Re: [R] Estimating parameters of 3 parameters lognormal distribution
It has to do with setting a to large upper bound for the gamma parameter.
However, we are estimating the value of gamma because we do not know it so how
can we set an upper bound for gamma??
Perhaps other optimization algorithms in optim() can be used (Nelder-Mead works
for this x at least).
Here is my findings:
N - 100
mu - 1
sigma - 2
gamma - 3
set.seed(42)
x - rlnorm(N, mu, sigma) + gamma
min(x)
[1] 3.006832
foo - function(x, mu, sigma, gamma)
+ {dlnorm(x - gamma, mu, sigma)}
## min(x) is not the right upper bound to use for gamma
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0.9, sigma = 1.9, gamma = 2.9),
+ lower = c(-Inf, 0, -Inf),
+ upper=c(Inf, Inf, min(x)))
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
## now upper bound for gamma is not to large
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0.9, sigma = 1.9, gamma = 2.9),
+ lower = c(-Inf, 0, -Inf),
+ upper=c(Inf, Inf, 3))
fit
mu sigmagamma
1.08694600 2.02546254 2.99393215
(0.20902998) (0.17736860) (0.01663611)
## now upper bound for gamma is not to large
## still able to find correct estimates for parameters for other start values
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0, sigma = 1, gamma = 0),
+ lower = c(-Inf, 0.0001, -Inf),
+ upper=c(Inf, Inf, 3))
fit
mu sigmagamma
1.08671881 2.02579806 2.99394094
(0.20907178) (0.17745706) (0.01663591)
## Nelder-Mead method seems to do the job
fit - fitdistr(x, densfun = foo, method = Nelder-Mead,
start = list(mu = 0, sigma = 1, gamma = 0))
fit
mu sigmagamma
1.08678997 2.02571374 2.99393924
(0.20906041) (0.17743258) (0.01663503)
## these two functions are defined and used in the code for fitdistr
## myfn is the objective function to be optimized in optim()
dens - function(parm, ...) foo(x, parm, ...)
myfn - function(parm, ...) -sum(log(dens(parm, ...)))
## for gamma 3. Here min(x) = 3.006832
## have at least one illegal value of log-normal
table(dens(1, 2, min(x)) 0)
FALSE TRUE
199
## resulting in
myfn(0.9, 1.9, min(x))
[1] Inf
## but for gamma = 3 we get finite likelihood
myfn(0.9, 1.9, 3.00)
[1] 322.4375
## also finite likelihood
myfn(0.9, 1.9, -3.00)
[1] 456.7857
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
http://www.vestas.com
Company reg. name: Vestas Wind Systems A/S
This e-mail is subject to our e-mail disclaimer statement.
Please refer to www.vestas.com/legal/notice
If you have received this e-mail in error please contact the sender.
-Original Message-
From: Rolf Turner [mailto:r.tur...@auckland.ac.nz]
Sent: 17. januar 2014 22:47
To: Vito Ricci
Cc: Frede Aakmann Tøgersen; Göran Broström; r-h...@stat.math.ethz.ch
Subject: Re: [R] Estimating parameters of 3 parameters lognormal
distribution
Can you please tell us (me!) how you chose starting values?
Out of curiosity I tried the following:
set.seed(42)
x - rlnorm(100,1,2) + 3
require(MASS)
strt - list(mu=1,sigma=2,gamma=3)
fit - fitdistr(x,densfun=function(x,mu,sigma,gamma)
{dlnorm(x-gamma,mu,sigma)
},
start=strt,lower=c(0,0,-Inf),
upper=c(Inf,Inf,min(x)))
and it ran just fine and gave sensible answers. But when I took:
strt - list(mu=0.9,sigma=1.9,gamma=2.9)
(not very different from the previous starting values)
I got an error:
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
On 18/01/14 01:00, Vito Ricci wrote:
OK. It runs fine! Many thanks Frede.
Regards.
Vito
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Il Venerdì 17 Gennaio 2014 12:38, Frede Aakmann Tøgersen
fr...@vestas.com ha scritto:
In package MASS there is the fitdistr function using maximum likelihood
estimation to infer on the parameters of distributions based on observed
data.
One of the arguments of fitdistr () allows you to specify the probability
density function.
You only know how the parametrization of your three parameters
lognormal distribution is defined since you really haven't told us much.
Please have a look at fitdistr() and tell us if fit your needs.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented,
Re: [R] Estimating parameters of 3 parameters lognormal distribution
On 18/01/2014 09:06, Frede Aakmann Tøgersen wrote:
It has to do with setting a to large upper bound for the gamma
parameter. However, we are estimating the value of gamma because we
do not know it so how can we set an upper bound for gamma??
You have starting values, so have some idea of the MLE
Perhaps other optimization algorithms in optim() can be used
(Nelder-Mead works for this x at least).
The parameters could be transformed here, and probably should. But most
optimization methods are going to cope with such simple constraints
provided foo() returns NA outside the feasible region.
Here is my findings:
N - 100
mu - 1
sigma - 2
gamma - 3
set.seed(42)
x - rlnorm(N, mu, sigma) + gamma
min(x)
[1] 3.006832
foo - function(x, mu, sigma, gamma)
+ {dlnorm(x - gamma, mu, sigma)}
## min(x) is not the right upper bound to use for gamma
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0.9, sigma = 1.9, gamma = 2.9),
+ lower = c(-Inf, 0, -Inf),
+ upper=c(Inf, Inf, min(x)))
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
## now upper bound for gamma is not to large
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0.9, sigma = 1.9, gamma = 2.9),
+ lower = c(-Inf, 0, -Inf),
+ upper=c(Inf, Inf, 3))
fit
mu sigmagamma
1.08694600 2.02546254 2.99393215
(0.20902998) (0.17736860) (0.01663611)
## now upper bound for gamma is not to large
## still able to find correct estimates for parameters for other start values
fit - fitdistr(x, densfun = foo,
+ start = list(mu = 0, sigma = 1, gamma = 0),
+ lower = c(-Inf, 0.0001, -Inf),
+ upper=c(Inf, Inf, 3))
fit
mu sigmagamma
1.08671881 2.02579806 2.99394094
(0.20907178) (0.17745706) (0.01663591)
## Nelder-Mead method seems to do the job
fit - fitdistr(x, densfun = foo, method = Nelder-Mead,
start = list(mu = 0, sigma = 1, gamma = 0))
fit
mu sigmagamma
1.08678997 2.02571374 2.99393924
(0.20906041) (0.17743258) (0.01663503)
## these two functions are defined and used in the code for fitdistr
## myfn is the objective function to be optimized in optim()
dens - function(parm, ...) foo(x, parm, ...)
myfn - function(parm, ...) -sum(log(dens(parm, ...)))
