[R] subset function unexpected behavior
I was surprised to see this unexpected behavior of subset in a for loop. I
looked in subset.data.frame and it seemed to me that both versions should
work, since the subset call should be evaluated in the global environment -
but perhaps I don't understand environments well enough. Can someone
enlighten me? In any case, this is a bit of a gotcha for naive users of
subset.
input.data -
data.frame(sch=c(1,1,2,2),
pop=c(100,200,300,400))
school.var - sch
school.list - 1:2
for(sch in school.list){
print(sch)
#do this before subset!:
right.sch.p -
input.data[,school.var] == sch
print( subset(input.data,right.sch.p)) #this is what I expected
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## [1] 2
## sch pop
## 3 2 300
## 4 2 400
for(sch in school.list){
print(sch)
print(subset(input.data,input.data[,school.var] == sch)) #note - compact
version fails!
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
## [1] 2
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
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Re: [R] Odp: How to repeat for function?
Hi r-help-boun...@r-project.org napsal dne 01.02.2010 18:14:04: Dear Petr, The intention is to get a ratio value for every observations (1400 obs) for every variable (4 variables). Well, maybe that mat-matrix(runif(12),4,3) ratiomat-t(t(mat)/vec) And from the ratios, I would like to rank the variables based on how many times the variable being the highest among 4 of them (or the total number of being the highest ratios, out of 1400 times). Eh. I am stupid enough not to understand what you mean. Which variables? Do you mean maximum in each column? apply(ratiomat,2, max) [1] 0.9686974 0.4097230 0.3085468 From the 1400 by 4 dataset, I ran a principal component analysis and I'll use only the first PC for the calculation of the ratios. The PC (a vector) will has 4 values for each variables. The ratios actually representing the contribution of a variable divide by the total contribution from the first PC. I can not say much about principal components analysis above what is described in help page. Does it make sense to you now? Thank you very much for giving a response. I really appreciate it. Not much. Reproducible code could be more informative then describing a problem by your words. Regards Petr Best regards, ayu From: Petr Pikal [via R] ml-node+1458931-701777...@n4.nabble.com Sent: Mon, 1 February, 2010 15:58:50 Subject: Odp: How to repeat for function? Hi few comments [hidden email] napsal dne 01.02.2010 14:51:17: Dear Users, I have one problem here, I tried many time and even read a few notes on writing function but still. Can anyone help me on how to simplify Part B (please refer the programming below), so that I don't have to repeat the calculation of num and r ? Thank you very much..every help is very much appreciated... ## Part A n=1400 m=matrix(c(0,0,0,0),4,1) m2=matrix(c(0,2,0,0),4,1) c4-matrix(c(1.0,0.2,-0.5,0.3,0.2,1,0.2,-0.5,-0.5,0.2,1,0.2,0.3,-0.5,0.2,1), 4,4,byrow=T) set.seed(428) X=mvrnorm(n,m2,c4) X.pca-princomp(X) loadings(X.pca) pc=X.pca$loadings[,1] ##  Part B num1=rep(1:n) is same as num1 - 1:n for(i in 1:n)num1[i]=pc[1]%*%X[i,1] If I understand what the above code does I am a bit surprised pc[1] is pc[1] [1] 0.525037 for each number in cycle X[i,1] is again a number X[1,1] [1] 0.7512862 X[2,1] [1] 0.5020333 so it basically  results in pc[1] * X[,1] which you can compute for all 4 columns by kronecker(pc, X, *) But I wonder if this is what you really want. Maybe you shall think it over again and try with some smaller manageable subset lit smallX - X[1:5,] do all your computation and if it does not produce what you want, specify what you want. Regards Petr num2=rep(1:n) for(i in 1:n)num2[i]=pc[2]%*%X[i,2] num3=rep(1:n) for(i in 1:n)num3[i]=pc[3]%*%X[i,3] num4=rep(1:n) for(i in 1:n)num4[i]=pc[4]%*%X[i,4] den=rep(1:n) for(i in 1:n)den[i]=pc%*%X[i,] r1=num1/den r2=num2/den r3=num3/den r4=num4/den MLAV=sum(r2r1 r2r3 r2r4) MLAV Best regards, ayu -- View this message in context: http://n4.nabble.com/How-to-repeat-for-function- tp1458806p1458806.html Sent from the R help mailing list archive at Nabble.com. __ [hidden email] 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. __ [hidden email] 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. View message @ http://n4.nabble.com/How-to-repeat-for-function-tp1458806p1458931.html To unsubscribe from How to repeat for function?, click here. -- View this message in context: http://n4.nabble.com/How-to-repeat-for-function- tp1458806p1459009.html Sent from the R help mailing list archive at Nabble.com. [[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] subset function unexpected behavior
Hi:
Try this for your second loop instead:
for(s in school.list){
print(s)
print(subset(input.data, sch == s))
}
[1] 1
sch pop
1 1 100
2 1 200
[1] 2
sch pop
3 2 300
4 2 400
Don't confound the 'sch' variable in your data frame with the
index in your loop :)
HTH,
Dennis
On Mon, Feb 1, 2010 at 8:17 PM, David Katz da...@davidkatzconsulting.comwrote:
I was surprised to see this unexpected behavior of subset in a for loop. I
looked in subset.data.frame and it seemed to me that both versions should
work, since the subset call should be evaluated in the global environment -
but perhaps I don't understand environments well enough. Can someone
enlighten me? In any case, this is a bit of a gotcha for naive users of
subset.
input.data -
data.frame(sch=c(1,1,2,2),
pop=c(100,200,300,400))
school.var - sch
school.list - 1:2
for(sch in school.list){
print(sch)
#do this before subset!:
right.sch.p -
input.data[,school.var] == sch
print( subset(input.data,right.sch.p)) #this is what I expected
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## [1] 2
## sch pop
## 3 2 300
## 4 2 400
for(sch in school.list){
print(sch)
print(subset(input.data,input.data[,school.var] == sch)) #note - compact
version fails!
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
## [1] 2
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
--
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Re: [R] Weighted SD
There is 'cov.wt'. (If you have univariate data, then you need to use 'as.matrix'.) On 01/02/2010 20:32, Thomas Lumley wrote: On Mon, 1 Feb 2010, David Winsemius wrote: On Feb 1, 2010, at 12:37 PM, Антон Морковин wrote: Dear all, what function can be used to calculate weighted SD or/and SE¿ There's no such stuff in 'base' package. Seems as though lm with a weights argument could be used fairly simply. As I have pointed out from time to time, it depends on what the weights are for -- are they precision weights, which lm() can handle, or sampling weights, which it can't? -tomas Thomas Lumley Assoc. Professor, Biostatistics tlum...@u.washington.edu University of Washington, Seattle __ 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. -- Patrick Burns pbu...@pburns.seanet.com http://www.burns-stat.com (home of 'The R Inferno' and 'A Guide for the Unwilling S User') __ 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] character variables in substitute()
Hi:
I played around with this earlier, but had only limited success. This is
what I
got; perhaps others can embellish this with more efficient (and correct)
code.
I couldn't get more than one expression in a line without overplotting.
Here's
an example:
plot(c(0, 1), c(0, 1))
text(0.5, 0.5, expression(R == 13.34859, P[m] == 2.53071))
# This makes sense, of course, because each expression is centered at
# the same point.
# You can put the text() statements along a row (pardon the crudity,
# and yes, there are no commas; it just makes the expressions more
# complicated):
plot(c(0, 1), c(0, 1))
text(0, 0.5, expression(R == 13.34859), adj = 0)
text(0.28, 0.5, expression(P[m] == 2.53071), adj = 0)
text(0.55, 0.5, expression(k[a] == 4.06000), adj = 0)
text(0.75, 0.5, expression(alpha[r] == 0.00719), adj = 0)
# However, one expression per line worked out OK (except for the
# desired five digit accuracy in k[a]...)
plot(c(0, 1), c(0, 1))
text(0.1, 0.9, expression(R == 13.34859), adj = 0)
text(0.1, 0.8, expression(P[m] == 2.53071), adj = 0)
text(0.1, 0.7, expression(k[a] == 4.06000), adj = 0)
text(0.1, 0.6, expression(alpha[r] == 0.00719), adj = 0)
# Trying multiple expressions in a plot title is equally challenging...here,
# only the first is rendered:
plot(c(0, 1), c(0, 1),
main = expression(R == 13.34859, P[m] == 2.53071, k[a] == 4.06000,
alpha[r] == 0.00719))
# The best I could do was paste text strings, but that's less
# than what we were hoping for...
plot(c(0, 1), c(0, 1),
main = paste('R = 13.34859, ', 'P[m] = 2.53071, ', 'k[a] = 4.06000, ',
'alpha[r] = 0.00719'), cex.main = 0.8)
I went through the plotmath demo and example and tried to find some
previous posts, but came up empty. There has to be a better way...
HTH,
Dennis
On Mon, Feb 1, 2010 at 9:50 PM, dkStevens david.stev...@usu.edu wrote:
In trying to create a plotmath expression for plot labeling, such as
R = 6, beta = 15
where I want beta to be the Greek beta and, possibly, R in italics (like
one
would get in an explicit expression. The reason for this is that I want to
write a string builder function that takes vectors of variable names and
their values and return a plotmath expression for labeling a plot. One
approach I tried is
pname = c(R,beta)
params = c(6,15)
substitute(p == v,list(p = pname[i],v = format(params[i],digits=4))
but this just copies the character strings in pname into the expression.
Can
anyone help me do what I want? How would I manage passing the resulting
string back to the calling routine? Any help will be much appreciated.
