[R] loops over regression models
Dear R help listers,
I am trying to replicate results in Gelman and Hill's book (Chapter 3
in regressions and multilevel models). Below I estimated two models
(chp3.1 and chp3.3 in R codes) with the same data and dependent
variable but different independent variables. I have been using Stata
for quite a while, and I know I can use foreach to build a loop to
condense the codes (especially if I have a large number of models to
run).
In Stata, it would be something like:
// read in data
use kidiq, clear
// run two regression
reg kid_score mom_hs
reg kid_score mom_iq
// the next three lines are equivalent of the previous two lines
foreach var in mom_hs mom_iq {
reg kid_score `var'
}
***
So I want to figure out how to use R to do this. Below are my codes:
library(foreign)
# read in stata data file
kidiq -data.frame(read.dta('kidiq.dta', convert.factor=FALSE))
# bivariate regressions
chp3.1 - lm(kid_score ~ mom_hs, data=kidiq)
summary(chp3.1)
chp3.3 - lm(kid_score ~ mom_iq, data=kidiq)
summary(chp3.3)
clist - c(mom_iq, mom_hs)
for (x in clist) {
lm(kid_score ~ x, data = kidiq)
}
Error in model.frame.default(formula = kid_score ~ x, data = kidiq,
drop.unused.levels = TRUE) :
variable lengths differ (found for 'x')
##
But I got an error message that says variable length differ. I tried
various ways to work around this, for example, I tried:
clist - c(mom_iq, mom_hs)
for (x in 1:length(clist)) {
lm(kid_score ~ clist[x], data = kidiq)
}
But none of these work. So I am wondering if anyone could give me
some hint. Thanks a lot
Jun Xu, PhD
Assistant Professor
Department of Sociology
Ball State University
Muncie, IN
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] subset
Many thanks for your elaborated explaina. At 2012-01-13 11:34:51,R. Michael Weylandt michael.weyla...@gmail.com wrote: As Jorge noted, the fix is to use %in%: a fuller explanation of why `==` didn't work is that it implicitly used vector recycling: look at with(data, id == c(a, c)) implicitly, this expands to id == c(a,c, a, c) to get the lengths to match. Obviously only the first elements work here. But when you had c(a, d) it expanded to c(a,d, a, d) and you get TRUE for the 1st and 4th slot. This, however, was just a lucky coincidence. Had you used c(d, a) there would have been no matches. Anyways, definitely use %in% but hopefully this clarifies things. Michael On Thu, Jan 12, 2012 at 9:50 PM, Jorge I Velez jorgeivanve...@gmail.com wrote: Hi, Use %in% instead of ==. HTH, Jorge.- On Thu, Jan 12, 2012 at 9:36 PM, ÃÏÐÀ wrote: Hi all I have a question about subset function. dat id x1 x2 x3 1 a 1 11 111 2 b 2 22 222 3 c 3 33 333 4 d 4 44 444 subset(dat,id==c(a,c)) id x1 x2 x3 1 a 1 11 111 subset(dat,id==c(a,d)) id x1 x2 x3 1 a 1 11 111 4 d 4 44 444 From the above, if I choose id=a,c, the result is wrong,but if I choose id=a,d, the result is right. What's the reason for it? Many thanks! My best [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] subset
Thanks£¡ ÔÚ 2012-01-13 10:51:14£¬Jorge I Velez jorgeivanve...@gmail.com дµÀ£º Hi, Use %in% instead of ==. HTH, Jorge.- On Thu, Jan 12, 2012 at 9:36 PM, ÃÏÐÀ wrote: Hi all I have a question about subset function. dat id x1 x2 x3 1 a 1 11 111 2 b 2 22 222 3 c 3 33 333 4 d 4 44 444 subset(dat,id==c(a,c)) id x1 x2 x3 1 a 1 11 111 subset(dat,id==c(a,d)) id x1 x2 x3 1 a 1 11 111 4 d 4 44 444 From the above, if I choose id=a,c, the result is wrong,but if I choose id=a,d, the result is right. What's the reason for it? Many thanks! My best [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] analytical solution of partial differential equation
i am trying to solve a partial differential equation analytically(PDE) in R . i have found some functions that do the stuff numerically. But that will not meet my purpose. is there any function to solve PDE analytically. please help. -- View this message in context: http://r.789695.n4.nabble.com/analytical-solution-of-partial-differential-equation-tp4291618p4291618.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] loops over regression models
Hi
Dear R help listers,
I am trying to replicate results in Gelman and Hill's book (Chapter 3
in regressions and multilevel models). Below I estimated two models
(chp3.1 and chp3.3 in R codes) with the same data and dependent
variable but different independent variables. I have been using Stata
for quite a while, and I know I can use foreach to build a loop to
condense the codes (especially if I have a large number of models to
run).
In Stata, it would be something like:
// read in data
use kidiq, clear
// run two regression
reg kid_score mom_hs
reg kid_score mom_iq
// the next three lines are equivalent of the previous two lines
foreach var in mom_hs mom_iq {
reg kid_score `var'
}
***
So I want to figure out how to use R to do this. Below are my codes:
library(foreign)
# read in stata data file
kidiq -data.frame(read.dta('kidiq.dta', convert.factor=FALSE))
# bivariate regressions
chp3.1 - lm(kid_score ~ mom_hs, data=kidiq)
summary(chp3.1)
chp3.3 - lm(kid_score ~ mom_iq, data=kidiq)
summary(chp3.3)
clist - c(mom_iq, mom_hs)
for (x in clist) {
lm(kid_score ~ x, data = kidiq)
use
as.formula(paste(kid_score ~ , eval(x)))
as I understand x is unevaluated and you need to evaluate it inside your
data.
And you also need to assign values of lm inside cycle or explicitly print
them.
Regards
Petr
}
Error in model.frame.default(formula = kid_score ~ x, data = kidiq,
drop.unused.levels = TRUE) :
variable lengths differ (found for 'x')
##
But I got an error message that says variable length differ. I tried
various ways to work around this, for example, I tried:
clist - c(mom_iq, mom_hs)
for (x in 1:length(clist)) {
lm(kid_score ~ clist[x], data = kidiq)
}
But none of these work. So I am wondering if anyone could give me
some hint. Thanks a lot
Jun Xu, PhD
Assistant Professor
Department of Sociology
Ball State University
Muncie, IN
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] beanplot-Error: sample is too sparse to find TD
Hi all, Since two days I am trying to find a solution to be able to create beanplots for my data. When I call the beanplot function the following error appears: beanplot(y1 ~ x1, log=, what=c(1,1,1,0), ylim=c(0,1)) Error in bw.SJ(x, method = dpi) : sample is too sparse to find TD What is really strange: I have 32 different vectors and the problem occurs for about 20 of them. All of the 32 vectors contain the same number of data values (about 20.000), within the same range, and without any NA values. The only thing which is slightly different sometimes is the distribution. I also had a look at the source code of the function which causes the error: TD - -TDh(cnt, b, n, d) if (!is.finite(TD) || TD = 0) stop(sample is too sparse to find TD) TDh - function(x, h, n, d) .C(R_band_phi6_bin, as.integer(n), as.integer(length(x)), as.double(d), x, as.double(h), u = double(1))$u But this also doesn't really help me. Has anyone of you an idea what's causing my problem and how I can avoid it? The only thing I found so far was this thread: http://r.789695.n4.nabble.com/bandwidth-estimation-using-bw-SJ-td851441.html Greets, Enomis [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Help needed in interpreting linear models
Dear members of the R-help list, I have sent the email below to the R-SIG-ME list to ask for help in interpreting some R output of fitted linear models. Unfortunately, I haven't yet received any answers. As I am not sure if my email was sent successfully to the mailing list I am asking for help here: Dear members of the R-SIG-ME list, I am new to linear models and struggling with interpreting some of the R output but hope to get some advice from here. I created the following dummy data set: scores - c(2,6,10,12,14,20) weight - c(60,70,80,75,80,85) height - c(180,180,190,180,180,180) The scores of a game/match should be dependent on the weight of the player but not on the height. For me the output of the following two linear models make sense: (lm1 - summary(lm(scores ~ weight))) Call: lm(formula = scores ~ weight) Residuals: 123456 1.08333 -1.41667 -3.91667 1.3 0.08333 2.8 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -38.083310.0394 -3.793 0.01921 * weight0.6500 0.1331 4.885 0.00813 ** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 2.661 on 4 degrees of freedom Multiple R-squared: 0.8564, Adjusted R-squared: 0.8205 F-statistic: 23.86 on 1 and 4 DF, p-value: 0.008134 (lm2 - summary(lm(scores ~ height))) Call: lm(formula = scores ~ height) Residuals: 1 2 3 4 5 6 -8.800e+00 -4.800e+00 1.377e-14 1.200e+00 3.200e+00 9.200e+00 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 25.2000 139.6175 0.1800.866 height -0.0800 0.7684 -0.1040.922 Residual standard error: 7.014 on 4 degrees of freedom Multiple R-squared: 0.002703, Adjusted R-squared: -0.2466 F-statistic: 0.01084 on 1 and 4 DF, p-value: 0.9221 The p-value of the first output is 0.008134 which makes sense as scores and weight have a high correlation and therefore, the scores can be explained by the explanatory variable/factor weight very well. Hence, the R-squared value is close to 1. For the second example it also makes sense that the p-value is almost 1 (p=0.9221) as there is hardly any correlation between scores and height. What is not clear to me is shown in my 3rd linear model which includes both weight and height. (lm3 - summary(lm(scores ~ weight + height))) Call: lm(formula = scores ~ weight + height) Residuals: 1 2 3 4 5 6 1.189e+00 -1.946e+00 -2.165e-15 4.865e-01 -1.081e+00 1.351e+00 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 49.45946 33.50261 1.476 0.23635 weight 0.713510.08716 8.186 0.00381 ** height -0.508110.19096 -2.661 0.07628 . --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 1.677 on 3 degrees of freedom Multiple R-squared: 0.9573, Adjusted R-squared: 0.9288 F-statistic: 33.6 on 2 and 3 DF, p-value: 0.008833 It makes sense that the R-squared value is higher when one adds both explanatory variables/factors to the linear model as the more variables are added the more variance is explained and therefore the fit of the model will be better. However, I do NOT understand why the p-value of height (Pr( | t |) = 0.07628) is now almost significant? And also, I do NOT understand why the overall p-value of 0.008833 is less significant as compared to the one from model lm1 which was p-value: 0.008134. The p-value of weight being low (p=0.00381) makes sense as this factor explains the scores very well. After fitting the 3 models (lm1, lm2 and lm3) I wanted to compare model lm1 with lm3 using the anova function to check whether the factor height significantly improves the model. In other words I wanted to check if adding height to the model helps explaining the scores of the players. The output of the anova looks as follows: lm1 - lm(scores ~ weight) lm2 - lm(scores ~ weight + height) anova(lm1,lm2) Analysis of Variance Table Model 1: scores ~ weight Model 2: scores ~ weight + height Res.Df RSS Df Sum of Sq F Pr(F) 1 4 28. 2 3 8.4324 119.901 7.0801 0.07628 . --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 In my opinion the p-value should be almost 1 and not close to significance (0.07) as we have seen from model lm2 height does not at all explain the scores. Here, I thought that a significant p-value means that the factor height adds significant value to the model. I would be very grateful if anyone could help me in interpreting the R output. Best regards -- View this message in context: http://r.789695.n4.nabble.com/Help-needed-in-interpreting-linear-models-tp4291670p4291670.html Sent from the R help mailing list archive at Nabble.com.
