Displaying 20 results from an estimated 7000 matches similar to: "Fitting the negative hypergeometric distribution"
2011 May 03
3
fitting distributions using fitdistr (MASS)
Please guide me through to resolve the error message that I get
this is what i have done.
>x1<- rnorm(100,2,1)
>x1fitbeta<-fitdistr(x1,"beta")
Error in fitdistr(x1, "beta") : 'start' must be a named list
Yes, I do understand that sometime for the distribution to converge to the
given set of data, it requires initial parameters of the distribution, to
2013 Jan 22
2
Assistant
Good-day Sir,
I am R.Language users but am try to? estimate parameter of beta distribution particular dataset but give this error, which is not clear to me: (Initial value in "vmmin" is not finite)
beta.fit <- fitdistr(data,densfun=dbeta,shape1=value , shape2=value)
kindly assist.
expecting your reply:
2009 Nov 20
2
How to use results of distribution fitting for further processing?
This is probably simple, but I have a hard time finding the solution. Any help greatly appreciated.
I would like to use the results of fitdistr(z,densfun=dweibull,start=list(scale=1,shape=1)) for further processing. How do I assign the values of scale and shape to b and a without manually entering the numbers?
TIA
__________________________________________________________________
2013 Oct 28
0
"Optimization fail" error from fitdistr (Weibull distribution)
Hello everyone,
This is Kangmin.
I am trying to produce shape and scale of my wind data. My data is based on
wind speed frequency with 1km/hr increment. data is described below.
Windspeed (km/h) Frequency
1 351
2 147
3 317
4 378
5 527
6 667
7 865
8 970
9 987
10 907
11 905
12 642
13 1000
14 983
15 847
16 842
17 757
18 698
19 632
20 626
21 599
22 529
23 325
24 391
2001 Sep 08
0
R-function available for noncentral hypergeometric distribution
For those who are interested, I have made available a R function for
noncentral hypergeometric distribution at
http://www.geocities.com/jg_liao/software/Hypergeometric/hypergeometric_in_R.txt
The paper that describes the algorithm will appear in The American
Statistician.
The function does not run on S-plus as the R's scoping rule is used.
Here is how the function can be used:
> n1
2008 Sep 25
1
What distribution is related to hypergeometric?
I have been reading, in various sources, that a poisson distribution is
related to binomial, extending the idea to include numbers of events in a
given period of time.
In my case, the hypergeometric distribution seems more appropriate, but I
need a temporal dimension to the distribution.
I have weekly samples of two kinds of events: call them A and B. I have a
count of A events. These change
2005 Sep 19
3
Extended Hypergeometric Distribution
Dear R Users,
There exists a non-central hypergeometric distribution function in the (MCMCpack) package, and a hypergeometric distribution function in the (stats) package.
Is there a function for sampling from an extended hypergeometric distribution?
Thanks,
Narcyz
This message is intended for the addressee named and may con...{{dropped}}
2008 Sep 19
0
Re lative Novice ? "Can I get some explanation of the docs for fitdistr(MASS)?"
In the docs I see:
Usage
fitdistr(x, densfun, start, ...)
Arguments
x A numeric vector.
densfun Either a character string or a function returning a density
evaluated at its first argument.
Distributions "beta", "cauchy", "chi-squared", "exponential", "f", "gamma",
"geometric", "log-normal", "lognormal",
2010 Mar 30
1
Multivariate hypergeometric distribution version of phyper()
Dear R Users,
I employed the phyper() function to estimate the likelihood that the
number of genes overlapping between 2 different lists of genes is due to
chance. This appears to work appropriately.
Now i want to try this with 3 lists of genes which phyper() does not
appear to support.
Some googling suggests i can utilize the Multivariate hypergeometric
distribution to achieve this. eg.:
2013 Apr 16
2
Strange error with log-normal models
Hi,
I have some data, that when plotted looks very close to a log-normal distribution. My goal is to build a regression model to test how this variable responds to several independent variables.
To do this, I want to use the fitdistr tool from the MASS package to see how well my data fits the actual distribution, and also build a generalized linear model using the glm command.