## for gamma 3. Here min(x) = 3.006832
## have at least one illegal value of log-normal
table(dens(1, 2, min(x)) 0)
FALSE TRUE
199
## resulting in
myfn(0.9, 1.9, min(x))
[1] Inf
## but for gamma = 3 we get finite likelihood
myfn(0.9, 1.9, 3.00)
[1] 322.4375
## also finite likelihood
myfn(0.9, 1.9, -3.00)
[1] 456.7857
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730 5135
M +45 2547 6050
fr...@vestas.com
http://www.vestas.com
Company reg. name: Vestas Wind Systems A/S
This e-mail is subject to our e-mail disclaimer statement.
Please refer to www.vestas.com/legal/notice
If you have received this e-mail in error please contact the sender.
-Original Message-
From: Rolf Turner [mailto:r.tur...@auckland.ac.nz]
Sent: 17. januar 2014 22:47
To: Vito Ricci
Cc: Frede Aakmann Tøgersen; Göran Broström; r-h...@stat.math.ethz.ch
Subject: Re: [R] Estimating parameters of 3 parameters lognormal
distribution
Can you please tell us (me!) how you chose starting values?
Out of curiosity I tried the following:
set.seed(42)
x - rlnorm(100,1,2) + 3
require(MASS)
strt - list(mu=1,sigma=2,gamma=3)
fit - fitdistr(x,densfun=function(x,mu,sigma,gamma)
{dlnorm(x-gamma,mu,sigma)
},
start=strt,lower=c(0,0,-Inf),
upper=c(Inf,Inf,min(x)))
and it ran just fine and gave sensible answers. But when I took:
strt - list(mu=0.9,sigma=1.9,gamma=2.9)
(not very different from the previous starting values)
I got an error:
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
On 18/01/14 01:00, Vito Ricci wrote:
OK. It runs fine! Many thanks Frede.
Regards.
Vito
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Il Venerdì 17 Gennaio 2014 12:38, Frede Aakmann Tøgersen
fr...@vestas.com ha scritto:
In package MASS there is the fitdistr function using maximum likelihood
estimation to infer on the parameters of distributions based on observed
data.
One of the arguments of fitdistr () allows you to specify the probability
density function.
You only know how the parametrization of your three parameters
lognormal distribution is defined since you really haven't told us much.
Please have a look at fitdistr() and tell us if fit your
Re: [R] *** caught segfault *** :: a common problem in 3.0.2
I think you are trying to access memory that you are not allowed to. Or did you specify the type of all of your data to be of the right ones? The segfault message at least say so. *** caught segfault *** address 0x2464943a08, cause 'memory not mapped' Traceback: 1: .Call(gyroscope, ncol(Y), nrow(Y), t(Y), as.integer(as.numeric(l) - 1), as.character(lnames), as.integer(perms), as.numeric(coloring), as.integer(useDensity), as.integer(showProfile), as.integer(showPatch)) 2: gyro2(data) Why don't you have as.integer for ncol(Y), nrow(Y)?? So are you having ncol(Y), nrow(Y) in the right order when you use t(Y)?? Why don't you have as.double for t(Y)?? Is this correct: as.integer(as.numeric(l) - 1)?? Should as.numeric be as.double? Have you checked that all indexes for your arrays in the C/C++ code are defined correctly with the arguments of your .Call()?? Yours sincerely / Med venlig hilsen Frede Aakmann Tøgersen Specialist, M.Sc., Ph.D. Plant Performance Modeling Technology Service Solutions T +45 9730 5135 M +45 2547 6050 fr...@vestas.com http://www.vestas.com Company reg. name: Vestas Wind Systems A/S This e-mail is subject to our e-mail disclaimer statement. Please refer to www.vestas.com/legal/notice If you have received this e-mail in error please contact the sender. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Christopher T Gregg Sent: 17. januar 2014 19:48 To: r-help@R-project.org Subject: [R] *** caught segfault *** :: a common problem in 3.0.2 Hi, I have struggled recently with R crashing with the following error in various contexts. It has occurred when running the R package RCytoscape, a package we generated internally and when running the base system. I have seen that this is an issue with the Mavericks OS (http://citizen- statistician.org/2013/10/27/warning-mac-os-10-9-mavericks-and-r-dont-play- nicely/), but I don't run Mavericks. I have tried reinstalling the latest R version without any improvements and I am seeking a solution for this frustating problem. The error and sessionInfo() are below. Thank you very much for any suggestions to overcome this problem. best wishes, Chris Here is the error: *** caught segfault *** address 0x2464943a08, cause 'memory not mapped' Traceback: 1: .Call(gyroscope, ncol(Y), nrow(Y), t(Y), as.integer(as.numeric(l) - 1), as.character(lnames), as.integer(perms), as.numeric(coloring), as.integer(useDensity), as.integer(showProfile), as.integer(showPatch)) 2: gyro2(data) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace sessionInfo() R version 3.0.2 (2013-09-25) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] gyroscope_0.9 Christopher Gregg, PhD. New York Stem Cell Foundation-Robertson Investigator Assistant Professor, Neurobiology and Anatomy Adjunct Assistant Professor, Human Genetics 323 Wintrobe Bldg 530 University of Utah, School of Medicine 20 North 1900 East Salt Lake City, Utah 84132-3401 phone: (801) 581-8212 fax: (801) 585-9736 Gregg Lab Website: www.neuro.utah.edu/labs/gregg/index.htmlhttp://www.neuro.utah.edu/l abs/gregg/index.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] For loop on column names