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Hi. I run a latent class analysis with polymotous variables. Because of small sample size, I have to use bootstrap method in order to select a proper model. Is there any package or way that I can run a bootstrap method after runing latent class model with polytomous variables? Thank KeunBok Lee -- Keun Bok Leeï¼ì´ê·¼ë³µï¼ PhD Student Department of Sociology University of California Berkeley, CA 94720, USA [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
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Re: [R] remoting ESS/R with tramp
p...@potis.org writes: When I move R and Rscript out of my path, as expected, I also have problems. If I use tramp to connect to remote, open a shell there with M-x eshell, and type 'env' , my path is the path on local. This is the correct behaviour of eshell: /scp:slbps0:/root $ which env eshell/env is a compiled Lisp function in `esh-var.el' If you want the environment of your remote eshell session, you shall call /bin/env instead. Before starting Emacs, try adding /usr/local/R-2.14.0/bin to your local path, e.g.: export PATH= $PATH:/usr/local/R-2.14.0/bin With Tramp means, you could do (add-to-list 'tramp-remote-path /usr/local/R-2.14.0/bin) Alternatively, you could instruct Tramp to preserve the path settings of your remote account: (add-to-list 'tramp-remote-path 'tramp-own-remote-path) Best regards, Michael. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] how to create stratified (cross-validation) partitions according to numerical features
Hi all, I want to fragment a dataset into k-cross-validation partitions (folds). The content of the folds should be stratified, but not according to a single (categorical) feature, but according to a range of features (numeric, if possible numeric and categorical). Does anybody know a way to do this? I only found a way to do this for a single split (training-test split) with the package sampling. I will paste the example code for the training-test split below to make clear what I am looking for. With best regards, Martin example code: library(sampling) data - as.matrix( iris[1:4] ) # skipping iris class column as this method only works for numerical features, but thats ok prob - 0.3 # probability to be selected into test set samplecube(data, pik=rep(prob, times=nrow(data)), order=2) [...] QUALITY OF BALANCING TOTALS HorvitzThompson_estimators Relative_deviation Sepal.Length 876.5 874.6667-0.20916524 Sepal.Width 458.6 458.-0.05814799 Petal.Length 563.7 563.-0.06504642 Petal.Width 179.9 178.6667-0.68556606 [1] 0 1 0 0 1 0 0 0 1 0 1 1 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 1 0 0 0 1 0 0 1 [38] 0 0 1 0 1 1 0 0 0 1 0 1 0 0 1 0 0 1 0 0 0 1 0 0 1 1 0 0 0 1 0 0 0 0 0 0 0 [75] 0 0 1 1 1 0 0 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 1 0 1 [112] 0 0 0 1 0 0 1 0 1 0 0 0 0 1 0 1 0 1 0 1 1 0 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 [149] 0 0 -- Dipl-Inf. Martin Gütlein Phone: +49 (0)761 203 7633 (office) +49 (0)177 623 9499 (mobile) Email: guetl...@informatik.uni-freiburg.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] syntax for reading into R
we haven't had time to create the file again, yet, but thanx also for the great tip about this special mailing list! marion 2012/1/3 Barry Rowlingson b.rowling...@lancaster.ac.uk On Tue, Jan 3, 2012 at 10:54 AM, Marion Wenty marion.we...@gmail.com wrote: hello barry, thank you very much for your help! we managed to do what you said and it basicaly worked - there only is a problem with our file, but we will create the file again and then it should work completely. i will keep you posted how it turned out. thanks again! Great - you might get more help from the R-sig-geo mailing list, where we tend to talk about spatial data a bit more than on R-help! Barry [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help needed in interpreting linear models
Hi It seems to me quite like a homework for which the policy of this list is not to respond. But far from being an expert in statistics I only express my opinion. It seems to me that your height variable behaves like a two level factor and the 190 value points to rather suspicious value in weight if I look at the plot plot(scores, weight) Regards Petr Dear members of the R-help list, I have sent the email below to the R-SIG-ME list to ask for help in interpreting some R output of fitted linear models. Unfortunately, I haven't yet received any answers. As I am not sure if my email was sent successfully to the mailing list I am asking for help here: Dear members of the R-SIG-ME list, I am new to linear models and struggling with interpreting some of the R output but hope to get some advice from here. I created the following dummy data set: scores - c(2,6,10,12,14,20) weight - c(60,70,80,75,80,85) height - c(180,180,190,180,180,180) The scores of a game/match should be dependent on the weight of the player but not on the height. For me the output of the following two linear models make sense: (lm1 - summary(lm(scores ~ weight))) Call: lm(formula = scores ~ weight) Residuals: 123456 1.08333 -1.41667 -3.91667 1.3 0.08333 2.8 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) -38.083310.0394 -3.793 0.01921 * weight0.6500 0.1331 4.885 0.00813 ** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 2.661 on 4 degrees of freedom Multiple R-squared: 0.8564, Adjusted R-squared: 0.8205 F-statistic: 23.86 on 1 and 4 DF, p-value: 0.008134 (lm2 - summary(lm(scores ~ height))) Call: lm(formula = scores ~ height) Residuals: 1 2 3 4 5 6 -8.800e+00 -4.800e+00 1.377e-14 1.200e+00 3.200e+00 9.200e+00 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 25.2000 139.6175 0.1800.866 height -0.0800 0.7684 -0.1040.922 Residual standard error: 7.014 on 4 degrees of freedom Multiple R-squared: 0.002703, Adjusted R-squared: -0.2466 F-statistic: 0.01084 on 1 and 4 DF, p-value: 0.9221 The p-value of the first output is 0.008134 which makes sense as scores and weight have a high correlation and therefore, the scores can be explained by the explanatory variable/factor weight very well. Hence, the R-squared value is close to 1. For the second example it also makes sense that the p-value is almost 1 (p=0.9221) as there is hardly any correlation between scores and height. What is not clear to me is shown in my 3rd linear model which includes both weight and height. (lm3 - summary(lm(scores ~ weight + height))) Call: lm(formula = scores ~ weight + height) Residuals: 1 2 3 4 5 6 1.189e+00 -1.946e+00 -2.165e-15 4.865e-01 -1.081e+00 1.351e+00 Coefficients: Estimate Std. Error t value Pr(|t|) (Intercept) 49.45946 33.50261 1.476 0.23635 weight 0.713510.08716 8.186 0.00381 ** height -0.508110.19096 -2.661 0.07628 . --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 1.677 on 3 degrees of freedom Multiple R-squared: 0.9573, Adjusted R-squared: 0.9288 F-statistic: 33.6 on 2 and 3 DF, p-value: 0.008833 It makes sense that the R-squared value is higher when one adds both explanatory variables/factors to the linear model as the more variables are added the more variance is explained and therefore the fit of the model will be better. However, I do NOT understand why the p-value of height (Pr( | t |) = 0.07628) is now almost significant? And also, I do NOT understand why the overall p-value of 0.008833 is less significant as compared to the one from model lm1 which was p-value: 0.008134. The p-value of weight being low (p=0.00381) makes sense as this factor explains the scores very well. After fitting the 3 models (lm1, lm2 and lm3) I wanted to compare model lm1 with lm3 using the anova function to check whether the factor height significantly improves the model. In other words I wanted to check if adding height to the model helps explaining the scores of the players. The output of the anova looks as follows: lm1 - lm(scores ~ weight) lm2 - lm(scores ~ weight + height) anova(lm1,lm2) Analysis of Variance Table Model 1: scores ~ weight Model 2: scores ~ weight + height Res.Df RSS Df Sum of Sq F Pr(F) 1 4 28. 2 3 8.4324 119.901 7.0801 0.07628 . --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 In my opinion the p-value should be almost 1 and not close to significance (0.07) as
[R] how to find the number of iterations kmeans used to converge?
Dear all, I need to know in which number of iterations the kmeans converge each time I run it. Any idea how to do it? Thank you for your attention, Rui __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] find inflexion point of discrete value list with R
d2y - diff(dy) which(dy==0) ## critical values sign(s2y)[which(dy==0)] ## test for max/min/saddle which(d2y==0) ## inflection points I would think that testing for d2y==0 would be akin to the error in numeric analysis warned about in FAQ 7.31. Seems unlikely that in real data that there would always be three points in a row with equal differences at a true inflection and even then, many of the ones you did find satisfying that criterion would not be in fact inflection points. Wouldn't it be better to fit a spline and then do your testing on the spline approximation? Counter-example: x=1:10 y=c(1,2,3,5,7,10,13,16,20,24) dy - diff(y) d2y - diff(dy) which(d2y==0) [1] 1 3 5 6 8 And actually the original data was a pretty good counter-example as well. The original post wasn't entirely clear, but I thought the data were indeed integers and that the discrete-state version of min/max/inflection point was indeed what was wanted. Yes, if the underlying variable is continuous you might want to use splinefun(), with its deriv= argument, and uniroot(), to find maxima and minima. Might be a little tricky in general, although with an interpolation spline between a finite set of points you can at least deal with it exhaustively. my real data is not limited to integer. Do you know a ready to use code example for this? Would it be a good idea to create a function and make it public to the community? And if yes as single .R file, or as a library? kind regards, -- Jonas Stein n...@jonasstein.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Remove space from string
Dear R users, I have some trivial query. I have a string, I want to remove space from the string. For eg. Input: a - Remove space Output required: Removespace I tried using str_trim but only removes end spaces. library(stringr). Regards Vikram [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Remove space from string
gsub( ,,a) /Gustaf On Fri, Jan 13, 2012 at 12:24 PM, Vikram Bahure economics.vik...@gmail.comwrote: Dear R users, I have some trivial query. I have a string, I want to remove space from the string. For eg. Input: a - Remove space Output required: Removespace I tried using str_trim but only removes end spaces. library(stringr). Regards Vikram [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Gustaf Rydevik, M.Sci. tel: +44(0)74 253 760 42 address:St John's hill 18/5 EH8 9UQ Edinburgh, UK skype:gustaf_rydevik [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] remoting ESS/R with tramp
Tom, what happens with: (Emacs) M-x ssh t (you should have the remote shell buffer now) R (once R is started) M-x ess-remote r ? Claudia -- Claudia Beleites Spectroscopy/Imaging Institute of Photonic Technology Albert-Einstein-Str. 9 07745 Jena Germany email: claudia.belei...@ipht-jena.de phone: +49 3641 206-133 fax: +49 2641 206-399 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Odp: Remove space from string
Hi Dear R users, I have some trivial query. I have a string, I want to remove space from the string. For eg. Input: a - Remove space Output required: Removespace It seems to be simple. Even myself with very poor knowledge of regexpr can suggest solution. gsub( , , a) Regards Petr I tried using str_trim but only removes end spaces. library(stringr). Regards Vikram [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to find the number of iterations kmeans used to converge?
On Jan 13, 2012, at 5:53 AM, Rui Esteves wrote: Dear all, I need to know in which number of iterations the kmeans converge each time I run it. Any idea how to do it? Look at the help page (to see that it is not part of the returned object) and then look at the code (to see that the object returned from the .C() call is immediately checked to see if the number of iterations exceeded the maximum set by the user. Search for this code: if (Z$iter iter.max) Then you should be able to see your way forward. You can either create a modified return object or you can insert a line that prints that value. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] second order cone programmming, linear inequalities constraints
Hi all, I can see two packages implementing second order cone programming in R: CLSOCP and DWD. Both of them allow to specify only equality linear constraints. Do you know if there is a library to solve this problem: second order cone programming with inequality liner constraints excluding mosek and cplex. thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Averaging within a range of values
Hello all. I have two data frames. Group Start End G1 200 700 G2 500 1000 G3 20003000 G4 40006000 G5 70008000 and Pos C0 C1 200 0.9 0.6 500 0.8 0.8 800 0.9 0.7 1000 0.7 0.6 20000.6 0.4 25001.2 0.8 30000.6 1.5 35000.7 0.7 40000.8 0.8 45000.6 0.6 5000 0.9 0.9 55000.7 0.8 60000.8 0.7 65000.4 0.4 7000 0.5 0.8 75000.7 0.9 80000.9 0.5 85000.8 0.6 90000.9 0.8 I need to conditionally average all values in columns C0 and C1 based upon the bins I defined in the first data frame. For example, for the bin G1 in the first dataframe, the values are 200 to 700 so i would average the value at pos 200 (0.9) and 500 (0.8) for C0 and then perform the same thing for C1. I can do this in excel with array formulas but I'm relatively new to R and would like know if there is a function that will perform the same action. I don't know if this will help, but the excel array function I used was average(if(range=start)*(range=end),range)). Where the range is the entire pos column. Initially I looked at the aggregate function. I can use aggregate when I give a single vector to be used for grouping such as (A,B,C) but I'm not sure how to define grouping as the bin 200-500 and the second bin as 500-1000 etc. and use that as my grouping vector. Any help would be greatly appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Averaging-within-a-range-of-values-tp4291958p4291958.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] simulating stable VAR process
Hello Paul Thanks for the answer but my point is not how to simulate a VAR(p) process and check that it is stable. My question is more how can I generate a VAR(p) such that I already know that it is stable. We know a condition that assure that it is stable (see first message) but this is not a condition on coefficients etc... What I want is generate say a 1000 random VAR(3) processes over say 500 time periods that will be STABLE (meaning If I run stability() all will pass the test) When I try to do that it seems that none of the VAR I am generating pass this test, so I assume that the class of stable VAR(p) is very small compared to the whole VAR(p) process. -- View this message in context: http://r.789695.n4.nabble.com/simulating-stable-VAR-process-tp4261177p4291835.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] GLHT in multcomp: Two similar models, one doesn't work
i ran this model model2-glm(rojos~ageandsex+sector+season+sector:season,quasipoisson) glht(model2,linfct=mcp(ageandsex=Tukey)) General Linear Hypotheses Multiple Comparisons of Means: Tukey Contrasts Linear Hypotheses: Estimate M - H == 0 0.2898 SUB - H == 0 -0.2261 SUB - M == 0 -0.5159 I tried to do the same changing factor season (with 2 levels) for month (with four levels), and i get an error and no results monthmodel2-glm(rojos~ageandsex+sector+month+sector:month,quasipoisson) glht(monthmodel2,linfct=mcp(ageandsex=Tukey)) Error en modelparm.default(model, ...) : dimensions of coefficients and covariance matrix don't match I understand nothing, Why this is happen. I beg for your help - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/GLHT-in-multcomp-Two-similar-models-one-doesn-t-work-tp4291875p4291875.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] colored outliers
Hi Justin, it still does not work. All points become red. I use this skript with your modifications: TOC_NI-read.csv2(C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv, sep=;, dec=,, encoding=UTF-8) circ-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,] plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3) Maybe the the Sourcefile will help to solve the prob? http://r.789695.n4.nabble.com/file/n4291954/TOC-NI.csv TOC-NI.csv Thank you so muich! GeO -- View this message in context: http://r.789695.n4.nabble.com/colored-outliers-tp4282207p4291954.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help needed in interpreting linear models
Hi Petr, thanks for your answer. First of all it's not homework I am a student and need to analyse cancer data using linear models. I looked into that topic since a week now and still struggling in interpreting some of the R output that is why I was asking for help here. I don't quite understand your answer because the 180/190 values belong to height and not to weight. What do you want to show with plot(scores,weight). What I can see from the plot is that there is a correlation between the two variables and therefore weight explains scores. Regards -- View this message in context: http://r.789695.n4.nabble.com/Help-needed-in-interpreting-linear-models-tp4291670p4291894.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] apply transformation
Hello All,
I have the following dataset:
Year 2006 2007
Jan Jan 0.0204 0.0065
Feb Feb 0.0145 0.0082
Mar Mar 0.0027 0.0122
dput(d_tmp)
structure(list(Year = c(Jan, Feb, Mar), `2006` = c(0.0204,
0.0145, 0.0027), `2007` = c(0.0065, 0.0082, 0.0122)), .Names =
c(Year,
2006, 2007), row.names = c(Jan, Feb, Mar), class =
data.frame)
I am trying to use the apply function but the values seem to be
getting coerced to characters. I could recast in my function ... but I
suspect there should be an easier way.