--
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[R] Suppressing scientific notation on plot axis tick labels
Is there a better alternative to x = c(1e7, 2e7) x.lb = c(0,1e7,2e7) s.lb = format(x.lb, scientific = FALSE, big.mark = ,) barplot(x, yaxt = n, ylab = ) axis(side = 2, at = x.lb, labels = s.lb) (I am sure there is a better alternative to line 2 :)). Thank you. -- View this message in context: http://n4.nabble.com/Suppressing-scientific-notation-on-plot-axis-tick-labels-tp1459697p1459697.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] how to use optim() or nlm() to solve three nonlinear equations
Dear all, I just know how to solve an eaquation by using optim() or nlm(). But, now, I have three nonlinear equations, how could we use optim() or nlm() to solve a system of nonlinear equations in R? Thank you so much. Sincerely, Joe ___ æ¨ççæ´»å³æé ï¼ æºéãå¨æ¨ãçæ´»ãå·¥ä½ä¸æ¬¡æå®ï¼ [[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] character variables in substitute()
Hi David, there is a solution using bquote instead of substitute expr-bquote(italic(.(pname[1]))==.(params[1])~, ~.(as.name(pname[2]))==.(params[2])) plot(1,1,main=expr) hth. dkStevens schrieb: In trying to create a plotmath expression for plot labeling, such as R = 6, beta = 15 where I want beta to be the Greek beta and, possibly, R in italics (like one would get in an explicit expression. The reason for this is that I want to write a string builder function that takes vectors of variable names and their values and return a plotmath expression for labeling a plot. One approach I tried is pname = c(R,beta) params = c(6,15) substitute(p == v,list(p = pname[i],v = format(params[i],digits=4)) but this just copies the character strings in pname into the expression. Can anyone help me do what I want? How would I manage passing the resulting string back to the calling routine? Any help will be much appreciated. -- Eik Vettorazzi Institut für Medizinische Biometrie und Epidemiologie Universitätsklinikum Hamburg-Eppendorf Martinistr. 52 20246 Hamburg T ++49/40/7410-58243 F ++49/40/7410-57790 __ 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] Yield to Maturity using R
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The
query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at
101. The yield to maturity can be calculated after solving the equation -
1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
This can be solved by trial and error method s.t. ytm = 9.601%. I wanted to
find out how to solve this equation in R.
## SOLUTION
Mr. Elisson had given me following wonderful solution
f.ytm-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
And I got the answer as 9.601.
## _
I was just trying to generalize this solution to any equation and accordingly
written a code as given below.
The following input I will be reading using csv file and thus my equation will
change if tenure or no_comp etc. changes. So taking into account the variable
nature of the input, I am trying to write a generalized code.
## Input
price = 101 # Price of bond
tenure = 3
no_comp = 1 # no of times coupon paid in a year.
coupon_rate = 0.10 # i.e. 10%
face_value = 100
# Computations
coupon_payment = face_value * coupon_rate
cash_flow = c(rep(c(coupon_payemnt), (no_comp * tenure - 1)), face_value +
coupon_payment)
cash_flow
## I am trying to customize the code as given by Mr Ellison.
f.ytm = function(ytm)
{
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
}
solution = uniroot(f.ytm, interval=c(0,25))
ytm = solution$root
However, when I execute this code I get following error.
solution = uniroot(f.ytm, interval=c(0,25))
Error in uniroot(f.ytm, interval = c(0, 25)) : f.lower = f(lower) is NA
Please guide. ytm should be 0.09601 (i.e. 9.601%)
with regards
Madhavi Bhave
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Re: [R] character variables in substitute()
That's the trick, Eik! Nice job.
To return to the problem I was struggling with, it works with the following
incantation:
names - c('R', 'P', 'k', 'alpha')
vals - c(13.34859, 2.53071, 4.06000, 0.00719)
expr - bquote(.(names[1])==.(vals[1])~,
~.(as.name(names[2]))[m]==.(vals[2])~,
~.(as.name(names[3]))[a]==.(vals[3])~, ~.(as.name
(names[4]))[r]==.(vals[4]))
plot(c(0, 1), c(0, 1))
text(0.5, 0.5, expr)
Thanks for the valuable tip!
Dennis
On Tue, Feb 2, 2010 at 2:37 AM, Eik Vettorazzi
e.vettora...@uke.uni-hamburg.de wrote:
Hi David,
there is a solution using bquote instead of substitute
expr-bquote(italic(.(pname[1]))==.(params[1])~, ~.(as.name
(pname[2]))==.(params[2]))
plot(1,1,main=expr)
hth.
dkStevens schrieb:
In trying to create a plotmath expression for plot labeling, such as
R = 6, beta = 15
where I want beta to be the Greek beta and, possibly, R in italics (like
one
would get in an explicit expression. The reason for this is that I want to
write a string builder function that takes vectors of variable names and
their values and return a plotmath expression for labeling a plot. One
approach I tried is
pname = c(R,beta)
params = c(6,15)
substitute(p == v,list(p = pname[i],v = format(params[i],digits=4))
but this just copies the character strings in pname into the expression.
Can
anyone help me do what I want? How would I manage passing the resulting
string back to the calling routine? Any help will be much appreciated.
--
Eik Vettorazzi
Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
Martinistr. 52
20246 Hamburg
T ++49/40/7410-58243
F ++49/40/7410-57790
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[R] Memory Problem
Hi, When I run the repeat loop in R for large dataset, I got Memory problem. How can I solve these problem. -- View this message in context: http://n4.nabble.com/Memory-Problem-tp1459740p1459740.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] Abstract submission deadline, useR! 2010
The abstract submission deadline for the R Users Conference, useR! 2010, is just one month away. Submission deadline: 2010-03-01 Early registration will also close on 2010-03-01. Now is the perfect time to prepare an abstract presenting innovations or exciting applications of R. The call for abstracts along with the link for submission is available at: http://www.R-project.org/useR-2010/#Call This meeting of the R user community will take place at the Gaithersburg, Maryland, USA campus of the National Institute of Standards and Technology (NIST), July 21-23, 2010. The conference schedule is comprised of invited lectures and user-contributed sessions as well as half-day tutorials presented by R experts on July 20, 2010, prior to the conference. Invited speakers will include Mark Handcock, Frank Harrell Jr, Friedrich Leisch, Michael Meyer, Richard Stallman, Luke Tierney, and Diethelm Wuertz. The full spectrum of conference information is available from the webpage: http://www.R-project.org/useR-2010/ We hope to meet you in Gaithersburg! Kate Mullen, for the organizing committee ___ r-annou...@stat.math.ethz.ch mailing list https://stat.ethz.ch/mailman/listinfo/r-announce __ 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] [R-pkgs] Major update: mboost 2.0-0 released
Dear useRs, we are happy to announce the release of mboost 2.0-0 on CRAN: http://cran.r-project.org/package=mboost This version contains major updates and changes to the implementation of the main algorithm. Some slight changes to the user-interface where necessary. Please consult the manual and the list of CHANGES below. The package 'mboost' (Model-based Boosting) implements boosting for optimizing general risk functions utilizing component-wise (penalized) least squares estimates or regression trees as base-learners for fitting generalized linear, additive and interaction models to potentially high-dimensional data. A big variety of models can be investigated using 'mboost' including survival models, expectile regression models, ordinal regression models as well as standard models such as simple linear models. In all cases, the predictor can be specified very flexible using linear, smooth, random and spatial effects as well as decision trees or an arbitrary combination suiting the intention of the researcher. For more details and a nice example showing some of the functionality of 'mboost' see ?mboost-package. We appreciate any feedback. mboost development team ___ CHANGES in `mboost' VERSION 2.0-0 (2010-02-01) o generic implementation of component-wise functional gradient boosting in `mboost_fit', specialized code for linear, additive and interaction models removed o new families available for ordinal, expectile and censored regression o computations potentially based on package Matrix (reduces memory usage) o various speed improvements o added interface to extract selected base-learners (selected()) o added interface for parallel computations in cvrisk with arbitrary packages (e.g. multicore, snow) o added which argument in predict and coef functions and improved usability of which in plot-function. Users can specify which as numeric value or as a character string o added function cv() to generate matrices for k-fold cross-validation, subsampling and bootstrap o new function stabsel() for stability selection with error control o added function model.weights() to extract the weights o added interface to expand model by increasing mstop in model[mstop] o alternative definition of degrees of freedom available o Interface changes: - class definition / Family() arguments changed - changed behavior of subset method (model[mstop]). Object is directly altered and not duplicated - argument center in bols replaced with intercept - argument z in base-learners replaced with by - bns and bss deprecated -- ** Dipl.-Stat. Benjamin Hofner Institut für Medizininformatik, Biometrie und Epidemiologie Friedrich-Alexander-Universität Erlangen-Nürnberg Waldstr. 6 - 91054 Erlangen - Germany benjamin.hof...@imbe.med.uni-erlangen.de http://www.imbe.med.uni-erlangen.de/~hofnerb/ http://www.benjaminhofner.de ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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] Suppressing scientific notation on plot axis tick labels
Ruben Roa has kindly suggested using 'scipen' option - cf. fixed notation will be preferred unless it is more than ‘scipen’ digits wider. However, options(scipen = 50) x = c(1e7, 2e7) barplot(x) still does not produce the desired result. -- View this message in context: http://n4.nabble.com/Suppressing-scientific-notation-on-plot-axis-tick-labels-tp1459697p1459789.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] 3D plot of following data
Hello R-experts, I am having difficulties with 3D plotting (i.e. the evolution of various forward curves through time). I have two comma seperated files both ordered by date (in the first column) one containing contracts (meaning forward delivery months from YEAR_ Letter F ... January through letter Z ... December) and the other holding the closing price of the respective contract on the day also defined in the first column (see attachments). What I would like to do is plot a three dimensional figure with trade day (date) on the X-axis, contract on the Y-axis and the price of the forward contract being the z-value. I am quite a newbie and did not manage to merge these two files in a logic way, so that R could do a 3D plot. Any help would be appreciated. -- WD__ 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 use optim() or nlm() to solve three nonlinear equations
Nai-Wei Chen wrote: I just know how to solve an eaquation by using optim() or nlm(). But, now, I have three nonlinear equations, how could we use optim() or nlm() to solve a system of nonlinear equations in R? Thank you so much. You don't use general optimize routines to solve a set of nonlinear equations. It is not efficient and it is not recommended. There are two packages: nleqslv and BB, that you can use to solve non linear equation systems. Both are available on CRAN. /B -- View this message in context: http://n4.nabble.com/how-to-use-optim-or-nlm-to-solve-three-nonlinear-equations-tp1459716p1459803.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] Solving an optimization problem: selecting an quot; optimalquot; subset
Erwin Kalvelagen-2 wrote: Hans W Borchers hwborchers at googlemail.com writes: # Prepare inputs for MILP solver obj - c(rep(0, n), 0, 1, 1, 0) typ - c(rep(B, n), B, C, C, B) mat - matrix(c(s, -z, -1, 1, 0,# a = a_p + a_m rep(0, n), 1, 0, 0, 0, # constant term rep(0, n), 0, 1, 0, -M, # a_p = M * d0 rep(0, n), 0, 0, 1, M, # a_m = M * (d0-1) rep(1, n), 0, 0, 0, 0), # subset size = k nrow=5, byrow=T) dir - c(==, ==, =, =, =) rhs - c(0, 1, 0, M, k) max - FALSE You can drop the binary variable d0. The condition one of a_p,a_m is zero holds automatically as we are minimizing a_p+a_m. Erwin Kalvelagen Right; for me adding M and d0 is kind of a reflex, but that is only necessary when a_p + a_m is used as an intermediate result. One more remark: I saw some spurious behavior with both solvers (Rsymphony, Rglpk) -- that is, slightly different results in different runs. It could relate to the tolerance with which these solvers compare and identify solutions. At the moment I don't know how to change the tolerance as a parameter. I guess this will not happen when using a more powerful solver such as CPLEX. Hans Werner -- View this message in context: http://n4.nabble.com/Solving-an-optimization-problem-selecting-an-optimal-subset-tp1446084p1459843.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] Yield to Maturity using R
Hi Madhavi,
the error message means, that your function returns NA evaluated at the
lower limit of the search interval.
try f.ytm(0) to check that.
