Hi,
I have a data which is stored in sql table and in every minute data
inserting to this table .Now my requirement is I need to plot the live
streaming graph as per below link or PFA. Can anyone help out me how to do
that in R ?
link : http://www.highcharts.com/studies/live-server.htm
T
Let's take a different view of the problem.
Given f=(f_1,...,f_m):R^n -> R^m we want to minimize ||f(x)||.
What distinguishes this from a general minimization problem is that you know
the structure of the
objective function F(x)=||f(x)||² and have the individual constituents f_j.
Make use of that
`stringi::stri_count()`
I know that the `stringr` pkg saves some typing (it wraps the
`stringi` pkg), but you should really just use the `stringi` package.
It has many more very useful functions with not too much more typing.
On Thu, Oct 20, 2016 at 5:47 PM, Jan Kacaba wrote:
> Hello dear R-help
Jan,
Within the stringr package you can find the function str_count().
Mark
R. Mark Sharp, Ph.D.
Director of Primate Records Database
Southwest National Primate Research Center
Texas Biomedical Research Institute
P.O. Box 760549
San Antonio, TX 78245-0549
Telephone: (210)258-9476
e-mail: msh...
Hi again,
Sorry, the text command should read:
text(x=dta$age[samp010],y=dta$fit5[samp010],labels=dta$exposure[samp010],
col=expcol[samp010])
Jim
On Fri, Oct 21, 2016 at 8:50 AM, Jim Lemon wrote:
> Hi mviljama,
> Without knowing what "dta" contains, it's a bit difficult. Here is an example:
>
Hi mviljama,
Without knowing what "dta" contains, it's a bit difficult. Here is an example:
set.seed(2345)
dta<-data.frame(age=sample(20:80,50),skin=sample(0:1,50,TRUE),
gender=sample(0:1,50,TRUE),trt=sample(0:1,50,TRUE),
exposure=sample(1:21,50,TRUE),fit5=runif(50))
# define your subset here fo
Hello dear R-help
I tried to find function which returns number of occurrences of a pattern
in string. The closest match I've found is str_locate_all in stringr
package. I can use str_locate_all but write my function but I don't want
reinvent wheel.
JK
[[alternative HTML version deleted]
On Thu, 20 Oct 2016, David Winsemius wrote:
Why are you creating this factor? The `date` column has the desirable
properties associated with the "Date" class. Axis labeling will be
correct.
David, et al.:
Because originally I mis-understood the parameters; it's gone now.
rainbyday2 <- xyp
> On Oct 20, 2016, at 9:20 AM, Rich Shepard wrote:
>
> On Wed, 19 Oct 2016, David Winsemius wrote:
>
>> I am getting annoyed, exhausted, and frustrated reading code like that.
>> Never, ever, ... ever, use the "$" operator in a formula. Use the 'data'
>> argument and the 'formula' as they are s
On Wed, Oct 19, 2016 at 10:31 AM, David Winsemius
wrote:
>
>> On Oct 19, 2016, at 4:54 AM, Kevin E. Thorpe
>> wrote:
>>
>> Hello.
>>
>> I am posting this on behalf of one of my students who is getting error
>> messages when installing some packages. I have not seen this before nor have
>> I be
I have a slight doubt with using text() with the label parameter having
to contain a vector of of integers (specifically integers in the range
[1, 21] corresponding to factors of my categorical variable that I want
to numbers to tell).
What I'm currently plotting is the following command:
tex
optimr/optimrx are for unconstrained and bounds constrained problems.
I think you are going to have to do the masking yourself. It's more messy and
tedious than
difficult. The optimization is on a new set of parameters newpar that has
fewer components, so in your objective function and constraint
1. Your par(), opar() business is junk -- lattice displays do not use or
modify these. See ?trellis.par.set .
2. You do *not* need rain$ in your formula with the data=rain argument; you
*do* need it in the at argument and elsewhere. The data argument only
controls where the variables in the formul
Thanks.
It really seemed to be a bug.
Rui Barradas
Em 20-10-2016 11:33, Peter Dalgaard escreveu:
It's a bug (left.open=FALSE code gets executed in some cases). Hoping to have a
fix tested and in place before 3.3.2.
