Please help, it possible to make an r means barplot in R using base, where
there are more than 2 factors and not with lattice, Franklin.
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OK, it seems that i misunderstand something, i forget how and when i pick up
the monition in my mind that " as possible as avoid using for loop".
TKS for all your suggestions!
But i still want the way to break sapply, if not exsits now, create it .
such as:
sapply<-function(...){
out<-FA
On Thu, 9 Oct 2014 11:18:43 PM Franklin Mairura wrote:
> Please help, it possible to make an r means barplot in R using base,
where
> there are more than 2 factors and not with lattice, Franklin.
>
Hi Franklin,
Have a look at the barNest function in the plotrix package.
Jim
Hi
But It is not doing what you wanted. It does not extend 365 days year to 366
days. Instead it will incorrectly rename all days after 28th February in non
leap years
> format(strptime(59, format = "%j"), "%d-%b")
[1] "28-II"
> format(strptime(60, format = "%j"), "%d-%b")
[1] "01-III"
> forma
Hi R helpers,
I want to write a function which will
1. Count the number of fridays in the current month ( to extract month from
given date) and also the number of fridays in the preceeding month
2. Calculate the ratio of the number of fridays in current month to the
number of fridays in the prec
On 10/10/2014, 7:28 AM, Abhinaba Roy wrote:
> Hi R helpers,
>
> I want to write a function which will
>
> 1. Count the number of fridays in the current month ( to extract month from
> given date) and also the number of fridays in the preceeding month
>
> 2. Calculate the ratio of the number of f
Doing as much as possible with vectors instead of loops of a good thing.
Fooling yourself that apply functions are vectorized is, well, not a good thing.
If you want to write a function to use instead of sapply, fine, but don't call
it *apply because those functions always give you one result fo
Hi Duncan,
I have converted the string to a POSIXIt object using
> strptime('31-may-2014',format="%d-%b-%Y")
But could not figure out the way forward.
Could you please elaborate a bit?
On Fri, Oct 10, 2014 at 5:14 PM, Duncan Murdoch
wrote:
> On 10/10/2014, 7:28 AM, Abhinaba Roy wrote:
> > Hi
On 10/10/2014 8:10 AM, Abhinaba Roy wrote:
Hi Duncan,
I have converted the string to a POSIXIt object using
> strptime('31-may-2014',format="%d-%b-%Y")
But could not figure out the way forward.
Could you please elaborate a bit?
Try this:
?POSIXlt
Duncan Murdoch
On Fri, Oct 10, 2014 at
Or try this:
fridays <- function(the_day){
require(lubridate)
day(the_day) = 1
the_month = seq(the_day, the_day+months(1)-days(1),1)
sum(wday(the_month) == 6)
}
That returns the number of Fridays in the month of a Date object given as arg:
> fridays(as.Date("2012-01-01"))
[1] 4
>
On 10 October 2014 08:04, pari hesabi wrote:
> Hello,As an example for Exponential distribution the MLE is got by this
> structure:t <- rexp(100, 2)loglik <- function(theta){ log(theta) - theta*t}a
> <- maxLik(loglik, start=1)print(a)Exponential distribution has a simple
> loglikelihood functio
I want to get rid of this thread. what to do?
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On 10 October 2014 16:16, Tasnuva Tabassum wrote:
> I want to get rid of this thread. what
On Fri, Oct 10, 2014 at 1:32 AM, Frederic Ntirenganya wrote:
Please try to not post in HTML. It is one of the forum "rules".
> Dear All,
>
> The following function gives me what I wanted.
> The input of function is one value and this push me to use sapply function
> to call it.
>
> Is there a be
maximum likelihood estimation
pari hesabi
6:04 AM
To: r-help@r-project.org
Hello,
As an example for Exponential distribution the MLE is got by this structure:
t <- rexp(100, 2)
loglik <- function(theta){ log(theta) - theta*t}
a <- maxLik(loglik, start=1)
print(a)
Exponential distribution has
Here is one way of doing it:
> require(lubridate)
> now <- as.Date('2014-10-10') # some date
> # get first of month
> first_mon <- now - day(now) + 1
> # create sequence of days in the month so you can count Fridays
> x <- seq(first_mon, by = '1 day', length = days_in_month(first_mon))
> first_f
On Fri, Oct 10, 2014 at 7:28 AM, Abhinaba Roy wrote:
> Hi R helpers,
>
> I want to write a function which will
>
> 1. Count the number of fridays in the current month ( to extract month from
> given date) and also the number of fridays in the preceeding month
>
> 2. Calculate the ratio of the numb
Dear friends,
The addHeader function in the rtf package always generates two rows of texts
even if a subtitle was not specified. Is there any way to stop the function
from generating the second blank line?
FYI, I cannot use the other functions, such as the addParagram and the
addText, because the
In my Impression, there seems exsits a function (let just call weekday) which
can return a POSIXlt format date 's weekday. That means, weekday( 31-may-2014
) may return the right weekday maybe 5, so a clumsy method is start to judge
from 1-may-2014 to 31-may-2014 , you get a vector x, sum(w
Dear all R users
I am trying to find the bayesian analysis using R2winBUGS but i have errors
in initial values with two groups.
the R-code
#Initial values for the MCMC in WinBUGS
init1<-list(uby1=rep(0.0,10),lam1=c(0.0,0.0,0.0,0.0,0.0,0.0),
gam1=c(1.0,1.0,1.0,1.0,1.0,1.0),psd1=1.0,phi1=matrix(data
Thanks Jim :)
Regards
Abhinaba
On Fri, Oct 10, 2014 at 9:02 PM, jim holtman wrote:
> Here is one way of doing it:
>
>
> > require(lubridate)
> > now <- as.Date('2014-10-10') # some date
> > # get first of month
> > first_mon <- now - day(now) + 1
> > # create sequence of days in the month so y
Hello,
First please keep in mind I am not a programmer and know very little about R. I
am running the 64bit version of R on a Windows 8.1 machine. I am trying to run
a script (which I have successfully run in the past) to download some weather
data from a NOAA ftp site.
When I attempt to run the
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