Hi Try for (cname in colnames(mydf)) print((percent(length(is.null(mydf [, cname]) / lines)) Br. Frede Oprindelig meddelelse Fra: Jeff Johnson Dato:18/01/2014 02.10 (GMT+01:00) Til: R help Emne: [R] For loop on column names I'm trying to find a more efficient to calculate the percent a field is populated and repeat it for each field (column). First, I'm counting the number of lines: lines - as.integer(countLines(extract) - 1) dput(lines) 10L extract - 'C:/Users/jeffjohn/Desktop/batchextract_100k_sample.csv' mydf - read.csv(file = extract, header = TRUE) Here's the list of columns in my file: dput(colnames(mydf)) c(PERSONPROFILE_POS, PARTY_ID, PERSON_FIRST_NAME, PERSON_LAST_NAME, PERSON_MIDDLE_NAME, PARTY_NUMBER, ACCOUNT_NUMBER, ABILITEC_LINK, ADDRESS1, ADDRESS2, ADDRESS3, ADDRESS4, CITY, COUNTY, STATE, PROVINCE, POSTAL_CODE, COUNTRY, PRIMARY_PER_TYPE, SELLTOADDR_LOS, LOCATION_ID, SELLTOADDR_SOS, PARTY_SITE_ID, PRIMARYPHONE_CPOS, CONTACT_POINT_ID_PCP, CONTACT_POINT_PURPOSE_PCP, PHONE_LINE_TYPE, PRIMARY_FLAG_PCP, PHONE_COUNTRY_CODE, PHONE_AREA_CODE, PHONE_NUMBER, EMAIL_CPOS, CONTACT_POINT_ID_ECP, CONTACT_POINT_PURPOSE_ECP, PRIMARY_FLAG_ECP, EMAIL_ADDRESS, BB_PARTY_ID) I want to count the percentage populated for each field. Rather than do: percent(length(is.null(mydf$PERSONPROFILE_POS)) / lines) percent(length(is.null(mydf$PARTY_ID)) / lines) etc. and repeat for each field manually, I want to use a for loop. I am trying the following: a - length(colnames(mydf)) # this is to get the total number of columns for (i in 1:a) print((percent(length(is.null(a)) / lines)) which isn't correct. I'm new to programming, so I don't quite know how to deal with this. Any suggestions? Thanks much. -- Jeff [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] eliminating white space in lattice plot
I forgot to mention for multiple plots ? print.trellis Duncan -Original Message- From: Gerrit Draisma [mailto:gdrai...@xs4all.nl] Sent: Saturday, 18 January 2014 21:49 To: arun; Duncan Mackay Subject: Re: [R] eliminating white space in lattice plot Ha Arun and Duncan, Thanks for your answers. They helped to solve my problem. I found that the width and height parameters can be specified directly in the pdf() command, and more useful for me in the R-code chunk in a Sweave source file, as = pdf(wideplot.pdf,width=10,height=3) xyplot(y~x,data=X,type=l,aspect=0.25) dev.off() = or = fig=TRUE,echo=FALSE,height=10,width=7= xyplot(y~x,data=X,type=l,aspect=0.25) @ = Have a good day. Mine is already. Gerrit. op 1/17/2014 5:37 PM arun schreef: Hi, May be this helps: x11(width=10,height=3) xyplot(y~x,data=X,type=l,aspect=0.25) dev.copy2pdf(file=wideplot.pdf) dev.off() A.K. On Friday, January 17, 2014 11:26 AM, Gerrit Draisma gdrai...@xs4all.nl wrote: Dear R-users, How do I get rid of the white space above and under a plot made with lattice xyplot? I searched the documentation but could not find how to do it. Standard lattice plots are square, but sometimes I want a wide but low graph and use aspect parameter to obtain such a graph. But then the plot has a lot of white space above and below the graph. That makes it more difficult to include in a LaTeX document. Thanks, Gerrit Draisma = library(lattice) X-data.frame(x=1:100,y=runif(100)) # pdf(wideplot.pdf) xyplot(y~x,data=X,type=l,aspect=0.25) # dev.off() __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] MULTINOMIAL LOGIT WITH ENDOGENEITY
My aim is to estimate a multinomial model with some endogenous variables. I tried the control function approach but the results are rather bad. I am thinking to ML estimation. Is there a package I can use to estimate this model? Thanks a lot Erasmo Papagni Department of economics University of Naples II __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generate Variable Length Strings from Various Sources
Hi Rainer,
Thanks for the tip.
Your suggestion works perfectly, however as per the R Mantra of avoiding
for loops, I propose the following this alternate:
# number of strings to be created
n - 50
# random length of each string
v.length = sample( c( 2:4), n, rep = TRUE )
# letter sources
src.1 = LETTERS[ 1:10 ]
src.2 = LETTERS[ 11:20 ]
src.3 = z
src.4 = c( 1, 2 )
# turn into a list
src - list( src.1, src.2, src.3, src.4 )
my.g = function(len,src)
{
my.s = src[[ sample( 1:4, 1 ) ]]
tmp = sample(my.s,len,rep=TRUE)
n1 = paste(tmp,collapse=)
n1
} # end
sapply(v.length,my.g,src)
// Cheers.