I think,
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
will not return what you expected, since the code evaluates solely E
=NULL (tenure * no_comp - 1) times. (the opening curly bracket is misplaced)
hth
Madhavi Bhave schrieb:
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following characteristics.
Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at 101. The yield to maturity can be calculated after solving the equation -
1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
This can be solved by trial and error method s.t. ytm = 9.601%. I wanted to find out how to solve this equation in R.
## SOLUTION
Mr. Elisson had given me following wonderful solution
f.ytm-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
And I got the answer as 9.601.
## _
I was just trying to generalize this solution to any equation and accordingly written a code as given below.
The following input I will be reading using csv file and thus my equation will change if tenure or no_comp etc. changes. So taking into account the variable nature of the input, I am trying to write a generalized code.
## Input
price = 101 # Price of bond
tenure = 3
no_comp = 1 # no of times coupon paid in a year.
coupon_rate = 0.10 # i.e. 10%
face_value = 100
# Computations
coupon_payment = face_value * coupon_rate
cash_flow = c(rep(c(coupon_payemnt), (no_comp * tenure - 1)), face_value +
coupon_payment)
cash_flow
## I am trying to customize the code as given by Mr Ellison.
f.ytm = function(ytm)
{
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
}
solution = uniroot(f.ytm, interval=c(0,25))
ytm = solution$root
However, when I execute this code I get following error.
solution = uniroot(f.ytm, interval=c(0,25))
Error in uniroot(f.ytm, interval = c(0, 25)) : f.lower = f(lower) is NA
Please guide. ytm should be 0.09601 (i.e. 9.601%)
with regards
Madhavi Bhave
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--
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Institut für Medizinische Biometrie und Epidemiologie
Universitätsklinikum Hamburg-Eppendorf
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20246 Hamburg
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F ++49/40/7410-57790
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[R] hvcluster() with distance method from vegdist(), package = vegan
hello, i'd be happy if someone could provide help with the following problem: i have a dist.matrix that comes from vegdist() function of the vegan package. the used method = horn is not accepted as argument in hvcluster(...,dist.method=...). is there a way to incorporate the method horn in hvcluster()? thanks in advance! yours, kay -- View this message in context: http://n4.nabble.com/hvcluster-with-distance-method-from-vegdist-package-vegan-tp1459859p1459859.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] hvcluster() with distance method from vegdist(), package = vegan
sorry, it should say pvclust(), of course... -- View this message in context: http://n4.nabble.com/hvcluster-with-distance-method-from-vegdist-package-vegan-tp1459859p1459865.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] Yield to Maturity using R
Dear Sir,
Thank you for valuable guidance. Though I have been using R occassionally, it
was limited to some basics and that way I am new to R. As suggested by you, I
have gone through the said chapter of Introduction to R manual, though I have
some urgent comittments to meet.
I have tried writing function as given below.
f = function(price, tenure, no_comp, coupon_rate, face_value)
{
coupon_payment = face_value * coupon_rate / no_comp
cash_flow = c(rep(c(coupon_payment), (no_comp * tenure - 1)), face_value +
coupon_payment)
E = NULL
for (i in 1 : (tenure * no_comp - 1))
{
E[i] = cash_flow[i]/(1+ytm)^i
}
F = NULL
{
F = sum(E) + ((face_value + coupon_payment)/(1+ytm)^(no_comp * tenure)) -
price
}
return(data.frame(S = uniroot.all(F, interval=c(0,25
}
output = f(1010, 3, 1, 0.10, 1000)
## End of code
However, when I try to execute the same, I get following error.
Error: object 'ytm' not found
My objective is to find ytm itself and I am not able to figure out where I am
going wrong and how to overcome the same.
Regards
Madhavi
--- On Tue, 2/2/10, Dennis Murphy djmu...@gmail.com wrote:
From: Dennis Murphy djmu...@gmail.com
Subject: Re: [R] Yield to Maturity using R
To: Madhavi Bhave madhavi_bh...@yahoo.com
Date: Tuesday, 2 February, 2010, 3:49 AM
Hi:
On Tue, Feb 2, 2010 at 3:01 AM, Madhavi Bhave madhavi_bh...@yahoo.com wrote:
Dear R helpers,
Yesterday I had raised following query which was addressed by Mr Ellison. The
query and the wonderful solution as provided by Mr. Ellison are as given below.
## PROBLEM
I am calculating the 'Yield to Maturity' for the Bond with following
characteristics.
Its a $1000 face value, 3 year bond with 10% annual coupon and is priced at
101. The yield to maturity can be calculated after solving the equation -
1010 = [100 / (1+ytm)] + [100 / (1+ytm)^2] + [ 1100 / (1 + ytm)^3]
This can be solved by trial and error method s.t. ytm = 9.601%. I wanted to
find out how to solve this equation in R.
## SOLUTION
Mr. Elisson had given me following wonderful solution
f.ytm-function(ytm) 100 / (1+ytm) +100 / ((1+ytm)^2) + 1100 / ((1 +
ytm)^3) -1010
uniroot(f.ytm, interval=c(0,25))
#$root has the answer
And I got the answer as 9.601.
## _
I was just trying to generalize this solution to any equation and accordingly
written a code as given below.
The following input I will be reading using csv file and thus my equation will
change if tenure or no_comp etc. changes. So taking into account the variable
nature of the input, I am trying to write a generalized code.
## Input
price = 101 # Price of bond
tenure = 3
no_comp = 1 # no of times coupon paid in a year.
coupon_rate = 0.10 # i.e. 10%
face_value = 100
# Computations
coupon_payment = face_value * coupon_rate
cash_flow = c(rep(c(coupon_payemnt), (no_comp * tenure - 1)), face_value +
coupon_payment)
cash_flow
## I am trying to customize the code as given by Mr Ellison.
f.ytm = function(ytm)
{
for (i in 1 : (tenure * no_comp - 1))
E = NULL
F = NULL
{
E[i] = cash_flow[i]/(1+ytm)^i
F = (sum(E) + (face_value + coupon_payment)/((1+ytm)^(tenure * no_comp))) -
price
}
}
For this to work, tenure, no_comp, cash_flow, face_value and coupon_payment
have to be
visible to the function - i.e., they either have to be in the function's
calling environment
or in the global environment. These are called 'free variables' under the
lexical
scoping rules of R. (Welcome to function writing :)
You might want to look a little more closely at uniroot(), especially the
conventions
it requires for the *functions* it can evaluate. You want the body of what you
send
to uniroot() for evaluation to be a function of a single variable. Your
previously supplied
solution meets that requirement. This one doesn't (yet).
Moreover, it appears that E[i] is never used and nothing is returned from
f.ytm. I'd
suggest an excursion into R function writing fundamentals. Start with Ch. 10 of
the Introduction to R manual.
HTH,
Dennis
solution = uniroot(f.ytm, interval=c(0,25))
ytm = solution$root
However, when I execute this code I get following error.
solution = uniroot(f.ytm, interval=c(0,25))
Error in uniroot(f.ytm, interval = c(0, 25)) : f.lower = f(lower) is NA
Please guide. ytm should be 0.09601 (i.e. 9.601%)
with regards
Madhavi Bhave
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Re: [R] Memory Problem
On 02.02.2010 11:33, Meenakshi wrote: Hi, When I run the repeat loop in R for large dataset, I got Memory problem. How can I solve these problem. buy more memory, bigger machine, more efficient programming, import of only relevant data, use of specific tools, .. or in other words: Depends on your problem and please do read the posting guide. Uwe Ligges __ 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] 3D plot of following data
Since your files did not come through the mail list, I'd suggest to put them on some webside and provide a link. Uwe Ligges On 02.02.2010 13:01, walter.dju...@chello.at wrote: Hello R-experts, I am having difficulties with 3D plotting (i.e. the evolution of various forward curves through time). I have two comma seperated files both ordered by date (in the first column) one containing contracts (meaning forward delivery months from YEAR_ Letter F ... January through letter Z ... December) and the other holding the closing price of the respective contract on the day also defined in the first column (see attachments). What I would like to do is plot a three dimensional figure with trade day (date) on the X-axis, contract on the Y-axis and the price of the forward contract being the z-value. I am quite a newbie and did not manage to merge these two files in a logic way, so that R could do a 3D plot. Any help would be appreciated. -- WD __ 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] Biclustering / Co-clustering in more than 2 dimensions
Hi, I was wondering if there existed a package in R that would bicluster / co-cluster in more than 2 dimensions. thanks! [[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] Subtracting time series
Hi all, I've got a time series object (xts) called table. Table contains closing price and volume for each market day of 2009 on a given equity. I'd like to get another xts object, say table2, that contains for each market day holds the close of that day minus the close of the day before and the volume of that day minus the volume of the day before. So table2 - table[x,] - table[(x-1),] But I can't quit get the syntax, any help would be appreciated. Nick [[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] strange behaviour: recognition of decimal numbers by 'which'
It is a strange behaviour in that I did not expect it... but I am sure
there is a simple explanation for it and it'll have to do with the way
numbers are stored in R, but it's caught me by surprise and I don't
find it obvious.
Here's a simplified example reproducing the behaviour I encountered:
I create an empty vector, and I fill it with a sequence of numbers: 0,
0.005, 0.01, 0.015, 0.02, etc... until 0.1.
