> On May 17, 2017, at 6:43 AM, ลห พงพง wrote:
>
> I have written a user written splitting function recently, but I do not know
> how to use my splitting criteria to predict the test data. If I use the
> function ‘predict’, there will return an error message, please give me some
> helpful advi
Hi Omar,
You may want to try subsetting your data and then passing each morsel
to be plotted as a violin plot, either as separate calls to ggplot or
directly to a violin plotting routine.
vioplot (vioplot)
violin_plot (plotrix)
Jim
On Thu, May 18, 2017 at 5:59 AM, Abdelrahman, Omar (RER)
wrote:
Thank you, curly quotes got me! I was able to subset the data and produce the
violin plot. Now, is there a way to generate multiple plots separately (no
facets)? With so many levels of each variable, I am trying to avoid doing it
iteratively. Neither ggplot2 books nor web searches have yielded a
Thanks again
RE: "so all the more reason to give us an example that we can run to trigger
the same error." Are you asking for an example of the data? Below is a "small"
example, but with so many levels of the different variables I am not sure it
can be useful.
STATION Geo Wshed DATE
I have written a user written splitting function recently, but I do not know
how to use my splitting criteria to predict the test data. If I use the
function ‘predict’, there will return an error message, please give me some
helpful advice.
__
R-help@r
Hi All,
I've a data-set on product sub-product matrix on which I'm doing multiple
calculation, but unfortunately using nested loops, the programme is taking
long time to execute. Can anyone help me how to get rid of the following
jungle? Any direction would be helpful.
GA <- "India"
verticle <- "
Hi Thambu,
You must have an internet connection, otherwise you couldn't access
CRAN. Have you tried using "install.packages" (see the help page) from
within R? Your message isn't clear whether you have used that function
or tried to download the package from a Web browser and then install
it from t
> On May 17, 2017, at 1:01 PM, Davide Piffer wrote:
>
> Thanks! This gets closer to the solution but a small problem remains.
> I get 2 rows and only one column, whereas I need a 2x2 matrix (like a
> contingency table for Fisher's exact test).Also another issue is it
> repeats the first number o
Thanks! This gets closer to the solution but a small problem remains.
I get 2 rows and only one column, whereas I need a 2x2 matrix (like a
contingency table for Fisher's exact test).Also another issue is it
repeats the first number of the column, instead of using all 4.
For example, first vector o
Not really enough info here since you don't specify much about the data frame
or how the results should be provided, but maybe something like this:
y <- data.frame(matrix(1:100, 10, 10))
y.mat <- lapply(y, matrix, nrow=2)
str(y.mat)
List of 10
$ X1 : int [1:2, 1:5] 1 2 3 4 5 6 7 8 9 10
$ X2 : i
> On May 17, 2017, at 12:18 PM, Davide Piffer wrote:
>
> I need to convert each vector of a dataframe into a matrix with 2 rows
> and 2 columns (i.e. contingency table).
> Note I don't want to convert the entire df into a matrix! I want to
> apply a function that converts each 4 elements vector
I need to convert each vector of a dataframe into a matrix with 2 rows
and 2 columns (i.e. contingency table).
Note I don't want to convert the entire df into a matrix! I want to
apply a function that converts each 4 elements vector of a df into a 2
x 2 matrix.
I wrote something like this, but it
"follow-on" is one of the main reasons I stopped work on optimx and refactored
to optimr/optimrx, where I
separated this functionality into the polyopt() function. optimr has just a few
solvers, while optimrx is used to
add them as I get round to doing it, but it's on R-forge. Mainly a matter of
Hi,
I would like to know if some of you have a solution for this problem:
I use optimx (from package optimx) to fit the parameters of a model
(complex model based on several imbricated exponential functions).
I use the two methods : method = c("Nelder-Mead", "BFGS") with the options:
control
You should consult a linear models text, but, assuming I have
correctly understood your post, the procedure is this (the translation
to R code is trivial, and I leave it to you):
Let y be the response variable, P1 be the first set of predictors and
z be your new predictor to be added.
1. regress
Here is an example that works... a reproducible example always includes
code AND enough sample data to exercise the code:
dta <- read.table( text=
"STATIONGeo Wshed DATEPARAMETER RESULT
BB36Bay C-100 1/10/2013 'Phosphorus
Dear Members & Experts,
Since the Dictionary () function is no longer available with the tm package.
How do I use other functions to do the same as below? I want to capture a list
of specific terms from a corpus. By example, if my corpus has 102 files. I want
to see a list with occurrences of
> On 17 May 2017, at 12:31 , Henric Winell wrote:
>
> On 2017-05-17 09:42, Patrick Connolly wrote:
>
>> After installing R-3.4.0 I ran 'make check' which halted here:
>> $ > tail reg-tests-1d.Rout.fail -n 16
>
> This problem was brought up on the R-devel list early this morning. See
> https
On 2017-05-17 09:42, Patrick Connolly wrote:
After installing R-3.4.0 I ran 'make check' which halted here:
$ > tail reg-tests-1d.Rout.fail -n 16
This problem was brought up on the R-devel list early this morning. See
https://stat.ethz.ch/pipermail/r-devel/2017-May/074275.html
Henric Wi
Hello, Urs,
you may have seen Wolfgang Viechtbauer's answer already which offers
an R-technical solution, but this may leave the mathematical grounds
of linear models. See inline below for my concern and a hint.
Am 17.05.2017 um 09:12 schrieb Urs Kleinholdermann:
Dear list members,
I want to a
After installing R-3.4.0 I ran 'make check' which halted here:
$ > tail reg-tests-1d.Rout.fail -n 16
> ## format()ing invalid hand-constructed POSIXlt objects
> d <- as.POSIXlt("2016-12-06"); d$zone <- 1
> tools::assertError(format(d))
> d$zone <- NULL
> stopifnot(identical(format(d),"2016-12-0
You could use an offset term. An example:
n <- 100
x1 <- rnorm(n)
x2 <- rnorm(n)
x3 <- rnorm(n)
y <- 0 + .2 * x1 - .5 * x2 + .3 * x3 + rnorm(n)
res1 <- lm(y ~ x1 + x2)
summary(res1)
res2 <- lm(y ~ 1 + offset(coef(res1)[2] * x1 + coef(res1)[3] * x2))
summary(res2) ### identical intercept as in res1
Dear list members,
I want to add a predictor to a linear model without changing the
coefficients of the existing model. How is that done with R?
So if I have a response y and predictors x1, x2, x3 I want to make a model lm1
like
lm1 = lm(y~x1+x2)
After this model is computed I want to add x3 l
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