Re: [R] stringr::str_split_fixed query
Dear David,
str_split_fixed calls str_locate_all, which gives
str_locate_all(ab, )
## [[1]]
## start end
## [1,] 1 0
## [2,] 2 1
##
in your example, since is a character of length 1. substring() is
probably more intuitive to get your expected result:
substring(ab, 1:2, 1:2)
## [1] a b
David Barron dnbar...@gmail.com schrieb am Wed, 14. Jan 18:47:
I'm puzzled as to why I get this behaviour with str_split_fixed in the
stringr package.
stringr::str_split_fixed('ab','',2)
[,1] [,2]
[1,]ab
stringr::str_split_fixed('ab','',3)
[,1] [,2] [,3]
[1,]a b
In the first example, I was expecting to get
[,1] [,2]
[1,] a b
Can someone explain?
Thanks,
David
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Re: [R] Approximation of a function from R^2 to R
Thanks for the reply. I have datapoints. What I actually want to do is to estimate a joint density function. I used npudens() to estimate the kernel density of two vectors and got the densities at the evaluation points. Now I want an approximation for this so I can have an estimate for different points. -- View this message in context: http://r.789695.n4.nabble.com/Approximation-of-a-function-from-R-2-to-R-tp4701805p4701828.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] R 3.1.2 mle2() function on Windows 7 Error and multiple solutions
Hi,
I tried your problem with optimx package. I found a better solution than that
found by mle2.
?library(optimx)
# the objective function needs to be re-written
LL2 - function(par,y) {
lambda - par[1]
alpha - par[2]
beta - par[3]
R = Nweibull(y,lambda,alpha,beta)
-sum(log(R))
}
optimx(fn=LL2, par=c(.01,325,.8),y=y, lower=c(.1,.1,.1),upper =
c(Inf, Inf,Inf),control=list(all.methods=TRUE))
# Look at the solution found by `nlminb' and `nmkb'. This is the optimal one.
This log-likelihood is larger than that of mle2 and other optimizers in optimx.
If this solution is not what you are looking for, your problem may be poorly
scaled. First, make sure that the likelihood is coded correctly. If it is
correct, then you may need to improve the scaling of the problem.
Hope this is helpful,
Ravi
[[alternative HTML version deleted]]
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Re: [R] stringr::str_split_fixed query
FWIW this is fixed in the dev version of stringr which uses stringi
under the hood:
stringr::str_split_fixed('ab','',2)
[,1] [,2]
[1,] a b
stringr::str_split_fixed('ab','',3)
[,1] [,2] [,3]
[1,] a b
Hadley
On Wed, Jan 14, 2015 at 12:47 PM, David Barron dnbar...@gmail.com wrote:
I'm puzzled as to why I get this behaviour with str_split_fixed in the
stringr package.
stringr::str_split_fixed('ab','',2)
[,1] [,2]
[1,]ab
stringr::str_split_fixed('ab','',3)
[,1] [,2] [,3]
[1,]a b
In the first example, I was expecting to get
[,1] [,2]
[1,] a b
Can someone explain?
Thanks,
David
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--
http://had.co.nz/
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Re: [R] R 3.1.2 mle2() function on Windows 7 Error and multiple solutions
A more important point that want to make is that I find few people taking
advantage of the comparative evaluation or benchmarking ability of optimx.
There is no uniformly best optimizer for all problems. Different ones turn
out to perform better for different problems and it is quite difficult to know
a priori which one would be the best for my problem. Thus, the benchmarking
capability provided by optimx is a powerful feature.
Ravi
From: Ben Bolker bbol...@gmail.com
Sent: Thursday, January 15, 2015 9:29 AM
To: Ravi Varadhan; R-Help
Cc: malqura...@ksu.edu.sa
Subject: Re: R 3.1.2 mle2() function on Windows 7 Error and multiple solutions
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
For what it's worth, you can use either nlminb (directly) or optimx
within the mle2 wrapper by specifying the 'optimizer' parameter ...
this gives you flexibility in optimization along with the convenience
of mle2 (likelihood ratio tests via anova(), likelihood profiling, etc.)
On 15-01-15 09:26 AM, Ravi Varadhan wrote:
Hi,
I tried your problem with optimx package. I found a better
solution than that found by mle2.
?library(optimx)
# the objective function needs to be re-written
LL2 - function(par,y) {
lambda - par[1] alpha - par[2] beta - par[3] R =
Nweibull(y,lambda,alpha,beta)
-sum(log(R)) }
optimx(fn=LL2, par=c(.01,325,.8),y=y,
lower=c(.1,.1,.1),upper = c(Inf,
Inf,Inf),control=list(all.methods=TRUE))
# Look at the solution found by `nlminb' and `nmkb'. This is the
optimal one. This log-likelihood is larger than that of mle2 and
other optimizers in optimx.
If this solution is not what you are looking for, your problem may
be poorly scaled. First, make sure that the likelihood is coded
correctly. If it is correct, then you may need to improve the
scaling of the problem.
Hope this is helpful,
Ravi
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Re: [R] passing elements of named vector / named list to function
It x is a list...
do.call(fun,x)
You should keep external data input away from this code construct to avoid
intentional or unintentional misuse of your fun.
If your toy example were your actual usage I would suggest the collapse
argument of paste.
---
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 January 15, 2015 6:25:43 AM PST, Rainer M Krug rai...@krugs.de wrote:
Hi
Following scenario: I have a function fun
--8---cut here---start-8---
fun - function(A, B, C, ...){paste(A, B, C, ...)}
--8---cut here---end---8---
and x defined as follow
--8---cut here---start-8---
x - 1:5
names(x) - LETTERS[x]
--8---cut here---end---8---
now I want to pass the *elements* of x to fun as named arguments, i.e.
,
| fun(A=1, B=2, C=3, D=4, E=5)
| [1] 1 2 3 4 5
`
The below examples obviously do not work:
,
| fun(x)
| Error in paste(A, B, C, list(...)) :
| argument B is missing, with no default
| fun(unlist(x))
| Error in paste(A, B, C, list(...)) :
| argument B is missing, with no default
`
How can I extract from x the elements and pass them on to fun()?
I could easily change x to a list() if this would be easier.
--8---cut here---start-8---
x - list(A=1, B=2, C=3, D=4, E=5)
--8---cut here---end---8---
In my actual program, x can have different elements as well as fun -
this is decided programmatically.
Any suggestions how I can achieve this?
