Displaying 20 results from an estimated 2000 matches similar to: "multiplicate 2 functions"
2005 Jul 22
1
Generate a function
hi all,
I need to generate a function inside a loop:
tmp is an array
for (i in 1:10)
{
func<- func * function(beta1) dweibull(tmp[i],beta1,eta)
}
because then i need to integrate this function on beta.
I could have written this :
func<-function(beta1) prod(dweibull(tmp,beta1,eta)) (with eta and beta1 set)
but it is unplottable and no integrable... i could make it a bit different but
2005 Aug 02
1
plotting 3 functions on same graph
hi all,
I wish to draw on the same graphic device 3 functions.
But i don't want them to be on different graph, i want to compare them on the same
I don't need mfrow or mfcol, I need something else...
1 graph on 1 device inside this graph 3 ploted function.
I saw something unsing data.frame, but i think it's overkill, and something less complicated must exist, if not why?
why not
2005 Aug 02
1
(no subject)
hi all,
I wish to draw on the same graphic device 3 functions.
But i don't want them to be on different graph, i want to compare them on the same
I don't need mfrow or mfcol, I need something else...
1 graph on 1 device inside this graph 3 ploted function.
I saw something unsing data.frame, but i think it's overkill, and something less complicated must exist, if not why?
why not
2005 Jul 27
1
thks all
hi all
I wish to thanks every body on the R mailing list for answering very fast, directly in my mail box ;).
I've finish my work with R and i can say that it is very difficult at the beginning, and when you succeed you are stopped by a stack overflow when you call your nice recursive function (which was working with a tab of 100 element) with a tab of 900 elements, but R just do what you
2005 Jul 19
1
integrate fails with errors
Hi all,
i'm new to R,
I need to modelize in R a statistic algorithm,
This algo use Weibull, normal law, linear regression, normalisation, root mean square, to find eta and beta fitting the weibull model (to analyse few results) and further when we will get more information apply bayes model .
the problem is when When i try to integrate it fails with errors.
by the way i like to integrate
2005 Aug 02
1
multiple scale
Hi all
i need to put on one graph 2 functions who's x axis is the same and y not.
I mean on horizontal the time, and on vertical left: pressure, on vertical right: rpm of a motor, is R able to do that?
i've found this that i could adapt maybe (i don't need time series really?) :/ :
(http://tolstoy.newcastle.edu.au/R/help/04/03/1456.html)
##
## Description: A simple function which
2005 Dec 07
2
concatenate data frame
hi all
Here is a small part of my code:
tab_tmp<-tab[1:(no[off_set[i-1]+1]+(no[off_set[i]+1]-no[off_set[i-1]+1])),length(tab)];
tab_tmp1<-tab[(no[off_set[i-1]+1]+(no[off_set[i]+1]-no[off_set[i-1]+1])):length(TotalFillTimeHours),length(tab)];
tab<-c(tab_tmp,tab_tmp1);
attach(tab);
Here is the output:
Error in attach(tab) : attach only works for lists and data frames
Execution halted
2005 Dec 16
1
column name of a table
hy all,
I wish to switch in a part in my code that use "read.table" to "scan", actually i use this:
tab<-scan("data.dat",what=integer(),skip=1)
dim(tab)<-c(75,length(tab)/75)
tab<-t(tab)
It gives me a new tab with 75 columns, but i when i was using read.table with headers then attach i could use the columns names to access the data values, now how can i
2006 Jan 04
1
produce hours greater than 23
Hy all,
I wish to use the date function to draw againt the lifetime of a motor.
This lifetime is given to me in Hours (it can go over 5000 hours)
I'm unable to find how to convert this lifetime value to something like %H:%M:%S because when R see 24H it says 1 day, i don't want that, i just want %H:%M:%S with a value of %H higher than 24...
for example:
i've got this value in hours:
2005 Oct 18
2
hist of dates
Hi all
I wish to draw an histogram... with dates but the following append, i don't know where is the problem, help(hist.Date) works and i don't see any usefull information on what i'm doing wrong...
> hist.Date(dt_cycles)
Error: couldn't find function "hist.Date"
> hist.date(dt_cycles)
Error: couldn't find function "hist.date"
> cycles
[1] 7 1
2005 Nov 24
1
font size in legend
Hy all
I use barplot to draw frequencies by dates.
