Re: [R] sd, mean with a frequency distribution matrix
Hi, For the second one, sqrt((sum(p[,1]^2*p[,2])-(sum(p[,1]*p[,2]))^2/sum(p[,2]))/(sum(p[,2])-1)), please refer to the following link for an example to explain how it works. http://www.lboro.ac.uk/media/wwwlboroacuk/content/mlsc/downloads/var_stand_deviat_group.pdf For the first one: sd(unlist(sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2].\: sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2])) to get all in the first column of the matrix repeated the number of times in the second column. After that, make the resulting list to become a vector so that it can be executed with sd function. Here is some illustrative example. p [,1] [,2] [1,] 103 [2,] 204 [3,] 305 sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2])) [[1]] [1] 10 10 10 [[2]] [1] 20 20 20 20 [[3]] [1] 30 30 30 30 30 unlist(sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2]))) [1] 10 10 10 20 20 20 20 30 30 30 30 30 sd(unlist(sapply(1:dim(p)[1],function(i)rep(p[i,1],p[i,2] [1] 8.348471 -- View this message in context: http://r.789695.n4.nabble.com/sd-mean-with-a-frequency-distribution-matrix-tp4703218p4703338.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Loop over regression results
Or even easier is you use lmList() from the nlme package
library(nlme)
data(iris)
regression.list - lmList(Sepal.Width ~ Petal.Width | Species, data = iris)
summary(regression.list)
coef(regression.list)
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance
Kliniekstraat 25
1070 Anderlecht
Belgium
To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to say
what the experiment died of. ~ Sir Ronald Aylmer Fisher
The plural of anecdote is not data. ~ Roger Brinner
The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of data.
~ John Tukey
2015-02-16 16:17 GMT+01:00 David L Carlson dcarl...@tamu.edu:
Or for the slopes and t-values:
do.call(rbind, lapply(mod, function(x) summary(x)[[coefficients]][2,]))
Estimate Std. Error t value Pr(|t|)
setosa 0.8371922 0.5049134 1.658091 1.038211e-01
versicolor 1.0536478 0.1712595 6.152348 1.41e-07
virginica 0.6314052 0.1428938 4.418702 5.647610e-05
David C
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L
Carlson
Sent: Monday, February 16, 2015 8:52 AM
To: Ronald Kölpin; r-help@r-project.org
Subject: Re: [R] Loop over regression results
In R you would want to combine the results into a list. This could be done
when you create the regressions or afterwards. To repeat your example using
a list:
data(iris)
taxon - levels(iris$Species)
mod - lapply(taxon, function (x) lm(Sepal.Width ~ Petal.Width,
data=iris, subset=Species==x))
names(mod) - taxon
lapply(mod, summary)
coeffs - do.call(rbind, lapply(mod, coef, [1))
coeffs
# (Intercept) Petal.Width
# setosa3.222051 0.8371922
# versicolor1.372863 1.0536478
# virginica 1.694773 0.6314052
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ronald
Kölpin
Sent: Monday, February 16, 2015 7:37 AM
To: r-help@r-project.org
Subject: [R] Loop over regression results
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear all,
I have a problem when trying to present the results of several
regression. Say I have run several regressions on a dataset and saved
the different results (as in the mini example below). I then want to
loop over the regression results in order so save certain values to a
matrix (in order to put them into a paper or presentation).
Aside from the question of how to access certain information stored by
lm() (or printed by summary()) I can't seem to so loop over lm()
objects -- no matter whether they are stored in a vector or a list.
They are always evaluated immediately when called. I tried quote() or
substitute() but that didn't work either as Objects of type 'symbol'
cannot be indexed.
In Stata I would simply do something like
forvalues k = 1/3 {
quietly estimates restore mod`k'
// [...]
}
and I am looking for the R equivalent of that syntax.
Kind regard and thanks
RK
attach(iris)
mod1 - lm(Sepal.Width ~ Petal.Width, data=iris, subset=Species==setosa)
mod2 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==versicolor)
mod3 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==virginica)
summary(mod1); summary(mod2); summary(mod3)
mat - matrix(data=NA, nrow=3, ncol=5,
dimnames=list(1:3, c(Model, Intercept, p(T |T|),
Slope, R^2)))
mods - c(mod1, mod2, mod3)
for(k in 1:3)
{
mod - mods[k]
mat[2,k] - as.numeric(coef(mod))[1]
mat[3,k] - as.numeric(coef(mod))[1]
}
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and
Re: [R] list of list
On 16 Feb 2015, at 17:43 , Troels Ring tr...@gvdnet.dk wrote:
Dear friends - this is simple I know but I can figure it out without your
help.
I have for each of 2195 instances 10 variables measured at specific times
from 6 to several hundred, so if I just take one of the instances, I can make
a list of the 10 variables together with their variable times. But when I
have 2195 such instances I cannot get it how to make a list of these
individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is mean to
take the list of here 5 entries made in RES[[i]] and I write this (ignoring
the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
Your description doesn't quite make sense to me, but if you really want ASL to
be a list of lists, you want each member to be a list and the (i,j)th item
accessed as ASL[[i]][[j]]. So I'd expect to see something like
for (j {
ASL[[j]] - list()
for (i
ASL[[j]][[|]] -
}
You also really don't want to start with an empty list and extend it on every
iteration. If you need an n-element list, preallocate it using vector(n,
mode=list).
If the above doesn't make sense, rephrase the question
--
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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Re: [R] ggplot2 shifting bars to only overlap in groups
Again, I come to think about violin plots which is more informative than the error bars. But for some reason, the violin in the *west* became way too slimmer than the *east* one, though the density plot tells me that is not necessarily the case. I am still trying to figure that out, and that would be even more irrelevant as long as *shifting bars in gorups*. So maybe I will come up with another post later when I got the solution. 祝好， He who is worthy to receive his days and nights is worthy to receive* all else* from you (and me). The Prophet, Gibran Kahlil On Mon, Feb 16, 2015 at 9:59 PM, John Kane jrkrid...@inbox.com wrote: Lovely, a much more elegant solution. John Kane Kingston ON Canada -Original Message- From: hyil...@gmail.com Sent: Mon, 16 Feb 2015 02:30:09 +0800 To: jrkrid...@inbox.com, djmu...@gmail.com Subject: Re: [R] ggplot2 shifting bars to only overlap in groups Thanks so much, John and Dennis (who did not respond in the mailing list for some reason). I feel quite obliged to keep you thinking about this. I do agree that not using the bar chart with error bars is a better option. And since *condition* is an important ordinal factor for me, it would be much better to have *condition* be positioned at a relative order. Thus, only color coding it as John's latest solution would not be optimal. It would have been better with the random data, but with my actual data, it does seem necessary to do a jitter for the *male* since it got clattered in the *west*. Here is the actual data along with the solution based on Dennis' code: ## data dat1 - structure(list(t = c(1.2454860689532, 0.627186899108052, 0.877176019393987, 1.26720638917869, 1.16906482219006, 0.889738853288831, 0.852034797572489, 1.30007600828822, 1.22896141479778, 0.820236562746995, 0.822197641624559, 1.39529772379005, 1.10479557445486, 0.760017179713665, 0.761340230517717, 1.11132156961026, 1.30042963441715, 0.811425854755042, 0.979421690403349, 1.3297658281305, 1.13377482477157, 0.895243910826397, 0.874181486658082, 1.15728885642541, 1.11121780853125, 0.703348405369258, 0.850897112058048, 1.14260584106012, 1.09383015337114, 0.911388765620587, 0.84622335453925, 1.09847968194129), condition = structure(c(4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c(c1, c2, c2, c4), class = factor), direction = structure(c(1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c(up, down), class = factor), location = structure(c(2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c(east, west), class = factor), gender = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(male, female), class = factor), ci = c(0.0307396796826649, 0.0302954863637637, 0.0400142340797275, 0.0527186825100342, 0.051675810189946, 0.0368383294010065, 0.0404823188495183, 0.0526312391852324, 0.0347332720922338, 0.0354587857740343, 0.0303368490163547, 0.0710445198259065, 0.0229339653012889, 0.0261217906562281, 0.0285673216713352, 0.0351642108247828, 0.0542657646932069, 0.0566816739316165, 0.0481239729953889, 0.0434272572423839, 0.0497366325101659, 0.0342004255233646, 0.0349733697554762, 0.0405364256564456, 0.0478372176424872, 0.0341294939361437, 0.0424566961614424, 0.0463489561778199, 0.0191707406475215, 0.0501106812754005, 0.0321562411182704, 0.0218613299095178)), .Names = c(t, condition, direction, location, gender, ci), row.names = c(NA, -32L ), class = data.frame) pp - ggplot(dat1, aes(x = condition, y = t, color = gender, linetype = direction)) + geom_errorbar(aes(ymin = t - ci, ymax = t + ci), position = position_dodge(width = 0.6), size = 1, width = 0.5) + geom_point(position = position_dodge(width = 0.6), size = 2.5) + facet_wrap(~ location) + scale_color_manual(values = c(blue, darkorange))+ theme_bw()+ scale_y_continuous(breaks=seq(0.6,1.5,0.1)) pp ## EOF I have also attached the output. Best ， He who is worthy to receive his days and nights is worthy to receive* all
Re: [R] list of list
You have two named objects when your goal is to have one that contains five
others.