Thanks,
Rainer
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[R] Mlogit: Error in model.frame.default/ Error in solve.default
Dear R users, 1) My problem in short: Mlogit cannot calculate certain conditional models. 2) My database: The target was a logistical regression analysis and a probability function which should include generic coefficients and alternative-specific ones. The database was a survey, the dependent variable consisted of six possible choices. I worked with two databases: First, the true survey data (out of the six choices, one is quite dominant with a share of around 40%, two have around 20% each, and the remaining are marginal). Second, the survey with randomized values for both choice and independent variables to avoid the possible problems with the less uniform distribution of the choice variable in the true set. Both datasets include 725 individuals with twelve variables due to receive a generic coefficient and 25 individual ones which should have an alternative specific coefficient. I have already to note that the problems above emerged when estimating with only one independent variable. Before the estimate, both datasets were brought into the required long format. The original data in the long format looked ok. 3) The estimates and the issues. I started in both datasets with an estimate with only one variable with a generic coefficient. I used the following order: mlogit(choice ~ variable, data = survey_long) R’s comment was “Error in solve.default(H, g[!fixed]) : system is computationally singular: reciprocal condition number = ...”. I was able to get an estimate for such a comparably simple model when using mlogit(choice ~ variable | 0, data = survey_long). However, the output did not contain estimates on the alternative specific coefficents for the intercept. I continued by gradually adding variables with generic coefficients. If it was more than five of them, R’s message was “Error in solve.default(H, g[!fixed]) : system is computationally singular: reciprocal condition number = ...”. One of the variables had the problem “Error in model.frame.default(terms(formula, lhs = lhs, rhs = rhs, data = data), : variable lengths differ (found for 'C15_CC_Dist')”. I compared the length for all the variables, and it was all the same. Another thing: It was impossible to add variables with an alternative specific coefficient: “Error in solve.default(H, g[!fixed]) :system is computationally singular: reciprocal condition number = ...”. These statements appeared both when using the true survey as well as when using the randomized data. Does anyone have any comment or help on that? What can I do? I can send the datasets if required. Any comment would be helpful. Best regards, LvC -- View this message in context: http://r.789695.n4.nabble.com/Mlogit-Error-in-model-frame-default-Error-in-solve-default-tp4701843.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.
[R] How to get items for both LHS and RHS for only specific columns in arules?
Hi all, I have a question about the arules package in R. I hope the example
tables are readable in your email, otherwise you can view it in the
question.txt in the attachment.Within the apriori function in the arules
package, I want the outcome to only contain these two variables in the LHS
HouseOwnerFlag=0 and HouseOwnerFlag=1. The RHS should only contain attributes
from the column Product. For instance: lhs rhs
support confidence lift1 {HouseOwnerFlag=0} =
{Product=SV 16xDVD M360 Black} 0.250 0.250 1.002
{HouseOwnerFlag=1} = {Product=Adventure Works 26 720p} 0.250
0.250 1.003 {HouseOwnerFlag=0} = {Product=Litware Wall Lamp E3015
Silver}0.167 0.333 1.334 {HouseOwnerFlag=1} = {Product=Contoso
Coffee Maker 5C E0900} 0.167 0.333 1.33So now I use the following:
rules - apriori(sales, parameter=list(support =0.01, confidence =0.8,
minlen=2), appearance = list(lhs=c(HouseOwnerFlag=0,
HouseOwnerFlag=1)))Then I use this to ensure that only the Product column is
on the RHS: inspect( subset( rules, subset = rhs %pin% Product= ) )The
outcome is like this (for the sake of readability, I omitted the colomns for
support, lift, confidence):lhs
rhs 1 {ProductKey=153, IncomeGroup=Moderate,
BrandName=Adventure Works } = {Product=SV 16xDVD M360 Black} 2
{ProductKey=176, MaritalStatus=M, ProductCategoryName=TV and Video } =
{Product=Adventure Works 26 720p} 3 {BrandName=Southridge Video,
NumberChildrenAtHome=0 }= {Product=Litware Wall Lamp E3015
Silver} 4 {HouseOwnerFlag=1, BrandName=Southridge Video, ProductKey=170 }
= {Product=Contoso Coffee Maker 5C E0900} So apparently the LHS is able to
contain every possible column, not just HouseOwnerFlag like I specified. I see
that I can put default=rhs in the apriori function to prevent this, like so:
rules - apriori(sales, parameter=list(support =0.001, confidence =0.5,
minlen=2), appearance = list(lhs=c(HouseOwnerFlag=0, HouseOwnerFlag=1),
default=rhs)) Then upon inspecting (without the subset part, just
inspect(rules), there are far less rules (7) than before but it does indeed
only containHouseOwnerFlag in the LHS:lhs rhs
support confidence lift1 {HouseOwnerFlag=0} =
{MaritalStatus=S}0.250 0.250 1.002
{HouseOwnerFlag=1} = {Gender=M} 0.250 0.250
1.003 {HouseOwnerFlag=0} = {NumberChildrenAtHome=0} 0.167
0.333 1.334 {HouseOwnerFlag=1} = {Gender=M}
0.167 0.333 1.33However on the RHS there's nothing from the column
Product in the RHS. So it has no use to inspect it with subset as ofcourse it
would return null. I tested it several times with different support numbers to
experiment and see if Product would appear or not, but the 7 same rules remain
the same.So my question is, how can I specify both the LHS (HouseOwnerFlag) and
RHS (Product)? What am I doing wrong?You can reproduce this problem by
downloading this testdataset from the attachment (testdf.txt) or via this
link:https://www.dropbox.com/s/tax5xalac5xgxtf/testdf.txt?dl=0 Mind you, I only
took the first 20 rows from a huge dataset (12 million), so the output here
won't have the same product names as the example I displayed above. But the
problem still remains the same. (if you would like to have the entire dataset I
can email it ofcourse). I want to be able to get only HouseOwnerFlag=0 and/or
HouseOwnerFlag=1 on the LHS and the column Product on the RHS. I asked this
question on other forum before, but no response at all unfortunately. Since
this mailinglist is dedicated to R only I thought you guys might be able to
help me. Thanks in advance! I look forward to hear from you.Kim
sales - structure(list(ProductCategoryName = structure(c(6L, 6L, 2L,
2L, 2L, 7L, 3L, 3L, 3L, 3L,
3L, 3L, 3L, 3L, 3L, 2L, 2L, 2L, 2L,
2L), .Label = c(Audio,
Cameras and camcorders , Cell phones,
Computers,
Games and Toys, Home Appliances, Music, Movies and Audio Books,
TV and
Video), class = factor), ProductSubcategory = structure(c(26L,
26L, 11L, 12L, 12L, 21L,
27L, 27L, 27L, 27L, 27L, 27L, 27L, 27L,
27L, 12L, 12L, 12L, 12L,