I can always use a for loop to get the output I need but just
wondering if there a way to get the same using apply or some other
function ... (the number of years can be changing in my requirement)
My final output needs to be as follows:
Year20062006-Lbl20072007-Lbl
Jan 0.0204 '2.04%' 0.0065 '0.65%'
Feb 0.0145 '1.45%' 0.0082 '0.82%'
Mar 0.0027 '0.27%' 0.0122 '1.22%'
i.e.
dput(d_final)
structure(list(Year = structure(c(2L, 1L, 3L), .Label = c(Feb,
Jan, Mar), class = factor), X2006 = c(0.0204, 0.0145, 0.0027
), X2006.Lbl = structure(c(3L, 2L, 1L), .Label = c('0.27%',
'1.45%', '2.04%'), class = factor), X2007 = c(0.0065, 0.0082,
0.0122), X2007.Lbl = structure(1:3, .Label = c('0.65%', '0.82%',
'1.22%'), class = factor)), .Names = c(Year, X2006,
X2006.Lbl,
X2007, X2007.Lbl), row.names = c(NA, -3L), class = data.frame)
Please advise.
Santosh
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[R] read.table as integer
Hello, I have a csv file with many variables, both characters and integers. I would like to load it on R and do some operations on integer variables, the problem is that R loads the entire dataset considering all variables as characters, instead I would like that R makes the distinction between the two types, because there are too many variables to do: x1-as.integer(x1) x2-as.integer(x2) x3-as.integer(x3) ... I tried to specify read.table(... stringsAsFactors=FALSE) but it doesn't work. Thanks, Best Regards __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problems with plotCI
Got problems with plotCI (plotrix) I only want to plot the upper part of the error bar in my barplot I had the exact same commands working two months ago Now I wanted to change the legend and when I ran it again plotCI stopped working. To me it seems like there must some bug in R library(plotrix) library (graphics) x = matrix(c( 13.75516276,3.944604404, 15.02280537, 80.27687015,91.35282561, 81.8232087, 5.96796709,3.625017815, 3.202591556), nrow=3,byrow=T) yx=matrix(c( 2.5,7.5,12.5,3.5,8.5,13.5,4.5,9.5,14.5), nrow=3,byrow=T) ste-c(4.870993623, 1.139221564,2.70870722,5.789702998,2.770116512,5.4600946821,1.2926938771,1.2 881562951,1.996090108) w-c(grey,light grey,white) r-c(Leioproctus, Lasioglossum,Bombus) barplot(x,ylab=Relative abundance (%, +SE), xlab=Land use, col=c(grey,light grey,white), beside=TRUE, space=c(0,2), ylim=c(0,107),cex.lab=1.3 ) legend(x=3.5, y=108, box.lty=0, legend=r, fill=w) box(bty='l' ) #original command, now not working # plotCI(x=yx,y=x,uiw=ste,liw=NA, pch=NA_integer_, col=black, lwd=1,add=TRUE) ## why is this following command working plotCI(x=yx,y=x,uiw=NA, liw=ste,pch=NA_integer_, col=black, lwd=1,add=TRUE) Cheers Lasse Bech Jacobsen Horticultural Master Student Copenhagen University Faculty of Life Sciences Department of Ecology and Zoology +4526829470 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] colored outliers
Hi what do you want to achieve? Hi Justin, it still does not work. All points become red. I use this skript with your modifications: TOC_NI-read.csv2(C:/Users/hilliges/Desktop/Master/Daten/Statistik/TOC-NI.csv, sep=;, dec=,, encoding=UTF-8) circ-TOC_NI[order(TOC_NI$NI,decreasing=T),][1:4,] plot(NI~TOC,data=TOC_NI,col=blue, pch=16, xlim=c(0,450)) Points are plotted as a small circles in blue, you can make the points bigger by let say cex=2 abline(lm(NI~TOC,data=TOC_NI),col = red,lwd=3) red line is plotted according to linear model points(NI~TOC,data=TOC_NI,col='red',pch=1,size=3) all points are newly plotted as red circles (and possibly overplot the former ones). There is no size parameter for plot command, therefore it is ommited. see ?plot.default and ?par for available options. You maybe want cex=3 to get those new points as bigger circles. Regards Petr Maybe the the Sourcefile will help to solve the prob? http://r.789695.n4.nabble.com/file/n4291954/TOC-NI.csv TOC-NI.csv Thank you so muich! GeO -- View this message in context: http://r.789695.n4.nabble.com/colored- outliers-tp4282207p4291954.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help needed in interpreting linear models
I forgot to post it to r help. Petr Hi Hi Petr, thanks for your answer. First of all it's not homework I am a student and need to analyse cancer data using linear models. I looked into that topic since a week now and still struggling in interpreting some of the R output that is why I was asking for help here. I don't quite understand your answer because the 180/190 values belong to height and not to weight. What do you want to show with plot(scores,weight). What I can see from the plot is that there is a correlation between the two variables and therefore weight explains scores. Yes, but as far as I remember (I do not keep mails so now I can not see the data you posted - Nabble is not available for me) I said that the height value 190 (which was unique, all others were 180 if I remember correctly) is pointing to scores/weight pair which is slightly out from the simple linear model lm(scores~weight) so it is kind of an outlier from the model. Therefore adding the variable (height) to the model improves it and therefore the height variable in the second model is slightly significant as you found from anova. You can also inspect your models by plot(predict(fit), y.variable) abline(0,1) The better is the model the more close are the points to 0,1 line. Of course you can use some more formal evaluation (residuals, hatvalues...) and you can find appropriate literature e.g. at CRAN web. Those two are my favourites, however there are plenty other sources. Using R for Data Analysis and Graphics - Introduction, Examples and Commentary” by John Maindonald (PDF, data sets and scripts are available at JM's homepage). “Practical Regression and Anova using R” by Julian Faraway (PDF, data sets and scripts are available at the book homepage). Regards Petr Regards -- View this message in context: http://r.789695.n4.nabble.com/Help-needed- in-interpreting-linear-models-tp4291670p4291894.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] analytical solution of partial differential equation
ATANU ata.sonu at gmail.com writes: i am trying to solve a partial differential equation analytically(PDE) in R . i have found some functions that do the stuff numerically. But that will not meet my purpose. is there any function to solve PDE analytically. please help. With extremely limited exceptions (D(), deriv()), R doesn't do analytical solutions. There is an interface to Yacas. You might try Mathematica, or Sage (search if you are looking for a free solution. Good luck, though: solving anything but the simplest PDEs analytically is very challenging ... __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] strange Sys.Date() side effect
Basically I want to create a string vector where first element is a
date.
Note that if you supply c() with objects of different types (as you
have),
the results will probably not be what you wanted.
This is the key. I assumed (not correctly)
that Sys.Date() generates a character
Sys.Date()
[1] 2012-01-13
but as you point out:
class(Sys.Date())
[1] Date
and it is the best to provide c() with arguments of the same type
so, in the end, explicitely casting Sys.Date() to character works for
me:
N - 2; c(as.character(Sys.Date()), sprintf('N=%d', N))
[1] 2012-01-13 N=2
Thanks!
Ryszard
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-Original Message-
From: MacQueen, Don [mailto:macque...@llnl.gov]
Sent: Thursday, January 12, 2012 5:27 PM
To: Czerminski, Ryszard; r-help@r-project.org
Subject: Re: [R] strange Sys.Date() side effect
My best guess is that you are misunderstanding what the c() function
does.
I'd suggest reading the help page for c, obtained by typing
?c
Note that if you supply c() with objects of different types (as you
have),
the results will probably not be what you wanted.
Given what c() does, your output and warning message make sense. But I'm
unable to figure out what you're really trying to do. Something like
this,
perhaps?
N - 2; sprintf('N = %d', N)
[1] N = 2
-Don
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 1/10/12 5:04 AM, Czerminski, Ryszard
ryszard.czermin...@astrazeneca.com wrote:
Any ideas what is the problem with this code?
N - 2; c(Sys.Date(), sprintf('N = %d', N))
[1] 2012-01-10 NA
Warning message:
In as.POSIXlt.Date(x) : NAs introduced by coercion
Best regards,
Ryszard
Ryszard Czerminski
AstraZeneca Pharmaceuticals LP
35 Gatehouse Drive
Waltham, MA 02451
USA
781-839-4304
ryszard.czermin...@astrazeneca.com
---
---
Confidentiality Notice: This message is private and may
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Re: [R] read.table as integer
On Jan 13, 2012, at 7:02 AM, Francisco wrote: Hello, I have a csv file with many variables, both characters and integers. I would like to load it on R and do some operations on integer variables, the problem is that R loads the entire dataset considering all variables as characters, instead I would like that R makes the distinction between the two types, because there are too many variables to do: x1-as.integer(x1) x2-as.integer(x2) x3-as.integer(x3) ... I tried to specify read.table(... stringsAsFactors=FALSE) but it doesn't work. You need to use colClasses -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read.table as integer
On Fri, Jan 13, 2012 at 7:02 AM, Francisco franciscororol...@google.com wrote: Hello, I have a csv file with many variables, both characters and integers. I would like to load it on R and do some operations on integer variables, the problem is that R loads the entire dataset considering all variables as characters, instead I would like that R makes the distinction between the two types, because there are too many variables to do: x1-as.integer(x1) x2-as.integer(x2) x3-as.integer(x3) ... I tried to specify read.table(... stringsAsFactors=FALSE) but it doesn't work. There must be non-integers in some of the columns that are supposed to be integer. Lets assume that the first row has no such garbage. Then we can get the desired classes from that row and apply it to the entire data frame. In this example the second column has such garbage: # test data Lines - a,b,c D,2,3 a,b,9 C,5,6 # read in just row 1 and read in all rows DF1 - read.csv(text = Lines, nrow = 1, as.is = TRUE) DF - DF0 - read.csv(text = Lines, as.is = TRUE) # there will warning as its converting garbage to NAs to.int - function(v, v1) if (inherits(v1, integer)) as.integer(v) else v DF - mapply(to.int, DF0, DF1, SIMPLIFY = FALSE) DF - as.data.frame(DF) As we see here the second column becomes integer despite garbage in it: str(DF0) # as read in 'data.frame': 3 obs. of 3 variables: $ a: chr D a C $ b: chr 2 b 5 $ c: int 3 9 6 str(DF) # as converted 'data.frame': 3 obs. of 3 variables: $ a: Factor w/ 3 levels a,C,D: 3 1 2 $ b: int 2 NA 5 $ c: int 3 9 6 -- Statistics Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] IF ELSE
Can somebody explain the problem in the following expression?
Thank you
/ if (species == 1){
+ fitness -
(1-b)*exp(-((microsites-niche.preference)/(niche.width.specialist+a.specialist)^2)*(1-a.specialist)
+ }else
Error: unexpected '}' in:
fitness -
(1-b)*exp(-((microsites-niche.preference)/(niche.width.specialist+a.specialist)^2)*(1-a.specialist)
}
if (species ==2){
+ fitness )
(1-b)*exp(-((microsites-niche.preference)/(niche.width.generalist+a.generalist)^2)*(1-a.generalist)
Error: unexpected ')' in:
if (species ==2){
fitness )
} else fitness - 0
Error: unexpected '}' in }
}
Error: unexpected '}' in }/
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[R] function to replace values doesn't work on vectors
I've got a numeric vector with values ranging from 1 to 5, I would like to catagorize these values like this: 1 becomes catagory 1 3 becomes catagory 3 And everything else in catagory 2. The simple function I wrote beneath works for single numeric data, but for some reason I am unable to feed it vectors. Any help would be appreciated, as I'm fairly new to R. -- View this message in context: http://r.789695.n4.nabble.com/function-to-replace-values-doesn-t-work-on-vectors-tp4292235p4292235.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] IF ELSE
Hi jost87,
Try using a good text editor with syntax highlighting and
parenthesis/brace matching (personal preference is Emacs + ESS, others
like Rstudio, and there are tons more). I find at least one
mismatched pair of parentheses/braces in those expressions.
Cheers,
Josh
On Fri, Jan 13, 2012 at 6:31 AM, jost87 jos...@hotmail.com wrote:
Can somebody explain the problem in the following expression?
Thank you
/ if (species == 1){
+ fitness -
(1-b)*exp(-((microsites-niche.preference)/(niche.width.specialist+a.specialist)^2)*(1-a.specialist)
+ }else
Error: unexpected '}' in:
fitness -
(1-b)*exp(-((microsites-niche.preference)/(niche.width.specialist+a.specialist)^2)*(1-a.specialist)
}
if (species ==2){
+ fitness )
(1-b)*exp(-((microsites-niche.preference)/(niche.width.generalist+a.generalist)^2)*(1-a.generalist)
Error: unexpected ')' in:
if (species ==2){
fitness )
} else fitness - 0
Error: unexpected '}' in }
}
Error: unexpected '}' in }/
--
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and provide commented, minimal, self-contained, reproducible code.
--
Joshua Wiley
Ph.D. Student, Health Psychology
Programmer Analyst II, Statistical Consulting Group
University of California, Los Angeles
https://joshuawiley.com/
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Re: [R] How can I prevent solve.QP from printing the solution progress ?
Many thanks all,
Tolga
-Original Message-
From: Chibisi Chima-Okereke [mailto:cchima-oker...@mango-solutions.com]
Sent: 12 January 2012 17:14
To: Uzuner, Tolga I
Cc: Jack Teall; Richard Pugh
Subject: RE: How can I prevent solve.QP from printing the solution progress ?
Try something like this ...
myFile - tempfile()
sink(file = myFile)
solve.Qp ()
sink()
file.remove(myFile) #or
unlink(myFile)
Kind regards
Chibisi
-Original Message-
From: Jack Teall
Sent: 12 January 2012 16:49
To: Matt Aldridge; Richard Pugh; Consultants
Subject: FW: How can I prevent solve.QP from printing the solution progress ?
tolga.i.uzu...@jpmorgan.com
Jack Teall
T:+44 (0)1249 766811
F: +44 (0)1249 767707
www.mango-solutions.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Uzuner, Tolga I
Sent: 12 January 2012 16:46
To: r-help@r-project.org
Subject: [R] How can I prevent solve.QP from printing the solution progress ?
Dear R Users,
How can I prevent solve.Qp from printing the solution progress ?