(In my real code this is just a way to store certain values when
certain conditions are met etc etc)
Then I ask which element contains teh value 0.01. Ok. Same with 0.02.
But when I ask about 0.03 or above... the answer is none. But I can
see 0.03 in my vector, it is there!
This must be because of the value stored internally. Indeed, if I do:
v-0.03, when I reach the value that appeared to contain 0.03, I don't
obtain zero, but 3.47e-18
why is this?
what is the recommended way to overcome this?
Interestingly, if I store the values as character, then I can match
them fine...
Below is the code, and my sessionInfo() output.
Thanks!
Jose de las Heras
code example:
#v-vector(mode=character)
v-c()
i-0
while (i = 0.1)
{
v-c(v,i)
i-i+0.005
}
v
which(v==0.01)
which(v==0.02)
which(v==0.03)
sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United Kingdom.1252
[2] LC_CTYPE=English_United Kingdom.1252
[3] LC_MONETARY=English_United Kingdom.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United Kingdom.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] biomaRt_2.2.0
loaded via a namespace (and not attached):
[1] RCurl_1.3-1 tools_2.10.0 XML_2.6-0
--
The University of Edinburgh is a charitable body, registered in
Scotland, with registration number SC005336.
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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] strange behaviour: recognition of decimal numbers by 'which'
FAQ 7.31
j.delashe...@ed.ac.uk wrote:
It is a strange behaviour in that I did not expect it... but I am sure
there is a simple explanation for it and it'll have to do with the way
numbers are stored in R, but it's caught me by surprise and I don't find
it obvious.
Here's a simplified example reproducing the behaviour I encountered:
I create an empty vector, and I fill it with a sequence of numbers: 0,
0.005, 0.01, 0.015, 0.02, etc... until 0.1.
(In my real code this is just a way to store certain values when certain
conditions are met etc etc)
Then I ask which element contains teh value 0.01. Ok. Same with 0.02.
But when I ask about 0.03 or above... the answer is none. But I can see
0.03 in my vector, it is there!
This must be because of the value stored internally. Indeed, if I do:
v-0.03, when I reach the value that appeared to contain 0.03, I don't
obtain zero, but 3.47e-18
why is this?
what is the recommended way to overcome this?
Interestingly, if I store the values as character, then I can match
them fine...
Below is the code, and my sessionInfo() output.
Thanks!
Jose de las Heras
code example:
#v-vector(mode=character)
v-c()
i-0
while (i = 0.1)
{
v-c(v,i)
i-i+0.005
}
v
which(v==0.01)
which(v==0.02)
which(v==0.03)
sessionInfo()
R version 2.10.0 (2009-10-26)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_United Kingdom.1252
[2] LC_CTYPE=English_United Kingdom.1252
[3] LC_MONETARY=English_United Kingdom.1252
[4] LC_NUMERIC=C
[5] LC_TIME=English_United Kingdom.1252
attached base packages:
[1] stats graphics grDevices utils datasets methods base
other attached packages:
[1] biomaRt_2.2.0
loaded via a namespace (and not attached):
[1] RCurl_1.3-1 tools_2.10.0 XML_2.6-0
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Biclustering / Co-clustering in more than 2 dimensions
Hi Tim, maybe you are looking for something like this: http://www.nature.com/nbt/journal/v26/n5/full/nbt1397.html It is relatively easy to implement in R, I think. Best, Gabor On Tue, Feb 2, 2010 at 3:44 PM, Tim Smith tim_smith_...@yahoo.com wrote: Hi, I was wondering if there existed a package in R that would bicluster / co-cluster in more than 2 dimensions. thanks! [[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. -- Gabor Csardi gabor.csa...@unil.ch UNIL DGM __ 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] Subtracting time series
On Tue, 2 Feb 2010, Nick Torenvliet wrote: Hi all, I've got a time series object (xts) called table. Table contains closing price and volume for each market day of 2009 on a given equity. I'd like to get another xts object, say table2, that contains for each market day holds the close of that day minus the close of the day before and the volume of that day minus the volume of the day before. So table2 - table[x,] - table[(x-1),] This does not do what you want because xts is smart and aligns along the time index. But I can't quit get the syntax, any help would be appreciated. Look at diff() and lag() and their respective methods for xts series. Z Nick [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] finding files whose name does NOT contain a given character
Unluckily I dela with miRNA files whose name may contain the character *. Because of the special meaning of * I have to remove it. I found out how to make list.files() extract only those file names which contain a * Namely: # list.files(pattern=\\*) Now I have to process all files whose name does NOT contain the character *. I cannot have list.files() extract all files whose name does NOT match pattern=\\* I tried using ^ in such a pattern but nothing is returned. Any suggestion is welcome. Thank you so much, Maura tutti i telefonini TIM! [[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] Plot - specification for grid of x axis
Dear all, I have a simple question for which I cannot find the answer. I need to make an easy plot, but for the x axis I need to be able to specify by myself the division of x axis from x[,1] either every single observation, or every 5th, or 10th or 20th x - matrix(data=NA, nrow=100, ncol=2) x[,1]-seq(1,100,1) x[,2]-rnorm(100) Thanks a lot for your help!!! [[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] finding files whose name does NOT contain a given character
mau...@alice.it wrote: Unluckily I dela with miRNA files whose name may contain the character *. Because of the special meaning of * I have to remove it. I found out how to make list.files() extract only those file names which contain a * Namely: # list.files(pattern=\\*) Now I have to process all files whose name does NOT contain the character *. I cannot have list.files() extract all files whose name does NOT match pattern=\\* I tried using ^ in such a pattern but nothing is returned. Any suggestion is welcome. That'll be something like pattern=^[^*]*$ (untested, I don't think I have any filenames with * inside...) Alternatively, you might try allfiles - list.files() withstar - allfiles[grepl(\\*, allfiles)] nostar - allfiles[!grepl(\\*, allfiles)] -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - (p.dalga...@biostat.ku.dk) FAX: (+45) 35327907 __ 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] finding files whose name does NOT contain a given character
mau...@alice.it wrote: Unluckily I dela with miRNA files whose name may contain the character *. Because of the special meaning of * I have to remove it. I found out how to make list.files() extract only those file names which contain a * Namely: # list.files(pattern=\\*) Now I have to process all files whose name does NOT contain the character *. I cannot have list.files() extract all files whose name does NOT match pattern=\\* I tried using ^ in such a pattern but nothing is returned. Any suggestion is welcome. Thank you so much, Maura Maybe ?setdiff could even help here ... tutti i telefonini TIM! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Finding the difference between two vectors
Hello everyone, I have two vectors having only one element different: vector1 vector2 vector1 TWC TWC TWC VFC TWX NA VIA/B VFC VFC WHR VIA/B VIA/B WPO WHR -WHR WYN WPO WPO WYNN WYN WYN YUMWYNN WYNN NA YUM YUM In that case, in the first vector, the value twx is missing . What I want to do is doing the following avoiding loop: putting the NA row in front of the TWX row as this would be the row where the value is missing. I don't know if there is any function doing this. I thought about using a which() function to retrieve all the missing elements index ( twx, 1 here) and insert them in the right index in the first vector but how can I insert in a vector, is there an appropriated function? thank you - Anna Lippel -- View this message in context: http://n4.nabble.com/Finding-the-difference-between-two-vectors-tp1460020p1460020.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] merging data frames gives all NAs
David,
Now the code is:
for (j in seq_along(rwy)) { # subset the data and merge them
ar4rw = ar4rw - subset(arrgnd, arrgnd$Runway==rwy[j])
if(j == 1) {
arrw = ar4rw
}
else {
arrw = merge(arrw, ar4rw)
}
}
I attach the data. I needed 500 rows to get both runways in rwy.
The suggestions did not help much, but did get rid of the row of NAs in
ar4rw. Why?
When I run through the loop for 2 runways, I get
# j = 1, Runway = 31L
Browse[1] arrw[1:3,]
DateTime Date month hour minute quarter weekday IATA ICAO Flight
552 1/1/09 23:03 2009-01-01 1 23 3 92 5 AA AAL AAL22
563 1/1/09 23:17 2009-01-01 1 23 17 93 5 DL DAL DAL242
565 1/1/09 23:24 2009-01-01 1 23 24 93 5 DL DAL DAL624
AircraftType Tail Arrived STA Runway FromTo Delay
552 B762 N329AA 23:03:35 23:10 * 31L* LAX /JFK 0
563 B763 N1611B 23:17:37 23:46 31L KATL /KJFK 0
565 B752 N654DL 23:24:04 23:48 31L LAS /JFK 0
Operator dq gw
552 AMERICAN AIRLINES 2009-01-01 92 1
563 DELTA AIR LINES 2009-01-01 93 1
565 DELTA AIR LINES 2009-01-01 93 1
# j = 2 Runway=31R
Browse[1] ar4rw[1:3,]
DateTime Date month hour minute quarter weekday IATA ICAO Flight
529 1/1/09 21:46 2009-01-01 1 21 46 87 5 TA TAI TAI570
530 1/1/09 21:48 2009-01-01 1 21 48 87 5 AA AAL AAL2018
531 1/1/09 21:50 2009-01-01 1 21 50 87 5 BA BAW BAW183
AircraftType Tail Arrived STA Runway FromTo Delay
529 A320 N496TA 21:46:58 22:30 * 31R* MSLP /KJFK 0
530 B752 N621AM 21:48:43 21:50 31R TLPL /JFK 0
531 B744 G-CIVI 21:50:26 22:50 31R EGLL /KJFK 0
Operator dq gw
529 TACA INTERNATIONAL AIRLINES 2009-01-01 87 1
530 AMERICAN AIRLINES 2009-01-01 87 1
531 BRITISH AIRWAYS 2009-01-01 87 1
# But the merge gives all NAs!
] arrw[1:3,]
DateTime Date month hour minute quarter weekday IATA ICAO Flight
NA NA NA NA NA NA NA NA NA NA NA
NA.1 NA NA NA NA NA NA NA NA NA NA
NA.2 NA NA NA NA NA NA NA NA NA NA
AircraftType Tail Arrived STA Runway FromTo Delay Operator dq gw
NA NA NA NA NA NA NA NA NA NA NA
NA.1 NA NA NA NA NA NA NA NA NA NA
NA.2 NA NA NA NA NA NA NA NA NA NA
Thanks,
Jim Rome
On Feb 1, 2010, at 5:30 PM, David Winsemius wrote:
On Feb 1, 2010, at 5:16 PM, James Rome wrote:
Dear kind R helpers,
I have a vector of runway names in rwy (31R, 31L,... the number
is user selectable)
arrgnd is a data frame with data for all flights and all runways,
with a Runway column.