On the x axis it shows only 1 date for 2 bars, i've understand why, because R cant put so much date on the same axis, it will get out of the graph.
that's why i tought about reducing the size of the font used to stamp the x axis.
anyone knows the right way? i've tryed par(cin) and barplot(cin=3) but it says i cannot do it at this
2005 Nov 30
1
about sorting table
hi all,
I load a table with headers that enable me to acces it by the column names:
tab<-read.table("blob/data.dat",h=T)
attach(tab)
everythings are OK, but i try to sort this table against one of his column like this:
tab<-tab[order(tab$IndexUI),];
It is still ok, the table is sorted, if i type "tab" i see a sorted table.
but, when i call the column by their names,
2005 Dec 19
1
change read.table by scan
Hi all,
Before the amount of data given has grown i was initially using read.table to load the values inside R.
It was feeding my needs because i could tell read.table h=T, then use attach to access the values by columns names.
Now it takes 20 seconds to load the data's and the first enhancement i could do is to win some time on the load of the data's...
How could i use the scan
2005 Dec 08
1
truncate/overwrite a data frame
hi all,
I've got a data frame, this data frame have 76 columns and 22600 rows.
The data inside can be redundant because the data can be captured simultaneously and overlap each other.
My aim is to supress these overlaps
I've test some solutions to do that but they all give a big cpu load and eat all of the memory then swap a lot, then killall R because it don't end.
actually
2005 Oct 13
1
drawing against a date
hy all
I wish to draw a graph against a date,
I have a set of date like this DD/MM/YYYY, corresponding to it a set of integer , i wish to draw on x side the dates (the space between the dates have to be constant, not based on the time between 2 dates but on the number of dates) and on y side the integers.
Do i have to make a tricky function under R ?
I've search the help for graphical
2009 Dec 01
2
Slightly OT - Oreka Call Recording
Greetings all-
I'd like to install Oreka on a Centos 5.x server for monitoring my Asterisk systems(using port mirroring) but I find I'm having problems with the version of libpcap installed. The latest is 0.9.4 but the orkaudio RPM (built for Centos 4.2) requires libpcap 0.8.3. I've tried making symlinks to overcome the issue but with no success. Without having to build Oreka from
2011 Oct 20
1
R code Error : Hybrid Censored Weibull Distribution
Dear Sir/madam,
I'm getting a problem with a R-code which calculate Fisher Information
Matrix for Hybrid Censored Weibull Distribution. My problem is that:
when I take weibull(scale=1,shape=2) { i.e shape>1} I got my desired
result but when I take weibull(scale=1,shape=0.5) { i.e shape<1} it gives
error : Error in integrate(int2, lower = 0, upper = t) : the integral is
probably
2008 Dec 06
2
Call Recording - Asterisk
Hello folks,
I wanted to setup Oreka to monitor calls on a trixbox box I have setup.
Oreka doesn't seem to be catching all of the calls though.... I have port
mirroring setup on the port that trixbox is connected to, mirrored to the
port Oreka is connected to.
I have read that Asterisk doesn't work as a SIP Proxy, so I wondered if this
meant that some phones, after checking in with
2012 Feb 21
5
help error: In dweibull(x, shape, scale, log) : NaNs produzidos
Guys,
I'm having an error when I use the command:
library(MASS)> dados<-read.table("inverno.txt",header=FALSE)> vento50<-fitdistr(dados[[1]],densfun="weibull")Mensagens de aviso perdidas:1: In dweibull(x, shape, scale, log) : NaNs produzidos2: In dweibull(x, shape, scale, log) : NaNs produzidos3: In dweibull(x, shape, scale, log) : NaNs produzidos4: In
2000 Jan 27
1
error in dweibull (PR#405)
Full_Name: R. Woodrow Setzer, Jr.
Version: 0.90.1
OS: linux - Redhat 6.1
Submission from: (NULL) (165.247.155.206)
dweibull(0,1,1) evaluates to 0; it should be 1.
Note that dweibull(.Machine$double.eps) evaluates to 1.
> dweibull(.01,1,1)
[1] 0.9900498
> dweibull(.00001,1,1)
[1] 0.99999
> dweibull(.Machine$double.eps,1,1)
[1] 1
> dweibull(0,1,1)
[1] 0