ASL - vector( list, 5 )
for (j in 1:5){
ASL[[j]] - vector( list, 5 )
for (i in 1:5) {
ASL[[j]][[i]] - i^j
}
}
---
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.
On February 16, 2015 8:43:51 AM PST, Troels Ring tr...@gvdnet.dk wrote:
Dear friends - this is simple I know but I can figure it out without
your help.
I have for each of 2195 instances 10 variables measured at specific
times from 6 to several hundred, so if I just take one of the
instances,
I can make a list of the 10 variables together with their variable
times. But when I have 2195 such instances I cannot get it how to make
a
list of these individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is
mean
to take the list of here 5 entries made in RES[[i]] and I write this
(ignoring the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
All best wishes
Troels Ring
Aalborg, Denmark
__
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[R] list of list
Dear friends - this is simple I know but I can figure it out without
your help.
I have for each of 2195 instances 10 variables measured at specific
times from 6 to several hundred, so if I just take one of the instances,
I can make a list of the 10 variables together with their variable
times. But when I have 2195 such instances I cannot get it how to make a
list of these individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is mean
to take the list of here 5 entries made in RES[[i]] and I write this
(ignoring the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
All best wishes
Troels Ring
Aalborg, Denmark
__
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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] list of list
Is this what you mean:
ASL - list()
for (j in 1:5){
RES - list()
for (i in 1:5) RES[[i]] - i ^ j # create list
ASL[[j]] - RES # store 'list of list'
}
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Mon, Feb 16, 2015 at 11:43 AM, Troels Ring tr...@gvdnet.dk wrote:
Dear friends - this is simple I know but I can figure it out without your
help.
I have for each of 2195 instances 10 variables measured at specific times
from 6 to several hundred, so if I just take one of the instances, I can
make a list of the 10 variables together with their variable times. But
when I have 2195 such instances I cannot get it how to make a list of these
individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is mean
to take the list of here 5 entries made in RES[[i]] and I write this
(ignoring the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
All best wishes
Troels Ring
Aalborg, Denmark
__
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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.
[[alternative HTML version deleted]]
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Re: [R] P-value from Matching
Hi Peter, That was my first port of call before I posted this thread. Unfortunately, it does not seem to explicitly state which test is used or how the p-value is calculated. Thanks, Rob. -- View this message in context: http://r.789695.n4.nabble.com/P-value-from-Matching-tp4703309p4703345.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Comparing gam models fitted using ti()
I wonder if this might be better asked on a statistical list -- e.g. stats.stackexchange.com -- as this seems to involve complex statistical model comparison issues, which are normally OT here. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Mon, Feb 16, 2015 at 10:13 AM, Arnaud Mosnier a.mosn...@gmail.com wrote: Dear R-Help list, I want to compare gam models including interaction with simpler models. For interaction models, I used gam(Y~ti(X1) + ti(X2) + ti(X1,X2)) removing the interaction, the models end as Y~ti(X1) + ti(X2) How those models compare with models with the form Y ~ s(X1) + s(X2) In my case, the former (i.e. ti() ) have a better AIC value than those with s(). Should I keep the ti() form or use only the s() form when I have no interaction ? Thanks for your help ! Arnaud [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] Comparing gam models fitted using ti()
Dear R-Help list, I want to compare gam models including interaction with simpler models. For interaction models, I used gam(Y~ti(X1) + ti(X2) + ti(X1,X2)) removing the interaction, the models end as Y~ti(X1) + ti(X2) How those models compare with models with the form Y ~ s(X1) + s(X2) In my case, the former (i.e. ti() ) have a better AIC value than those with s(). Should I keep the ti() form or use only the s() form when I have no interaction ? Thanks for your help ! Arnaud [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] list of list
Thanks to Jim, Peter and Jeff who all saw the solution!
Best wishes
Troels
Den 16-02-2015 kl. 18:46 skrev Jeff Newmiller:
You have two named objects when your goal is to have one that contains five
others.
ASL - vector( list, 5 )
for (j in 1:5){
ASL[[j]] - vector( list, 5 )
for (i in 1:5) {
ASL[[j]][[i]] - i^j
}
}
---
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.
On February 16, 2015 8:43:51 AM PST, Troels Ring tr...@gvdnet.dk wrote:
Dear friends - this is simple I know but I can figure it out without
your help.
I have for each of 2195 instances 10 variables measured at specific
times from 6 to several hundred, so if I just take one of the
instances,
I can make a list of the 10 variables together with their variable
times. But when I have 2195 such instances I cannot get it how to make
a
list of these individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is
mean
to take the list of here 5 entries made in RES[[i]] and I write this
(ignoring the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
All best wishes
Troels Ring
Aalborg, Denmark
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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 -- To UNSUBSCRIBE and more, see
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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.
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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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] list of list
You can let lapply() do the preallocation and the looping for you with
ASL - lapply(1:5, function(j) lapply(1:5, function(i) i^j))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Mon, Feb 16, 2015 at 9:46 AM, Jeff Newmiller jdnew...@dcn.davis.ca.us
wrote:
You have two named objects when your goal is to have one that contains
five others.
ASL - vector( list, 5 )
for (j in 1:5){
ASL[[j]] - vector( list, 5 )
for (i in 1:5) {
ASL[[j]][[i]] - i^j
}
}
---
Jeff NewmillerThe . . Go Live...
DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live
Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/BatteriesO.O#. #.O#. with
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---
Sent from my phone. Please excuse my brevity.
On February 16, 2015 8:43:51 AM PST, Troels Ring tr...@gvdnet.dk wrote:
Dear friends - this is simple I know but I can figure it out without
your help.
