested, I read the help for you.
Based on the examples, you need:
## set universe
sets_options("universe", seq(from = 0, to = 10, by = 0.1))
teste2 = fuzzy_variable( a = fuzzy_trapezoid( corners = c( 0, 1, 2, 3 ) ) )
comparing
plot(teste) # complete with Warning
and
pl
Please keep the list included in the thread (e.g. reply-all?).
I looked at the file and agree that it looks like xml with a utf8 byte order
mark and Unix line endings, which means it is not XLS and it is not XLSX (which
is a zipped directory of xml files with DOS line endings). Excel complains b
Sorry, I forgot to mention:
> install.packages("sets")
...
> library(sets)
> teste = fuzzy_variable( a = fuzzy_trapezoid( corners = c( 0, 1, 2, 3
) ), universe = seq( from = 0, to = 10, by = 0.1 ) )
> teste
A fuzzy variable with values: a, universe
The ideia is to set the universe group for th
I am not as familiar with those other languages as I should be, but
in both examples aren't you using 'map' to call 'reduce' on each
component of a list? In R the basic mapping function is lapply.
> data <- list(one = c(1, 1), three = c(3), two = c(2, 2))
> rData <- lapply(X=data,
FUN=functio
This is a good example to illustrate that R is a vectorized language, and
programming concepts that would work in fortran, C, or any of many other
languages are not always effective in R.
The vectorized method below has produced the same value for 'p' in every
example I¹ve tried, and is much faste
Yes, I meant confirmatory factor analysis, I'm sorry if I wasn't clear. I
was doing my analyses in lavaan and saw that there are several robust
options (MLM, MLMVS, MLMV, MLF, MLR), but I wasn't sure about their
specifics, so that was what I was actually asking about.
I will definitely consult the
I am trying to model data with multivariate Hawkes distribution. Take the
below example. I am able to compute likelihood but don't know how to
maximize it.
library(hawkes)
lambda0 <- c(0.2,0.2)
alpha <- matrix(c(0.5,0,0,0.5),byrow=TRUE,nrow=2)
beta<- c(0.7,0.7)
history <- simulateHawkes(lamb
XLS has nothing to do with XML. The shift from XLS to XLSX/XLSM formats was
where XML was introduced. You might occasionally find mislabelled files that
seem to work anyway, but there is a significant difference inside true XLS
files.
Use a package designed to handle your data format. There ar
This appears to be a question about a contributed package, though you have not
specified which one (so your example code is not reproducible).
Be warned that I have never seen discussion of fuzzy logic on this list, so any
help you get here is likely to be from someone reading the documentation
The ishares website has the S&P 500 stocks you can download as a XLS file,
which opens fine in Excel, but I am not able to open it in R due to what seems
to be invalid XML formatting. I tried using XLConnect and XML as shown below.
Does anyone know a workaround or can point out what I am doin
Hello, everyone!
Is there a way to set universe group for each fuzzy_variable? I try to do this:
> teste = fuzzy_variable( a = fuzzy_trapezoid( corners = c( 0, 1, 2, 3 ) ),
> universe = seq( from = 0, to= 10, by = 0.1 ) )
But when I print teste:
> teste
A fuzzy variable with values: a, univer
Hi Georg,
it is difficult to figure out what happens between your expectation and the
outcome if we cannot see a minimal dataset.
Based on your description I did this
library(tidyr)
library(dplyr)
test_df <- data_frame(channel = LETTERS[1:5], unit = letters[1:5], custID =
c(1:5), dummy = 1)
tes
On 28/07/2016 7:19 AM, Joanne Thomas wrote:
Hi,
Apologies if something similar has come up but I have been unable to find it in
the archive.
The company I work for has an issue with a package they've developed called
ESVP. This is dependent on OpenSM, which I'm told is already somewhat delica
Ulrik - many thanks for your reply.
I'm aware of many simple solutions as the one you suggest, both iterative
and functional style - but I'm trying to learn how to bend Reduce() for the
purpose of using it in more complex processing tasks. What I'm trying to
work out is how to have the accumulator
Hi,
Apologies if something similar has come up but I have been unable to find it in
the archive.
The company I work for has an issue with a package they've developed called
ESVP. This is dependent on OpenSM, which I'm told is already somewhat delicate
with respect to calls on rJava. When tryin
Hi Stefan,
in that case,lapply(data, length) should do the trick.
Best wishes,
Ulrik
On Thu, 28 Jul 2016 at 12:57 Stefan Kruger wrote:
> David - many thanks for your response.
>
> What I tried to do was to turn
>
> data <- list(one = c(1, 1), three = c(3), two = c(2, 2))
>
> into
>
> result <-
Hi Georg,
it's hard to tell without a reproducible example.
Should spread really merge elements? Does spread know anything about
CustID? Maybe you need to make a useful key of the CustIDs first and spread
on that?
Maybe I'm all off, because I'm really just guessing.
Best,
Ulrik
On Thu, 28 Jul
David - many thanks for your response.
What I tried to do was to turn
data <- list(one = c(1, 1), three = c(3), two = c(2, 2))
into
result <- list(one = 2, three = 1, two = 2)
that is creating a new list which has the same names as the first, but
where the values are the vector lengths.
I kno
Hi All,
I need to spread a data.frame on 2 variables, e. g. "channel" and "unit".
If I do it in two steps spreads keeps all cases that does not look like
the one before although it contains the same values for a specific case.
Here is what I have right now:
-- cut --
test1$dummy <- 1
test2 <-
I think you need someone who speaks your own language. Have a look at
https://www.facebook.com/groups/KoreaRUsers/
- Peter D.
> On 28 Jul 2016, at 08:54 , 박희국 wrote:
>
> Dear Sir/Madam
>
>
> Dear
>
> Hello, I am a student in South Korea, and I would like to analysis genome
> converge and
Dear R users:
I'm happy to announce the first CRAN release of the clubSandwich package:
https://cran.r-project.org/web/packages/clubSandwich
clubSandwich provides several variants of the cluster-robust variance
estimator for ordinary and weighted least squares linear regression models,
including
Dear Sir/Madam
Dear
Hello, I am a student in South Korea, and I would like to analysis genome
converge and fit curve using nls.lm in R. But I am a very beginner user R. So
this is not easy for me. I try to find and implement many example and
tutorial. But it dose not help for me.
And I fina
Hi,
I have a list of N categories that a user can click on. Lets say there are
K such users totally. I have the past 3 months data which tells which user
has clicked on which category on which date for how many times. For ex -
{20th June 2016 : [10,15,12,15]} this dict is for a particular user and
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