12L), .Label = c(Air Conditioners,
Re: [R] sparse matrix from vector outer product
thanks, that sounds good! Martin Maechler maech...@stat.math.ethz.ch schrieb am Thu Jan 15 2015 at 09:07:04: Philipp A flying-sh...@web.de on Wed, 14 Jan 2015 14:02:40 + writes: Hi, creating a matrix from two vectors a, b by multiplying each combination can be done e.g. via a %*% t(b) or via outer(a, b) # default for third argument is '*' really the best (most efficient) way would be tcrossprod(a, b) But this yields a normal matrix. of course. Please always use small self-contained example code, here, e.g., a - numeric(17); a[3*(1:5)] - 10*(5:1) b - numeric(12); b[c(2,3,7,11)] - 1:3 Is there an efficient way to create sparse matrices (from the Matrix package) like that? Right now i’m doing a.sparse = as(a, 'sparseVector') b.sparse = as(t(b), 'sparseMatrix') a.sparse %*% b.sparse but this strikes me as wasteful. not really wasteful I think. But there is a nicer and more efficient way : require(Matrix) tcrossprod(as(a, sparseVector), as(b, sparseVector)) now also gives 17 x 12 sparse Matrix of class dgCMatrix [1,] . . . . . . . . . . . . [2,] . . . . . . . . . . . . [3,] . 50 100 . . . 150 . . . 50 . [4,] . . . . . . . . . . . . [5,] . . . . . . . . . . . . [6,] . 40 80 . . . 120 . . . 40 . [7,] . . . . . . . . . . . . [8,] . . . . . . . . . . . . [9,] . 30 60 . . . 90 . . . 30 . [10,] . . . . . . . . . . . . [11,] . . . . . . . . . . . . [12,] . 20 40 . . . 60 . . . 20 . [13,] . . . . . . . . . . . . [14,] . . . . . . . . . . . . [15,] . 10 20 . . . 30 . . . 10 . [16,] . . . . . . . . . . . . [17,] . . . . . . . . . . . . [[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] Iteratively subsetting data by factor level across multiple variables
Hi R experts! I would like to have a scripted solution that will iteratively subset data across many variables per factor level of each variable. To illustrate, if I create a dataframe (df) by: variation - c(A,B,C,D) element1 - as.factor(c(0,1,0,1)) element2 - as.factor(c(0,0,1,1)) response - c(4,2,6,2) df - data.frame(variation,element1,element2,response) I would like a function that would allow me to subset the data into four groups and perform analysis across the groups. One group for each of the two factor levels across two variables. In this example its fairly easy because I only have two variables with two levels each, but would I would like this to be extendable across situations where I am dealing with more than 2 variables and/or more than two factor levels per variable. I am looking for a result that will mimic the output of the following: element1_level0 - subset(df,df$element1==0) element1_level1 - subset(df,df$element1==1) element2_level0 - subset(df,df$element2==0) element2_level1 - subset(df,df$element2==1) The purpose would be to perform analysis on the df across each subset. Simplistically this could be represented as follows: mean(element1_level0$response) mean(element1_level1$response) mean(element2_level0$response) mean(element2_level1$response) Thanks, Reid [[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] Latest version of Rtools is incompatible with latest version of R !!
I have installed R version 3.1.2. I tried to install RTools32.exe which is the latest version (using this link - http://cran.r-project.org/bin/windows/Rtools/) ! However on using the function find_rtools() an error message was displayed which said --- Rtools is required to build R packages, but no version of Rtools compatible with R 3.1.2 was found. (Only the following incompatible version(s) of Rtools were found:3.2)Please download and install Rtools 3.1 from http://cran.r-project.org/bin/windows/Rtools/ and then run find_rtools(). I want to know why the latest version of R is not supporting the latest Rtools! Any suggestions? Thanks for your time,Prameet [[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] Iteratively subsetting data by factor level across multiple variables
There are lots of ways to do this. You have to decide on how you want to organize the results. Here are two ways that use only core R packages. Many people like the plyr package for this split-data/analyze-parts/combine-results sort of thing. df - data.frame(x=1:27,response=log2(1:27), g1=rep(letters[1:2],len=27),g2=rep(LETTERS[24:26],c(10,10,7))) s - split(seq_len(nrow(df)), df[c(g1,g2)]) mean(subset(df, df$g1==a df$g2==Z)$response) [1] 4.578656 vapply(s, function(si)mean(df$response[si]), FUN.VALUE=0) # a.Z part is previous result a.X b.X a.Y b.Y a.Z b.Z 1.976834 2.381378 3.880430 3.976834 4.578656 4.581611 coef(lm(response~x, data=subset(df, df$g1==a df$g2==Z))) # regression example (Intercept) x 3.12905040 0.06040022 vapply(s, function(si)coef(lm(response ~ x, data=df[si,])), FUN.VALUE=rep(0,2)) a.X b.Xa.Yb.Ya.Zb.Z (Intercept) 0.0862735 0.6882213 2.40741927 2.50763309 3.12905040 3.13556268 x 0.3781121 0.2821928 0.09820075 0.09182506 0.06040022 0.06025202 For the particular case of computing means of a partition of the data you can use lm() once, which gives the same numbers organized in a different way: coef(lm(response ~ x * (g1:g2) - x - 1, data=df)) g1a:g2Xg1b:g2Xg1a:g2Yg1b:g2Yg1a:g2Zg1b:g2Z 0.08627350 0.68822126 2.40741927 2.50763309 3.12905040 3.13556268 x:g1a:g2X x:g1b:g2X x:g1a:g2Y x:g1b:g2Y x:g1a:g2Z x:g1b:g2Z 0.37811212 0.28219281 0.09820075 0.09182506 0.06040022 0.06025202 Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Jan 15, 2015 at 11:42 AM, Reid Bryant reidbry...@gmail.com wrote: Hi R experts! I would like to have a scripted solution that will iteratively subset data across many variables per factor level of each variable. To illustrate, if I create a dataframe (df) by: variation - c(A,B,C,D) element1 - as.factor(c(0,1,0,1)) element2 - as.factor(c(0,0,1,1)) response - c(4,2,6,2) df - data.frame(variation,element1,element2,response) I would like a function that would allow me to subset the data into four groups and perform analysis across the groups. One group for each of the two factor levels across two variables. In this example its fairly easy because I only have two variables with two levels each, but would I would like this to be extendable across situations where I am dealing with more than 2 variables and/or more than two factor levels per variable. I am looking for a result that will mimic the output of the following: element1_level0 - subset(df,df$element1==0) element1_level1 - subset(df,df$element1==1) element2_level0 - subset(df,df$element2==0) element2_level1 - subset(df,df$element2==1) The purpose would be to perform analysis on the df across each subset. Simplistically this could be represented as follows: mean(element1_level0$response) mean(element1_level1$response) mean(element2_level0$response) mean(element2_level1$response) Thanks, Reid [[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. [[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] Psych package: why am I receiving NA for many of the factor scores?