I have for each of 2195 instances 10 variables measured at specific
times from 6 to several hundred, so if I just take one of the
instances,
I can make a list of the 10 variables together with their variable
times. But when I have 2195 such instances I cannot get it how to make
a
list of these individual lists
As a toy example demonstrating mercilessly my problem, if ASL[j] is
mean
to take the list of here 5 entries made in RES[[i]] and I write this
(ignoring the times) it certainly doesn't work
ASL - list()
RES - list()
for (j in 1:5){
for (i in 1:5)
ASL[[j]] -
RES[[i]] - i^j }
All best wishes
Troels Ring
Aalborg, Denmark
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[R] plm function plm r.squared error
DearGiovanni,Congratulationsfor the truly helpful plm package! Being new to R, I have a problem with the plm function for financial markets timeseries data: After having defined a large, unbalanced panel pdata.frame (https://www.dropbox.com/s/2r9t1cu9v65gobk/Database_CN_2004.csv?dl=0)and running a simple OLS model of two variables regressing company returns(Perf) on index returns (PerfIn)synch -plm (Perf ~ PerfIn , data=pCN04, na.action = na.omit, index=c(Unit, Week), model=pooling) I keepreceiving the following error:Error in model.matrix.pFormula(formula, data, rhs =1, model = model, : NA in theindividual index variableIn addition:Warning message:In `[.data.frame`(index, as.numeric(rownames(mf)),) : NAsintroduced by coercion There aremuch more NAs in y (Perf) than in x (PerfIn) and I have tried to get rid ofthem with na.omit. Another error I have is with the same model for the r.squaredfunction r.squared(pCN04,model=NULL, Perf ~ PerfIn, type= ess)Error in tss.default(object, model = model) : unused argument (model = model)In addition: Warning message:In mean.default(haty) : argument is not numeric or logical: returning NAMany thanks in advance,your help is very precious to me!Constanze [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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-es] consultas formularios web
Estimado Carlos Ortega Gracias, lo voy a mirar, estaba casi seguro que en el grupo esto se había tratado. Javier Marcuzzi De: Carlos Ortega Enviado el: lunes, 16 de febrero de 2015 06:59 p.m. Para: Javier Ruben Marcuzzi CC: R-help-es@r-project.org Hola, Tienes una presentación sobre esto que hizo Gregorio Serrano en el Grupo de Madrid: http://madrid.r-es.org/martes-20-mayo-de-2014/ Webscrapping with el paquete Relenium Y otra alternativa es utilizar RSelenium: http://cran.r-project.org/web/packages/RSelenium/vignettes/RSelenium-basics.html Saludos, Carlos Ortega www.qualityexcellence.es El 16 de febrero de 2015, 20:13, javier.ruben.marcu...@gmail.com escribió: Estimados Les consulto por lo siguiente, incluso creo que de esto se habló en una oportunidad en esta lista, por ese motivo cualquier sugerencia es bienvenida. Hay algo de información que me hace falta para un trabajo, pero esta no es de una única fuente, y desconozco si brindan los registros, pero lo que es accesible son los sitios web donde estas fuentes publican un formulario HTML simple, la respuesta es otro HTML simple, luego puede ser otro HTML simple con algunas tablas. Es decir, algo de información hay pero necesita de trabajo, nada que un script y consultas automatizadas junto a una base de datos y un excelente diseño no pueda lograr. Una alternativa podría ser el siguiente paquete http://cran.r-project.org/web/packages/httr/index.html (como otros tanto que tendría que estudiar), o simplemente curl, tengo que organizar desde el inicio. ¿Alguna sugerencia o recomendación? Recuerdo que yo mismo participe en debates sobre como tomar información desde internet a R, pero como hace un tiempo no uso esa tecnología, puedo estar muy desactualizado. Javier Marcuzzi [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R-es] consultas formularios web
Hola, Tienes una presentación sobre esto que hizo Gregorio Serrano en el Grupo de Madrid: http://madrid.r-es.org/martes-20-mayo-de-2014/ Webscrapping with el paquete Relenium Y otra alternativa es utilizar RSelenium: http://cran.r-project.org/web/packages/RSelenium/vignettes/RSelenium-basics.html Saludos, Carlos Ortega www.qualityexcellence.es El 16 de febrero de 2015, 20:13, javier.ruben.marcu...@gmail.com escribió: Estimados Les consulto por lo siguiente, incluso creo que de esto se habló en una oportunidad en esta lista, por ese motivo cualquier sugerencia es bienvenida. Hay algo de información que me hace falta para un trabajo, pero esta no es de una única fuente, y desconozco si brindan los registros, pero lo que es accesible son los sitios web donde estas fuentes publican un formulario HTML simple, la respuesta es otro HTML simple, luego puede ser otro HTML simple con algunas tablas. Es decir, algo de información hay pero necesita de trabajo, nada que un script y consultas automatizadas junto a una base de datos y un excelente diseño no pueda lograr. Una alternativa podría ser el siguiente paquete http://cran.r-project.org/web/packages/httr/index.html (como otros tanto que tendría que estudiar), o simplemente curl, tengo que organizar desde el inicio. ¿Alguna sugerencia o recomendación? Recuerdo que yo mismo participe en debates sobre como tomar información desde internet a R, pero como hace un tiempo no uso esa tecnología, puedo estar muy desactualizado. Javier Marcuzzi [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R] method of moments estimation
Hi, I'm trying to use method of moments estimation to estimate 3 unkown paramters delta,k and alpha. so I had system of 3 non linear equations: 1) [delta^(1/alpha) *gamma (k-(1/alpha)) ]/gamma(k) = xbar 2) [delta^(2/alpha) *gamma (k-(2/alpha)) ]/gamma(k) = 1/n *sum (x^2) 3) [delta^(3/alpha) *gamma (k-(3/alpha)) ]/gamma(k) = 1/n *sum (x^3) where gamma is a gamma function and n is the sample size How can I solve these system , Is there a package on R can give me MOM ?? Thank you , Sara [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Fitting gamma distribution
On 17/02/15 12:59, smart hendsome wrote:
I'm very new to r-programming. I have rainfall data. I have tried to fit gamma
into my data but there is error. Anyone can help me please.
My rainfall data as I uploaded. When I try run the coding:
library(MASS)
KLT1-read.csv('C:/Users/User/Dropbox/PhD
Materials/Coding_PhD_Thesis/Kelantan_Average/K1.csv')
KLT-KLT1$Amount
fd_g - fitdistr(KLT, gamma)
Error in stats::optim(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, :
initial value in 'vmmin' is not finite
Anyone can help me? Thanks.
No. No one can help. Not unless you provided a self-contained
reproducible example as the posting guide requests.
cheers,
Rolf Turner
P. S. I am not sure, but the error may be indicating that your data
begin with a long string of zeros. (In which case fitting a gamma
distribution is simply idiotic.) Rainfall data often contain many
zeroes. (There are, contrary to popular belief, many dry days.)
What you *may* want to fit is a mixture of a gamma distribution and a
point-mass (atom) at 0.
R. T.
--
Rolf Turner
Technical Editor ANZJS
Department of Statistics
University of Auckland
Phone: +64-9-373-7599 ext. 88276
Home phone: +64-9-480-4619
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Re: [R] split dataframe to several dataframes in R
Hi,
I think you need not split the data.frame to get the desired result.
You can work with your list lst4 itself.
#Convert the vectors in the list to data.frames.
lst4 - lapply(lst4, function(x) {as.data.frame(t(iris1.csv))})
#Get the data.frames in the list to the global environment
list2env(lst4 ,.GlobalEnv)
Regards
On 17 February 2015 at 09:23, Zilefac Elvis via R-help
r-help@r-project.org wrote:
Hi All,I have a dataframe called 'means' as shown below:iris1.csv - iris
iris2.csv - iris
names - c(iris1.csv, iris2.csv)
dat - mget(names)
lst4 - lapply(dat, function(x) apply(x[,-5], 2, mean))
# Build the new data frame
means - as.data.frame(do.call(rbind, lst4))
means$source - names(lst4)
means
# Sepal.Length Sepal.Width Petal.Length Petal.Width isv
source
# iris1.csv 5.843.0573333.7581.199333 0.333
iris1.csv
# iris2.csv 5.843.0573333.7581.199333 0.333
iris2.csvQUESTION: How can I split 'means' such that there are two files
(dataframes) on my workspace:datframe 1# Sepal.Length Sepal.Width
Petal.Length Petal.Width isv
# iris1.csv 5.843.0573333.7581.199333
0.333dataframe 2:# Sepal.Length Sepal.Width Petal.Length
Petal.Width isv# iris2.csv 5.843.0573333.758
1.199333 0.333
Many thanks,Asong.