[[alternative HTML version deleted]]
__
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] For loop on column names
I doubt it. 1. The OP failed to specify how populatedness is defined. Is it NULL, NA, , ,...? 2. What is percent() ? Is this the OP's function or one from a package or pseudocode or ... ? 3. lapply(df,function) is generally preferable in R to: for(name in colnames(df)) function(df[ ,name]) The former packages everything neatly in a list, while with the latter you are stuck mucking about with canonical naming schemes and/or assignments that may clutter up your workspace. The plyR package may also be helpful her, especially for a novice. Given the OP's admitted ignorance to both programming and R, it seems to me that the obvious advice is to stop knocking around in the dark this way and spend time with some R tutorials. A good R book, perhaps tuned to his/her discipline, would probably also be a worthwhile purchase. Cheers, Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Sat, Jan 18, 2014 at 2:40 AM, Frede Aakmann Tøgersen fr...@vestas.com wrote: Hi Try for (cname in colnames(mydf)) print((percent(length(is.null(mydf [, cname]) / lines)) Br. Frede Oprindelig meddelelse Fra: Jeff Johnson Dato:18/01/2014 02.10 (GMT+01:00) Til: R help Emne: [R] For loop on column names I'm trying to find a more efficient to calculate the percent a field is populated and repeat it for each field (column). First, I'm counting the number of lines: lines - as.integer(countLines(extract) - 1) dput(lines) 10L extract - 'C:/Users/jeffjohn/Desktop/batchextract_100k_sample.csv' mydf - read.csv(file = extract, header = TRUE) Here's the list of columns in my file: dput(colnames(mydf)) c(PERSONPROFILE_POS, PARTY_ID, PERSON_FIRST_NAME, PERSON_LAST_NAME, PERSON_MIDDLE_NAME, PARTY_NUMBER, ACCOUNT_NUMBER, ABILITEC_LINK, ADDRESS1, ADDRESS2, ADDRESS3, ADDRESS4, CITY, COUNTY, STATE, PROVINCE, POSTAL_CODE, COUNTRY, PRIMARY_PER_TYPE, SELLTOADDR_LOS, LOCATION_ID, SELLTOADDR_SOS, PARTY_SITE_ID, PRIMARYPHONE_CPOS, CONTACT_POINT_ID_PCP, CONTACT_POINT_PURPOSE_PCP, PHONE_LINE_TYPE, PRIMARY_FLAG_PCP, PHONE_COUNTRY_CODE, PHONE_AREA_CODE, PHONE_NUMBER, EMAIL_CPOS, CONTACT_POINT_ID_ECP, CONTACT_POINT_PURPOSE_ECP, PRIMARY_FLAG_ECP, EMAIL_ADDRESS, BB_PARTY_ID) I want to count the percentage populated for each field. Rather than do: percent(length(is.null(mydf$PERSONPROFILE_POS)) / lines) percent(length(is.null(mydf$PARTY_ID)) / lines) etc. and repeat for each field manually, I want to use a for loop. I am trying the following: a - length(colnames(mydf)) # this is to get the total number of columns for (i in 1:a) print((percent(length(is.null(a)) / lines)) which isn't correct. I'm new to programming, so I don't quite know how to deal with this. Any suggestions? Thanks much. -- Jeff [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Estimating parameters of 3 parameters lognormal distribution
Hi Rolf,
I fit 3 parameters lognormal in this way
myf-function(x,m,s,t)(1/((x-t)*s*(2*pi)^0.5))*exp(((-(log(x-t)-m)^2)/(2*s^2)))
then fitdistr(y, myf, list(m=1,s=1, t=2))
I choose initial parameters value seeing data, I noted that a goog estimate of
t would be 2...Me too I've some errors, having NA in log function.
Best regards.
Vito
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Il Venerdì 17 Gennaio 2014 22:46, Rolf Turner r.tur...@auckland.ac.nz ha
scritto:
Can you please tell us (me!) how you chose starting values?
Out of curiosity I tried the following:
set.seed(42)
x - rlnorm(100,1,2) + 3
require(MASS)
strt - list(mu=1,sigma=2,gamma=3)
fit - fitdistr(x,densfun=function(x,mu,sigma,gamma)
{dlnorm(x-gamma,mu,sigma)
},
start=strt,lower=c(0,0,-Inf),
upper=c(Inf,Inf,min(x)))
and it ran just fine and gave sensible answers. But when I took:
strt - list(mu=0.9,sigma=1.9,gamma=2.9)
(not very different from the previous starting values)
I got an error:
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
On 18/01/14 01:00, Vito Ricci wrote:
OK. It runs fine! Many thanks Frede.
Regards.
Vito
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Il Venerdì 17 Gennaio 2014 12:38, Frede Aakmann Tøgersen fr...@vestas.com
ha scritto:
In package MASS there is the fitdistr function using maximum likelihood
estimation to infer on the parameters of distributions based on observed
data.
One of the arguments of fitdistr () allows you to specify the probability
density function.
You only know how the parametrization of your three parameters lognormal
distribution is defined since you really haven't told us much.
Please have a look at fitdistr() and tell us if fit your needs.
[[alternative HTML version deleted]]
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Re: [R] xts error: number of items to replace is not a multiple of replacement length
This is particularly complicated and a bit of an edge case, so I wouldn't try too hard to understand what's going on, but I'll try to explain anyway. Your code creates two calls to the subset-replacement function (xts:::`[-.xts`). The first call replaces the value of Close in a temporary copy of the first row of the object on the left-hand-side of the assignment. This works fine. The second call tries to replace the first *element* of the object on the left-hand-side of the assignment with the modified temporary copy of the first row. This is the problem. This throws a warning because x[1] - 1:4 is different from x[1,] - 1:4. The former replaces the first element, the latter replaces the first row. If you want your original command to work, you need to use the comma in your first subset call on the left-hand-side. sample.xts[2007-01-02::2007-01-04,]$Close - sample.xts[2007-01-02::2007-01-04]$Close+1 That said, I wouldn't encourage you to do this, because the code isn't clear and it's likely you'll forget the comma and re-confuse yourself at a later date. Simply remember to subset by column first, then row, if you insist on making two calls to the subset-replacement function. I would encourage you to use Arun's solution. It's cleaner and faster because it's only one function call and it avoids the $ function (which is marginally slower on xts objects). Best, -- Joshua Ulrich | about.me/joshuaulrich FOSS Trading | www.fosstrading.com On Thu, Jan 16, 2014 at 7:53 PM, ce zadi...