I am trying to subset arrgnd into a dat frame for each selected
runway, and then combine them back together using the following code:
for (j in 1:nr) { # nr = number of user-selected runways
Safer would be:
for (j in seq_along(rwy) {
ar4rw = arrgnd[arrgnd$Runway==rwy[j],]
Clearer would be :
ar4rw - subset(arrgnd, Runway= j) # and I think the NA line's will
also disappear.
^ == ^
if (j == 1) {
arrw = ar4rw
}
else {
arrw = merge(arrw, ar4rw)
}
}
You really should give us something like:
dput(rwy)
dput( head(arrgnd, 10) )
but, the merge step gives me a data frame with all NAs. In addition,
ar4rw always gets a row with NAs at the start, which I do not
understand. There are no rows with all NAs in the arrgnd data frame.
ar4rw[1:2,] # first time through for 31R
DateTime Date month hour minute quarter weekday IATA ICAO Flight
NA NA NA NA NA NA NA NA NA NA NA
529 1/1/09 21:46 2009-01-01 1 21 46 87 5 TA TAI TAI570
AircraftType Tail Arrived STA Runway FromTo Delay
NA NA NA NA NA NA NA NA
529 A320 N496TA 21:46:58 22:30 31R MSLP /KJFK 0
Operator dq gw
NA NA NA NA
529 TACA INTERNATIONAL AIRLINES 2009-01-01 87 1
ar4rw[1:2,] # second time through for 31L
DateTime Date month hour minute quarter weekday IATA ICAO Flight
NA NA NA NA NA NA NA NA NA NA NA
552 1/1/09 23:03 2009-01-01 1 23 3 92 5 AA AAL AAL22
AircraftType Tail Arrived STA Runway FromTo Delay Operator
NA NA NA NA NA NA NA NA NA
552 B762 N329AA 23:03:35 23:10 31L LAX /JFK 0 AMERICAN AIRLINES
dq gw
NA NA NA
But after the merge, I get all NAs. What am I doing wrong?
The data layout gets mangled and I cannot tell what rows are being
matched to what. Use dput to convey an unambiguous, and easily
replicated example.
Thanks,
Jim Rome
552 2009-01-01 92 1
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David Winsemius, MD
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David Winsemius, MD
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West Hartford, CT
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Re: [R] 3D plot of following data
On 02.02.2010 16:01, walter.dju...@chello.at wrote:
Here they are for now for you.
And I will put them somewhere in a few minutes and poost the link.
No need to do so. If you really want to use 3D plots (which I doubt you
really want since you could do better in 2D using dotcharts or so):
setwd('d:/')
#READ IN Contracts
NG.contracts - read.csv('NG_contracts.csv', header=FALSE, sep=;,
na.string=)
NG.contracts$V1 - as.Date(NG.contracts$V1,format='%d.%m.%Y')
#READ IN Values
NG.values - read.csv('NG_values.csv', header=FALSE, sep=;, na.string=?)
NG.values$V1 - as.Date(NG.values$V1, format='%d.%m.%Y')
## new stuff by UL:
NG.valuesl - reshape(NG.values, direction=long, varying=list(2:32),
idvar=V1)
NG.contractsl - reshape(NG.contracts, direction=long,
varying=list(2:32), idvar=V1)
NG - NG.valuesl[,c(V1,V2)]
colnames(NG) - c(Date, value)
rownames(NG) - NULL
NG$contract - NG.contractsl[,V2]
## two ways, one for nice viewing, one for easier printout:
library(rgl)
plot3d(NG$Date, NG$contract, NG$value)
library(scatterplot3d)
scatterplot3d(NG$Date, NG$contract, NG$value, pch=20)
Now you need to prettify exes etc.
Uwe Ligges
Greetings,
Walter
Uwe Liggeslig...@statistik.tu-dortmund.de schrieb:
Since your files did not come through the mail list, I'd suggest to put
them on some webside and provide a link.
Uwe Ligges
On 02.02.2010 13:01, walter.dju...@chello.at wrote:
Hello R-experts,
I am having difficulties with 3D plotting (i.e. the evolution of various
forward curves through time).
I have two comma seperated files both ordered by date (in the first column) one containing contracts
(meaning forward delivery months from YEAR_ Letter F ... January through letter
Z ... December) and the other holding the closing price of the respective contract on the
day also defined in the first column (see attachments).
What I would like to do is plot a three dimensional figure with trade day
(date) on the X-axis, contract on the Y-axis and the price of the forward
contract being the z-value.
I am quite a newbie and did not manage to merge these two files in a logic way,
so that R could do a 3D plot.
Any help would be appreciated.
--
WD
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--
mfg
WD
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Re: [R] finding files whose name does NOT contain a given charac
On 02-Feb-10 15:18:28, Peter Dalgaard wrote: mau...@alice.it wrote: Unluckily I dela with miRNA files whose name may contain the character *. Because of the special meaning of * I have to remove it. I found out how to make list.files() extract only those file names which contain a * Namely: # list.files(pattern=\\*) Now I have to process all files whose name does NOT contain the character *. I cannot have list.files() extract all files whose name does NOT match pattern=\\* I tried using ^ in such a pattern but nothing is returned. Any suggestion is welcome. That'll be something like pattern=^[^*]*$ (untested, I don't think I have any filenames with * inside...) Alternatively, you might try allfiles - list.files() withstar - allfiles[grepl(\\*, allfiles)] nostar - allfiles[!grepl(\\*, allfiles)] Peter's first suggestion works. As often with regeular expressions, [...] is your friend! I have created dummy files abcde pq*st uvwxyz, and then list.files() # [1] abcde pq*st uvwxyz list.files(pattern=[*]) # [1] pq*st list.files(pattern=^[^*]*$) # [1] abcde uvwxyz Ted. E-Mail: (Ted Harding) ted.hard...@manchester.ac.uk Fax-to-email: +44 (0)870 094 0861 Date: 02-Feb-10 Time: 15:41:53 -- XFMail -- __ 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] Suppressing scientific notation on plot axis tick labels
options(scipen = 50, digits = 5) x = c(1e7, 2e7) barplot(x) Still scientific... -- View this message in context: http://n4.nabble.com/Suppressing-scientific-notation-on-plot-axis-tick-labels-tp1459697p1459828.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] R assistance
Dear Madam/Sir, I read carefully The R Installation and Administration paper. I downloaded Rtools11.exe and I run it. As a result there is a directory R on my PC. Could you please tell me what should I do next in order to run R? Thank you in advance. BR Dragomir Nedeltchev [[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] hiding/protecting utility functions in .Rprofile
On Mon, Feb 1, 2010 at 13:19, Michael Friendly frien...@yorku.ca wrote: [Env: WinXp, R 2.9.2] In my .Rprofile, I define a number of utility functions I'd like to have available in my R session, but don't want them to be *normally* listed by ls(), or more importantly, saved if I save my session variables/functions. How can I do this? -- Michael Friendly Email: friendly AT yorku DOT ca Professor, Psychology Dept. York University Voice: 416 736-5115 x66249 Fax: 416 736-5814 4700 Keele Street http://www.math.yorku.ca/SCS/friendly.html Toronto, ONT M3J 1P3 CANADA __ 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. My strategy for loading my own functions at startup is to keep my functions in a separate file (not .Rprofile) that I source into its own environment via .Rprofile. My .Rprofile looks like this: ### .diag - new.env() sys.source( /path/to/sourcefile/functions_to_load.R, envir = .diag ) attach( .diag ) ### I can't claim credit for this; I put it together from other posts on this mailing list, but I don't remember the source now. And I cannot speak to whether this is the best way, but it works fine for me. Hope that helps. __ 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] 3D plot of following data
a link to the 3D plot data files http://members.chello.at/gwd the files are NG_contracts and NG_values -- mfg WD __ 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] R assistance
Download the R installer from CRAN, and run that. A desktop icon should pop up, and you should be set. Stephen On Tue, Feb 2, 2010 at 8:16 AM, dragomir nedeltchev dnedelche...@yahoo.com wrote: Dear Madam/Sir, I read carefully The R Installation and Administration paper. I downloaded Rtools11.exe and I run it. As a result there is a directory R on my PC. Could you please tell me what should I do next in order to run R? Thank you in advance. BR Dragomir Nedeltchev [[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. -- Stephen Sefick Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ 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] Suppressing scientific notation on plot axis tick labels
Strange, the following works reproducibly on my machine (Windows 2000 Pro): options(scipen = 50, digits = 5) x = c(1e7, 2e7) ?barplot barplot(x) while I also get scientific with your code. After I called ?barplot once, I'm incapable of getting the scientific notation again, though... But I admit that I am outdated, I still run 2.8.1 (but not for long...) Michael -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Shvorob Sent: Dienstag, 2. Februar 2010 12:59 To: r-help@r-project.org Subject: Re: [R] Suppressing scientific notation on plot axis tick labels options(scipen = 50, digits = 5) x = c(1e7, 2e7) barplot(x) Still scientific... -- View this message in context: http://n4.nabble.com/Suppressing-scientific-notation-on-plot-a xis-tick-labels-tp1459697p1459828.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] Odp: Finding the difference between two vectors
Hi Option one vec1-sample(letters[1:10]) vec2-vec1[-5] vec2-c(vec2,NA) # missing position mis-which(!(vec1 %in% vec2)) c(vec2[1:(mis-1)], NA,vec2[mis:((length(vec2)-1))]) [1] f e g d NA j c i b h gives you desired vector. For two or more values missing it shall be rephrased probably by rle Option two mis-which(!(vec1 %in% vec2)) vec2 - vec1 vec2 [mis] -NA Which could be used in case of several values missing. But I wonder what you want to achieve. It seems to me like you want to do something for which merge can be used. See ?merge Regards Petr r-help-boun...@r-project.org napsal dne 02.02.2010 16:34:23: Hello everyone, I have two vectors having only one element different: vector1 vector2 vector1 TWC TWC TWC VFC TWX NA VIA/B VFC VFC WHR VIA/B VIA/B WPO WHR -WHR WYN WPO WPO WYNN WYN WYN YUMWYNN WYNN NA YUM YUM In that case, in the first vector, the value twx is missing . What I want to do is doing the following avoiding loop: putting the NA row in front of the TWX row as this would be the row where the value is missing. I don't know if there is any function doing this. I thought about using a which() function to retrieve all the missing elements index ( twx, 1 here) and insert them in the right index in the first vector but how can I insert in a vector, is there an appropriated function? thank you - Anna Lippel -- View this message in context: http://n4.nabble.com/Finding-the-difference- between-two-vectors-tp1460020p1460020.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] Plot - specification for grid of x axis
When you create your plot us the argument xaxt='n'. That will suppress the default axis labels and tick marks. Then use the axis function which lets you specify exactly where you want the tick marks and exactly what you want the labels to be. You can also specify tck=1 as an argument if you want grid lines (see ?par for details). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare greg.s...@imail.org 801.408.8111 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r- project.org] On Behalf Of Trafim Vanishek Sent: Tuesday, February 02, 2010 8:12 AM To: r-help@r-project.org Subject: [R] Plot - specification for grid of x axis Dear all, I have a simple question for which I cannot find the answer. I need to make an easy plot, but for the x axis I need to be able to specify by myself the division of x axis from x[,1] either every single observation, or every 5th, or 10th or 20th x - matrix(data=NA, nrow=100, ncol=2) x[,1]-seq(1,100,1) x[,2]-rnorm(100) Thanks a lot for your help!!! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Deleting many columns of a data frame with the same name in a row
Hi, I have a data frame datas with half of the columns with the same name A. I want to delete all those columns from the data frame so here is what I did: datas$A - NULL The problem is that it deleted only one column, I would have to do it as many times as there are A columns. Is there a way to do it in one time? thank you - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460078.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] create an object in a loop (v3)
Hi David,
Thanks for your answer.