Thanks in advance,
Tolga
This email is confidential and subject to important disclaimers and conditions
including on offers for the purchase or sale of securities, accuracy and
completeness of information, viruses, confidentiality, legal privilege, and
legal entity disclaimers, available at
http://www.jpmorgan.com/pages/disclosures/email.
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Re: [R] Averaging within a range of values
Regarding your last question, read ?cut --- 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. doggysaywhat chwh...@ucsd.edu wrote: Hello all. I have two data frames. Group Start End G1200 700 G2500 1000 G320003000 G440006000 G570008000 and PosC0 C1 2000.9 0.6 500 0.8 0.8 8000.9 0.7 1000 0.7 0.6 2000 0.6 0.4 2500 1.2 0.8 3000 0.6 1.5 3500 0.7 0.7 4000 0.8 0.8 4500 0.6 0.6 5000 0.9 0.9 5500 0.7 0.8 6000 0.8 0.7 6500 0.4 0.4 7000 0.5 0.8 7500 0.7 0.9 8000 0.9 0.5 8500 0.8 0.6 9000 0.9 0.8 I need to conditionally average all values in columns C0 and C1 based upon the bins I defined in the first data frame. For example, for the bin G1 in the first dataframe, the values are 200 to 700 so i would average the value at pos 200 (0.9) and 500 (0.8) for C0 and then perform the same thing for C1. I can do this in excel with array formulas but I'm relatively new to R and would like know if there is a function that will perform the same action. I don't know if this will help, but the excel array function I used was average(if(range=start)*(range=end),range)). Where the range is the entire pos column. Initially I looked at the aggregate function. I can use aggregate when I give a single vector to be used for grouping such as (A,B,C) but I'm not sure how to define grouping as the bin 200-500 and the second bin as 500-1000 etc. and use that as my grouping vector. Any help would be greatly appreciated. -- View this message in context: http://r.789695.n4.nabble.com/Averaging-within-a-range-of-values-tp4291958p4291958.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] function to replace values doesn't work on vectors
I don't see any signs of a simple function, but I'd use ifelse(). y - ifelse(x == 1, 1, ifelse(x == 3, 3, 2)) or some such. (Lack of reproducible example means lack of actual testing.) Sarah On Fri, Jan 13, 2012 at 9:11 AM, WoutH w.denhollan...@lumc.nl wrote: I've got a numeric vector with values ranging from 1 to 5, I would like to catagorize these values like this: 1 becomes catagory 1 3 becomes catagory 3 And everything else in catagory 2. The simple function I wrote beneath works for single numeric data, but for some reason I am unable to feed it vectors. Any help would be appreciated, as I'm fairly new to R. -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Troubles with stemming (tm + Snowball packages) under MacOS
Dear all, I have some troubles using the stemming algorithm provided by the tm (text mining) + Snowball packages. Here is my config: MacOS 10.5 R 2.12.0 / R 2.13.1 / R 2.14.1 (I have tried several versions) I have installed all the needed packages (tm, rJava, rWeka, Snowball) + dependencies. I have desactivated AWT (like written in http://r.789695.n4.nabble.com/Problem-with-Snowball-amp-RWeka-td3402126.html) with : Sys.setenv(NOAWT=TRUE) The command tm_map(reuters, stemDocument) gives the following errors : - First time: Error in .jnew(name) : java.lang.InternalError: Can't start the AWT because Java was started on the first thread. Make sure StartOnFirstThread is not specified in your application's Info.plist or on the command line Refreshing GOE props... - Second time: Stemmer 'porter' unknown! Stemmer 'english' unknown! Stemmer 'porter' unknown! Stemmer 'english' unknown! Stemmer 'porter' unknown! Stemmer 'english' unknown! Stemmer 'porter' unknown! Stemmer 'english' unknown! Stemmer 'porter' unknown! Stemmer 'english' unknown! (etc.) I have already search the Web for a solution, but I have found nothing useful. Here is the full source code (all the librairies are already loaded): -- Sys.setenv(NOAWT=TRUE) source - ReutersSource(reuters-21578.xml, encoding=UTF-8) reuters - Corpus(source) reuters - tm_map(reuters, as.PlainTextDocument) reuters - tm_map(reuters, removePunctuation) reuters - tm_map(reuters, tolower) reuters - tm_map(reuters, removeWords, stopwords(english)) reuters - tm_map(reuters, removeNumbers) reuters - tm_map(reuters, stripWhitespace) reuters - tm_map(reuters, stemDocument) -- Thank you for your help, Julien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Quantiles in boxplot
Hi, I have a simple question about quartiles in R, especially how they are calculated using the boxplot. Quartiles (.25 and .75) in boxplot are different from the summary function and also don't match with the 9 types in the quantile function. See attachment for details. Can you give me the details on how the boxplot function does calculate these values? Cheers, Rene Brinkhuis (Netherlands) Quartiles in R.pdf Description: Adobe PDF document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] access/row access/col access
Hello, I have a data.frame and I want to transfor it in a list of rows or columns. I can do apply(myDataFrame,MARGIN=1,FUN=???) I remember that there is a function which mean return or access column ... something like :: or ], or [, I can't remember can somebody refresh my memory? -- View this message in context: http://r.789695.n4.nabble.com/access-row-access-col-access-tp4292531p4292531.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quantiles in boxplot
The explanation is in ?boxplot.stats (as the help for boxplot states). Details: The two ‘hinges’ are versions of the first and third quartile, i.e., close to ‘quantile(x, c(1,3)/4)’. The hinges equal the quartiles for odd n (where ‘n - length(x)’) and differ for even n. Whereas the quartiles only equal observations for ‘n %% 4 == 1’ (n = 1 mod 4), the hinges do so _additionally_ for ‘n %% 4 == 2’ (n = 2 mod 4), and are in the middle of two observations otherwise. And so on, with references. Sarah 2012/1/13 René Brinkhuis rene.brinkh...@live.nl: Hi, I have a simple question about quartiles in R, especially how they are calculated using the boxplot. Quartiles (.25 and .75) in boxplot are different from the summary function and also don't match with the 9 types in the quantile function. See attachment for details. Can you give me the details on how the boxplot function does calculate these values? Cheers, Rene Brinkhuis (Netherlands) -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] access/row access/col access
I have to admit, I have very little idea what you are trying to do. Can you provide an example? In general, the i-th column of a data frame can be accessed with mydataframe[, i] but that doesn't help with whatever you want to do with apply(). Sarah On Fri, Jan 13, 2012 at 10:45 AM, statquant2 statqu...@gmail.com wrote: Hello, I have a data.frame and I want to transfor it in a list of rows or columns. I can do apply(myDataFrame,MARGIN=1,FUN=???) I remember that there is a function which mean return or access column ... something like :: or ], or [, I can't remember can somebody refresh my memory? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Odp: Remove space from string
I have NO knowledge of regexpr, but someone helped me out once and I put
it into a function I call trim. Here is the line I use:
function(x) gsub(^[[:space:]]+|[[:space:]]+$, , x)
One more thing you can try. Hope it helps, Roger
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On Behalf Of Petr PIKAL
Sent: Friday, January 13, 2012 7:40 AM
To: Vikram Bahure
Cc: r-help@r-project.org
Subject: [R] Odp: Remove space from string
Hi
Dear R users,
I have some trivial query.
I have a string, I want to remove space from the string.
For eg.
Input:
a - Remove space
Output required:
Removespace
It seems to be simple. Even myself with very poor knowledge of regexpr
can suggest solution.
gsub( , , a)
Regards
Petr
I tried using str_trim but only removes end spaces. library(stringr).
Regards
Vikram
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[R] plotting regression line in with lattice
#Dear All,
#I'm having a bit of a trouble here, please help me...
#I have this data
set.seed(4)
mydata - data.frame(var = rnorm(100),
temp = rnorm(100),
subj = as.factor(rep(c(1:10),5)),
trt = rep(c(A,B), 50))
#and this model that fits them
lm - lm(var ~ temp * subj, data = mydata)
#i want to plot the results with lattice anf fit the regression line, predicted
with my model, trough them
#to do so, I'm using this approach, outlined Lattice Tricks for the power
useR by D. Sarkar
temp_rng - range(mydata$temp, finite = TRUE)
grid - expand.grid(temp = do.breaks(temp_rng, 30),
subj = unique(mydata$subj),
trt = unique(mydata$trt))
model - cbind(grid, var = predict(lm, newdata = grid))
orig - mydata[c(var,temp,subj,trt)]
combined - make.groups(original = orig, model = model)
xyplot(var ~ temp | subj,
data = combined,
groups = which,
type = c(p, l),
distribute.type = TRUE
)
# so far every thing is fine, but, i also whant assign a filling to the data
points for the two treatments trt=1 and trt=2
# so I have written this piece of code, that works fine, but when it comes to
plot the regression line, it seems that type is not recognized by the panel
function...
my.fill - c(black, grey)
plot - with(combined,
xyplot(var ~ temp | subj,
data = combined,
group = combined$which,
type = c(p, l),
distribute.type = TRUE,
panel = function(x, y, ..., subscripts){
fill - my.fill[combined$trt[subscripts]]
panel.xyplot(x, y, pch = 21, fill = my.fill, col =
black)
},
key = list(space = right,
text = list(c(trt1, trt2), cex = 0.8),
points = list(pch = c(21), fill = c(black, grey)),
rep = FALSE)
)
)
plot
#I've also tried to move type and distribute type within panel.xyplot, as well
as subsseting the data in it panel.xyplot like this
plot - with(combined,
xyplot(var ~ temp | subj,
data = combined,
panel = function(x, y, ..., subscripts){
fill - my.fill[combined$trt[subscripts]]
panel.xyplot(x[combined$which==original],
y[combined$which==original], pch = 21, fill = my.fill, col = black)
panel.xyplot(x[combined$which==model],
y[combined$which==model], type = l, col = black)
},
key = list(space = right,
text = list(c(trt1, trt2), cex = 0.8),
points = list(pch = c(21), fill = c(black, grey)),
rep = FALSE)
)
)
plot
#but no success with that either...
#can anyone help me to get the predicted values plotted as a line instead of
being points?
#really appricieate
#matteo
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Re: [R] function to replace values doesn't work on vectors
WoutH wrote
I've got a numeric vector with values ranging from 1 to 5, I would like to
catagorize these values like this:
1 becomes catagory 1
3 becomes catagory 3
And everything else in catagory 2. The simple function I wrote beneath
works for single numeric data, but for some reason I am unable to feed it
vectors. Any help would be appreciated, as I'm fairly new to R.
function.123 - function(x)
{
x1 - ifelse(x == 1 | x == 3, return(x) ,return(2))
return(x1)
}
Don't use the return() in the ifelse since return() returns immediately from
the function.
Just
x1 - ifelse(x==1 | x==3, x, 2)
will do.
Berend
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Re: [R] access/row access/col access
http://cran.r-project.org/doc/manuals/R-intro.pdf --- 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. statquant2 statqu...@gmail.com wrote: Hello, I have a data.frame and I want to transfor it in a list of rows or columns. I can do apply(myDataFrame,MARGIN=1,FUN=???) I remember that there is a function which mean return or access column ... something like :: or ], or [, I can't remember can somebody refresh my memory? -- View this message in context: http://r.789695.n4.nabble.com/access-row-access-col-access-tp4292531p4292531.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem Installing R to SuSE 10 via RPM
Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Nabble? Was Re: function to replace values doesn't work on vectors
Interesting: The email I received through the R-help list didn't have
all the information that was apparently there:
On Fri, Jan 13, 2012 at 11:32 AM, Berend Hasselman b...@xs4all.nl wrote:
WoutH wrote
I've got a numeric vector with values ranging from 1 to 5, I would like to
catagorize these values like this:
1 becomes catagory 1
3 becomes catagory 3
And everything else in catagory 2. The simple function I wrote beneath
works for single numeric data, but for some reason I am unable to feed it
vectors. Any help would be appreciated, as I'm fairly new to R.
function.123 - function(x)
{
x1 - ifelse(x == 1 | x == 3, return(x) ,return(2))
return(x1)
}
This function? Not in the original email as I received it, although it showed
up in Berend's reply.
I hope that it was a momentary glitch; greater disagreement between Nabble
and the email list will cause all sorts of fun. If the interface,
whatever it is,
starts stripping out code? I'll have to quit answering Nabble queries entirely.
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
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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] Coloring counties on a full US map based on a certain criterion
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Thanks a lot for any hint!
--
Dimitri Liakhovitski
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem Installing R to SuSE 10 via RPM
I think it's saying you need to install R-base before R-base-devel. You'll need to add a cran repository as SUSE might not have the most up-to-date version of R. This is the code for Ubuntu I assume it's the same, just change the distro and keyserver: sudo apt-get update sudo add-apt-repository 'deb http://cran.ma.imperial.ac.uk/bin/linux/ubuntu oneiric/' gpg --keyserver keyserver.ubuntu.com --recv-key E084DAB9 gpg -a --export E084DAB9 | sudo apt-key add - You may also want to get Sun java, again, change distro and keyserver: sudo add-apt-repository ppa:ferramroberto/java sudo apt-key adv --recv-key --keyserver keyserver.ubuntu.com B725097B3ACC3965 sudo apt-get update sudo apt-get install sun-java6-jdk sun-java6-plugin Then run: sudo apt-get install r-base r-base-dev sudo R CMD javareconf Cheers, Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Matthew Pettis Sent: 13 January 2012 16:43 To: r-help@r-project.org Subject: [R] Problem Installing R to SuSE 10 via RPM Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem Installing R to SuSE 10 via RPM
Thanks, will do! I thought devel included base, but evidently, that's not the case... On Fri, Jan 13, 2012 at 10:58 AM, Gavin Blackburn gavin.blackb...@strath.ac.uk wrote: I think it's saying you need to install R-base before R-base-devel. You'll need to add a cran repository as SUSE might not have the most up-to-date version of R. This is the code for Ubuntu I assume it's the same, just change the distro and keyserver: sudo apt-get update sudo add-apt-repository 'deb http://cran.ma.imperial.ac.uk/bin/linux/ubuntu oneiric/' gpg --keyserver keyserver.ubuntu.com --recv-key E084DAB9 gpg -a --export E084DAB9 | sudo apt-key add - You may also want to get Sun java, again, change distro and keyserver: sudo add-apt-repository ppa:ferramroberto/java sudo apt-key adv --recv-key --keyserver keyserver.ubuntu.comB725097B3ACC3965 sudo apt-get update sudo apt-get install sun-java6-jdk sun-java6-plugin Then run: sudo apt-get install r-base r-base-dev sudo R CMD javareconf Cheers, Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Matthew Pettis Sent: 13 January 2012 16:43 To: r-help@r-project.org Subject: [R] Problem Installing R to SuSE 10 via RPM Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Do not seek to follow in the footsteps of the wise men of old. Seek what they sought. - Matsuo Munefusa (Basho) [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Hi,
You've just about got it. See below.