@excite.com wrote: Indeed it works ! Thanks a lot. But why? -Original Message- From: arun [smartpink...@yahoo.com] Date: 01/16/2014 08:44 PM To: r-help@r-project.org r-help@r-project.org Subject: Re: [R] xts error: number of items to replace is not a multiple of replacement length Hi, Try: sample.xts[2007-01-02::2007-01-04,Close] -sample.xts[2007-01-02::2007-01-04,Close] +1 sample.xts[2007-01-02::2007-01-04] # Open High LowClose #2007-01-02 50.03978 50.11778 49.95041 52.11778 #2007-01-03 50.23050 50.42188 50.23050 52.39767 #2007-01-04 50.42096 50.42096 50.26414 52.33236 A.K. On Thursday, January 16, 2014 8:34 PM, ce zadi...@excite.com wrote: Dear all , I am getting this error while trying to change columns of an xts object with a date range as index. library(xts) Loading required package: zoo Attaching package: ‘zoo’ The following object is masked from ‘package:base’: as.Date, as.Date.numeric data(sample_matrix) sample.xts - as.xts(sample_matrix, descr='my new xts object') head(sample.xts) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 sample.xts$Close - sample.xts$Close+1 head(sample.xts) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 51.11778 2007-01-03 50.23050 50.42188 50.23050 51.39767 2007-01-04 50.42096 50.42096 50.26414 51.33236 2007-01-05 50.37347 50.37347 50.22103 51.33459 2007-01-06 50.24433 50.24433 50.11121 51.18112 2007-01-07 50.13211 50.21561 49.99185 50.99185 sample.xts[2007-01-02::2007-01-04] Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 51.11778 2007-01-03 50.23050 50.42188 50.23050 51.39767 2007-01-04 50.42096 50.42096 50.26414 51.33236 sample.xts[2007-01-02::2007-01-04]$Close - sample.xts[2007-01-02::2007-01-04]$Close+1 Warning message: In NextMethod(.Generic) : number of items to replace is not a multiple of replacement length __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] shading and estimating area abetween two cdf curves
These examples do not apply to the objects generated by the ecdf function. Look at the objects generated by the ecdf function and you will see that they are functions, not vectors of numbers (the 'f' in ecdf stands for 'function'). You need to apply the functions to vectors of numbers to get what you want. plot(ecdf(x)) works because there is a special plotting method for the function objects made by ecdf() which essentially does plot(x, ecdf(x)(x)). To go beyond that you have to evaluate the function yourself, as in the second part of this example: data0 - log2(1:16) data1 - 10 / (1 + sqrt(1:15)) ecdf0 - ecdf(data0) ecdf1 - ecdf(data1) plot(ecdf0, xlim=range(data0,data1), col=blue) plot(ecdf1, add=TRUE, col=orange) tmpx - seq(min(data0,data1), max(data0,data1), len=2001) p0 - ecdf0(tmpx) p1 - ecdf1(tmpx) segments(x0=tmpx, y0=p0, y1=p1, col=adjustcolor(ifelse(p0p1, red, black), alpha.f=0.1)) (There are better ways to fill the area between two lines, but using segments to make lots of vertical lines makes for a simple example.) Bill Dunlap TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Peter Maclean Sent: Friday, January 17, 2014 11:37 PM To: r-help@r-project.org Subject: [R] shading and estimating area abetween two cdf curves I would like to shade the area between two curves generated by the ecdf function. There are examples such as those explained here http://www.alisonsinclair.ca/2011/03/shading-between-curves-in-r/ These examples do not apply to the objects generated by the ecdf function. # library(truncnorm) x - rtruncnorm(100, 3, 25, 15, 5) y - rtruncnorm(100, 5, 30, 17, 8) x - ecdf(x) y - ecdf(y) plot(x) lines(y) #This is not working polygon(x, y, xpd = xpd, col = orange, lty = 2, lwd = 2, border = red) #How to shade the area between the curve #calculate the area between the two curves Peter Maclean Department of Economics UDSM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Estimating parameters of 3 parameters lognormal distribution
Offhand, I'd say that the issue isn't so much the starting values as the risk
of hitting the upper bound on gamma. If you get gamma set to min(x) at any
point during the iteration, the density at the corresponding point is zero, the
log-likelihood is -Inf, and poof, the magic smoke escapes.
Perhaps try min(x)-1e-7 or so?
-pd
On 17 Jan 2014, at 22:46 , Rolf Turner r.tur...@auckland.ac.nz wrote:
Can you please tell us (me!) how you chose starting values?
Out of curiosity I tried the following:
set.seed(42)
x - rlnorm(100,1,2) + 3
require(MASS)
strt - list(mu=1,sigma=2,gamma=3)
fit - fitdistr(x,densfun=function(x,mu,sigma,gamma)
{dlnorm(x-gamma,mu,sigma)
},
start=strt,lower=c(0,0,-Inf),
upper=c(Inf,Inf,min(x)))
and it ran just fine and gave sensible answers. But when I took:
strt - list(mu=0.9,sigma=1.9,gamma=2.9)
(not very different from the previous starting values)
I got an error:
Error in stats::optim(x = c(45.178764955739, 3.87862565957867,
8.61957940802123, :
L-BFGS-B needs finite values of 'fn'
On 18/01/14 01:00, Vito Ricci wrote:
OK. It runs fine! Many thanks Frede.
Regards.
Vito
Se non ora, quando?
Se non qui, dove?
Se non tu, chi?
Il Venerdì 17 Gennaio 2014 12:38, Frede Aakmann Tøgersen fr...@vestas.com
ha scritto:
In package MASS there is the fitdistr function using maximum likelihood
estimation to infer on the parameters of distributions based on observed
data.
One of the arguments of fitdistr () allows you to specify the probability
density function.
You only know how the parametrization of your three parameters lognormal
distribution is defined since you really haven't told us much.
Please have a look at fitdistr() and tell us if fit your needs.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Generate Variable Length Strings from Various Sources
People who subscribe to an avoid-for-loops mantra are often missing the
point. Converting for loops to indexing can often yield speed improvements.
Converting for loops to apply functions does not improve speed. (If you think
it does, you are probably handling memory allocation differently.) Worse, in
some cases it can reduce readability, so doing it because of a mantra can
actually harm your code.
Also, posting in HTML is hardly the R-help way (see the Posting Guide).
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
Burhan ul haq ulh...@gmail.com wrote:
Hi Rainer,
Thanks for the tip.