But I don't really see how I can extend it with my real data.
The thing is that I have more than 3 names and 1 value for each name.
Moreover, each is different from one run to another. That is why I was
trying with a modification of names(). Also to be noted is that I
simplified the name in assign(); I actually have 2 other variables
that will be pasted to create the name.
Here is my code (I kept only what is important for that part):
library(WRS)
seq.num - seq(7,10,1)#column (variable) indexes to be used as
numerical variables
# fac2list() separates the data from file[,k] into groups from levels in
file[3] and store into list mode.
for(i in 1:length(seq.num)) {
k - seq.num[i]
name.num - names(file)[k]
assign(paste(names(file)[3], name.num, sep=_), fac2list(file[,k],
file[3]))
names(paste(names(file)[3], name.num, sep=_)) -
levels(factor(file[[3]])) #that line doesn't work, but I would like
something in this direction
}
Thanks in advance for your help.
Regards,
Ivan
Le 2/1/2010 18:47, David Winsemius a écrit :
On Feb 1, 2010, at 12:33 PM, Ivan Calandra wrote:
I have a follow-up question:
I use assign() to store some value in my paste()-created object as
suggested:
for (i in 1:3) {
assign(paste(object, i, sep=), c(a, b, c))
}
Then I would like to change the names of the elements of that object
within the loop. Since it is all in a loop, I cannot give the name of
the object manually by doing something like: names(object1) -
c(tooth, bone, species).
The only thing I can give to names() is paste(object, i, sep=),
which doesn't work.
Any idea of how to do it?
for (i in paste(object, 1:3, sep=)) {
+ assign(i, c(tooth=a, bone=b, species=c) )
+ }
object1
toothbone species
a b c
Thanks in advance
Ivan
Le 2/1/2010 17:14, David Winsemius a écrit :
Upon reading it yesterday, it appeared as it would have required
some serious testing and there was no data on which to do any work.
You were clearly not taking the time to isolate the problem and
construct a dataset. But who knows? When you say What I want to do
is. ... ,I would like the name of the list to be created in the loop
too, maybe all you needed was to be pointed to was:
?assign
But if that were the case, then you lost most of your audience along
the way with a bunch of unneeded and obscure code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] using rbind in a for loop
Hi, On Mon, Feb 1, 2010 at 11:16 PM, Eunjung Kim eunjung.haw...@gmail.com wrote: *Dear R users, *I'm facing a trivial problem but I cannot solve it. My question is: I want to make data set like this. i_lon1 i_lat1 i_lon2 i_lat2 i_lon3 i_lat3 i_lon4 i_lat4 i+1_lon1 i+1_lat1 i+1_lon2 i+1_lat2 i+1_lon3 i+1_lat3 i+1_lon4 i+1_lat4 i+2_lon1 i+2_lat1 i+2_lon2 i+2_lat2 i+2_lon3 i+2_lat3 i+2_lon4 i+2_lat4 . . . i+5_lon1 i+5_lat1 i+5_lon2 i+5_lat2 i+5_lon3 i+5_lat3 i+5_lon4 i+5_lat4 I'd like to help, but I have no idea what you have listed above is supposed to be. Can you try to give a better example/explanation of what you want? Are you trying to make one big matrix? 5 different matrices? -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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] Deleting many columns of a data frame with the same name in a row
This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A but it worked - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460091.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] character variables in substitute()
This is an interesting approach but it doesn't quite get what I want. The sample below is what I'm looking for where Eik's Title is compared with the text line within the plot. My text line was produced by the explicit text(0.6,1.2,expression(R *'= 13.35, ' *P[m] *'= 2.531, ' *alpha[r] *'= 0.007187'),pos=4) and the title was produced by expr-bquote(italic(.(pNames[1]))==.(vparams[1])~,~.((pNames[2]))==.(vparams[2])) plot(1,1,main=expr) in which pNames is a vector of character c(R,P[m],...) and vparams is a vector of characters representing the values with the # of digits limited to 4. What I'm struggling with is to get the pNames recognized as objects similar to those in the expression above, but using variables rather than explicitly. In addition, is is to all go into a function to build the expression for varying numbers of elements (e.g. maybe just R = or P[m] =, both, or even more. The return from this function would be the expr that goes into a text(... command or,as Eik used, a main= plot parameter. http://n4.nabble.com/file/n1460088/help.png -- View this message in context: http://n4.nabble.com/character-variables-in-substitute-tp1459566p1460088.html Sent from the R help mailing list archive at Nabble.com. [[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 function unexpected behavior
Thanks, that helps! Subset creates a new context where a name clash can
occur. So if I don't want to check for that possibility, I should use a
special kind of index like .sch, or avoid subset:
for(sch in school.list){
print(sch)
print(input.data[input.data[,school.var] == sch,])}
which works no matter what variable names I use. That seems like a
reasonable requirement for good code.
(Checking for a name clash would be at least theoretically needed since
school.var is a parameter that can be any character name.)
Although subset conveniently avoids extra typing in many cases (not here),
this suggests to me that it's not ideal for code that can be used in a
variety of contexts. Note that unlike attach, subset does not issue a
warning!
-
Hi:
Try this for your second loop instead:
for(s in school.list){
print(s)
print(subset(input.data, sch == s))
}
[1] 1
sch pop
1 1 100
2 1 200
[1] 2
sch pop
3 2 300
4 2 400
Don't confound the 'sch' variable in your data frame with the
index in your loop :)
HTH,
Dennis
On Mon, Feb 1, 2010 at 8:17 PM, David Katz [hidden email]wrote:
- Hide quoted text -
I was surprised to see this unexpected behavior of subset in a for loop. I
looked in subset.data.frame and it seemed to me that both versions should
work, since the subset call should be evaluated in the global environment
-
but perhaps I don't understand environments well enough. Can someone
enlighten me? In any case, this is a bit of a gotcha for naive users of
subset.
input.data -
data.frame(sch=c(1,1,2,2),
pop=c(100,200,300,400))
school.var - sch
school.list - 1:2
for(sch in school.list){
print(sch)
#do this before subset!:
right.sch.p -
input.data[,school.var] == sch
print( subset(input.data,right.sch.p)) #this is what I expected
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## [1] 2
## sch pop
## 3 2 300
## 4 2 400
for(sch in school.list){
print(sch)
print(subset(input.data,input.data[,school.var] == sch)) #note - compact
version fails!
}
## [1] 1
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
## [1] 2
## sch pop
## 1 1 100
## 2 1 200
## 3 2 300
## 4 2 400
--
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Re: [R] Plot - specification for grid of x axis
Have a look at ?par and check out the 'xaxp' parameter par(mfrow=c(1,2)) plot(x[,2] ~ x[,1], xaxp=c(0,100,5)) plot(x[,2] ~ x[,1], xaxp=c(0,100,10), cex.axis=0.5) -- View this message in context: http://n4.nabble.com/Plot-specification-for-grid-of-x-axis-tp1460005p1460083.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] Deleting many columns of a data frame with the same name in a row
Try: newdf - datas[names(datas) != A] On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A but it worked - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460091.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] character variables in substitute()
Try this: leg - function(pNames, vparams) parse(text = paste(pNames, vparams, sep = ==)) plot(0, main = leg(Sigma, 3.2)) legend(topleft, legend = leg(c(alpha[beta], delta), 1:2)) On Tue, Feb 2, 2010 at 11:44 AM, dkStevens david.stev...@usu.edu wrote: This is an interesting approach but it doesn't quite get what I want. The sample below is what I'm looking for where Eik's Title is compared with the text line within the plot. My text line was produced by the explicit text(0.6,1.2,expression(R *'= 13.35, ' *P[m] *'= 2.531, ' *alpha[r] *'= 0.007187'),pos=4) and the title was produced by expr-bquote(italic(.(pNames[1]))==.(vparams[1])~,~.((pNames[2]))==.(vparams[2])) plot(1,1,main=expr) in which pNames is a vector of character c(R,P[m],...) and vparams is a vector of characters representing the values with the # of digits limited to 4. What I'm struggling with is to get the pNames recognized as objects similar to those in the expression above, but using variables rather than explicitly. In addition, is is to all go into a function to build the expression for varying numbers of elements (e.g. maybe just R = or P[m] =, both, or even more. The return from this function would be the expr that goes into a text(... command or,as Eik used, a main= plot parameter. http://n4.nabble.com/file/n1460088/help.png -- View this message in context: http://n4.nabble.com/character-variables-in-substitute-tp1459566p1460088.html Sent from the R help mailing list archive at Nabble.com. [[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] Error with toString
Romain, to keep it simple, I reproduced the example that you can find in the toString function help and I still got the same error: x - c(a, b, aaa) toString(x) Error in toString(x) : could not find function .jcall - Anna Lippel -- View this message in context: http://n4.nabble.com/Error-with-toString-tp1290327p1460119.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] barplot y axis too short
Thanks for the replies.