On Fri, Jan 13, 2012 at 11:52 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Using a random sampling to give you the basic idea.
There are 99 counties in Iowa, so to construct the criterion:
countycol - sample(1:5, 99, replace=TRUE)
And to invent a set of colors (RColorBrewer is a better choice for
final maps):
classcolors - rainbow(5)
then you can use them in your map just as you would for any other
plotting command:
map('county', 'iowa', fill= TRUE, col = classcolors[countycol])
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Absolutely. The only thing you have to watch out for is that you put your
values in the same order as:
map('county', 'iowa', plot=FALSE)$names
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Sarah, this is amazing, thank you so much.
One question: I am trying to do for the whole US (on one map) what
you've helped me do for Iowa.
In other words, I would like to create a file of inputs like you
countycol with 1,000+ lines - for all US counties (probably without
Hawaii and Alaska, right?) and then somehow feed it into the map
command but so that the output is the whole map of the US, and not one
state. Is it at all possible?
Or maybe it's possible to create 48 colored state maps one by one -
the way you showed me - save them, and then somehow paste those
states onto the whole US map?
Thanks a lot for your help!
Dimitri
On Fri, Jan 13, 2012 at 12:05 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Hi,
You've just about got it. See below.
On Fri, Jan 13, 2012 at 11:52 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Using a random sampling to give you the basic idea.
There are 99 counties in Iowa, so to construct the criterion:
countycol - sample(1:5, 99, replace=TRUE)
And to invent a set of colors (RColorBrewer is a better choice for
final maps):
classcolors - rainbow(5)
then you can use them in your map just as you would for any other
plotting command:
map('county', 'iowa', fill= TRUE, col = classcolors[countycol])
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Absolutely. The only thing you have to watch out for is that you put your
values in the same order as:
map('county', 'iowa', plot=FALSE)$names
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] access/row access/col access
Like Sarah said, what you say below makes very little sense, but as a total shot in the dark, is this what you mean? lapply(1:nrow(df), function(i) x[i,] ) Michael On Jan 13, 2012, at 11:15 AM, Sarah Goslee sarah.gos...@gmail.com wrote: I have to admit, I have very little idea what you are trying to do. Can you provide an example? In general, the i-th column of a data frame can be accessed with mydataframe[, i] but that doesn't help with whatever you want to do with apply(). Sarah On Fri, Jan 13, 2012 at 10:45 AM, statquant2 statqu...@gmail.com wrote: Hello, I have a data.frame and I want to transfor it in a list of rows or columns. I can do apply(myDataFrame,MARGIN=1,FUN=???) I remember that there is a function which mean return or access column ... something like :: or ], or [, I can't remember can somebody refresh my memory? -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
On Fri, Jan 13, 2012 at 12:15 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Sarah, this is amazing, thank you so much.
One question: I am trying to do for the whole US (on one map) what
you've helped me do for Iowa.
In other words, I would like to create a file of inputs like you
countycol with 1,000+ lines - for all US counties (probably without
Hawaii and Alaska, right?) and then somehow feed it into the map
command but so that the output is the whole map of the US, and not one
state. Is it at all possible?
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
Or maybe it's possible to create 48 colored state maps one by one -
the way you showed me - save them, and then somehow paste those
states onto the whole US map?
Thanks a lot for your help!
Dimitri
On Fri, Jan 13, 2012 at 12:05 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Hi,
You've just about got it. See below.
On Fri, Jan 13, 2012 at 11:52 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Using a random sampling to give you the basic idea.
There are 99 counties in Iowa, so to construct the criterion:
countycol - sample(1:5, 99, replace=TRUE)
And to invent a set of colors (RColorBrewer is a better choice for
final maps):
classcolors - rainbow(5)
then you can use them in your map just as you would for any other
plotting command:
map('county', 'iowa', fill= TRUE, col = classcolors[countycol])
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Absolutely. The only thing you have to watch out for is that you put your
values in the same order as:
map('county', 'iowa', plot=FALSE)$names
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Portfolio Optimization
Hi, I'm an R newbie and I've been struggling with a optimization problem for the past couple of days now. Here's the problem - I have a matrix of expected payouts from different stock option strategies. Each column in my matrix represents a different stock and each row represents the return to the strategy given a certain market move. So the rows are not a time series of percentage returns but a dollar payout in different expected scenarios, i.e. Expected Return Matrix (ER) = stock1 stock2 stockn scenario1 $ $ $ scenario2 $ $ $ scenario3 $ $ $ ... I want to create an optimal portfolio of these strategies by applying a vector of weights. The weights will be the number of contracts of each to buy and won't be a percentage weighting. There are a few constraints I need it comply with: - The weights have to be integers - The minimum portfolio return (ER* Weights) across the scenarios has to be greater than some negative number I specify - There has to be a certain minimum number of stocks in the portfolio so length(weights)some number I specify. Any help is GREATLY appreciated since I have tried so many different functions and packages. Even if someone can just lead me to the correct function to use that would be a great help as I've looked at optim, solveLP, ROI package and many others. Thanks, S [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Brillouin index
Dear colleagues. I wonder if anybody knows about a procedure in R to calculate the Brillouin Diversity index. I searched the net but did not find anything about it. Thanks a lot for any help Best, Philipp *** Prof. Dr. Philipp Fischer Head of AWI Center for Scientific Diving Dept. In situ Ecology Section Shelf Sea Systems Alfred-Wegener-Institut Biologische Anstalt Helgoland Building A D-27498 Helgoland Phone: +49(4725)819-3344 Skype: fischer_philipp Fax: +49(4725)819-3369 http://www.awi.de/People/show?pfischer http://www.awi.de/en/infrastructure/underwater/scientific_diving/ http://www.awi.de/en/research/research_divisions/biosciences/shelf_sea_ecology/ http://www.forschungstauchen-deutschland.de __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] tm package, custom reader
I need help with creating custom xml reader for use with the tm package. The objective is to crate a corpus for analysis. Files that I'm working with come from solr and are in a funky XML format never the less I'm able to parse the XML files using solrDocs.R function provided by Duncan Temple Lang. The problem I'm having that once I parse the document I need to create a custom reader that would be compatible with the tm package. If someone build a custom reader for tm package, or has some ideas of how to go about this, I would greatly appreciate the help. Thanks -- View this message in context: http://r.789695.n4.nabble.com/tm-package-custom-reader-tp4292766p4292766.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] meta-analysis normal quantile plot metafor
At 15:53 11/01/2012, Ricc wrote: Hello, I once used the metawin software to perform a meta-analysis (see metawinsoft, Rosenberg et al.) and produced normal qqplot to test for a potential bias in the dataset. I now want to re-use the same dataset with the package metafor by W. Viechtbauer (great package btw). I run the qqnorm.rma.uni function. I use standardized effect sizes as in metawin. I think it would help if you said which parameters you used to control the envelope. Did you smooth it? Did you use the Bonferroni correction? QQplot generated with metafor differs from the plot obtained with metawin: most of the datapoint fall outside the confidence envelope (using the same confidence level). I don't understand very well how the pseudo confidence envelope was created in metafor. Is it more conservative than that from metawin or created using the package envelope ? Unfortunately I do not have access to metawin's code so that I cannot compare implementations but the manual let me think that metawin print classical confidence interval... Thanks for input ! Ricc More precisions: R version 2.13.1 (2011-07-08) Platform: x86_64-pc-linux-gnu (64-bit) metafor_1.6-0 Michael Dewey i...@aghmed.fsnet.co.uk http://www.aghmed.fsnet.co.uk/home.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] fUtilities removed
When setting up my new machine, I had the surprise to see that Package 'fUtilities' was removed from the CRAN repository. This is problematic for my work. I use many of its functions, and it will complicate things a lot if other programmers want to use my previous code in the future. Plus, nowhere can I find the justification for its removal. Thanks for any info on this [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] GLHT in multcomp: Two similar models, one doesn't work
Nobody? - Mario Garrido Escudero PhD student Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Universidad de Salamanca -- View this message in context: http://r.789695.n4.nabble.com/GLHT-in-multcomp-Two-similar-models-one-doesn-t-work-tp4291875p4292889.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem Installing R to SuSE 10 via RPM
Two things: 1. Looking at the sizes of the RPMs at the URL provided, the 'R-base-devel' RPM is only 84k. So as with other Linux distros, that is likely to contain R header files and other such things to enable compilation of CRAN packages during their installation using the source tarballs. The 'R-base' RPM is 33Mb, so that is your primary install for R. You will want both. 2. I don't know about the configuration of the RPMs on SUSE, but presume that as with RH/Fedora, the RPMs can be created in such a fashion that they are not relocatable. That means that they are configured to be only installed 'system-wide', not to a specific user's directory tree. You may have issues trying to install them without root access on your computer, thus may get other permission related errors when you get the correct RPM. Whether they are or are not relocatable would be a question for the SUSE maintainer for the R package. If they are, there are 'rpm' command line arguments that are relevant. So use 'man rpm' for more details. If the RPMs are not relocatable, you will be left with the option of building R from source in order to install R locally and will want to read the R Installation and Administration manual for details. HTH, Marc Schwartz On Jan 13, 2012, at 11:00 AM, Matthew Pettis wrote: Thanks, will do! I thought devel included base, but evidently, that's not the case... On Fri, Jan 13, 2012 at 10:58 AM, Gavin Blackburn gavin.blackb...@strath.ac.uk wrote: I think it's saying you need to install R-base before R-base-devel. You'll need to add a cran repository as SUSE might not have the most up-to-date version of R. This is the code for Ubuntu I assume it's the same, just change the distro and keyserver: sudo apt-get update sudo add-apt-repository 'deb http://cran.ma.imperial.ac.uk/bin/linux/ubuntu oneiric/' gpg --keyserver keyserver.ubuntu.com --recv-key E084DAB9 gpg -a --export E084DAB9 | sudo apt-key add - You may also want to get Sun java, again, change distro and keyserver: sudo add-apt-repository ppa:ferramroberto/java sudo apt-key adv --recv-key --keyserver keyserver.ubuntu.comB725097B3ACC3965 sudo apt-get update sudo apt-get install sun-java6-jdk sun-java6-plugin Then run: sudo apt-get install r-base r-base-dev sudo R CMD javareconf Cheers, Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Matthew Pettis Sent: 13 January 2012 16:43 To: r-help@r-project.org Subject: [R] Problem Installing R to SuSE 10 via RPM Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] multidimensional array calculation
Hello,
probably it is quite easy but I can get it: I have
mulitple numeric vectors and a function using
all of them to calculate a new value:
L - c(200,400,600)
AR - c(1.5)
SO - c(1,3,5)
T - c(30,365)
fun - function(L,AR,SO,T){
exp(L*AR+sqrt(SO)*log(T))
}
How can I get an array or dataframe where
all possible combinations of the factors are listed
and the new value is calculated.
I thought about an array like:
array(NA, dim = c(3,1,3,2),
dimnames=list(c(200,400,600),c(1.5),c(1,3,5),c(30,365)))
but how can I get the array populated according to the function?
As I want to get in the end a 2D dataframe I probably will use the melt.array()
function from the reshape package or is there another way to get simple such
a full-factorial dataframe with all possible combinations?
Best regards,
Johannes
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] access/row access/col access
On Jan 13, 2012, at 10:45 AM, statquant2 wrote: Hello, I have a data.frame and I want to transfor it in a list of rows or columns. I can do apply(myDataFrame,MARGIN=1,FUN=???) I remember that there is a function which mean return or access column ... something like :: or ], or [, I can't remember can somebody refresh my memory? It is definitely the long way around) to do this, since [,5] would get the same results, but if you want the fifth element in each row you could do this: dat id x1 x2 x3 y1 y2 y3 z1 z2 z3 v 1 1 2 4 5 10 20 15 200 150 170 2.5 2 2 3 7 6 25 35 40 300 350 400 4.2 apply(dat, 1, [, 5) [1] 10 25 When 'apply' is used, the trailing arguments are matched either by position or name to the arguments of the function. In this case the 5 gets matched to the i for the [ function. Because [ is primitive the name is actually ignored even if offered, so using any other name would not change the result: apply(dat, 1, [, j=5) [1] 10 25 Whereas if you were using quantile as your function, you might perhaps use prob=c(,25,,75), na.rm=TRUE or na.rm=TRUE prob=c(,25,,75), prob=c(, 25,,75) -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Thank you somuch, Sarah. I tried, and it's working just wonderfully
(code below).
One last question, if I may: is it possible to get rid of borders
between counties (just leave the fill)? I did not find this argument
in help...
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$group])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
Or maybe it's possible to create 48 colored state maps one by one -
the way you showed me - save them, and then somehow paste those
states onto the whole US map?
Thanks a lot for your help!
Dimitri
On Fri, Jan 13, 2012 at 12:05 PM, Sarah Goslee sarah.gos...@gmail.com
wrote:
Hi,
You've just about got it. See below.
On Fri, Jan 13, 2012 at 11:52 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Using a random sampling to give you the basic idea.