Your suggestion works perfectly, however as per the R Mantra of
avoiding
for loops, I propose the following this alternate:
# number of strings to be created
n - 50
# random length of each string
v.length = sample( c( 2:4), n, rep = TRUE )
# letter sources
src.1 = LETTERS[ 1:10 ]
src.2 = LETTERS[ 11:20 ]
src.3 = z
src.4 = c( 1, 2 )
# turn into a list
src - list( src.1, src.2, src.3, src.4 )
my.g = function(len,src)
{
my.s = src[[ sample( 1:4, 1 ) ]]
tmp = sample(my.s,len,rep=TRUE)
n1 = paste(tmp,collapse=)
n1
} # end
sapply(v.length,my.g,src)
// Cheers.
[[alternative HTML version deleted]]
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[R] accessing an object with a variable ?
I want access and object , its name is in an variable : require(xts) data(sample_matrix) head(sample_matrix) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 assign(a,sample_matrix) head(a) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 But following doesn't work: b=sample_matrix assign(a,b) a [1] sample_matrix how I can access sample_matrix data when I have only b variable ? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Estimating parameters of 3 parameters lognormal distribution
On 19/01/14 06:20, peter dalgaard wrote: Offhand, I'd say that the issue isn't so much the starting values as the risk of hitting the upper bound on gamma. If you get gamma set to min(x) at any point during the iteration, the density at the corresponding point is zero, the log-likelihood is -Inf, and poof, the magic smoke escapes. Perhaps try min(x)-1e-7 or so? Yep, that did it. Thanks Peter. (Actually I used min(x) - eps where I set eps equal to sqrt(.Machine$double.eps) --- which seems to be a kind of standard construction for a small-but-not-too-small number.) cheers, Rolf __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing an object with a variable ?
Yes, it works perfect, thanks a lot. -Original Message- From: Marius 't Hart [mar...@ai178182.ai.rug.nl] Date: 01/18/2014 04:58 PM To: ce zadi...@excite.com CC: r-help@r-project.org Subject: Re: [R] accessing an object with a variable ? Does this help you? http://stackoverflow.com/questions/3971844/access-variable-value-where-the-name-of-variable-is-stored-in-a-string-in-r On Sat, 18 Jan 2014, ce wrote: I want access and object , its name is in an variable : require(xts) data(sample_matrix) head(sample_matrix) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 assign(a,sample_matrix) head(a) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 But following doesn't work: b=sample_matrix assign(a,b) a [1] sample_matrix how I can access sample_matrix data when I have only b variable ? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing an object with a variable ?
May be this helps: assign(a,eval(parse(text=b))) A.K. On Saturday, January 18, 2014 4:24 PM, ce zadi...@excite.com wrote: I want access and object , its name is in an variable : require(xts) data(sample_matrix) head(sample_matrix) Open High Low Close 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 assign(a,sample_matrix) head(a) Open High Low Close 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 But following doesn't work: b=sample_matrix assign(a,b) a [1] sample_matrix how I can access sample_matrix data when I have only b variable ? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] remove axes on barp from plotrix
Hello guys, I'm using the barp function from the plotrix package, which works really really great. Unfortunately, I need a finer spacing of the labels on the y - axis. Is there any way to control this? Looking at the documentation of barp, it seems to be done rather automatically. An option would be to not plot axes at all from barp, and add them later by using axes(), which I was not able to get to work.. I tried axes=FALSE, and yaxt=n, and par settings: yaxp = c(a,b,n), all of which did not work. Any ideas?! Best regards Chrisee -- View this message in context: http://r.789695.n4.nabble.com/remove-axes-on-barp-from-plotrix-tp4683808.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] *** caught segfault *** :: a common problem in 3.0.2
Thank you, Frede. I will look into these suggestions. I did not specify the data type carefully…I will assess this and see if it cleans up the problem. best wishes, Chris On Jan 18, 2014, at 2:23 AM, Frede Aakmann Tøgersen fr...@vestas.com wrote: I think you are trying to access memory that you are not allowed to. Or did you specify the type of all of your data to be of the right ones? The segfault message at least say so. *** caught segfault *** address 0x2464943a08, cause 'memory not mapped' Traceback: 1: .Call(gyroscope, ncol(Y), nrow(Y), t(Y), as.integer(as.numeric(l) - 1), as.character(lnames), as.integer(perms), as.numeric(coloring), as.integer(useDensity), as.integer(showProfile), as.integer(showPatch)) 2: gyro2(data) Why don't you have as.integer for ncol(Y), nrow(Y)?? So are you having ncol(Y), nrow(Y) in the right order when you use t(Y)?? Why don't you have as.double for t(Y)?? Is this correct: as.integer(as.numeric(l) - 1)?? Should as.numeric be as.double? Have you checked that all indexes for your arrays in the C/C++ code are defined correctly with the arguments of your .Call()?? Yours sincerely / Med venlig hilsen Frede Aakmann Tøgersen Specialist, M.Sc., Ph.D. Plant Performance Modeling Technology Service Solutions T +45 9730 5135 M +45 2547 6050 fr...@vestas.com http://www.vestas.com Company reg. name: Vestas Wind Systems A/S This e-mail is subject to our e-mail disclaimer statement. Please refer to www.vestas.com/legal/notice If you have received this e-mail in error please contact the sender. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Christopher T Gregg Sent: 17. januar 2014 19:48 To: r-help@R-project.org Subject: [R] *** caught segfault *** :: a common problem in 3.0.2 Hi, I have struggled recently with R crashing with the following error in various contexts. It has occurred when running the R package RCytoscape, a package we generated internally and when running the base system. I have seen that this is an issue with the Mavericks OS (http://citizen- statistician.org/2013/10/27/warning-mac-os-10-9-mavericks-and-r-dont-play- nicely/), but I don't run Mavericks. I have tried reinstalling the latest R version without any improvements and I am seeking a solution for this frustating problem. The error and sessionInfo() are below. Thank you very much for any suggestions to overcome this problem. best wishes, Chris Here is the error: *** caught segfault *** address 0x2464943a08, cause 'memory not mapped' Traceback: 1: .Call(gyroscope, ncol(Y), nrow(Y), t(Y), as.integer(as.numeric(l) - 1), as.character(lnames), as.integer(perms), as.numeric(coloring), as.integer(useDensity), as.integer(showProfile), as.integer(showPatch)) 2: gyro2(data) Possible actions: 1: abort (with core dump, if enabled) 2: normal R exit 3: exit R without saving workspace 4: exit R saving workspace sessionInfo() R version 3.0.2 (2013-09-25) Platform: x86_64-apple-darwin10.8.0 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] stats graphics grDevices utils datasets methods base other attached packages: [1] gyroscope_0.9 Christopher Gregg, PhD. New York Stem Cell Foundation-Robertson Investigator Assistant Professor, Neurobiology and Anatomy Adjunct Assistant Professor, Human Genetics 323 Wintrobe Bldg 530 University of Utah, School of Medicine 20 North 1900 East Salt Lake City, Utah 84132-3401 phone: (801) 581-8212 fax: (801) 585-9736 Gregg Lab Website: www.neuro.utah.edu/labs/gregg/index.htmlhttp://www.neuro.utah.edu/l abs/gregg/index.html [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fwd: Implementing a Formula into a Column of a Data Frame