Apparently you cannot adjust the extension factor used by yaxs (which is set
at 4%), so what I did is
wrote a function (barplotCovered) to get y-axis limits based on the range of
the data and a user-defined axis expansion factor (axExFact).
I included an example below in case someone else has this problem.
# example data
data - c(.10, -.15, .52, .11)
# calculate a maximum and minimum for the y-scale:
# takes the biggest datum: max(data)
# then makes a buffer around it: max(data)*1.1
# then rounds that inflated number to one significant digit:
signif(max(data)*1.1, digits=1)
# then takes the max of that or 0 to allow for positive and negative numbers
axExFact -1.1
ymax - max(c(0, signif(max(data)*axExFact, digits=1)))
ymin - min(c(0, signif(min(data)*axExFact, digits=1)))
# here is the plot
barplot(data, ylim=c(ymin, ymax))
# another example
data2 - c(10, -15, 52, 11)
ymax2 - max(c(0, signif(max(data2)*1.1, digits=1)))
ymin2 - min(c(0, signif(min(data2)*1.1, digits=1)))
barplot(data2, ylim=c(ymin2, ymax2))
# here it is as a function
barplotCovered - function(data, axExFact, ...){
ymax - max(c(0, signif(max(data)*axExFact, digits=1)))
ymin - min(c(0, signif(min(data)*axExFact, digits=1)))
barplot(data, ylim=c(ymin, ymax), ...)
}
# example using the function
barplotCovered(data, axExFact=1.1, names.arg=c(a, b, c, d))
-
Jack Siegrist
Graduate Program in Ecology Evolution, and Department of Ecology,
Evolution, Natural Resources,
Rutgers University, New Brunswick, NJ 08901
Jack Siegrist wrote:
The function barplot automatically creates a y-axis that doesn't
necessarily cover the range of y-values to be plotted. I know how to
manually create my own y-axis so that it does cover the range, but I was
wondering if there is some parameter to change so that the scale of the
y-axis is automatically taller than the tallest bar.
I thought setting xpd=F would do it, since it says that xpd determines
whether bars will be plotted outside of the plotting region, but it had no
effect, so I guess it must be dealing with something different.
In the example below, the scale goes to 15 but the second bar goes to 16.
In this case I would like the scale to go to 20.
Thanks
#example data
data - c(12, 16)
#none of the following are any different
barplot(data)
barplot(data, xpd=T)
barplot(data, xpd=F)
--
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Re: [R] population variance and sample variance
On Mon, Oct 19, 2009 at 12:53 PM, Kingsford Jones kingsfordjo...@gmail.com wrote: sum((x-mean(x))^2)/(n) [1] 0.4894708 ((n-1)/n) * var(x) [1] 0.4894708 But this is not a built-in function in R to do so, right? hth, Kingsford On Mon, Oct 19, 2009 at 9:30 AM, Peng Yu pengyu...@gmail.com wrote: It seems that var() computes sample variance. It is straight forward to compute population variance from sample variance. However, I feel that it is still convenient to have a function that can compute population variance. Is there a population variance function available in R? $ Rscript var.R set.seed(0) n = 4 x = rnorm(n) var(x) [1] 0.6526278 sum((x-mean(x))^2)/(n-1) [1] 0.6526278 __ 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] Deleting many columns of a data frame with the same name in a row
Anna - You could also look at the problem from the other direction: data[,names(datas) != 'A'] - Phil Spector Statistical Computing Facility Department of Statistics UC Berkeley spec...@stat.berkeley.edu On Tue, 2 Feb 2010, anna wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A but it worked - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460091.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] Error with toString
What packages are loaded? On the basic system, it works fine:
toString(123)
[1] 123
toString(c('a', 'b', 'zz'))
[1] a, b, zz
On Tue, Feb 2, 2010 at 12:17 PM, anna lippelann...@hotmail.com wrote:
Romain, to keep it simple, I reproduced the example that you can find in the
toString function help and I still got the same error:
x - c(a, b, aaa)
toString(x)
Error in toString(x) : could not find function .jcall
-
Anna Lippel
--
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] Deleting many columns of a data frame with the same name in a row
Here is one way with an example: datas=data.frame(x=1:3,A=1:3,A=1:3) names(datas)=c(x,A,A) datas datas=datas[,names(datas)!=A,drop=FALSE] datas On 2/2/2010 8:35 AM, anna wrote: Hi, I have a data frame datas with half of the columns with the same name A. I want to delete all those columns from the data frame so here is what I did: datas$A- NULL The problem is that it deleted only one column, I would have to do it as many times as there are A columns. Is there a way to do it in one time? thank you - Anna Lippel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Deleting many columns of a data frame with the same name in a row
datas[ , A != colnames(datas)] Uwe Ligges On 02.02.2010 17:35, anna wrote: Hi, I have a data frame datas with half of the columns with the same name A. I want to delete all those columns from the data frame so here is what I did: datas$A- NULL The problem is that it deleted only one column, I would have to do it as many times as there are A columns. Is there a way to do it in one time? thank you - Anna Lippel __ 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 function unexpected behavior
On Tue, 2 Feb 2010, David Katz wrote:
Thanks, that helps! Subset creates a new context where a name clash can
occur. So if I don't want to check for that possibility, I should use a
special kind of index like .sch, or avoid subset:
for(sch in school.list){
print(sch)
print(input.data[input.data[,school.var] == sch,])}
which works no matter what variable names I use. That seems like a
reasonable requirement for good code.
(Checking for a name clash would be at least theoretically needed since
school.var is a parameter that can be any character name.)
Although subset conveniently avoids extra typing in many cases (not here),
this suggests to me that it's not ideal for code that can be used in a
variety of contexts. Note that unlike attach, subset does not issue a
warning!
attach() doesn't issue a warning for this situation, it warns when a variable
in the data set *fails to* mask one in the global environment. subset() can't
fail to mask variables further up the search path, so it doesn't ever warn.
-thomas
Thomas Lumley Assoc. Professor, Biostatistics
tlum...@u.washington.eduUniversity of Washington, Seattle
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[R] Doubt about cluster analysis
Dear R community, I'm a beginner with Cluster Analysis. I would like to know if there is a criterion to select the best set of clusters to do this analysis. Thanks in advance, [[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] Deleting many columns of a data frame with the same name in a row
Hi, On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A but it worked For what it's worth, you could also have done: clean - datas[,-which(names(datas)==A)] (note that indexing with a negative vector removes those rows/columns from your object (instead of picking them)). -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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] Error with toString
anna wrote: Romain, to keep it simple, I reproduced the example that you can find in the toString function help and I still got the same error: x - c(a, b, aaa) toString(x) Error in toString(x) : could not find function .jcall .jcall() is an rJava function. toString() is a base R function. I'm guessing that you're doing this as part of some code that involves rJava. In that case you will have to provide more details. The example you cite above works perfectly well for me. -Peter Ehlers - Anna Lippel -- Peter Ehlers University of Calgary __ 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] using rbind in a for loop
This might give you a hint as to how to do it since I have no idea of
what your data looks like:
lon - lat - matrix(1:100, nrow=5)
output - file('/tempxx.txt', open='w')
for (i in 1:5){
x - cbind(lon[,i], lat[,i])
write.table(x, row.names=FALSE, col.names=FALSE, file=output)
cat('\n', file=output)
}
close(output)
This gives a file like:
1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 14
15 15
16 16
17 17
18 18
19 19
20 20
21 21
22 22
23 23
24 24
25 25
On Mon, Feb 1, 2010 at 11:16 PM, Eunjung Kim eunjung.haw...@gmail.com wrote:
*Dear R users,
*I'm facing a trivial problem but I cannot solve it.
My question is:
I want to make data set like this.
i_lon1 i_lat1
i_lon2 i_lat2
i_lon3 i_lat3
i_lon4 i_lat4
i+1_lon1 i+1_lat1
i+1_lon2 i+1_lat2
i+1_lon3 i+1_lat3
i+1_lon4 i+1_lat4
i+2_lon1 i+2_lat1
i+2_lon2 i+2_lat2
i+2_lon3 i+2_lat3
i+2_lon4 i+2_lat4
.
.
.
i+5_lon1 i+5_lat1
i+5_lon2 i+5_lat2
i+5_lon3 i+5_lat3
i+5_lon4 i+5_lat4
My code is
for (i in 1:5) {
j - which(!is.na(lat[i,]))
p - cbind(lon[i,j],lat[i,j])
cat(c(,p),sep= ,file=data.dat,append=T)}
and what I got is even not close to what I want: many lines of empty and
whole bunch of numbers without line change.
How can I solve this?
Thanks in advance,
EJ
--
Eunjung Kim
Department of Oceanography
University of Hawaii at Manoa
1000 Pope Road
Honolulu, HI 96822
United States
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+1 513 646 9390
What is the problem that you are trying to solve?