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--
J.Aravind
Scientist
Germplasm Conservation Division
ICAR-National Bureau of Plant Genetic Resources
New Delhi - 110 012
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Re: [R] compute values by condition in DF by rownames
Hi,
I hope that someone can provide a better way to implement it. This is my
implementation.
data
X2 gbm_tcga lusc_tcga ucec_tcga_pub
1 gbm_tcga 1.0 0.14053719 -0.102847164
2 gbm_tcga 1.0 0.04413434 0.013568055
3 gbm_tcga 1.0 -0.20003971 0.038971817
4 gbm_tcga 1.0 0.14569916 0.009947045
5 lusc_tcga 0.140537191 1. 0.133080708
6 lusc_tcga 0.044134345 1. 0.062024713
7 lusc_tcga -0.200039712 1. -0.130239551
8 lusc_tcga 0.145699156 1. 0.041796670
9 ucec_tcga_pub -0.102847164 0.13308071 1.0
10 ucec_tcga_pub 0.013568055 0.06202471 1.0
11 ucec_tcga_pub 0.038971817 -0.13023955 1.0
12 ucec_tcga_pub 0.009947045 0.04179667 1.0
test
function(x)
{
tempMatrix - matrix(0,nrow=3,ncol=3)
lev - levels(x$X2)
for (i in 1:length(lev))
{
tempMatrix[i,1] = sum(ifelse(abs(x[x$X2==lev[i],2])0.2,1,0))
tempMatrix[i,2] = sum(ifelse(abs(x[x$X2==lev[i],3])0.2,1,0))
tempMatrix[i,3] = sum(ifelse(abs(x[x$X2==lev[i],4])0.2,1,0))
}
result - data.frame(lev,tempMatrix)
names(result) - c(X2,gbm_tcga,lusc_tcga,tcga_pub)
return(result)
}
test(data)
X2 gbm_tcga lusc_tcga tcga_pub
1 gbm_tcga4 10
2 lusc_tcga1 40
3 ucec_tcga_pub0 04
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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[R] split dataframe to several dataframes in R
quote author='R help mailing list-2' Hi All,I have a dataframe called 'means' as shown below:iris1.csv - iris iris2.csv - iris names - c(iris1.csv, iris2.csv) dat - mget(names) lst4 - lapply(dat, function(x) apply(x[,-5], 2, mean)) # Build the new data frame means - as.data.frame(do.call(rbind, lst4)) means$source - names(lst4) means # Sepal.Length Sepal.Width Petal.Length Petal.Width isv source # iris1.csv 5.843.0573333.7581.199333 0.333 iris1.csv # iris2.csv 5.843.0573333.7581.199333 0.333 iris2.csvQUESTION: How can I split 'means' such that there are two files (dataframes) on my workspace:datframe 1# Sepal.Length Sepal.Width Petal.Length Petal.Width isv # iris1.csv 5.843.0573333.7581.199333 0.333dataframe 2:# Sepal.Length Sepal.Width Petal.Length Petal.Width isv# iris2.csv 5.843.0573333.758 1.199333 0.333 Many thanks,Asong. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. /quote Quoted from: http://r.789695.n4.nabble.com/split-dataframe-to-several-dataframes-in-R-tp4703372.html _ Sent from http://r.789695.n4.nabble.com __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Loop over regression results
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear all,
I have a problem when trying to present the results of several
regression. Say I have run several regressions on a dataset and saved
the different results (as in the mini example below). I then want to
loop over the regression results in order so save certain values to a
matrix (in order to put them into a paper or presentation).
Aside from the question of how to access certain information stored by
lm() (or printed by summary()) I can't seem to so loop over lm()
objects -- no matter whether they are stored in a vector or a list.
They are always evaluated immediately when called. I tried quote() or
substitute() but that didn't work either as Objects of type 'symbol'
cannot be indexed.
In Stata I would simply do something like
forvalues k = 1/3 {
quietly estimates restore mod`k'
// [...]
}
and I am looking for the R equivalent of that syntax.
Kind regard and thanks
RK
attach(iris)
mod1 - lm(Sepal.Width ~ Petal.Width, data=iris, subset=Species==setosa)
mod2 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==versicolor)
mod3 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==virginica)
summary(mod1); summary(mod2); summary(mod3)
mat - matrix(data=NA, nrow=3, ncol=5,
dimnames=list(1:3, c(Model, Intercept, p(T |T|),
Slope, R^2)))
mods - c(mod1, mod2, mod3)
for(k in 1:3)
{
mod - mods[k]
mat[2,k] - as.numeric(coef(mod))[1]
mat[3,k] - as.numeric(coef(mod))[1]
}
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Re: [R] Nonlinear integer programming (again)
You may have good reason to distrust the Excel solver :) See below John Kane Kingston ON Canada -Original Message- From: rzw...@ets.org Sent: Sat, 14 Feb 2015 23:53:55 + To: r-help@r-project.org Subject: [R] Nonlinear integer programming (again) Oddly, Excel's Solver will produce a solution to such problems but (1) I don't trust it and (2) it cannot handle a large number of constraints. From IIRC a discussion on the R-help list but which I forgot to save the link. The idea that the Excel solver has a good reputation for being fast and accurate does not withstand an examination of the Excel solver's ability to solve the StRD nls test problems. Solver's ability is abysmal. 13 of 27 answers have zero accurate digits, and three more have fewer than two accurate digits --and this is after tuning the solver to get a good answer. ... Excel solver does have the virtue that it will always produce an answer, albeit one with zero accurate digits. Bruce McCullough FREE 3D MARINE AQUARIUM SCREENSAVER - Watch dolphins, sharks orcas on your desktop! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] How do I provide path character string to extdata content?