-pd
On 17 Oct 2016, at 21:48 , Rui Barradas wrote:
Hello,
Same on Windows
Indhira,
You have to assign cnt and new value in the loop if you want it to update in
the loop. However, to count the number of values of x > median(x), there are
multiple options. You are using a loop where none is needed in R, which has
many implicit vector functions that run with relational
On Wed, 19 Oct 2016, David Winsemius wrote:
I am getting annoyed, exhausted, and frustrated reading code like that.
Never, ever, ... ever, use the "$" operator in a formula. Use the 'data'
argument and the 'formula' as they are supposed to be used.
David,
I apologize for annoying, exhaustin
Your code,
cnt = 0
for(i in 1:length(x)){
ifelse(x[i] > median(x),cnt+1,cnt)
}
sets cnt to zero and never sets it to anything else. Hence it is zero
at the end of the loop. if you set cnt to the value of your call to ifelse
you should get the desired result
cnt <- ifelse(x[i] > media
Thanks Prof Nash.
The reason I used nlopr() in my problem is due to non linear
constraints. I wonder if optimrx/optim can model the below scenario. I
will be elated if it can.
The problem in hand goes like this:
There are 2 injectors and 2 producers. Consider these as some entity.
I have an actua
You do know that the median is defined as the point with half the values above
it and half below it? For even sample sizes it will always be 50%.
Your function is not working because you used the ifelse() function instead of
the programming command if() else:
> ?Control # Note the capital "C"
>
>From a statistician's point of view, "nonsense" may be OK, but there are other
>applications of R where
(partial or non-unique) solutions may be needed.
Yesterday I raised the question of how nonlinear least squares could be adapted
to underdetermined problems.
Many folk are unaware of such pos
Hi,
I would like to print percentage of points in a group which are greater than
median value in boxplot. I have tried below code but it always prints zero in
the graph. Can you let me know, how to modify code to get desired result?
perc.greaterthan.median <- function(x){
cnt = 0
for(i in
Dear list,
I want to label outliers in a ggplot box plot with the name of the subject for
which outlying data were observed.
I have proceeded by creating a simple function to identify outliers:
is_outlier <- function(x) {
return(x < quantile(x, 0.25) - 1.5 * IQR(x) | x > quantile(x, 0.75) +
One additional issue, since you are using logistic regression, you are
predicting a dichotomy (i.e. 0 and 1 or factor with 2 categories). The value
returned by predict() is a log odds ratio of belonging in the second category.
Alternatively if you use the type="response" argument with predict(),
I believe you have missing values and therefore you need to use
the argument
glm(formula, data, na.action=na.exclude, ...)
?na.exclude
The relevant line is
when 'na.exclude' is used the residuals and
predictions are padded to the correct length by inserting 'NA's
for cases omitted
Hi.
I am a bit puzzled. You do not get predicted values from variables but from
estimated model. AFAIK order of predicted values is the same as order of
original values.
> x<-1:10
> y<-5*x+3+rnorm(10)
> plot(x,y)
> a<-sample(letters[1:3], 10, replace=T)
> fit<-lm(y~x)
> points(x, predict(fit),
Some more context would help here but here goes anyway.
You should have a vector of predictions with length equal to the number
of rows in your original data-set so you can just use cbind. If that is
not true check the documentation for the correct setting of na.action.
If you used newdata = i
I'm using predict() for my glm() logistic model, but I'm having trouble
relating the predicted results to the rows that produced them.
I want to be able to plot predictions along some categorical variables.
So what can I do in order to get predicted values but also know what
variable values pr
> How do you reply to a specific post on this board instead of the thread?
You can reply to the individual, as I just did.
But I strongly suggest that you don't. You would be much better advised to
discontinue debate and follow the essential advice given by nls.lm, which - no
matter whether cou
It's a bug (left.open=FALSE code gets executed in some cases). Hoping to have a
fix tested and in place before 3.3.2.
-pd
> On 17 Oct 2016, at 21:48 , Rui Barradas wrote:
>
> Hello,
>
> Same on Windows 7.
>
> > findInterval(x=c(6, 1, 1, 1), vec=c(0, 1, 3, 5, 10), left.open=TRUE)
> [1] 4 2 1
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