There are 99 counties in Iowa, so to construct the criterion:
countycol - sample(1:5, 99, replace=TRUE)
And to invent a set of colors (RColorBrewer is a better choice for
final maps):
classcolors - rainbow(5)
then you can use them in your map just as you would for any other
plotting command:
map('county', 'iowa', fill= TRUE, col = classcolors[countycol])
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Absolutely. The only thing you have to watch out for is that you put your
values in the same order as:
map('county', 'iowa', plot=FALSE)$names
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Merging data XXXX
Hello everyone, I have 1 data frame (just a vector in the example below) with 12 individuals listed and a separate vector of 36 days (in week intervals). What is the best way to merge these together so that each individual (specialist here) has all 36 days matched with their specialist number (a one to many merge in SAS; essentially resulting in long format data). implement-as.Date(2012-4-30) start-implement-91 weeks-seq(start,by=weeks,length=36) weeks specialist-1:12 Thanks! Dan [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Just to clarify, according to help about the fill argument:
logical flag that says whether to draw lines or fill areas. If FALSE,
the lines bounding each region will be drawn (but only once, for
interior lines). If TRUE, each region will be filled using colors from
the col = argument, and bounding lines will not be drawn.
We have fill=TRUE - so why are the county borders still drawn?
Thank you!
Dimitri
On Fri, Jan 13, 2012 at 1:41 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you somuch, Sarah. I tried, and it's working just wonderfully
(code below).
One last question, if I may: is it possible to get rid of borders
between counties (just leave the fill)? I did not find this argument
in help...
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$group])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
Or maybe it's possible to create 48 colored state maps one by one -
the way you showed me - save them, and then somehow paste those
states onto the whole US map?
Thanks a lot for your help!
Dimitri
On Fri, Jan 13, 2012 at 12:05 PM, Sarah Goslee sarah.gos...@gmail.com
wrote:
Hi,
You've just about got it. See below.
On Fri, Jan 13, 2012 at 11:52 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Dear Rers,
is there a way to color counties on a full US map based on a criterion
one establishes (i.e., all counties I assign the same number should be
the same color)?
I explored a bit and looks like the package maps might be of help.
library(maps)
One could get a map of the US: map('usa')
One could get countries within a US state: map('county', 'iowa', fill
= TRUE, col = palette())
Using a random sampling to give you the basic idea.
There are 99 counties in Iowa, so to construct the criterion:
countycol - sample(1:5, 99, replace=TRUE)
And to invent a set of colors (RColorBrewer is a better choice for
final maps):
classcolors - rainbow(5)
then you can use them in your map just as you would for any other
plotting command:
map('county', 'iowa', fill= TRUE, col = classcolors[countycol])
Would it be possible to read in a file with counties and their
assignments (some counties have a 1, some counties have a 2, etc.) and
then have one map of the US with counties colored based on their
assignment?
Absolutely. The only thing you have to watch out for is that you put your
values in the same order as:
map('county', 'iowa', plot=FALSE)$names
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
On Fri, Jan 13, 2012 at 1:41 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Thank you somuch, Sarah. I tried, and it's working just wonderfully
(code below).
One last question, if I may: is it possible to get rid of borders
between counties (just leave the fill)? I did not find this argument
in help...
But the help does say that additional arguments are passed to lines(),
so you can use lty=0.
That can leave white bits between counties if the areas don't line up
precisely, so I think it looks better with the lines in black.
Sarah
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$group])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] fUtilities removed
On Jan 13, 2012, at 12:33 PM, Dominic Comtois wrote: When setting up my new machine, I had the surprise to see that Package 'fUtilities' was removed from the CRAN repository. https://stat.ethz.ch/pipermail/rmetrics-core/2012-January/000554.html https://stat.ethz.ch/pipermail/rmetrics-core/2011-November/000549.html This is problematic for my work. I use many of its functions, and it will complicate things a lot if other programmers want to use my previous code in the future. Plus, nowhere can I find the justification for its removal. You need to send your questions to the maintainers. They apparently did not respond to the requests to fix the errors. Thanks for any info on this You should perhaps subscribe to the list that is established for discussion on this and related packages. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Merging data XXXX
On Jan 13, 2012, at 12:42 PM, Dan Abner wrote: Hello everyone, I have 1 data frame (just a vector in the example below) with 12 individuals listed and a separate vector of 36 days (in week intervals). What is the best way to merge these together so that each individual (specialist here) has all 36 days matched with their specialist number (a one to many merge in SAS; essentially resulting in long format data). implement-as.Date(2012-4-30) start-implement-91 weeks-seq(start,by=weeks,length=36) weeks specialist-1:12 Thanks! Dan Two ways: merge(specialist, weeks) expand.grid(specialist, weeks) See ?merge which performs a SQL-like join operation and ?expand.grid which provides for all possible combinations of two or more vectors HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
On Fri, Jan 13, 2012 at 1:52 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Just to clarify, according to help about the fill argument:
logical flag that says whether to draw lines or fill areas. If FALSE,
the lines bounding each region will be drawn (but only once, for
interior lines). If TRUE, each region will be filled using colors from
the col = argument, and bounding lines will not be drawn.
We have fill=TRUE - so why are the county borders still drawn?
Thank you!
Dimitri
This prompted me to check the code:
if fill=TRUE, map() calls polygon()
if fill=FALSE, map() calls lines()
But polygon() draws borders by default.
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow)
To not draw borders, the border argument is provided:
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow, border=NA)
But that fails in map():
map('county', 'iowa', fill=TRUE, col=rainbow(20), border=NA)
Error in par(pin = p) :
invalid value specified for graphical parameter pin
because border is used as a named argument in map() already, for setting the
size of the plot area, so there's no way to alter the border argument
to polygon.
The work-around I suggested previous (lty=0) seems to be the only
way to deal with the problem.
Sarah
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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] Merging data XXXX
On Jan 13, 2012, at 1:42 PM, Dan Abner wrote: Hello everyone, I have 1 data frame (just a vector in the example below) with 12 individuals listed and a separate vector of 36 days (in week intervals). What is the best way to merge these together so that each individual (specialist here) has all 36 days matched with their specialist number (a one to many merge in SAS; essentially resulting in long format data). implement-as.Date(2012-4-30) start-implement-91 weeks-seq(start,by=weeks,length=36) weeks specialist-1:12 ?rep ?data.frame I would not think 'merge' is needed ... you just want to make the right number of copies and you should be paying attention to the 'each' and 'times' parameters. [[alternative HTML version deleted]] The above suggests failure adhere to the below PLEASE do read the posting guide http://www.R-project.org/posting-guide.html -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] grplasso
So does anyone use this package? - Original Message - From: Scott Raynaud scott.rayn...@yahoo.com To: r-help@r-project.org r-help@r-project.org Cc: Sent: Tuesday, January 10, 2012 1:40 PM Subject: grplasso I want to use the grplasso package on a data set where I want to fit a linear model. My interest is in identifying significant beta coefficients. The documentation is a bit cryptic so I'd appreciate some help. I know this is a strategy for large numbers of variables but consider a simple case for pedagogical puposes. Say I have two 3 category predictors (2 dummies each), a binary predictor and a continuous predictor with a continuous outcome: y x1 x2 x3 x4 x5 x6 rows of data here .. .. Naturally, I want to select x1 and x2 as a group and x3 and x4 as another group. The documentation has a couple of examples but it's not clear how they translate to the current problem. How do I specify my groups and run the lasso regression? Looks like this is the grouping part: index-c(NA,) but I'm not sure how to specify the df for the variables past the NA for the intercept. Once that's defined the penalty can be specified: lambda - lambdamax(x, y = y, index = index, penscale = sqrt, model = LogReg()) * 0.5^(0:5) In my case I'd use LinReg for the model. Then the model: fit - grplasso(x, y = y, index = index, lambda = lambda, model = LogReg(), penscale = sqrt, control = grpl.control(update.hess = lambda, trace = 0)) again using LinReg for the model. This can be plotted against lambda, but when I do lasso regression in other software I end up with a plot of the coefficients against the tuning parameter with a cutpoint or a table and graph that tells me what to include in the model based on some selected criterion. It's not clear from the example if there's a cross-validation or some other procedure to determine what variables to include. Plot(fit) produces a graph of coefficients against lambda but nothig to indicate what to include. What is used in the package, if anything, to make that determination? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] multidimensional array calculation
See
?expand.grid
For example,
df - expand.grid(L=L, AR=AR, SO=SO, T=T)
df$y - fun(df$L, df$AR, df$SO, df$T)
Jean
Johannes Radinger wrote on 01/13/2012 12:28:46 PM:
Hello,
probably it is quite easy but I can get it: I have
mulitple numeric vectors and a function using
all of them to calculate a new value:
L - c(200,400,600)
AR - c(1.5)
SO - c(1,3,5)
T - c(30,365)
fun - function(L,AR,SO,T){
exp(L*AR+sqrt(SO)*log(T))
}
How can I get an array or dataframe where
all possible combinations of the factors are listed
and the new value is calculated.
I thought about an array like:
array(NA, dim = c(3,1,3,2), dimnames=list(c(200,400,600),c(1.5),c(1,
3,5),c(30,365)))
but how can I get the array populated according to the function?
As I want to get in the end a 2D dataframe I probably will use the
melt.array()
function from the reshape package or is there another way to get simple
such
a full-factorial dataframe with all possible combinations?
Best regards,
Johannes
[[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] multidimensional array calculation
Perhaps repeated use of the outer() function. You could also write a
multi.outer() or adopt one of the solutions here:
http://stackoverflow.com/questions/6192848/how-to-generalize-outer-to-n-dimensions
Michael
On Fri, Jan 13, 2012 at 1:28 PM, Johannes Radinger jradin...@gmx.at wrote:
Hello,
probably it is quite easy but I can get it: I have
mulitple numeric vectors and a function using
all of them to calculate a new value:
L - c(200,400,600)
AR - c(1.5)
SO - c(1,3,5)
T - c(30,365)
fun - function(L,AR,SO,T){
exp(L*AR+sqrt(SO)*log(T))
}
How can I get an array or dataframe where
all possible combinations of the factors are listed
and the new value is calculated.
I thought about an array like:
array(NA, dim = c(3,1,3,2),
dimnames=list(c(200,400,600),c(1.5),c(1,3,5),c(30,365)))
but how can I get the array populated according to the function?
As I want to get in the end a 2D dataframe I probably will use the
melt.array()
function from the reshape package or is there another way to get simple such
a full-factorial dataframe with all possible combinations?
Best regards,
Johannes
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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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] plotting regression line in with lattice
Hi,
Since trt is a factor, you use it for indexing. try just delete in the code
fill - my.fill[combined$trt[subscripts]]
Weidong Gu
On Fri, Jan 13, 2012 at 11:30 AM, matteo dossena m.doss...@qmul.ac.uk wrote:
#Dear All,
#I'm having a bit of a trouble here, please help me...
#I have this data
set.seed(4)
mydata - data.frame(var = rnorm(100),
temp = rnorm(100),
subj = as.factor(rep(c(1:10),5)),
trt = rep(c(A,B), 50))
#and this model that fits them
lm - lm(var ~ temp * subj, data = mydata)
#i want to plot the results with lattice anf fit the regression line,
predicted with my model, trough them
#to do so, I'm using this approach, outlined Lattice Tricks for the power
useR by D. Sarkar
temp_rng - range(mydata$temp, finite = TRUE)
grid - expand.grid(temp = do.breaks(temp_rng, 30),
subj = unique(mydata$subj),
trt = unique(mydata$trt))
model - cbind(grid, var = predict(lm, newdata = grid))
orig - mydata[c(var,temp,subj,trt)]
combined - make.groups(original = orig, model = model)
xyplot(var ~ temp | subj,
data = combined,
groups = which,
type = c(p, l),
distribute.type = TRUE
)
# so far every thing is fine, but, i also whant assign a filling to the data
points for the two treatments trt=1 and trt=2
# so I have written this piece of code, that works fine, but when it comes to
plot the regression line, it seems that type is not recognized by the panel
function...
my.fill - c(black, grey)
plot - with(combined,
xyplot(var ~ temp | subj,
data = combined,
group = combined$which,
type = c(p, l),
distribute.type = TRUE,
panel = function(x, y, ..., subscripts){
fill - my.fill[combined$trt[subscripts]]
panel.xyplot(x, y, pch = 21, fill = my.fill, col =
black)
},
key = list(space = right,
text = list(c(trt1, trt2), cex = 0.8),
points = list(pch = c(21), fill = c(black, grey)),
rep = FALSE)
)
)
plot
#I've also tried to move type and distribute type within panel.xyplot, as
well as subsseting the data in it panel.xyplot like this
plot - with(combined,
xyplot(var ~ temp | subj,
data = combined,
panel = function(x, y, ..., subscripts){
fill - my.fill[combined$trt[subscripts]]
panel.xyplot(x[combined$which==original],
y[combined$which==original], pch = 21, fill = my.fill, col = black)
panel.xyplot(x[combined$which==model],
y[combined$which==model], type = l, col = black)
},
key = list(space = right,
text = list(c(trt1, trt2), cex = 0.8),
points = list(pch = c(21), fill = c(black, grey)),
rep = FALSE)
)
)
plot
#but no success with that either...
#can anyone help me to get the predicted values plotted as a line instead of
being points?