On 16/02/2015 6:37 AM, Ulrike Grömping wrote: Dear helpeRs, I have some png files in the inst/extdata directory of a package (e.g., man.png), and I want to provide character strings containing the paths to these files; of course, these path strings have to depend on the individual installation. So far, I have written a function that - if called - writes the correct strings to the global environment, but that is not CRAN compatible. The system.file() function does this. For example, if your package named foo contains inst/fig/man.png, you would use system.file(fig/man.png, package=foo). Ideally, I want the package namespace to export these text strings. I have played with .onLoad without luck. I do not have a good understanding of which hook is for which purpose. How can I go about this? Rather than exporting the strings, it's easier to export a function that produces the strings, e.g. fig - function(name) system.file(file.path(fig, name), package=foo) to be called as fig(man.png) Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Selection/filtering from Data
? subset. You have most of the command already written John Kane Kingston ON Canada -Original Message- From: pnsinh...@gmail.com Sent: Mon, 16 Feb 2015 13:31:33 +0530 To: r-help@r-project.org Subject: [R] Selection/filtering from Data From a dataset , I want select age =36 and income1. How to do ? parth __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. Can't remember your password? Do you need a strong and secure password? Use Password manager! It stores your passwords protects your account. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] P-value from Matching
On 16 Feb 2015, at 00:31 , Rob Wood robert.w...@adelphigroup.com wrote: Hi all, When using the match command from the matching package, the output reports the treatment effect, standard error, t-statistic and a p-value. Which test is used to generate this p-value, or how us it generated? I would assume this information to be available via the references on the help page for Match() (sic) in the Matching (sic) package. Peter D. Thanks, Rob. -- View this message in context: http://r.789695.n4.nabble.com/P-value-from-Matching-tp4703309.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] ggplot2 shifting bars to only overlap in groups
Lovely, a much more elegant solution. John Kane Kingston ON Canada -Original Message- From: hyil...@gmail.com Sent: Mon, 16 Feb 2015 02:30:09 +0800 To: jrkrid...@inbox.com, djmu...@gmail.com Subject: Re: [R] ggplot2 shifting bars to only overlap in groups Thanks so much, John and Dennis (who did not respond in the mailing list for some reason). I feel quite obliged to keep you thinking about this. I do agree that not using the bar chart with error bars is a better option. And since *condition* is an important ordinal factor for me, it would be much better to have *condition* be positioned at a relative order. Thus, only color coding it as John's latest solution would not be optimal. It would have been better with the random data, but with my actual data, it does seem necessary to do a jitter for the *male* since it got clattered in the *west*. Here is the actual data along with the solution based on Dennis' code: ## data dat1 - structure(list(t = c(1.2454860689532, 0.627186899108052, 0.877176019393987, 1.26720638917869, 1.16906482219006, 0.889738853288831, 0.852034797572489, 1.30007600828822, 1.22896141479778, 0.820236562746995, 0.822197641624559, 1.39529772379005, 1.10479557445486, 0.760017179713665, 0.761340230517717, 1.11132156961026, 1.30042963441715, 0.811425854755042, 0.979421690403349, 1.3297658281305, 1.13377482477157, 0.895243910826397, 0.874181486658082, 1.15728885642541, 1.11121780853125, 0.703348405369258, 0.850897112058048, 1.14260584106012, 1.09383015337114, 0.911388765620587, 0.84622335453925, 1.09847968194129), condition = structure(c(4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L, 4L, 4L, 4L, 4L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 3L, 3L, 3L, 3L), .Label = c(c1, c2, c2, c4), class = factor), direction = structure(c(1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 2L), .Label = c(up, down), class = factor), location = structure(c(2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L, 2L, 1L), .Label = c(east, west), class = factor), gender = structure(c(2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), .Label = c(male, female), class = factor), ci = c(0.0307396796826649, 0.0302954863637637, 0.0400142340797275, 0.0527186825100342, 0.051675810189946, 0.0368383294010065, 0.0404823188495183, 0.0526312391852324, 0.0347332720922338, 0.0354587857740343, 0.0303368490163547, 0.0710445198259065, 0.0229339653012889, 0.0261217906562281, 0.0285673216713352, 0.0351642108247828, 0.0542657646932069, 0.0566816739316165, 0.0481239729953889, 0.0434272572423839, 0.0497366325101659, 0.0342004255233646, 0.0349733697554762, 0.0405364256564456, 0.0478372176424872, 0.0341294939361437, 0.0424566961614424, 0.0463489561778199, 0.0191707406475215, 0.0501106812754005, 0.0321562411182704, 0.0218613299095178)), .Names = c(t, condition, direction, location, gender, ci), row.names = c(NA, -32L ), class = data.frame) pp - ggplot(dat1, aes(x = condition, y = t, color = gender, linetype = direction)) + geom_errorbar(aes(ymin = t - ci, ymax = t + ci), position = position_dodge(width = 0.6), size = 1, width = 0.5) + geom_point(position = position_dodge(width = 0.6), size = 2.5) + facet_wrap(~ location) + scale_color_manual(values = c(blue, darkorange))+ theme_bw()+ scale_y_continuous(breaks=seq(0.6,1.5,0.1)) pp ## EOF I have also attached the output. Best ， He who is worthy to receive his days and nights is worthy to receive* all else* from you (and me). The Prophet, Gibran Kahlil FREE 3D EARTH SCREENSAVER - Watch the Earth right on your desktop! __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] split dataframe to several dataframes in R
Hi All,I have a dataframe called 'means' as shown below:iris1.csv - iris iris2.csv - iris names - c(iris1.csv, iris2.csv) dat - mget(names) lst4 - lapply(dat, function(x) apply(x[,-5], 2, mean)) # Build the new data frame means - as.data.frame(do.call(rbind, lst4)) means$source - names(lst4) means # Sepal.Length Sepal.Width Petal.Length Petal.Width isv source # iris1.csv 5.843.0573333.7581.199333 0.333 iris1.csv # iris2.csv 5.843.0573333.7581.199333 0.333 iris2.csvQUESTION: How can I split 'means' such that there are two files (dataframes) on my workspace:datframe 1# Sepal.Length Sepal.Width Petal.Length Petal.Width isv # iris1.csv 5.843.0573333.7581.199333 0.333dataframe 2:# Sepal.Length Sepal.Width Petal.Length Petal.Width isv# iris2.csv 5.843.0573333.758 1.199333 0.333 Many thanks,Asong. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Fitting gamma distribution
I'm very new to r-programming. I have rainfall data. I have tried to fit gamma
into my data but there is error. Anyone can help me please.
My rainfall data as I uploaded. When I try run the coding:
library(MASS)
KLT1-read.csv('C:/Users/User/Dropbox/PhD
Materials/Coding_PhD_Thesis/Kelantan_Average/K1.csv')
KLT-KLT1$Amount
fd_g - fitdistr(KLT, gamma)
Error in stats::optim(x = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, :
initial value in 'vmmin' is not finite
Anyone can help me? Thanks.
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Re: [R-es] Gráfico de serie temporal de datos
Estimado Rafa Poquet En lo personal creo que los datos están en una forma confusa, lo claro sería data.frame(fecha, medición), en cambio usted tiene la misma fecha en dos columnas. Javier Marcuzzi De: Rafa Poquet - gmail Enviado el: lunes, 16 de febrero de 2015 08:41 a.m. Para: R-help-es@r-project.org Hola a tod@s, Tengo un problema al tratar de elaborar un gr�fico de serie temporal con unos datos de consumo de agua. El caso es que se trata de lecturas de contador de la compa��a suministradora, realizadas en fechas concretas. A continuaci�n expongo la tabla de datos base: Inicio Fin agua_m3 1 23/11/2011 20/01/2012 593 2 20/01/2012 15/03/2012 774 3 15/03/2012 17/05/2012 853 4 17/05/2012 11/07/2012 932 5 11/07/2012 04/09/2012 4192 6 04/09/2012 30/10/2012 1348 7 30/10/2012 08/01/2013 777 8 08/01/2013 01/03/2013 630 9 01/03/2013 06/05/2013 980 10 06/05/2013 03/07/2013 1083 11 03/07/2013 02/09/2013 342 12 02/09/2013 29/10/2013 874 Lo que pretendo conseguir es un gr�fico en el que observar la evoluci�n temporal del consumo de agua, de modo que tenga: � En el eje �y�: consumo de agua (m3) en el periodo de lectura correspondiente (agua_m3) � En el eje �x�: fecha de lectura de contador (Inicio) Por ahora, si utilizo la funci�n plot.ts() el problema me surge a la hora de hacer que los valores temporales los coja directamente de la columna �Inicio� y que no los tenga que definir como una serie temporal regular (con observaciones cada mes, cada dos meses,�). Soy nuevo en la comunidad R por lo que si necesit�is de m�s datos o aclaraciones me lo dec�s. Gracias de antemano por vuestra colaboraci�n, Rafa --- El software de antivirus Avast ha analizado este correo electr�nico en busca de virus. [[alternative HTML version deleted]] [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] difference between max in summary table and max function
Thanks for the clarification. The basic error on my side was that I had
misunderstood the digits option in the summary, I had not understood that
even integer numbers might end up being rounded. Problem is solved now.
-Original Message-
From: Allen Bingham [mailto:aebingh...@gmail.com]
Sent: zaterdag 14 februari 2015 8:16
To: 'Martyn Byng'; r-help@r-project.org
Cc: Franckx Laurent
Subject: RE: [R] difference between max in summary table and max function
I thought I'd chime in ... submitting the following:
?summary
Provides the following documentation for the default for generalized argument
(other than class=data.frame, factor, or matrix):
## Default S3 method:
summary(object, ..., digits = max(3, getOption(digits)-3))
so passing along the object testrow w/o a corresponding argument for digits
... defaults to digits=4 (assuming your system has the same default option of
digits = 7 that mine does).
... and since later in the documentation it indicates that digits is an:
integer, used for number formatting with signif()
so noting that all of the values you reported from summary(testrow) all have
4 significant digits (including the Max. of 131500) (excepting the min value of
1), summary() is doing what it is documented to do.