Dear Elizabeth, A correction to my suggestion: scaled - scale(mydata) wts - f4$weights scores -t( apply(scaled,1,function(x) colSums(x*wts,na.rm=TRUE))) #you need the colSums not the sum function Also, your confusion in getting the NAs with missing data was due to a bug in the fa function in the way it just ignored the missing statement. Thanks for catching that. It is now fixed and should be on CRAN real soon. Bill On Jan 14, 2015, at 9:39 AM, William Revelle li...@revelle.net wrote: Dear Elizabeth, Factor scores in the fa function are found by multiplying the standardized data by the factor weights using matrix multiplication. This will give scores only for subjects with complete data. However, if you want, you can create them yourself by standardizing your data and then multiplying them by the weights: mydata - rProjectSurveyDataJustVariables f4 - fa(my.data,4) #modify this to match your call wts - f4$wts scaleddata - scale(mydata) scores - apply(scaleddata,1,function(x) sum(x * wts,na.rm=TRUE)) #this will work with complete data, and impute factor scores for those cases with incomplete data. If the data are missing completely at random, this should give a reasonable answer. However, if the missingness has some structure to it, the imputed scores will be biased. This is a reasonable option to add to the fa function and I will do so. A side note. If you need help with a package, e.g., psych, you get faster responses by writing to the package author. I just happened to be browsing R-help when your question came in. Let me know if this solution works for you. Bill On Jan 13, 2015, at 7:46 PM, Elizabeth Barrett-Cheetham ebarrettcheet...@gmail.com wrote: Hello R Psych package users, Why am I receiving NA for many of the factor scores for individual observations? I'm assuming it is because there is quite a bit of missing data (denoted by NA). Are there any tricks in the psych package for getting a complete set of factor scores? My input is: rProjectSurveyDataJustVariables = read.csv(R Project Survey Data Just Variables.csv, header = TRUE) solution - fa(r = rProjectSurveyDataJustVariables, nfactors = 4, rotate = oblimin, fm = ml, scores = tenBerge, warnings = TRUE, oblique.scores = TRUE) solution Thank you. __ 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. William Revelle http://personality-project.org/revelle.html Professorhttp://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern University http://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 5 minutes to midnight http://www.thebulletin.org __ 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. William Revellehttp://personality-project.org/revelle.html Professor http://personality-project.org Department of Psychology http://www.wcas.northwestern.edu/psych/ Northwestern Universityhttp://www.northwestern.edu/ Use R for psychology http://personality-project.org/r It is 5 minutes to midnighthttp://www.thebulletin.org __ 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] Problem with latex when checking a package
Hello everyone. I'm checking a package in Windows 8.1, and when the
program starts to create the PDF manual, exits with status 1, printing
the following message:
# End(Not run)
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
! You can't use `macro parameter character #' in vertical mode.
l.164 ##
I think the program is having problems with the \dontrun{} sections
when creating the PDF. Does anyone know how to solve this problem?
Thanks in advance,
--
Lic. Leandro Gabriel Roser
Laboratorio de Genética
Dto. de Ecología, Genética y Evolución,
F.C.E.N., U.B.A.,
Ciudad Universitaria, PB II, 4to piso,
Nuñez, Cdad. Autónoma de Buenos Aires,
Argentina.
tel ++54 +11 4576-3300 (ext 219)
fax ++54 +11 4576-3384
[[alternative HTML version deleted]]
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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] Problem with latex when checking a package
You didn't show your code, so this is a guess.
My guess is that the \dontrun{} is outside the \examples{} section.
It must be inside the \examples{} section.
This is right
\examples{
abc - 123
\dontrun{
def - 456
}
ghi - 789
}
This is my guess as to what you did.
\examples{
abc - 123
ghi - 789
}
\dontrun{
def - 456
}
Rich
On Thu, Jan 15, 2015 at 10:31 PM, Leandro Roser learo...@gmail.com wrote:
Hello everyone. I'm checking a package in Windows 8.1, and when the
program starts to create the PDF manual, exits with status 1, printing
the following message:
# End(Not run)
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
! You can't use `macro parameter character #' in vertical mode.
l.164 ##
I think the program is having problems with the \dontrun{} sections
when creating the PDF. Does anyone know how to solve this problem?
Thanks in advance,
--
Lic. Leandro Gabriel Roser
Laboratorio de Genética
Dto. de Ecología, Genética y Evolución,
F.C.E.N., U.B.A.,
Ciudad Universitaria, PB II, 4to piso,
Nuñez, Cdad. Autónoma de Buenos Aires,
Argentina.
tel ++54 +11 4576-3300 (ext 219)
fax ++54 +11 4576-3384
[[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
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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] Problem with latex when checking a package
Hi, Richard. I was putting some piece of code in a wrong section
(description). After correcting that, I have no more errors.
Many tanks for your help,
Leandro.
2015-01-16 1:59 GMT-03:00 Richard M. Heiberger r...@temple.edu:
You didn't show your code, so this is a guess.
My guess is that the \dontrun{} is outside the \examples{} section.
It must be inside the \examples{} section.
This is right
\examples{
abc - 123
\dontrun{
def - 456
}
ghi - 789
}
This is my guess as to what you did.
\examples{
abc - 123
ghi - 789
}
\dontrun{
def - 456
}
Rich
On Thu, Jan 15, 2015 at 10:31 PM, Leandro Roser learo...@gmail.com wrote:
Hello everyone. I'm checking a package in Windows 8.1, and when the
program starts to create the PDF manual, exits with status 1, printing
the following message:
# End(Not run)
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
! You can't use `macro parameter character #' in vertical mode.
l.164 ##
I think the program is having problems with the \dontrun{} sections
when creating the PDF. Does anyone know how to solve this problem?