The summary
2012 Feb 21
3
HELP ERROR Weibull values must be > 0
GUYS,
I NEED HELP WITH ERROR:
library(MASS)
> dados<-read.table("mediaRGinverno.txt",header=FALSE)
> vento50<-fitdistr(dados[[1]],densfun="weibull")
Erro em fitdistr(dados[[1]], densfun = "weibull") :
Weibull values must be > 0
WHY RETURN THIS ERROR? WHAT CAN I DO?
BEST REGARDS
[[alternative HTML version deleted]]
2005 Sep 06
2
fitting distributions with R
Dear all
I've got the dataset
data:2743;4678;21427;6194;10286;1505;12811;2161;6853;2625;14542;694;11491;
?? ?? ?? ?? ?? 14924;28640;17097;2136;5308;3477;91301;11488;3860;64114;14334
I know from other testing that it should be possible to fit the data with the
exponentialdistribution. I tried to get parameterestimates for the
exponentialdistribution with R, but as the values
of the parameter
1997 May 12
1
R-alpha: Hypergeometric Distribution
A cut and paste typo has crept in and is rendering all values returned
for the hypergeometric distribution incorrect. The problem is in
src/main/arithmetic.c in the function "math4". The lines
PROTECT(sy = allocVector(REALSXP, n));
a = REAL(sa);
b = REAL(sb);
c = REAL(sc);
d = REAL(sc); /* <-- change this line */
y = REAL(sy);
should
2010 Mar 08
1
lapply and list indexing basics (after realizing I wasn't previously subscribed...sorry)
I have split my original dataframe to generate a list of dataframes each of
which has 3 columns of factors and a 4th column of numeric data.
I would like to use lapply to apply the fitdistr() function to only the 4th
column (x$isi) of the dataframes in the list.
Is there a way to do this or am I misusing lapply?
As a second solution I tried splitting only the numeric data column to yield
a
2010 Mar 08
1
lapply and list indexing basics
I have split my original dataframe to generate a list of dataframes each of
which has 3 columns of factors and a 4th column of numeric data.
I would like to use lapply to apply the fitdistr() function to only the 4th
column (x$isi) of the dataframes in the list.
Is there a way to do this or am I misusing lapply?
As a second solution I tried splitting only the numeric data column to yield
a list
2000 Mar 24
3
quantiles of the hypergeometric distribution (PR#502)
Hello!
I use R-version 1.0.0
To get the 0.95 quantile of the hypergeometric
distribution with the parameters m=45000,n=5000 and
k=600 I use the R-command
> qhyper(0.95,45000,5000,600).
The value obtained is 600. However, the true value
is 552. The latter can be obtained for example by
calling the corresponding distribution function
with the R commands
> x<-540:580
>
2005 Nov 17
1
Problem with fitdistr for gamma in R 2.2.0
Dear R developers,
I have encountered strange behaviour of fitdistr for gamma in recent R
build i.e. 2.2.0. I have attached the code for data at the end of this mail
so you can reproduce the problem. In short, I am able to run fitdistr under
2.1.0 without problems, while I get the following error under 2.2.0
(Version 2.2.0 Patched (2005-11-15 r36348))
> fitdistr(otm, "gamma")
Error
2010 Jan 12
1
Strange behavior when trying to piggyback off of "fitdistr"
Hello.
I am not certain even how to search the archives for this particular question, so if there is an obvious answer, please smack me with a large halibut and send me to the URLs.
I have been experimenting with fitting curves by using both maximum likelihood and maximum spacing estimation techniques. Originally, I have been writing distribution-specific functions in 'R' which work
2005 Sep 06
2
(no subject)
my problem actually arised with fitting the data to the weibulldistribution,
where it is hard to see, if the proposed parameterestimates make sense.
data1:2743;4678;21427;6194;10286;1505;12811;2161;6853;2625;14542;694;11491;
?? ?? ?? ?? ?? 14924;28640;17097;2136;5308;3477;91301;11488;3860;64114;14334
how am I supposed to know what starting values i have to take?
i get different
2008 Feb 10
1
Error in optim while using fitdistr() function for estimation of parameters
Hello,
I am trying to fit distribution for data consisting of 421 readings.It is
basically no of requests arrived per minute.It contains many 0 entries as no
of requests.When i use
fd<-fitdistr(V2,"gamma")
I get following error:
Error in optim(x = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, :
initial value in 'vmmin' is not finite
What should I do ? I need