-- Forwarded message -- From: Norman McBride npmcbride0...@gmail.com Date: Fri, Jan 17, 2014 at 9:54 AM Subject: Implementing a Formula into a Column of a Data Frame To: r-help@r-project.org Now, if I were to move the data over from an excel doc would that change anything. I do understand that that I would have to change it to a csv document and create a variable for it. Just to give you a more visual understanding of what I am doing I a portion of the head of the data is shown below. So the formula would specifically be: Battery Heat (Watts) / ( Surface Area (Square Meters) * ( Battery Temperature (Celsius) - Air Temperature (Celsius) ) and that would be the value put into the Cooling Coefficient Column currently set at one. One final note, when using function() should I input variable such as a,b, c and then change then change them when using the do.call function or should I just do it with the column names the first time? Thank you for the help I really appreciate it. structure(list(`Time (Seconds)` = c(20.318162, 21.316219, 22.316277, 23.316334, 24.316391, 25.316448, 26.316505, 27.315562, 28.315619, 29.315677, 30.315734, 31.315791, 32.315848, 33.315906, 34.315963, 35.31602, 36.316077, 37.316134, 38.316192, 39.316249), `Battery Temperature (Celsius)` = c(12.290895, 12.331806, 12.373601, 12.424348, 12.467361, 12.522784, 12.581674, 12.642473, 12.713441, 12.760907, 12.832591, 12.90363, 12.953117, 13.037519, 13.103664, 13.193626, 13.267647, 13.340661, 13.439084, 13.512353), `Air Temperature (Celsius)` = c(12.109872, 12.130341, 12.127966, 12.126647, 12.138111, 12.13635, 12.133192, 12.137278, 12.144941, 12.140494, 12.145511, 12.143312, 12.136964, 12.15125, 12.1607, 12.162753, 12.152606, 12.148803, 12.16181, 12.151917 ), `Battery Heat (Watts)` = c(1.634698, 1.634698, 1.634882, 1.634703, 1.634614, 1.634435, 1.634435, 1.634252, 1.634252, 1.634163, 1.633435, 1.633801, 1.633444, 1.63344, 1.804659, 2.013801, 2.013905, 2.013905, 2.0139, 2.013107), `Surface Area (Square Meters)` = c(0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115, 0.00637115), `Cooling Coefficient` = c(1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1)), .Names = c(Time (Seconds), Battery Temperature (Celsius), Air Temperature (Celsius), Battery Heat (Watts), Surface Area (Square Meters), Cooling Coefficient ), row.names = c(NA, 20L), class = data.frame) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] 2 phase coxian phase distribution
Can you pliz  send me the R code for estimating parameters of a 2 phase distribution given some data to fit. Im nelson sibanda , a final year student at Nust [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing an object with a variable ?
Does this help you? http://stackoverflow.com/questions/3971844/access-variable-value-where-the-name-of-variable-is-stored-in-a-string-in-r On Sat, 18 Jan 2014, ce wrote: I want access and object , its name is in an variable : require(xts) data(sample_matrix) head(sample_matrix) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 assign(a,sample_matrix) head(a) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 But following doesn't work: b=sample_matrix assign(a,b) a [1] sample_matrix how I can access sample_matrix data when I have only b variable ? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lm(y ~ group/x ) + predict.lm(...,type=terms)
Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y - (group==1)*fun1(x)+(group==2)*fun2(x) + rnorm(400) mod1 - lm(y ~ group/(x-1) ) # without intercetp mod2 - lm(y ~ group/x ) # with intercetp #data to predict new - data.frame(x=rep(0:2,each=2), + group=factor(rep(1:2,3))) new x group 1 0 1 2 0 2 3 1 1 4 1 2 5 2 1 6 2 2 coef(mod1) # checking coefficients, both make sense to me. group1 group2 group1:x group2:x 7.864981 -18.098424 -2.963931 15.051449 coef(mod2) (Intercept) group2group1:xgroup2:x 7.864981 -25.963405 -2.963931 15.051449 predict(mod1, new,type = c(response)) # two response type predictions are the same, make sense to me. 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod2, new,type = c(response)) 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod1, new,type = c(terms)) # make sense to me group group:x 1 7.864981 0.00 2 -18.098424 0.00 3 7.864981 -2.963931 4 -18.098424 15.051449 5 7.864981 -5.927861 6 -18.098424 30.102898 attr(,constant) [1] 0 # I want to know the computation process for group:x below??? this is what I am interested in predict(mod2, new,type = c(terms)) groupgroup:x 1 12.9817 0.5209069 2 -12.9817 0.5209069 3 12.9817 -2.4430237 4 -12.9817 15.5723560 5 12.9817 -5.4069544 6 -12.9817 30.6238052 attr(,constant) [1] -5.637629 Thanks in advance Xing __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lm(y ~ group/x ) + predict.lm(...,type=terms)
Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y - (group==1)*fun1(x)+(group==2)*fun2(x) + rnorm(400) mod1 - lm(y ~ group/(x-1) ) # without intercetp mod2 - lm(y ~ group/x ) # with intercetp #data to predict new - data.frame(x=rep(0:2,each=2), + group=factor(rep(1:2,3))) new x group 1 0 1 2 0 2 3 1 1 4 1 2 5 2 1 6 2 2 coef(mod1) # checking coefficients, both make sense to me. group1 group2 group1:x group2:x 7.864981 -18.098424 -2.963931 15.051449 coef(mod2) (Intercept) group2group1:xgroup2:x 7.864981 -25.963405 -2.963931 15.051449 predict(mod1, new,type = c(response)) # two response type predictions are the same, make sense to me. 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod2, new,type = c(response)) 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod1, new,type = c(terms)) # make sense to me group group:x 1 7.864981 0.00 2 -18.098424 0.00 3 7.864981 -2.963931 4 -18.098424 15.051449 5 7.864981 -5.927861 6 -18.098424 30.102898 attr(,constant) [1] 0 # I want to know the computation process for group:x below??? this is what I am interested in predict(mod2, new,type = c(terms)) groupgroup:x 1 12.9817 0.5209069 2 -12.9817 0.5209069 3 12.9817 -2.4430237 4 -12.9817 15.5723560 5 12.9817 -5.4069544 6 -12.9817 30.6238052 attr(,constant) [1] -5.637629 Thanks in advance Xing [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] accessing an object with a variable ?