Thanks in advance,
--
Lic. Leandro Gabriel Roser
Laboratorio de Genética
Dto. de Ecología, Genética y Evolución,
F.C.E.N., U.B.A.,
Ciudad Universitaria, PB II, 4to piso,
Nuñez, Cdad. Autónoma de Buenos Aires,
Argentina.
tel ++54 +11 4576-3300 (ext 219)
fax ++54 +11 4576-3384
[[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.
--
Lic. Leandro Gabriel Roser
Laboratorio de Genética
Dto. de Ecología, Genética y Evolución,
F.C.E.N., U.B.A.,
Ciudad Universitaria, PB II, 4to piso,
Nuñez, Cdad. Autónoma de Buenos Aires,
Argentina.
tel ++54 +11 4576-3300 (ext 219)
fax ++54 +11 4576-3384
__
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] Help installing packages with dependencies for R, behind
Hello all, Sorry about the first post, I forgot to mention that I am using R on Linux. As Brian suggested I looked closely into the R help and tried the verbose option ( install.packages(ggplot2, verbose = 'true')) but it wasn't very helpful. But after checking 'help(download.file)' I found than my format of the variable http_proxy was wrong and needed to be like this (protocol and port): http_proxy=http://myproxy.com:80 Downloads now work fine, I hope this saves some time to the next person dealing with this issue. Thanks for the help! --Jose -- Message: 26 Date: Wed, 14 Jan 2015 16:37:17 -0500 From: jose.nunez-zul...@barclays.com To: r-help@r-project.org Subject: [R] Help installing packages with dependencies for R, behind corporate firewall Message-ID: 34922d8098cb7048a99568d009af262d0916cd1...@nykpcmmgmb05.intranet.barcapint.com Content-Type: text/plain; charset=us-ascii Hello R-users, I have no practical experience with the R language itself but I've been tasked to install it behind a corporate firewall. Basic installation seems sane but when my user tries to install a custom library like this: install.packages(ggplot2) Installing package into '/home/myuser/rlibs' (as 'lib' is unspecified) Warning: unable to access index for repository http://cran.us.r-project.org/src/contrib Warning message: package 'ggplot2' is not available (for R version 3.1.2) See no progress and eventually nothing gets downloaded into my custom directory. My question is, there is a way to add verbosity to R to see if the network proxy setting are working correctly (I can get files using 'wget' without problems under the same account)? More details about my installation below: 1) I'm behind a firewall with http proxy access. I have no root and made a local installation 2) Contents of my ~/.Renviron: R_LIBS=/home/myuser/rlibs 3) Contents of ~/.Rprofile: r - getOption(repos) # hard code the US repo for CRAN r[CRAN] - http://cran.us.r-project.org; options(repos = r) rm(r) 4) Http environment variable proxy is set (like export http_proxy=proxy..com. I can see it if I do 'Sys.getenv(http_proxy)' from inside the R prompt) NOTE: I managed to install libraries 'by hand' but for module that have dependencies this doesn't work: cd /home/$USER/rlibs/; wget http://cran.us.r-project.org/src/contrib/timeDate_3011.99.tar.gz; /mylocal/R-3.1.2/bin/R CMD INSTALL -l /localrdir timeDate_3011.99.tar.gz I apologize if this is not the correct list (went through all of them, WIKI and other groups looking for an answer to this issue without much luck). Thanks, --Jose ___ This message is for information purposes only, it is not a recommendation, advice, offer or solicitation to buy or sell a product or service nor an official confirmation of any transaction. It is directed at persons who are professionals and is not intended for retail customer use. Intended for recipient only. This message is subject to the terms at: www.barclays.com/emaildisclaimer. For important disclosures, please see: www.barclays.com/salesandtradingdisclaimer regarding market commentary from Barclays Sales and/or Trading, who are active market participants; and in respect of Barclays Research, including disclosures relating to specific issuers, please see http://publicresearch.barclays.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] Latest version of Rtools is incompatible with latest version of R !!
On Thu, Jan 15, 2015 at 9:27 AM, PRAMEET LAHIRI prameet_lah...@yahoo.co.in wrote: I have installed R version 3.1.2. I tried to install RTools32.exe which is the latest version (using this link - http://cran.r-project.org/bin/windows/Rtools/) ! However on using the function find_rtools() an error message was displayed which said --- Rtools is required to build R packages, but no version of Rtools compatible with R 3.1.2 was found. (Only the following incompatible version(s) of Rtools were found:3.2)Please download and install Rtools 3.1 from http://cran.r-project.org/bin/windows/Rtools/ and then run find_rtools(). I want to know why the latest version of R is not supporting the latest Rtools! Any suggestions? find_rtools() is a function of the 'devtools' package. Maybe it's an issue with that package and not R, and I'm pretty sure Duncan Murdoch put great efforts in asserting Rtools is working well with R (that's been my experience for the last 5-10 years). I also know that devtools 1.7.0 have been submitted to CRAN, so if/when that becomes available you problem might be solved. You can of course also grab it earlier from the devtools GitHub page. My $.02 Henrik Thanks for your time,Prameet [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to speed up the simulation when processing time and date
I don't have time to look at your example in detail, but there are couple
of things that caught my eye.
Use as.POSIXct() instead of as.POSIXlt()
I don't see anything that requires POSIXlt, and POSIXct is simpler.
If everything in
Total_Zone1
is numeric, then leave it as a matrix, do not convert to data frame.
If you use as.POSIXct() then the times are actually the number of seconds
since an origin, and thus can be treated as numeric, making it possible to
leave Total_Zone1 as a matrix.
If it is a matrix, you can refer to the times using
Total_Zone1[,'time'] instead of Total_Zone1$time
Either of these might help speed things up, though I can't be sure without
trying it.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/15/15, 2:38 AM, Faranak Golestaneh faranak.golesta...@gmail.com
wrote:
Dear Friends,
I am trying to program a forecasting method in R. The predictors are
weather variables in addition to lag measured Power values. The accuracy
of
data is one minute and their corresponding time and date are available.
To add lag values of power to the predictors list, I am aiming to consider
last ten minutes values. If I was sure that the database is perfect and
the
values for all minutes throughout the year are available I could simply
shift the Power columns but as it may not be always the case, I have used
the following codes for each time t to check if all its corresponding ten
minutes lag values are available and extract them and store in a matrix.