#really appricieate
#matteo
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Re: [R] Problem Installing R to SuSE 10 via RPM
apt-get wont run on Suse unless you have aptitude, also your distribution won't compile a deb without some intervention. You won't be able to query the package regardless of key if your local proxy settings won't let you because of admin. Perhaps see if there is a tool to convert the deb package to rpm locally to install after downloading the deb, although building from source shouldn't be that challenging unless you have ridiculous Fortran compiler problems like I do with Linux. Hope that helps, Ken On Jan 13, 2012, at 1:13 PM, Marc Schwartz marc_schwa...@me.com wrote: Two things: 1. Looking at the sizes of the RPMs at the URL provided, the 'R-base-devel' RPM is only 84k. So as with other Linux distros, that is likely to contain R header files and other such things to enable compilation of CRAN packages during their installation using the source tarballs. The 'R-base' RPM is 33Mb, so that is your primary install for R. You will want both. 2. I don't know about the configuration of the RPMs on SUSE, but presume that as with RH/Fedora, the RPMs can be created in such a fashion that they are not relocatable. That means that they are configured to be only installed 'system-wide', not to a specific user's directory tree. You may have issues trying to install them without root access on your computer, thus may get other permission related errors when you get the correct RPM. Whether they are or are not relocatable would be a question for the SUSE maintainer for the R package. If they are, there are 'rpm' command line arguments that are relevant. So use 'man rpm' for more details. If the RPMs are not relocatable, you will be left with the option of building R from source in order to install R locally and will want to read the R Installation and Administration manual for details. HTH, Marc Schwartz On Jan 13, 2012, at 11:00 AM, Matthew Pettis wrote: Thanks, will do! I thought devel included base, but evidently, that's not the case... On Fri, Jan 13, 2012 at 10:58 AM, Gavin Blackburn gavin.blackb...@strath.ac.uk wrote: I think it's saying you need to install R-base before R-base-devel. You'll need to add a cran repository as SUSE might not have the most up-to-date version of R. This is the code for Ubuntu I assume it's the same, just change the distro and keyserver: sudo apt-get update sudo add-apt-repository 'deb http://cran.ma.imperial.ac.uk/bin/linux/ubuntu oneiric/' gpg --keyserver keyserver.ubuntu.com --recv-key E084DAB9 gpg -a --export E084DAB9 | sudo apt-key add - You may also want to get Sun java, again, change distro and keyserver: sudo add-apt-repository ppa:ferramroberto/java sudo apt-key adv --recv-key --keyserver keyserver.ubuntu.comB725097B3ACC3965 sudo apt-get update sudo apt-get install sun-java6-jdk sun-java6-plugin Then run: sudo apt-get install r-base r-base-dev sudo R CMD javareconf Cheers, Gavin. -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Matthew Pettis Sent: 13 January 2012 16:43 To: r-help@r-project.org Subject: [R] Problem Installing R to SuSE 10 via RPM Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide
Re: [R] Coloring counties on a full US map based on a certain criterion
But the help does say that additional arguments are passed to lines(),
so you can use lty=0.
That can leave white bits between counties if the areas don't line up
precisely, so I think it looks better with the lines in black.
I agree, indeed it leaves white bits. Of course, I could try to change
the type of the lines using lty...
Sorry for one more question: can one change the color of these lines
(borders between counties) from black?
Thank you!
D.
Sarah
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$group])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply transformation
Try this:
d_tmp2 - data.frame(apply(round(100*d_tmp[, -1], 2), 2, paste, %,
sep=))
names(d_tmp2) - paste(names(d_tmp[, -1]), Lbl, sep=-)
d_final - cbind(d_tmp, d_tmp2)
d_final[, 1+c(0, order(names(d_final[, -1])))]
Jean
santosh wrote on 01/13/2012 04:55:51 AM:
Hello All,
I have the following dataset:
Year 2006 2007
Jan Jan 0.0204 0.0065
Feb Feb 0.0145 0.0082
Mar Mar 0.0027 0.0122
dput(d_tmp)
structure(list(Year = c(Jan, Feb, Mar), `2006` = c(0.0204,
0.0145, 0.0027), `2007` = c(0.0065, 0.0082, 0.0122)), .Names =
c(Year,
2006, 2007), row.names = c(Jan, Feb, Mar), class =
data.frame)
I am trying to use the apply function but the values seem to be
getting coerced to characters. I could recast in my function ... but I
suspect there should be an easier way.
I can always use a for loop to get the output I need but just
wondering if there a way to get the same using apply or some other
function ... (the number of years can be changing in my requirement)
My final output needs to be as follows:
Year 2006 2006-Lbl 2007 2007-Lbl
Jan 0.0204 '2.04%' 0.0065 '0.65%'
Feb 0.0145 '1.45%' 0.0082 '0.82%'
Mar 0.0027 '0.27%' 0.0122 '1.22%'
i.e.
dput(d_final)
structure(list(Year = structure(c(2L, 1L, 3L), .Label = c(Feb,
Jan, Mar), class = factor), X2006 = c(0.0204, 0.0145, 0.0027
), X2006.Lbl = structure(c(3L, 2L, 1L), .Label = c('0.27%',
'1.45%', '2.04%'), class = factor), X2007 = c(0.0065, 0.0082,
0.0122), X2007.Lbl = structure(1:3, .Label = c('0.65%', '0.82%',
'1.22%'), class = factor)), .Names = c(Year, X2006,
X2006.Lbl,
X2007, X2007.Lbl), row.names = c(NA, -3L), class = data.frame)
Please advise.
Santosh
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Re: [R] Coloring counties on a full US map based on a certain criterion
Changing the color of the borders is what the border argument to
polygon() would do, if only it hadn't been overridden.
So no, you can't easily change the line color. It would be an easy
tweak to the code to add a polyborder argument that is passed
to polygon() as border, though. That would solve both your
problems.
Sarah
On Fri, Jan 13, 2012 at 3:13 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
But the help does say that additional arguments are passed to lines(),
so you can use lty=0.
That can leave white bits between counties if the areas don't line up
precisely, so I think it looks better with the lines in black.
I agree, indeed it leaves white bits. Of course, I could try to change
the type of the lines using lty...
Sorry for one more question: can one change the color of these lines
(borders between counties) from black?
Thank you!
D.
Sarah
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$group])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
You can set the fg graphics parameter, for example
oldpar - par(fg='white') # change the default fg (foreground color) to
white
map('county', 'iowa', fill=TRUE, col='light gray')
oldpar # reset fg to the default - black
map('state', 'iowa', lwd=3, add=TRUE)
Assuming you want the state outline in black with the county boundaries in
white. Otherwise just eliminate the last map command.
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Dimitri Liakhovitski
Sent: Friday, January 13, 2012 2:14 PM
To: Sarah Goslee
Cc: r-help
Subject: Re: [R] Coloring counties on a full US map based on a certain
criterion
But the help does say that additional arguments are passed to lines(),
so you can use lty=0.
That can leave white bits between counties if the areas don't line up
precisely, so I think it looks better with the lines in black.
I agree, indeed it leaves white bits. Of course, I could try to change
the type of the lines using lty...
Sorry for one more question: can one change the color of these lines
(borders between counties) from black?
Thank you!
D.
Sarah
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$gr
oup])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
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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.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Sarah, David, thank you very much for your help! It all works now:
### My criterion for all counties:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
### For gray border color:
oldpar - par(fg='gray') # change the default fg (foreground color) to white
### My US map:
map('county',fill=TRUE,col=classcolors[allcounties$group],lty=3,bg =
transparent)
### For state borders:
map('state', lwd=1, add=TRUE)
Dimitri
On Fri, Jan 13, 2012 at 3:24 PM, David L Carlson dcarl...@tamu.edu wrote:
You can set the fg graphics parameter, for example
oldpar - par(fg='white') # change the default fg (foreground color) to
white
map('county', 'iowa', fill=TRUE, col='light gray')
oldpar # reset fg to the default - black
map('state', 'iowa', lwd=3, add=TRUE)
Assuming you want the state outline in black with the county boundaries in
white. Otherwise just eliminate the last map command.
--
David L Carlson
Associate Professor of Anthropology
Texas AM University
College Station, TX 77843-4352
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Dimitri Liakhovitski
Sent: Friday, January 13, 2012 2:14 PM
To: Sarah Goslee
Cc: r-help
Subject: Re: [R] Coloring counties on a full US map based on a certain
criterion
But the help does say that additional arguments are passed to lines(),
so you can use lty=0.
That can leave white bits between counties if the areas don't line up
precisely, so I think it looks better with the lines in black.
I agree, indeed it leaves white bits. Of course, I could try to change
the type of the lines using lty...
Sorry for one more question: can one change the color of these lines
(borders between counties) from black?
Thank you!
D.
Sarah
Thank you!
Dimitri
### My criterion for all counties.:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
map('county',fill=TRUE,col=classcolors[allcounties$group])
map('state', lwd=2, add=TRUE)
Sure. Just start with
map('county')
instead.
I like to add something like:
map('state', lwd=3, add=TRUE)
I am trying:
### My criterion for all counties in the US:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-sample(1:5,3082,replace=TRUE)
### My colors:
classcolors - rainbow(5)
### Trying to build the map - not working:
map(database='usa',regions='county',fill=TRUE,col=classcolors[allcounties$gr
oup])
You'll need to instead coordinate with the names of the entire US:
length(map('county', plot=FALSE)$names)
[1] 3082
Sarah
--
Sarah Goslee
http://www.functionaldiversity.org
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Dimitri Liakhovitski
marketfusionanalytics.com
__
R-help@r-project.org mailing list
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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] Coloring counties on a full US map based on a certain criterion
On 14/01/2012 8:04 a.m., Sarah Goslee wrote:
On Fri, Jan 13, 2012 at 1:52 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Just to clarify, according to help about the fill argument:
logical flag that says whether to draw lines or fill areas. If FALSE,
the lines bounding each region will be drawn (but only once, for
interior lines). If TRUE, each region will be filled using colors from
the col = argument, and bounding lines will not be drawn.
We have fill=TRUE - so why are the county borders still drawn?
Thank you!
Dimitri
This prompted me to check the code:
if fill=TRUE, map() calls polygon()
if fill=FALSE, map() calls lines()
But polygon() draws borders by default.
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow)
To not draw borders, the border argument is provided:
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow, border=NA)
But that fails in map():
map('county', 'iowa', fill=TRUE, col=rainbow(20), border=NA)
Error in par(pin = p) :
invalid value specified for graphical parameter pin
because border is used as a named argument in map() already, for setting the
size of the plot area, so there's no way to alter the border argument
to polygon.
Coincidentally, I became aware of this just recently. When the maps
package was created (way back in the 'new' S era), polygon() didn't
add borders, and that is why ?map states that fill does not add
borders. A workaround is to change the map() option border= to
myborder= (it is then used twice in map()).
The work-around I suggested previous (lty=0) seems to be the only
way to deal with the problem.
In fact I believe there is another workaround if you don't want to
modify the code; use the option resolution=0 in the map() call. I.e. try
(in Sarah's original Iowa example):
map('county', 'iowa', fill= TRUE, col = classcolors[countycol],
resolution=0, lty=0)
This ensures that the polygon boundaries match up.
I'll fix the border issue in the next version of maps (*not* the one
just uploaded to CRAN, which was to add Cibola County to NM).
Ray Brownrigg
Sarah
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Re: [R] Coloring counties on a full US map based on a certain criterion
Hi Ray,
I'm glad to see you here. I was going to write this up a bit more
clearly and email it to you, but now I don't have to bother. :)
Coincidentally, I became aware of this just recently. When the maps package
was created (way back in the 'new' S era), polygon() didn't add borders,
and that is why ?map states that fill does not add borders. A workaround is
to change the map() option border= to myborder= (it is then used twice in
map()).
I though it was probably a legacy code issue.
In fact I believe there is another workaround if you don't want to modify
the code; use the option resolution=0 in the map() call. I.e. try (in
Sarah's original Iowa example):
map('county', 'iowa', fill= TRUE, col = classcolors[countycol],
resolution=0, lty=0)
This ensures that the polygon boundaries match up.
Ah! That works nicely, and wasn't clear to me from the help that it
would do so.
Thanks!
--
Sarah Goslee
http://www.functionaldiversity.org
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Portfolio Optimization
I would be biased towards using a heuristic, for instance Threshold Accepting (TA), for solving such a problem. (TA is implemented in package NMOF. Disclosure: I am the author of that package.) But you will not find a ready-to-use solution there. (1) you need an objective function, ie, a function that maps a given vector of holdings (and data like your scenario matrix) into a real number; the better the portfolio, the lower the number. (2) For TA, you need a so-called neighbourhood function. That is a function that changes one portfolio vector into another, by changing some elements. Examples for simple neighbourhoods are in the package vignettes. Do you have a budget constraint? If yes, and you want to work with integers, I would suggest using a cash variable. (See, eg, Algorithm 3 in http://www.swissfinanceinstitute.ch/rp20.pdf ) (3) The constraints can, at least in a first round, be included through penalties. Regards, Enrico PS. There is a mailing list dedicated to finance-with-R questions, and you may get better answers there. https://stat.ethz.ch/mailman/listinfo/r-sig-finance -- Enrico Schumann Lucerne, Switzerland http://nmof.net/ Am 13.01.2012 17:06, schrieb Sal Pellettieri: Hi, I'm an R newbie and I've been struggling with a optimization problem for the past couple of days now. Here's the problem - I have a matrix of expected payouts from different stock option strategies. Each column in my matrix represents a different stock and each row represents the return to the strategy given a certain market move. So the rows are not a time series of percentage returns but a dollar payout in different expected scenarios, i.e. Expected Return Matrix (ER) = stock1 stock2 stockn scenario1 $ $ $ scenario2 $ $ $ scenario3 $ $ $ ... I want to create an optimal portfolio of these strategies by applying a vector of weights. The weights will be the number of contracts of each to buy and won't be a percentage weighting. There are a few constraints I need it comply with: - The weights have to be integers - The minimum portfolio return (ER* Weights) across the scenarios has to be greater than some negative number I specify - There has to be a certain minimum number of stocks in the portfolio so length(weights)some number I specify. Any help is GREATLY appreciated since I have tried so many different functions and packages. Even if someone can just lead me to the correct function to use that would be a great help as I've looked at optim, solveLP, ROI package and many others. Thanks, S __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging within a range of values
Hello Jeff, thank you for the reply. I tried the cut function and I had two questions. How do I have the cut function take the first position in the start column in df1 as the first cut point and the first position in column 2 as the second cut point. The break variable seems to want a single vector. I tried compressing both vectors into one where I had say 200 700 500 1000 etc then cut gives me the 200-500 range, 500-700, and 700-1000. In this case I wanted the range, 200-700, and 500-1000. Is there a way to define the first point of each cut as positions along the START vector and all second points of the cut as positions along the END vector? I also had one additional question. When playing around with this, I noticed that I had to do this for the Pos column in the second data frame. But, when I get the ranges, how do I have it return the values in C0 or C1 in df2 that are in the same rows as those of the ranges? Thanks again for the help. -- View this message in context: http://r.789695.n4.nabble.com/Averaging-within-a-range-of-values-tp4291958p4293410.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Coloring counties on a full US map based on a certain criterion
Everyone, thanks a lot - this is great:
### My criterion for all counties:
allcounties-data.frame(county=map('county', plot=FALSE)$names)
allcounties$group-c(rep(1:6,513),rep(1,4))[order(c(rep(1:6,513),rep(1,4)))]
### My colors:
classcolors - rainbow(6)
### 1. If I want to have no borders between counties:
map('county',fill=TRUE,col=classcolors[allcounties$group],resolution=0,lty=0,bg
= transparent)
map('state', lwd=1, add=TRUE)
# 2. If I want to see borders between counties (of a desired
color, e.g., gray):
### For line color:
oldpar - par(fg='gray') # change the default fg (foreground color) to white
### My US map:
map('county',fill=TRUE,col=classcolors[allcounties$group],lty=1,bg =
transparent)
par(oldpar)
Dimitri
On Fri, Jan 13, 2012 at 3:48 PM, Sarah Goslee sarah.gos...@gmail.com wrote:
Hi Ray,
I'm glad to see you here. I was going to write this up a bit more
clearly and email it to you, but now I don't have to bother. :)
Coincidentally, I became aware of this just recently. When the maps package
was created (way back in the 'new' S era), polygon() didn't add borders,
and that is why ?map states that fill does not add borders. A workaround is
to change the map() option border= to myborder= (it is then used twice in
map()).