... sorry for being pedantic --- but doing so to point out how helpful the ?
command can be sometimes.
Hope this helps.
__
Allen Bingham
Bingham Statistical Consulting
aebingh...@gmail.com
LinkedIn Profile: www.linkedin.com/pub/allen-bingham/3b/556/325
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Martyn Byng
Sent: Friday, February 13, 2015 3:15 AM
To: Franckx Laurent; r-help@r-project.org
Subject: Re: [R] difference between max in summary table and max function
Its a formatting thing, try
summary(testrow,digits=20)
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Franckx Laurent
Sent: 13 February 2015 11:00
To: r-help@r-project.org
Subject: [R] difference between max in summary table and max function
Dear all
I have found out that the max in the summary of an integer vector is not always
equal to the actual maximum of that vector. For example:
testrow - c(1:131509)
summary(testrow)
Min. 1st Qu. MedianMean 3rd Qu.Max.
1 32880 65760 65760 98630 131500
max(testrow)
[1] 131509
This has occurred both in a Windows and in a Linux environment.
Does this mean that the max value in the summary is only an approximation?
Best regards
Laurent Franckx, PhD
Senior researcher sustainable mobility
VITO NV | Boeretang 200 | 2400 Mol
Tel. ++ 32 14 33 58 22| mob. +32 479 25 59 07 | Skype: laurent.franckx |
laurent.fran...@vito.be | Twitter @LaurentFranckx
VITO Disclaimer: http://www.vito.be/e-maildisclaimer
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Re: [R] Loop over regression results
In R you would want to combine the results into a list. This could be done when
you create the regressions or afterwards. To repeat your example using a list:
data(iris)
taxon - levels(iris$Species)
mod - lapply(taxon, function (x) lm(Sepal.Width ~ Petal.Width,
data=iris, subset=Species==x))
names(mod) - taxon
lapply(mod, summary)
coeffs - do.call(rbind, lapply(mod, coef, [1))
coeffs
# (Intercept) Petal.Width
# setosa3.222051 0.8371922
# versicolor1.372863 1.0536478
# virginica 1.694773 0.6314052
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ronald Kölpin
Sent: Monday, February 16, 2015 7:37 AM
To: r-help@r-project.org
Subject: [R] Loop over regression results
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear all,
I have a problem when trying to present the results of several
regression. Say I have run several regressions on a dataset and saved
the different results (as in the mini example below). I then want to
loop over the regression results in order so save certain values to a
matrix (in order to put them into a paper or presentation).
Aside from the question of how to access certain information stored by
lm() (or printed by summary()) I can't seem to so loop over lm()
objects -- no matter whether they are stored in a vector or a list.
They are always evaluated immediately when called. I tried quote() or
substitute() but that didn't work either as Objects of type 'symbol'
cannot be indexed.
In Stata I would simply do something like
forvalues k = 1/3 {
quietly estimates restore mod`k'
// [...]
}
and I am looking for the R equivalent of that syntax.
Kind regard and thanks
RK
attach(iris)
mod1 - lm(Sepal.Width ~ Petal.Width, data=iris, subset=Species==setosa)
mod2 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==versicolor)
mod3 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==virginica)
summary(mod1); summary(mod2); summary(mod3)
mat - matrix(data=NA, nrow=3, ncol=5,
dimnames=list(1:3, c(Model, Intercept, p(T |T|),
Slope, R^2)))
mods - c(mod1, mod2, mod3)
for(k in 1:3)
{
mod - mods[k]
mat[2,k] - as.numeric(coef(mod))[1]
mat[3,k] - as.numeric(coef(mod))[1]
}
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[R] compute values by condition in DF by rownames
Hi All, how to compute only abs(value) 0.2 of DF X2 gbm_tcga lusc_tcga ucec_tcga_pub gbm_tcga 1.0 0.1405371906 -0.1028471639 gbm_tcga 1.0 0.0441343447 0.0135680550 gbm_tcga 1.0 -0.2000397119 0.0389718175 gbm_tcga 1.0 0.1456991556 0.0099470445 lusc_tcga 0.140537191 1.00 0.1330807083 lusc_tcga 0.044134345 1.00 0.0620247126 lusc_tcga -0.200039712 1.00 -0.1302395515 lusc_tcga 0.145699156 1.00 0.0417966700 ucec_tcga_pub -0.102847164 0.1330807083 1.00 ucec_tcga_pub 0.013568055 0.0620247126 1.00 ucec_tcga_pub 0.038971817 -0.1302395515 1.00 ucec_tcga_pub 0.009947045 0.0417966700 1.00 and return X2 gbm_tcga lusc_tcga ucec_tcga_pub gbm_tcga 4 1 0 lusc_tcga1 4 0 ucec_tcga_pub 0 0 4 Thanks Karim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Loop over regression results
Or for the slopes and t-values:
do.call(rbind, lapply(mod, function(x) summary(x)[[coefficients]][2,]))
Estimate Std. Error t value Pr(|t|)
setosa 0.8371922 0.5049134 1.658091 1.038211e-01
versicolor 1.0536478 0.1712595 6.152348 1.41e-07
virginica 0.6314052 0.1428938 4.418702 5.647610e-05
David C
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David L Carlson
Sent: Monday, February 16, 2015 8:52 AM
To: Ronald Kölpin; r-help@r-project.org
Subject: Re: [R] Loop over regression results
In R you would want to combine the results into a list. This could be done when
you create the regressions or afterwards. To repeat your example using a list:
data(iris)
taxon - levels(iris$Species)
mod - lapply(taxon, function (x) lm(Sepal.Width ~ Petal.Width,
data=iris, subset=Species==x))
names(mod) - taxon
lapply(mod, summary)
coeffs - do.call(rbind, lapply(mod, coef, [1))
coeffs
# (Intercept) Petal.Width
# setosa3.222051 0.8371922
# versicolor1.372863 1.0536478
# virginica 1.694773 0.6314052
-
David L Carlson
Department of Anthropology
Texas AM University
College Station, TX 77840-4352
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Ronald Kölpin
Sent: Monday, February 16, 2015 7:37 AM
To: r-help@r-project.org
Subject: [R] Loop over regression results
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
Dear all,
I have a problem when trying to present the results of several
regression. Say I have run several regressions on a dataset and saved
the different results (as in the mini example below). I then want to
loop over the regression results in order so save certain values to a
matrix (in order to put them into a paper or presentation).
Aside from the question of how to access certain information stored by
lm() (or printed by summary()) I can't seem to so loop over lm()
objects -- no matter whether they are stored in a vector or a list.
They are always evaluated immediately when called. I tried quote() or
substitute() but that didn't work either as Objects of type 'symbol'
cannot be indexed.
In Stata I would simply do something like
forvalues k = 1/3 {
quietly estimates restore mod`k'
// [...]
}
and I am looking for the R equivalent of that syntax.