The problem is that, the process is highly time consuming and it takes a
long time to be simulated. Here I ve given reproducible example. I was
wondering any of you can suggest a better approach. Thank you.
rm(list = ls())
cat(\014)
st=2012/01/01
et=2012/02/27
st - as.POSIXlt(as.Date(st))
et - as.POSIXlt(as.Date(et))
time= seq(from=st, to=et,by=60)
timeas.POSIXlt(time)
#Window is the number of lag values
#leadTime is look-ahead time (forecast horizon)
leadTime=10;
Window=15;
=time[1:8000]
Total_Zone1=abind(matrix(rnorm(4000*2),4000*2,1),
matrix(rnorm(4000*2),4000*2,1),
matrix(rnorm(4000*2),4000*2,1),time[1:8000])
N_Train=nrow(Total_Zone1);
lag_Power=matrix(0,N_Train,Window)
colnames(Total_Zone1) - c( airtemp,humidity, Power, time)
Total_Zone1- as.data.frame(Total_Zone1)
for (tt in 4000:N_Train){
Statlag=Total_Zone1$time[tt]-(leadTime+Window)*60
EndLag=Total_Zone1$time[tt]-(leadTime)*60
Index_lags=which((Total_Zone1$timeStatlag)(Total_Zone1$time=EndLag))
if (size(Index_lags)[2]Window) {
Statlag2=Total_Zone1$time[tt]-24*60*60
Index_lags2=which(Total_Zone1$time==Statlag2)
tem1=rep(Total_Zone1[Index_lags2,c(Power)],Window-size(Index_lags)[2])
lag_Power[tt,]=t(c(Total_Zone1[Index_lags,c(Power)],tem1))
}else{
lag_Power[tt,]=t(Total_Zone1[Index_lags,c(Power)])
}
}
[[alternative HTML version deleted]]
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http://www.R-project.org/posting-guide.html
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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] Interactive Decion Trees in R
How can we have the Interactive Decision in R, I want a variable which is of Business importance to be injected in the Tree (Rpart specifically). R tree is not picking up the variable, we have this functionality in SAS and Statistica where we can force a variable to be injected in a branch and further split of tree is done by machine. Thanks Krishna [[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] two-sample KS test: data becomes significantly different after normalization
Thank you, Chris and Martin!
On Wed Jan 14 2015 at 7:31:12 AM Andrews, Chris chri...@med.umich.edu
wrote:
Your definition of p-value is not correct. See, for example,
http://en.wikipedia.org/wiki/P-value#Misunderstandings
-Original Message-
From: Monnand [mailto:monn...@gmail.com]
Sent: Wednesday, January 14, 2015 2:17 AM
To: Andrews, Chris
Cc: r-help@r-project.org
Subject: Re: [R] two-sample KS test: data becomes significantly different
after normalization
I know this must be a wrong method, but I cannot help to ask: Can I only
use the p-value from KS test, saying if p-value is greater than \beta, then
two samples are from the same distribution. If the definition of p-value is
the probability that the null hypothesis is true, then why there's little
people uses p-value as a true probability. e.g. normally, people will not
multiply or add p-values to get the probability that two independent null
hypothesis are both true or one of them is true. I had this question for
very long time.
-Monnand
On Tue Jan 13 2015 at 2:47:30 PM Andrews, Chris chri...@med.umich.edu
wrote:
This sounds more like quality control than hypothesis testing. Rather
than statistical significance, you want to determine what is an
acceptable
difference (an 'equivalence margin', if you will). And that is a
question
about the application, not a statistical one.
From: Monnand [monn...@gmail.com]
Sent: Monday, January 12, 2015 10:14 PM
To: Andrews, Chris
Cc: r-help@r-project.org
Subject: Re: [R] two-sample KS test: data becomes significantly different
after normalization
Thank you, Chris!
I think it is exactly the problem you mentioned. I did consider
1000-point data is a large one at first.
I down-sampled the data from 1000 points to 100 points and ran KS test
again. It worked as expected. Is there any typical method to compare
two large samples? I also tried KL diverge, but it only gives me some
number but does not tell me how large the distance is should be
considered as significantly different.
Regards,
-Monnand
On Mon, Jan 12, 2015 at 9:32 AM, Andrews, Chris chri...@med.umich.edu
wrote:
The main issue is that the original distributions are the same, you
shift the two samples *by different amounts* (about 0.01 SD), and you
have
a large (n=1000) sample size. Thus the new distributions are not the
same.
This is a problem with testing for equality of distributions. With
large samples, even a small deviation is significant.
Chris
-Original Message-
From: Monnand [mailto:monn...@gmail.com]
Sent: Sunday, January 11, 2015 10:13 PM
To: r-help@r-project.org
Subject: [R] two-sample KS test: data becomes significantly different
after normalization
Hi all,
This question is sort of related to R (I'm not sure if I used an R
function
correctly), but also related to stats in general. I'm sorry if this is
considered as off-topic.
I'm currently working on a data set with two sets of samples. The csv
file
of the data could be found here: http://pastebin.com/200v10py
I would like to use KS test to see if these two sets of samples are
from
different distributions.
I ran the following R script:
# read data from the file
data = read.csv('data.csv')
ks.test(data[[1]], data[[2]])
Two-sample Kolmogorov-Smirnov test
data: data[[1]] and data[[2]]
D = 0.025, p-value = 0.9132
alternative hypothesis: two-sided
The KS test shows that these two samples are very similar. (In fact,
they
should come from same distribution.)
However, due to some reasons, instead of the raw values, the actual
data
that I will get will be normalized (zero mean, unit variance). So I
tried
to normalize the raw data I have and run the KS test again:
ks.test(scale(data[[1]]), scale(data[[2]]))
Two-sample Kolmogorov-Smirnov test
data: scale(data[[1]]) and scale(data[[2]])
D = 0.3273, p-value 2.2e-16
alternative hypothesis: two-sided
The p-value becomes almost zero after normalization indicating these
two
samples are significantly different (from different distributions).
My question is: How the normalization could make two similar samples
becomes different from each other? I can see that if two samples are
different, then normalization could make them similar. However, if two
sets
of data are similar, then intuitively, applying same operation onto
them
should make them still similar, at least not different from each other
too
much.