I though it was probably a legacy code issue.
In fact I believe there is another workaround if you don't want to modify
the code; use the option resolution=0 in the map() call. I.e. try (in
Sarah's original Iowa example):
map('county', 'iowa', fill= TRUE, col = classcolors[countycol],
resolution=0, lty=0)
This ensures that the polygon boundaries match up.
Ah! That works nicely, and wasn't clear to me from the help that it
would do so.
Thanks!
--
Sarah Goslee
http://www.functionaldiversity.org
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--
Dimitri Liakhovitski
marketfusionanalytics.com
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Re: [R] Coloring counties on a full US map based on a certain criterion
Somewhat related question out of curiousity:
Does anyone know how often the list of the counties and county names
is updated in this package? Or is it done centrally for all packages
that deal with US counties?
Thanks!
Dimitri
On Fri, Jan 13, 2012 at 3:41 PM, Ray Brownrigg
ray.brownr...@ecs.vuw.ac.nz wrote:
On 14/01/2012 8:04 a.m., Sarah Goslee wrote:
On Fri, Jan 13, 2012 at 1:52 PM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com wrote:
Just to clarify, according to help about the fill argument:
logical flag that says whether to draw lines or fill areas. If FALSE,
the lines bounding each region will be drawn (but only once, for
interior lines). If TRUE, each region will be filled using colors from
the col = argument, and bounding lines will not be drawn.
We have fill=TRUE - so why are the county borders still drawn?
Thank you!
Dimitri
This prompted me to check the code:
if fill=TRUE, map() calls polygon()
if fill=FALSE, map() calls lines()
But polygon() draws borders by default.
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow)
To not draw borders, the border argument is provided:
plot(c(0,1), c(0,1), type=n)
polygon(c(0,0,1,1), c(0,1,1,0), col=yellow, border=NA)
But that fails in map():
map('county', 'iowa', fill=TRUE, col=rainbow(20), border=NA)
Error in par(pin = p) :
invalid value specified for graphical parameter pin
because border is used as a named argument in map() already, for setting
the
size of the plot area, so there's no way to alter the border argument
to polygon.
Coincidentally, I became aware of this just recently. When the maps package
was created (way back in the 'new' S era), polygon() didn't add borders,
and that is why ?map states that fill does not add borders. A workaround is
to change the map() option border= to myborder= (it is then used twice in
map()).
The work-around I suggested previous (lty=0) seems to be the only
way to deal with the problem.
In fact I believe there is another workaround if you don't want to modify
the code; use the option resolution=0 in the map() call. I.e. try (in
Sarah's original Iowa example):
map('county', 'iowa', fill= TRUE, col = classcolors[countycol],
resolution=0, lty=0)
This ensures that the polygon boundaries match up.
I'll fix the border issue in the next version of maps (*not* the one just
uploaded to CRAN, which was to add Cibola County to NM).
Ray Brownrigg
Sarah
--
Dimitri Liakhovitski
marketfusionanalytics.com
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Re: [R] Problem Installing R to SuSE 10 via RPM
On Jan 13, 2012, at 17:43 , Matthew Pettis wrote: Hi, I'm trying to install R from an rpm locally to my account (the reason I'm not doing it through yast/yast2/zypper is that the sys admin isn't yet willing to install it, and doesn't want to support it, but will help me support it if I install it locally -- in short, policy problems rather than technical). Below is the SuSE version, Kernel version, and rpm install error I'm getting, as well as the error... As others have said, installing from RPM outside the system folders could be a no-go. Another option is to install from source, this works fine if you have all the relevant packages installed (compilers, libraries, header files...). If you don't already have them get your sysadm to install them (say, one per day over a couple of weeks, until he sees the error of his ways...). But really, the painless option for both of you is just to install it from the official SUSE sources with yast, keep it updated automagically, and keep all add-on packages in your own home directory. The sysadmin workload for R on SUSE (and most other Linuxen) should be essentially nil. Can anyone help me with the error? I'm trying to install R-base 2.14.1, but it is telling me that I need R-base version 2.14.1 as a dependency. Am I using the wrong rpm for an installation starting from scratch? I got the rpm from: http://download.opensuse.org/repositories/devel:/languages:/R:/base/SLE_10/x86_64/ Thanks, Matt pettis@swat:~/bin cat /etc/*-release SUSE Linux Enterprise Server 10 (x86_64) VERSION = 10 PATCHLEVEL = 2 pettis@swat:~/bin uname -a Linux swat 2.6.16.60-0.34-smp #1 SMP Fri Jan 16 14:59:01 UTC 2009 x86_64 x86_64 x86_64 GNU/Linux pettis@swat:~/bin rpm -ivh R-base-devel-2.14.1-30.1.x86_64.rpm warning: R-base-devel-2.14.1-30.1.x86_64.rpm: Header V3 DSA signature: NOKEY, key ID 793371fe error: Failed dependencies: R-base = 2.14.1 is needed by R-base-devel-2.14.1-30.1.x86_64 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- 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 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] deviance and variance - GAM models
Hi all, This is pretty basic but I am not an expert and I couldn't find anything in the forum or my statistics book about it. I was reading a paper and the authors were using both explained deviance and explained variance as synonyms. They were describing a GAM regression. Is that right? I performed an analysis in R to take a look to the output of GAM regression and I think that: - 'R-sq. (adj)' is the percentage of variance explained by the regression, i.e., I can write The regression explains xx% of variance. - 'Deviance explained' is a simple measure of the quality of the fit but it is not related to the percentage of variance that is explained by the regression. Am I right? Thank you so much Ramón -- View this message in context: http://r.789695.n4.nabble.com/deviance-and-variance-GAM-models-tp4293293p4293293.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] outputs from command by
Hello R experts,
I have generated a data set below. I tried to export the object - my.pvalues
(class is by)
as .txt (or excel file), so the output in matrix form line up nicely in .xls
file. Do I have to convert to some other data formats before using write.table?
I am having difficulty to get this out.
I really appreciate it if anyone here could kindly guide me.
Kevin
# Data
a - c(rnorm(6,mean=0.5,sd=0.1),rnorm(18,mean=4,sd=1))
b - factor(rep(LETTERS[1:4], each=3, lenth=24))
grp - rep(factor(c(WT,TG)), each=12)
d - as.data.frame(cbind(a,b,grp))
my.pvalues -by(d,
INDICES=d[,grp],
{
FUN=function(x)
pairwise.t.test(x$a,x$b)$p.value
}
)
class(my.pvalues)
[1] by
### Results ###
my.pvalues
d[, grp]: 1
1 2 3
2 1 NA NA
3 1 1 NA
4 1 1 1
d[, grp]: 2
1 2 3
2 0.7629610481 NA NA
3 0.0006906049 0.0006059707 NA
4 0.0164179404 0.0141134949 0.03267941
[[alternative HTML version deleted]]
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[R] Help with t student test
Dear all, I've the following data.frame GENDER MEASURE01 MEASURE02 MEASURE03 MEASURE04 MEASURE05 MEASURE06 R. Rafuse MALE 91 6 8 12 T. LeikerMALE 6 7 1 800 E. Bizot FEMALE 9 8 2 9 88 K. French MALE 7 9 059 9 E. Van LanduytMALE 7 1 628 9 K. Harrell FEMALE 6 0 0831 W. Noren FEMALE 74 3 2 57 W. WilldenMALE 99 2 6 68 S. Kohut FEMALE 78 2 3 6 9 I'd like to run an indipendent t test with the MEASURE01-MEASURE06 as dipendent variable and GENDER as grouping variable. Namely, I wish to compare means all variables in the gorup male and female. Could you help me to find the right R command? Bests -- View this message in context: http://r.789695.n4.nabble.com/Help-with-t-student-test-tp4293203p4293203.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] The Future of R | API to Public Databases
Dear R Users - R is a wonderful software package. CRAN provides a variety of tools to work on your data. But R is not apt to utilize all the public databases in an efficient manner. I observed the most tedious part with R is searching and downloading the data from public databases and putting it into the right format. I could not find a package on CRAN which offers exactly this fundamental capability. Imagine R is the unified interface to access (and analyze) all public data in the easiest way possible. That would create a real impact, would put R a big leap forward and would enable us to see the world with different eyes. There is a lack of a direct connection to the API of these databases, to name a few: - Eurostat - OECD - IMF - Worldbank - UN - FAO - data.gov - ... The ease of access to the data is the key of information processing with R. How can we handle the flow of information noise? R has to give an answer to that with an extensive API to public databases. I would love your comments and ideas as a contribution in a vital discussion. Benjamin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Averaging over data sets
Here is a solution that works for your small example. It might be difficult to prepare your larger data sets to use the same method. db -rbind(d1,d2) aggregate(subset(db,select=-c(subject,trt)), by=list(subject=db$subject),mean) ## or, for example, aggregate(subset(db,select=-c(subject,trt)), by=list(subject=db$subject, trt=db$trt),mean) In order for aggregate() to work, its first argument must have only numeric columns. That is what subset(db,select=-c(subject,trt)) does for you. (d1 + d2)/2 did not work because d1 and d2 are data frames, not numbers. Much more complicated, you could have done your averages one at a time, (d1$eat1[d1$subject=='Felipe'] + d2$eat1[d2$subjedt=='Felipe'])/2 and similarly for eat3 and John. But that is of course not practical for larger data sets. -Don -- Don MacQueen Lawrence Livermore National Laboratory 7000 East Ave., L-627 Livermore, CA 94550 925-423-1062 On 1/12/12 10:16 PM, Felipe Nunes felipnu...@gmail.com wrote: Hi all, after using Amelia II to create 10 imputed data sets I need to average them to have one unique data that includes the average for each cell of the variables imputed, in addition to the values for the variables not imputed. Such data has many variables (some numeric, other factors), and more than 2 observations. I do not know how to average them out. Any help? Below I provide a small example: Suppose Amelia provided two datasets: d1 - data.frame(subject = c(Felipe, John), eat1 = 1:2, eat3 = 5:6, trt = c(t1, t2)) d2 - data.frame(subject = c(Felipe, John), eat1 = 3:4, eat3 = 6:7, trt = c(t1, t2)) I tried (d1 + d2)/2 but I lose my factors. mean() did not work either. The result I'd like is: subject eat1 eat3 trt 1 Felipe 2 5.5 t1 2 John 3 6.5 t2 thanks, *Felipe Nunes* CAPES/Fulbright Fellow PhD Student Political Science - UCLA Web: felipenunes.bol.ucla.edu [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] The Future of R | API to Public Databases
R is Open Source. You're welcome to write tools, and submit your package to CRAN. I think some part of this has been done, based on questions to the list asking about those parts. Personally, I've been using S-Plus and then R for 18 years, and never required data from any of them. Which doesn't make it not important, but suggests that public databases aren't the be-all and end-all for R use. Sarah On Fri, Jan 13, 2012 at 4:14 PM, Benjamin Weber m...@bwe.im wrote: Dear R Users - R is a wonderful software package. CRAN provides a variety of tools to work on your data. But R is not apt to utilize all the public databases in an efficient manner. I observed the most tedious part with R is searching and downloading the data from public databases and putting it into the right format. I could not find a package on CRAN which offers exactly this fundamental capability. Imagine R is the unified interface to access (and analyze) all public data in the easiest way possible. That would create a real impact, would put R a big leap forward and would enable us to see the world with different eyes. There is a lack of a direct connection to the API of these databases, to name a few: - Eurostat - OECD - IMF - Worldbank - UN - FAO - data.gov - ... The ease of access to the data is the key of information processing with R. How can we handle the flow of information noise? R has to give an answer to that with an extensive API to public databases. I would love your comments and ideas as a contribution in a vital discussion. Benjamin -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Regression Modeling Strategies 3-Day Short Course March 2012
*RMS Short Course 2012* Frank E. Harrell, Jr., Ph.D., Professor and Chair Department of Biostatistics, Vanderbilt University School of Medicine *March 7, 8 9, 2012* 8:00am - 4:30pm Vanderbilt University Nashville Tennessee USA http://biostat.mc.vanderbilt.edu/RMSShortCourse This course covers a variety of regression modeling and model validation methods as well as the R rms package. Please email interest to eve.a.ander...@vanderbilt.edu - Frank Harrell Department of Biostatistics, Vanderbilt University -- View this message in context: http://r.789695.n4.nabble.com/Regression-Modeling-Strategies-3-Day-Short-Course-March-2012-tp4293555p4293555.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