Kind regard and thanks
RK
attach(iris)
mod1 - lm(Sepal.Width ~ Petal.Width, data=iris, subset=Species==setosa)
mod2 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==versicolor)
mod3 - lm(Sepal.Width ~ Petal.Width, data=iris,
subset=Species==virginica)
summary(mod1); summary(mod2); summary(mod3)
mat - matrix(data=NA, nrow=3, ncol=5,
dimnames=list(1:3, c(Model, Intercept, p(T |T|),
Slope, R^2)))
mods - c(mod1, mod2, mod3)
for(k in 1:3)
{
mod - mods[k]
mat[2,k] - as.numeric(coef(mod))[1]
mat[3,k] - as.numeric(coef(mod))[1]
}
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Picking Best Discriminant Function Variables
Look at the function stepclass() in package klaR. - David L Carlson Department of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of David Moskowitz Sent: Sunday, February 15, 2015 11:34 AM To: n omranian via R-help Subject: [R] Picking Best Discriminant Function Variables Is there a way to have the LDA function give me the best 3 (or 4) predictor variables. When I put in all the variables, LDA uses all the variables, but I would like to know what would be the 3 (or 4) best to use out all the available variables and the coefficients for those. Here is the code I am using for Linear Discriminant Function library(MASS) results - lda(data$V1 ~ data$V2 + data$V3 + data$V4 + data$V5 + data$V6 + data$V7 + data$V8 + data$V9 + data$V10 + data$V11 + data$V12 + data$V13 + data$V14) Output: Coefficients of linear discriminants: LD1 LD2 data$V2 -0.4033997810.8717930699 data$V3 0.165254596 0.3053797325 data$V4 -0.3690752562.3458497486 data$V5 0.154797889 -0.1463807654 data$V6 -0.002163496-0.0004627565 data$V7 0.618052068 -0.0322128171 data$V8 -1.661191235 -0.4919980543 data$V9 -1.495818440 -1.6309537953 data$V10 0.134092628 -0.3070875776 data$V11 0.3550557100.2532306865 data$V12 -0.818036073-1.5156344987 data$V13 -1.1575593760.0511839665 data$V14 -0.0026912060.0028529846 So in the above example, I would like the LDA to return to me the 3 best predictors out of the 13 available. Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] compute values by condition in DF by rownames
I named your data frame temp aggregate(temp[,-1], list(temp[,1]), function(x) sum(abs(x).2)) Group.1 gbm_tcga lusc_tcga ucec_tcga_pub 1 gbm_tcga4 1 0 2 lusc_tcga1 4 0 3 ucec_tcga_pub0 0 4 Cheers Petr -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Karim Mezhoud Sent: Monday, February 16, 2015 4:10 PM To: r-help@r-project.org Subject: [R] compute values by condition in DF by rownames Hi All, how to compute only abs(value) 0.2 of DF X2 gbm_tcga lusc_tcga ucec_tcga_pub gbm_tcga 1.0 0.1405371906 -0.1028471639 gbm_tcga 1.0 0.0441343447 0.0135680550 gbm_tcga 1.0 -0.2000397119 0.0389718175 gbm_tcga 1.0 0.1456991556 0.0099470445 lusc_tcga 0.140537191 1.00 0.1330807083 lusc_tcga 0.044134345 1.00 0.0620247126 lusc_tcga -0.200039712 1.00 -0.1302395515 lusc_tcga 0.145699156 1.00 0.0417966700 ucec_tcga_pub -0.102847164 0.1330807083 1.00 ucec_tcga_pub 0.013568055 0.0620247126 1.00 ucec_tcga_pub 0.038971817 -0.1302395515 1.00 ucec_tcga_pub 0.009947045 0.0417966700 1.00 and return X2 gbm_tcga lusc_tcga ucec_tcga_pub gbm_tcga 4 1 0 lusc_tcga1 4 0 ucec_tcga_pub 0 0 4 Thanks Karim [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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. Tento e-mail a jakékoliv k němu připojené dokumenty jsou důvěrné a jsou určeny pouze jeho adresátům. Jestliže jste obdržel(a) tento e-mail omylem, informujte laskavě neprodleně jeho odesílatele. Obsah tohoto emailu i s přílohami a jeho kopie vymažte ze svého systému. Nejste-li zamýšleným adresátem tohoto emailu, nejste oprávněni tento email jakkoliv užívat, rozšiřovat, kopírovat či zveřejňovat. Odesílatel e-mailu neodpovídá za eventuální škodu způsobenou modifikacemi či zpožděním přenosu e-mailu. V případě, že je tento e-mail součástí obchodního jednání: - vyhrazuje si odesílatel právo ukončit kdykoliv jednání o uzavření smlouvy, a to z jakéhokoliv důvodu i bez uvedení důvodu. - a obsahuje-li nabídku, je adresát oprávněn nabídku bezodkladně přijmout; Odesílatel tohoto e-mailu (nabídky) vylučuje přijetí nabídky ze strany příjemce s dodatkem či odchylkou. - trvá odesílatel na tom, že příslušná smlouva je uzavřena teprve výslovným dosažením shody na všech jejích náležitostech. - odesílatel tohoto emailu informuje, že není oprávněn uzavírat za společnost žádné smlouvy s výjimkou případů, kdy k tomu byl písemně zmocněn nebo písemně pověřen a takové pověření nebo plná moc byly adresátovi tohoto emailu případně osobě, kterou adresát zastupuje, předloženy nebo jejich existence je adresátovi či osobě jím zastoupené známá. This e-mail and any documents attached to it may be confidential and are intended only for its intended recipients. If you received this e-mail by mistake, please immediately inform its sender. Delete the contents of this e-mail with all attachments and its copies from your system. If you are not the intended recipient of this e-mail, you are not authorized to use, disseminate, copy or disclose this e-mail in any manner. The sender of this e-mail shall not be liable for any possible damage caused by modifications of the e-mail or by delay with transfer of the email. In case that this e-mail forms part of business dealings: - the sender reserves the right to end negotiations about entering into a contract in any time, for any reason, and without stating any reasoning. - if the e-mail contains an offer, the recipient is entitled to immediately accept such offer; The sender of this e-mail (offer) excludes any acceptance of the offer on the part of the recipient containing any amendment or variation. - the sender insists on that the respective contract is concluded only upon an express mutual agreement on all its aspects. - the sender of this e-mail informs that he/she is not authorized to enter into any contracts on behalf of the company except for cases in which he/she is expressly authorized to do so in writing, and such authorization or power of attorney is submitted to the recipient or the person represented by the recipient, or the existence of such authorization is known to the recipient of the person represented by the recipient. __
Re: [R] Selection/filtering from Data
Hi Parth, Assume that your dataset is in the form of a data frame, named psdf. psdf-data.frame(age=sample(0:100,50),income=sample(8000:12000,50)) selectdf-subset(psdf,age = 36 income 1) Jim On Mon, Feb 16, 2015 at 7:01 PM, Partha Sinha pnsinh...@gmail.com wrote: From a dataset , I want select age =36 and income1. How to do ? parth __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] problems with packages installation
On 16/02/2015 07:49, PIKAL Petr wrote: Hi Jeff -Original Message- From: Jeff Newmiller [mailto:jdnew...@dcn.davis.ca.us] Sent: Friday, February 13, 2015 3:56 PM To: PIKAL Petr; r-help@r-project.org Cc: Richard M. Heiberger Subject: RE: [R] problems with packages installation I agree that the PG muddies the water a bit on this topic. However, the web page from which you downloaded the patched version warns: This is not an official release of R. Please check bugs in this version against the official release before reporting them. When a new version is under development, the number of bugs often increases temporarily by quite a bit. Any use you make of a patched or development version is at your own risk, and should be undertaken for two reasons only: to address a specific problem you are having with the released version, or because you want to help test the unreleased version out of the goodness of your heart (bless you). In either case, Yes, that is why I use devel versions but I was not able to find a bug (yet). chatter about bugs you encounter that are not in the released version belongs on R-devel. It was hard to tell what is the problem. It could be that our IT people silently changed a firewall or something, which would prevent package installation from menu. Far too often I am not able to open some web pages e.g. (https://stat.ethz.ch/mailman/listinfo/r-devel) and I do not have power to change it. However it seems to be that this particular problem is in R menu procedure for installation as I wrote in my previous mail. If you can reproduce the problem in a released version, posting on R- help would be more appropriate, but if you are sure it is a bug then filling a report is the right thing to do. It is hard to reproduce and I believe that the problem is probably already known by R core (I thank them for their marvellous work). AFAIK it was already fixed in R-devel at the time of posting. R-devel is 'under development', and at least until it reaches 'alpha' status the R-devel list is the place to report problems (if they persist for a few days and after checking with the current version). For me it is solved by install.packages function. Actually that is what the menu calls. The difference is the arguments you used (an explicit package rather than NULL for a menu). Thanks Best regards Petr -- Brian D. Ripley, rip...@stats.ox.ac.uk Emeritus Professor of Applied Statistics, University of Oxford 1 South Parks Road, Oxford OX1 3TG, UK __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Selection/filtering from Data