I did some further analysis about the data. I also tried to normalize
the
data into [0,1] range (using the formula (x-min(x))/(max(x)-min(x))),
but
same thing happened. At first, I thought it might be outliers caused
this
problem (I can see that an outlier may cause this problem if I
normalize
the data
Re: [R] sparse matrix from vector outer product
Philipp A flying-sh...@web.de on Wed, 14 Jan 2015 14:02:40 + writes: Hi, creating a matrix from two vectors a, b by multiplying each combination can be done e.g. via a %*% t(b) or via outer(a, b) # default for third argument is '*' really the best (most efficient) way would be tcrossprod(a, b) But this yields a normal matrix. of course. Please always use small self-contained example code, here, e.g., a - numeric(17); a[3*(1:5)] - 10*(5:1) b - numeric(12); b[c(2,3,7,11)] - 1:3 Is there an efficient way to create sparse matrices (from the Matrix package) like that? Right now i’m doing a.sparse = as(a, 'sparseVector') b.sparse = as(t(b), 'sparseMatrix') a.sparse %*% b.sparse but this strikes me as wasteful. not really wasteful I think. But there is a nicer and more efficient way : require(Matrix) tcrossprod(as(a, sparseVector), as(b, sparseVector)) now also gives 17 x 12 sparse Matrix of class dgCMatrix [1,] . . . . . . . . . . . . [2,] . . . . . . . . . . . . [3,] . 50 100 . . . 150 . . . 50 . [4,] . . . . . . . . . . . . [5,] . . . . . . . . . . . . [6,] . 40 80 . . . 120 . . . 40 . [7,] . . . . . . . . . . . . [8,] . . . . . . . . . . . . [9,] . 30 60 . . . 90 . . . 30 . [10,] . . . . . . . . . . . . [11,] . . . . . . . . . . . . [12,] . 20 40 . . . 60 . . . 20 . [13,] . . . . . . . . . . . . [14,] . . . . . . . . . . . . [15,] . 10 20 . . . 30 . . . 10 . [16,] . . . . . . . . . . . . [17,] . . . . . . . . . . . . __ 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] using poly() to predict
Thanks Prof Ripley.
For anybody else wondering about this, see:
http://stackoverflow.com/questions/26728289/extracting-orthogonal-polynomial-coefficients-from-rs-poly-function
=
The polynomials are defined recursively using the alpha and norm2
coefficients of the poly object you've created. Let's look at an example:
z - poly(1:10, 3)
attributes(z)$coefs# $alpha# [1] 5.5 5.5 5.5# $norm2# [1]1.0
10.0 82.5 528.0 3088.8
For notation, let's call a_d the element in index d of alpha and let's call
n_d the element in index d of norm2. F_d(x) will be the orthogonal
polynomial of degree d that is generated. For some base cases we have:
F_0(x) = 1 / sqrt(n_2)
F_1(x) = (x-a_1) / sqrt(n_3)
The rest of the polynomials are recursively defined:
F_d(x) = [(x-a_d) * sqrt(n_{d+1}) * F_{d-1}(x) - n_{d+1} / sqrt(n_d) *
F_{d-2}(x)] / sqrt(n_{d+2})
To confirm with x=2.1:
x - 2.1
predict(z, newdata=x)# 1 2 3# [1,]
-0.3743277 0.1440493 0.1890351# ...
a - attributes(z)$coefs$alpha
n - attributes(z)$coefs$norm2
f0 - 1 / sqrt(n[2])(f1 - (x-a[1]) / sqrt(n[3]))# [1] -0.3743277(f2
- ((x-a[2]) * sqrt(n[3]) * f1 - n[3] / sqrt(n[2]) * f0) /
sqrt(n[4]))# [1] 0.1440493(f3 - ((x-a[3]) * sqrt(n[4]) * f2 - n[4] /
sqrt(n[3]) * f1) / sqrt(n[5]))# [1] 0.1890351
The most compact way to export your polynomials to your C++ code would
probably be to export attributes(z)$coefs$alpha and
attributes(z)$coefs$norm2 and then use the recursive formula in C++ to
evaluate your polynomials.
On Wed, Jan 14, 2015 at 2:38 PM, Prof Brian Ripley rip...@stats.ox.ac.uk
wrote:
On 14/01/2015 14:20, Stanislav Aggerwal wrote:
This method of finding yhat as x %*% b works when I use raw polynomials:
x-1:8
y- 1+ 1*x + .5*x^2
fit-lm(y~poly(x,2,raw=T))
b-coef(fit)
xfit-seq(min(x),max(x),length=20)
yfit-b[1] + poly(xfit,2,raw=T) %*% b[-1]
plot(x,y)
lines(xfit,yfit)
But it doesn't work when I use orthogonal polynomials:
fit-lm(y~poly(x,2))
b-coef(fit)
yfit-b[1] + poly(xfit,2) %*% b[-1]
plot(x,y)
lines(xfit,yfit,col='red')
I have a feeling that the second version needs to incorporate poly() coefs
(alpha and norm2) somehow. If so, please tell me how.
I do know how to use predict() for this. I just want to understand how
poly() works.
What matters is how lm() and predict() use poly(): see ?makepredictcall
and its code.
str(fit) might also help.
Thanks very much for any help
Stan
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Emeritus Professor of Applied Statistics, University of Oxford
1 South Parks Road, Oxford OX1 3TG, UK
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Re: [R] RStudio connection to server: Safari cannot open page because it could not connect to the server
Hi John,
This isn't a question about R, and so is off-topic here. Try a web search
for home server port forwarding or similar.
Best,
Ista
On Jan 15, 2015 7:30 AM, John Sorkin jsor...@grecc.umaryland.edu wrote:
I set up Rstudio, and can access it from within my lan using
http:/192.168.108:8787.
I looked up my external IP address using one of the websites that returns
an ip addresses and tried to connect from outside my LAN using
http://73.213.144.65:8787
and received a message:
Safari cannot open the page because the page because it could not connect
to the server.
I can ping the 73.213.144.65
My LAN connects to the WWW through a wireless router which connects to a
cable model.
Can anyone help me connect?
Thank you,
John
John David Sorkin M.D., Ph.D.