Does this help you? http://stackoverflow.com/questions/3971844/access-variable-value-where-the-name-of-variable-is-stored-in-a-string-in-r On 14-01-18 04:21 PM, ce wrote: I want access and object , its name is in an variable : require(xts) data(sample_matrix) head(sample_matrix) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 assign(a,sample_matrix) head(a) Open High LowClose 2007-01-02 50.03978 50.11778 49.95041 50.11778 2007-01-03 50.23050 50.42188 50.23050 50.39767 2007-01-04 50.42096 50.42096 50.26414 50.33236 2007-01-05 50.37347 50.37347 50.22103 50.33459 2007-01-06 50.24433 50.24433 50.11121 50.18112 2007-01-07 50.13211 50.21561 49.99185 49.99185 But following doesn't work: b=sample_matrix assign(a,b) a [1] sample_matrix how I can access sample_matrix data when I have only b variable ? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lm(y ~ group/x ) + predict.lm(...,type=terms)
Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y - (group==1)*fun1(x)+(group==2)*fun2(x) + rnorm(400) mod1 - lm(y ~ group/(x-1) ) # without intercetp mod2 - lm(y ~ group/x ) # with intercetp #data to predict new - data.frame(x=rep(0:2,each=2), + group=factor(rep(1:2,3))) new x group 1 0 1 2 0 2 3 1 1 4 1 2 5 2 1 6 2 2 coef(mod1) # checking coefficients, both make sense to me. group1 group2 group1:x group2:x 7.864981 -18.098424 -2.963931 15.051449 coef(mod2) (Intercept) group2group1:xgroup2:x 7.864981 -25.963405 -2.963931 15.051449 predict(mod1, new,type = c(response)) # two response type predictions are the same, make sense to me. 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod2, new,type = c(response)) 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod1, new,type = c(terms)) # make sense to me group group:x 1 7.864981 0.00 2 -18.098424 0.00 3 7.864981 -2.963931 4 -18.098424 15.051449 5 7.864981 -5.927861 6 -18.098424 30.102898 attr(,constant) [1] 0 # I want to know the computation process for group:x below??? this is what I am interested in predict(mod2, new,type = c(terms)) groupgroup:x 1 12.9817 0.5209069 2 -12.9817 0.5209069 3 12.9817 -2.4430237 4 -12.9817 15.5723560 5 12.9817 -5.4069544 6 -12.9817 30.6238052 attr(,constant) [1] -5.637629 Thanks in advance Xing [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lm(y ~ group/x ) + predict.lm(...,type=terms)
Please see the posting guide: posting 3 times only reduces your chance of an informative answer (as does posting in HTML). Your posting is too vague for others to know what it is you do not understand. R is Open Source: please read the sources for the definitive answer to the 'know the computation process' query. If (as I suspect) your difficulty is rather with interpreting terms in formulae, study chapter 6 of MASS (Venables Ripley, 2002). On 18/01/2014 22:03, Xing Zhao wrote: Hi, all I am trying to figure out the computation result for predict.lm(...,type=terms) when the original fitting model has a nesting term, lm(y ~ group/x ). 'nested' A example, set.seed(731) group - factor(rep(1:2, 200)) x - rnorm(400) fun1 - function(x) -3*x+8 fun2 - function(x) 15*x-18 y - (group==1)*fun1(x)+(group==2)*fun2(x) + rnorm(400) mod1 - lm(y ~ group/(x-1) ) # without intercetp mod2 - lm(y ~ group/x ) # with intercetp #data to predict new - data.frame(x=rep(0:2,each=2), + group=factor(rep(1:2,3))) new x group 1 0 1 2 0 2 3 1 1 4 1 2 5 2 1 6 2 2 coef(mod1) # checking coefficients, both make sense to me. group1 group2 group1:x group2:x 7.864981 -18.098424 -2.963931 15.051449 coef(mod2) (Intercept) group2group1:xgroup2:x 7.864981 -25.963405 -2.963931 15.051449 predict(mod1, new,type = c(response)) # two response type predictions are the same, make sense to me. 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod2, new,type = c(response)) 1 2 3 4 5 6 7.864981 -18.098424 4.901050 -3.046975 1.937120 12.004474 predict(mod1, new,type = c(terms)) # make sense to me group group:x 1 7.864981 0.00 2 -18.098424 0.00 3 7.864981 -2.963931 4 -18.098424 15.051449 5 7.864981 -5.927861 6 -18.098424 30.102898 attr(,constant) [1] 0 # I want to know the computation process for group:x below??? this is what I am interested in predict(mod2, new,type = c(terms)) groupgroup:x 1 12.9817 0.5209069 2 -12.9817 0.5209069 3 12.9817 -2.4430237 4 -12.9817 15.5723560 5 12.9817 -5.4069544 6 -12.9817 30.6238052 attr(,constant) [1] -5.637629 Thanks in advance Xing [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