From a dataset , I want select age =36 and income1. How to do ? parth __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Pass additional arguments to do.call(grid.arrange, plots)
On 16/02/15 13:36, Bingzhang Chen wrote: Hi R users, I have a problem on how to pass an extra argument to do. call: The example codes are: # require(ggplot2) plots = lapply(1:5, function(.x) qplot(1:10,rnorm(10), main=paste(plot,.x))) require(gridExtra) do.call(grid.arrange, plots) #- I want to force the composite figures into 2 rows by adding 'now = 2' You mean 'nrow = 2', but I wasn't fooled for an instant! :-) in the function 'grid.arrange'. How can I do it? I searched on google but could not find a workable solution. I am working on RStudio 0.98.1102 on OSX Yosemite 10.10.2. Try: do.call(grid.arrange,c(plots,list(nrow=2)) The second argument to do.call() consists of a list constituting the arguments to the first argument. So just append the argument you want to use, to this list. Easy, once you know. :-) cheers, Rolf Turner P. S. Thank you for providing a clear and reproducible example. R. T. -- Rolf Turner Technical Editor ANZJS Department of Statistics University of Auckland Phone: +64-9-373-7599 ext. 88276 Home phone: +64-9-480-4619 __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] Nonlinear integer programming (again)
Zwick, Rebecca J RZwick at ETS.ORG writes: Oddly, Excel's Solver will produce a solution to such problems but (1) I don't trust it and (2) it cannot handle a large number of constraints. [...] My question is whether there is an R package that can handle this problem. There are not many free integer (nonlinear) programming (IP, INLP) solvers available. You could send your problem to one of the MINLP solvers at NEOS: http://neos.mcs.anl.gov/neos/solvers/ [See the list of relevant NEOS solvers (commercial and free) on this page: http://www.neos-guide.org/content/mixed-integer-nonlinear-programming] You can also use the 'rneos' package to send your request to one of these Web services, but most of the time I find it easier to directly fill the solver template. Please note that you have to format your problem and data according to the solver's needs, i.e. likely in AMLP or GAMS syntax. If you intend to solve such problems more often, I'd suggest to download one of the commercial solvers with academic licenses (e.g., Knitro, Gurobi, ...) and to install a corresponding R package to access the solver. For more information see the Optimization task view. I would *never* trust Excel for these kinds of problems... By the way, I may not correctly understand your objective, but perhaps you can formulate it as maximizing a quadratic function (with constraints). __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] #library(party) - Compare predicted results for ctree
Hello, I've created a ctree model called fit using 15 input variables for a factor predicted variable Response (YES/NO). When I run the following : table(predict(fit2), training_data$response) I get the following result: NO YES NO 48694 480 YES 0 0 It appears that the NO responses are predicted with 100% accuracy and the YES response are predicted with 0% accuracy. Why is this happening? It's because of my data or it's something in ctree algorithm? Thanks! Rodica __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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] #library(party) - Compare predicted results for ctree
On Mon, 16 Feb 2015, Rodica Coderie via R-help wrote: Hello, I've created a ctree model called fit using 15 input variables for a factor predicted variable Response (YES/NO). When I run the following : table(predict(fit2), training_data$response) I get the following result: NO YES NO 48694 480 YES 0 0 It appears that the NO responses are predicted with 100% accuracy and the YES response are predicted with 0% accuracy. Why is this happening? It's because of my data or it's something in ctree algorithm? Your data has less than 1% of YES observations and I would guess that the tree cannot separate these in a way such that majority voting gives a YES prediction. You might consider a different cutoff (other than 50%) or downsampling the NO observations. Thanks! Rodica __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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-es] Gráfico de serie temporal de datos
Hola a tod@s, Tengo un problema al tratar de elaborar un gr�fico de serie temporal con unos datos de consumo de agua. El caso es que se trata de lecturas de contador de la compa��a suministradora, realizadas en fechas concretas. A continuaci�n expongo la tabla de datos base: Inicio Fin agua_m3 1 23/11/2011 20/01/2012 593 2 20/01/2012 15/03/2012 774 3 15/03/2012 17/05/2012 853 4 17/05/2012 11/07/2012 932 5 11/07/2012 04/09/2012 4192 6 04/09/2012 30/10/2012 1348 7 30/10/2012 08/01/2013 777 8 08/01/2013 01/03/2013 630 9 01/03/2013 06/05/2013 980 10 06/05/2013 03/07/2013 1083 11 03/07/2013 02/09/2013 342 12 02/09/2013 29/10/2013 874 Lo que pretendo conseguir es un gr�fico en el que observar la evoluci�n temporal del consumo de agua, de modo que tenga: � En el eje �y�: consumo de agua (m3) en el periodo de lectura correspondiente (agua_m3) � En el eje �x�: fecha de lectura de contador (Inicio) Por ahora, si utilizo la funci�n plot.ts() el problema me surge a la hora de hacer que los valores temporales los coja directamente de la columna �Inicio� y que no los tenga que definir como una serie temporal regular (con observaciones cada mes, cada dos meses,�). Soy nuevo en la comunidad R por lo que si necesit�is de m�s datos o aclaraciones me lo dec�s. Gracias de antemano por vuestra colaboraci�n, Rafa --- El software de antivirus Avast ha analizado este correo electr�nico en busca de virus. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
Re: [R] How to get the error and error variance after HB using bayesm
Hi Arnab, Actually, I don´t think so as in Bayesian Regression the estimation converges towards the prior distribution and not a point estimate like in the frequentist approach. But I’m not very sure honestly. I used a Bayesian Logit Model in R (bayesm, rhierBinLogit by Rossi) to calculate individual pert-worth. I reviewer asked me to control for error variance as my results highly depend on it. But I do not know how to deal with using the Bayesian approach. Cheers, Michael From: arnabkrma...@gmail.com [mailto:arnabkrma...@gmail.com] On Behalf Of ARNAB KR MAITY Sent: Freitag, 13. Februar 2015 20:46 To: Michael Langmaack Cc: r-help@r-project.org Subject: Re: [R] How to get the error and error variance after HB using bayesm Hello Michael, I have a question here. Does Bayesian paradigm deal with MSE kind of stuff? Thanks Regards, Arnab Arnab Kumar Maity Graduate Teaching Assistant Division of Statistics Northern Illinois University DeKalb, IL 60115 Email: ma...@math.niu.edumailto:ma...@math.niu.edu Ph: 779-777-3428 On Fri, Feb 13, 2015 at 4:45 AM, Michael Langmaack michael.langma...@the-klu.orgmailto:michael.langma...@the-klu.org wrote: Hello all, I have a question concerning bayesian regression in R using the package bayesm (version 2.2-5) by Peter Rossi. I have binary choice data and estimated individual coefficients using the command rhierBinLogit(Data=Data,Mcmc=Mcmc). That worked out properly, conversion plots, histograms, parameter are fine. No a have to compute the errors and the error variance or something like the MSE. But I do not know how to do. I did not find a hint so far. I would be more than happy if anybody can help me. Thanks in advance! Best regards, Michael [[alternative HTML version deleted]] __ R-help@r-project.orgmailto:R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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 -- To UNSUBSCRIBE and more, see 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] How do I provide path character string to extdata content?
Dear helpeRs, I have some png files in the inst/extdata directory of a package (e.g., man.png), and I want to provide character strings containing the paths to these files; of course, these path strings have to depend on the individual installation. So far, I have written a function that - if called - writes the correct strings to the global environment, but that is not CRAN compatible. Ideally, I want the package namespace to export these text strings. I have played with .onLoad without luck. I do not have a good understanding of which hook is for which purpose. How can I go about this? Best, Ulrike __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