Professor of Medicine
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology and
Geriatric Medicine
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:16}}
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[R] RStudio connection to server: Safari cannot open page because it could not connect to the server
I set up Rstudio, and can access it from within my lan using http:/192.168.108:8787. I looked up my external IP address using one of the websites that returns an ip addresses and tried to connect from outside my LAN using http://73.213.144.65:8787 and received a message: Safari cannot open the page because the page because it could not connect to the server. I can ping the 73.213.144.65 My LAN connects to the WWW through a wireless router which connects to a cable model. Can anyone help me connect? Thank you, John John David Sorkin M.D., Ph.D. Professor of Medicine Chief, Biostatistics and Informatics University of Maryland School of Medicine Division of Gerontology and Geriatric Medicine Baltimore VA Medical Center 10 North Greene Street GRECC (BT/18/GR) Baltimore, MD 21201-1524 (Phone) 410-605-7119 (Fax) 410-605-7913 (Please call phone number above prior to faxing) Confidentiality Statement: This email message, including any attachments, is for the sole use of the intended recipient(s) and may contain confidential and privileged information. Any unauthorized use, disclosure or distribution is prohibited. If you are not the intended recipient, please contact the sender by reply email and destroy all copies of the original message. __ 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] Probit regression with misclassified binary dependent variable
Is there a generalized linear mixed-effects model implementation for R that could handle misclassified binary data? Unless I have overlooked something in the documentation, glmer with family=binomial(link=probit) does not handle misclassification. Closest to what I have found was misclass, which implements Hausman et al. 1998, in an old McSpatial package version 1.1.1, but it is no longer available and does not handle random effects. Hausman, J. A., Abrevaya, J., Scott-Morton, F. M. (1998). Misclassification of the dependent variable in a discrete-response setting. Journal of Econometrics, 87(2), 239-269. [[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 to speed up the simulation when processing time and date
Dear Friends,
I am trying to program a forecasting method in R. The predictors are
weather variables in addition to lag measured Power values. The accuracy of
data is one minute and their corresponding time and date are available.
To add lag values of power to the predictors list, I am aiming to consider
last ten minutes values. If I was sure that the database is perfect and the
values for all minutes throughout the year are available I could simply
shift the Power columns but as it may not be always the case, I have used
the following codes for each time t to check if all its corresponding ten
minutes lag values are available and extract them and store in a matrix.
The problem is that, the process is highly time consuming and it takes a
long time to be simulated. Here I ve given reproducible example. I was
wondering any of you can suggest a better approach. Thank you.
rm(list = ls())
cat(\014)
st=2012/01/01
et=2012/02/27
st - as.POSIXlt(as.Date(st))
et - as.POSIXlt(as.Date(et))
time= seq(from=st, to=et,by=60)
timeas.POSIXlt(time)
#Window is the number of lag values
#leadTime is look-ahead time (forecast horizon)
leadTime=10;
Window=15;
=time[1:8000]
Total_Zone1=abind(matrix(rnorm(4000*2),4000*2,1),
matrix(rnorm(4000*2),4000*2,1), matrix(rnorm(4000*2),4000*2,1),time[1:8000])
N_Train=nrow(Total_Zone1);
lag_Power=matrix(0,N_Train,Window)
colnames(Total_Zone1) - c( airtemp,humidity, Power, time)
Total_Zone1- as.data.frame(Total_Zone1)
for (tt in 4000:N_Train){
Statlag=Total_Zone1$time[tt]-(leadTime+Window)*60
EndLag=Total_Zone1$time[tt]-(leadTime)*60
Index_lags=which((Total_Zone1$timeStatlag)(Total_Zone1$time=EndLag))
if (size(Index_lags)[2]Window) {
Statlag2=Total_Zone1$time[tt]-24*60*60
Index_lags2=which(Total_Zone1$time==Statlag2)
tem1=rep(Total_Zone1[Index_lags2,c(Power)],Window-size(Index_lags)[2])
lag_Power[tt,]=t(c(Total_Zone1[Index_lags,c(Power)],tem1))
}else{
lag_Power[tt,]=t(Total_Zone1[Index_lags,c(Power)])
}
}
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[R] HOW TO DOWNLOAD INTRADAY DATA AT ONE TIME
Is there any way VIA R to download all available intraday data for stocks at once (for example, all the data available at the Indian stock exchange)? I need to make a comparative analysis and downloading the data by ticker is too time consuming, besides I want to know if there is any website that store the historical intraday data. Other sites delete the data gradually. [[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] passing elements of named vector / named list to function
Hi
Following scenario: I have a function fun
--8---cut here---start-8---
fun - function(A, B, C, ...){paste(A, B, C, ...)}
--8---cut here---end---8---
and x defined as follow
--8---cut here---start-8---
x - 1:5
names(x) - LETTERS[x]
--8---cut here---end---8---
now I want to pass the *elements* of x to fun as named arguments, i.e.
,
| fun(A=1, B=2, C=3, D=4, E=5)
| [1] 1 2 3 4 5
`
The below examples obviously do not work:
,
| fun(x)
| Error in paste(A, B, C, list(...)) :
| argument B is missing, with no default
| fun(unlist(x))
| Error in paste(A, B, C, list(...)) :
| argument B is missing, with no default
`
How can I extract from x the elements and pass them on to fun()?
I could easily change x to a list() if this would be easier.
--8---cut here---start-8---
x - list(A=1, B=2, C=3, D=4, E=5)
--8---cut here---end---8---
In my actual program, x can have different elements as well as fun -
this is decided programmatically.
Any suggestions how I can achieve this?
Thanks,
Rainer
--
Rainer M. Krug
email: Raineratkrugsdotde
PGP: 0x0F52F982
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Re: [R] R 3.1.2 mle2() function on Windows 7 Error and multiple solutions
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Hash: SHA1
For what it's worth, you can use either nlminb (directly) or optimx
within the mle2 wrapper by specifying the 'optimizer' parameter ...
this gives you flexibility in optimization along with the convenience
of mle2 (likelihood ratio tests via anova(), likelihood profiling, etc.)
On 15-01-15 09:26 AM, Ravi Varadhan wrote:
Hi,
I tried your problem with optimx package. I found a better
solution than that found by mle2.
?library(optimx)
# the objective function needs to be re-written
LL2 - function(par,y) {
lambda - par[1] alpha - par[2] beta - par[3] R =
Nweibull(y,lambda,alpha,beta)
-sum(log(R)) }
optimx(fn=LL2, par=c(.01,325,.8),y=y,
lower=c(.1,.1,.1),upper = c(Inf,
Inf,Inf),control=list(all.methods=TRUE))
# Look at the solution found by `nlminb' and `nmkb'. This is the
optimal one. This log-likelihood is larger than that of mle2 and
other optimizers in optimx.
If this solution is not what you are looking for, your problem may
be poorly scaled. First, make sure that the likelihood is coded
correctly. If it is correct, then you may need to improve the
scaling of the problem.
Hope this is helpful,
Ravi
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