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[R] Doubt about cluster analysis... y más
Hola, ¿qué tal? No sé si la conoces, pero te invito a participar en la lista oficial de ayuda de R en español: https://stat.ethz.ch/mailman/listinfo/r-help-es A través de ella, además de atender dudas de usuarios, tratamos de crear una comunidad. De hecho, hace no mucho organizamos las primeras jornadas de usuarios de R, etc. Un cordial saludo, Carlos J. Gil Bellosta http://www.datanalytics.com Francisco Javier Santos Alamillos wrote: Dear R community, I'm a beginner with Cluster Analysis. I would like to know if there is a criterion to select the best set of clusters to do this analysis. Thanks in advance, [[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] Error with toString
Here is the list of the loaded packages: package:lattice package:fSeries [5] package:fCalendarpackage:fEcofin package:fUtilities package:MASS [9] package:robustbase package:caTools package:PerformanceAnalytics package:xts [13] package:xlsReadWrite package:RBloomberg package:RUnitpackage:bitops [17] package:zoo package:stats package:graphics package:grDevices [21] package:datasets package:rcom package:rscproxy package:utils [25] package:methods package:base - Anna Lippel -- View this message in context: http://n4.nabble.com/Error-with-toString-tp1290327p1460190.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] Error with toString
No I just did it on the R console, I built the vector just before. - Anna Lippel -- View this message in context: http://n4.nabble.com/Error-with-toString-tp1290327p1460192.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] Deleting many columns of a data frame with the same name in a row
Try this: subset(DF, select = setdiff(names(DF), A)) On Tue, Feb 2, 2010 at 2:47 PM, anna lippelann...@hotmail.com wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(names(datas)==A but it worked - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460091.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. -- Henrique Dallazuanna Curitiba-Paraná-Brasil 25° 25' 40 S 49° 16' 22 O __ 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] List of object properties
Dear all, I have a simple question: How can I retrieve a list with all properties of an object? Example: If I fit a regression with model - coxph(Surv()~...) I know that I can access the coefficients with model$coefficients[...] afterwards, but how do I know which other variables are available and what are their names? Thanks and all the best Philipp __ 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] Deleting many columns of a data frame with the same name ina row
-Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Steve Lianoglou Sent: Tuesday, February 02, 2010 9:05 AM To: anna Cc: r-help@r-project.org Subject: Re: [R] Deleting many columns of a data frame with the same name ina row Hi, On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]-list(rep(NULL,length(which(na mes(datas)==A but it worked For what it's worth, you could also have done: clean - datas[,-which(names(datas)==A)] (note that indexing with a negative vector removes those rows/columns from your object (instead of picking them)). I would recommend replacing datas[, -which(names(data)==A)] with datas[, names(data)!=A] because the former returns a zero-column data.frame if there are columns of datas named A. (which(rep(F,n))-integer(0) and -integer(0) is identical to integer(0), which selects no elements). The latter returns the entire data.frame in the case. Logical subscripts are your friend. Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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] Error with write.table
I was trying to save a data frame to an excel file using the following command: write.table(myData, file=myData.csv,sep=,, row.names=F) The command works for some data frames, but for other data frames, I get the following error: Error in if (inherits(X[[j]], data.frame) ncol(xj) 1L) X[[j]] - as.matrix(X[[j]]) : missing value where TRUE/FALSE needed Is there something I'm doing wrong in trying to save the data frame to a .csv file? Many thanks in advance! John Woodard -- View this message in context: http://n4.nabble.com/Error-with-write-table-tp1460198p1460198.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] Subset and point plot
OK, I need help plotting. I have column headings of Day, Wgt, Foodin, Rep, Grp and Tanks. Rep=c(1,2,3) and Tanks=c(a1,a2,a3,a4,a5,a6, c1,c2,c3,c4,c5,c6, h1,h2,h3,h4,h5,h6). I created a subset where I only would like Rep=2, and Tanks=c(a4,c4,h4) and would like to graph (points) of Wgt and Day. I would think that I only need 3 colors, but when I run with only 3, only 2 lines show up. When I add a 4th color (pink in this case), I get the 3rd line. If I subset for c(a1,a2,a3), it works using three colors. Should be simple, but I dont see it. rm(list=ls()) daily-subset(raw, subset=Foodin1 Rep==2 Grp==fed Tanks==a4| Foodin1 Rep==2 Grp==fed Tanks==c4|Foodin1 Rep==2 Grp==fed Tanks==h4) attach(daily) x11() par(cex=1.4) plot(Day, Wgt col=c(red,blue,purple,pink)[Tanks]) detach(daily) -- M. Keith Cox, Ph.D. Alaska NOAA Fisheries, National Marine Fisheries Service Auke Bay Laboratories 17109 Pt. Lena Loop Rd. Juneau, AK 99801 keith@noaa.gov marlink...@gmail.com U.S. (907) 789-6603 [[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] Retrieve distinct values within a whole data frame
Hello everyone, I am trying to retrieve the list of distinct values within a whole data frame. I tried to use unique() function but it retrieves the distinct values within each column or row, I want it for the entire data frame, any idea? - Anna Lippel -- View this message in context: http://n4.nabble.com/Retrieve-distinct-values-within-a-whole-data-frame-tp1460205p1460205.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] Deleting many columns of a data frame with the same name in a row
thanks this is actually shorter :) - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460208.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] Deleting many columns of a data frame with the same name in a row
yes it looks really simpler, thank you! - Anna Lippel -- View this message in context: http://n4.nabble.com/Deleting-many-columns-of-a-data-frame-with-the-same-name-in-a-row-tp1460078p1460213.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] Retrieve distinct values within a whole data frame
Hi, On Tue, Feb 2, 2010 at 1:19 PM, anna lippelann...@hotmail.com wrote: Hello everyone, I am trying to retrieve the list of distinct values within a whole data frame. I tried to use unique() function but it retrieves the distinct values within each column or row, I want it for the entire data frame, any idea? Here's one way: R df - data.frame(a=c(1,2,3,4,3,2), b=c(4,5,6,6,4,3)) R unique(unlist(df)) [1] 1 2 3 4 5 6 I guess funny things might happen when the columns of your data.frame are of different types, though ... -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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] Error with write.table
How about a MINIMAL example of a data.frame that produces the error message you see? That will help track down what's going on... jlwoodard wrote: I was trying to save a data frame to an excel file using the following command: write.table(myData, file=myData.csv,sep=,, row.names=F) The command works for some data frames, but for other data frames, I get the following error: Error in if (inherits(X[[j]], data.frame) ncol(xj) 1L) X[[j]] - as.matrix(X[[j]]) : missing value where TRUE/FALSE needed Is there something I'm doing wrong in trying to save the data frame to a .csv file? Many thanks in advance! John Woodard __ 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] Retrieve distinct values within a whole data frame
ok it worked so the key is unlist() from what I see :working: - Anna Lippel -- View this message in context: http://n4.nabble.com/Retrieve-distinct-values-within-a-whole-data-frame-tp1460205p1460217.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] create an object in a loop (v3)
On Tue, Feb 2, 2010 at 11:44, Ivan Calandra
ivan.calan...@uni-hamburg.de wrote:
Hi David,
Thanks for your answer.
But I don't really see how I can extend it with my real data.
The thing is that I have more than 3 names and 1 value for each name.
Moreover, each is different from one run to another. That is why I was
trying with a modification of names(). Also to be noted is that I simplified
the name in assign(); I actually have 2 other variables that will be
pasted to create the name.
Here is my code (I kept only what is important for that part):
library(WRS)
seq.num - seq(7,10,1) #column (variable) indexes to be used as
numerical variables
# fac2list() separates the data from file[,k] into groups from levels in
file[3] and store into list mode.
for(i in 1:length(seq.num)) {
k - seq.num[i]
name.num - names(file)[k]
assign(paste(names(file)[3], name.num, sep=_), fac2list(file[,k],
file[3]))
names(paste(names(file)[3], name.num, sep=_)) -
levels(factor(file[[3]])) #that line doesn't work, but I would like
something in this direction
}
Sounds like a job for 'get'. Try this (untested):
names(get(paste(names(file)[3], name.num, sep=_)))
Good luck
Thanks in advance for your help.
Regards,
Ivan
Le 2/1/2010 18:47, David Winsemius a écrit :
On Feb 1, 2010, at 12:33 PM, Ivan Calandra wrote:
I have a follow-up question:
I use assign() to store some value in my paste()-created object as
suggested:
for (i in 1:3) {
assign(paste(object, i, sep=), c(a, b, c))
}
Then I would like to change the names of the elements of that object
within the loop. Since it is all in a loop, I cannot give the name of the
object manually by doing something like: names(object1) - c(tooth,
bone, species).
The only thing I can give to names() is paste(object, i, sep=), which
doesn't work.
Any idea of how to do it?
for (i in paste(object, 1:3, sep=)) {
+ assign(i, c(tooth=a, bone=b, species=c) )
+ }
object1
tooth bone species
a b c
Thanks in advance
Ivan
Le 2/1/2010 17:14, David Winsemius a écrit :
Upon reading it yesterday, it appeared as it would have required some
serious testing and there was no data on which to do any work. You were
clearly not taking the time to isolate the problem and construct a dataset.
But who knows? When you say What I want to do is. ... ,I would like the
name of the list to be created in the loop too, maybe all you needed was
to
be pointed to was:
?assign
But if that were the case, then you lost most of your audience along the
way with a bunch of unneeded and obscure code.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
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Re: [R] Error with toString
Anna, Try omitting pkg:RBloomberg. If you really need to use that package, you will have to install the non-CRAN package RDCOMClient from omegahat. I still don't see why toString() wouldn't do its job, even with RBloomberg loaded. -Peter Ehlers anna wrote: Here is the list of the loaded packages: package:lattice package:fSeries [5] package:fCalendarpackage:fEcofin package:fUtilities package:MASS [9] package:robustbase package:caTools package:PerformanceAnalytics package:xts [13] package:xlsReadWrite package:RBloomberg package:RUnitpackage:bitops [17] package:zoo package:stats package:graphics package:grDevices [21] package:datasets package:rcom package:rscproxy package:utils [25] package:methods package:base - Anna Lippel -- Peter Ehlers University of Calgary __ 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] character variables in substitute()
Hi:
Taking David's and Gabor's contributions, I came up with a function that I
hope
satisfies David's needs:
parText - function(names, pars) {
p1 - paste(* '= , pars[-length(pars)], , '*, sep = )
p2 - paste(* '= , pars[length(pars)], ')
p - c(p1, p2)
txt - paste(names, p, collapse = )
parse(text = txt)
}
Tests:
names - c('R', 'P', 'K', 'alpha')
vals - c(13.35, 2.53, 4.06, 0.0072)
# Different length inputs
vars - c('beta', 'K[m]')
pars - c(12, 0.6)
plot(c(0, 2), c(0, 2))
text(0, 1, parText(names, vals), adj = 0)
text(0, 0.5, parText(vars, pars))
title(parText(vars, pars))
Does that work?
Dennis
On Tue, Feb 2, 2010 at 8:44 AM, dkStevens david.stev...@usu.edu wrote:
This is an interesting approach but it doesn't quite get what I want. The
sample below is what I'm looking for where Eik's Title is compared with the
text line within the plot. My text line was produced by the explicit
text(0.6,1.2,expression(R *'= 13.35, ' *P[m] *'= 2.531, ' *alpha[r] *'=
0.007187'),pos=4)
and the title was produced by
expr-bquote(italic(.(pNames[1]))==.(vparams[1])~,~.((pNames[2]))==.(vparams[2]))
plot(1,1,main=expr)
in which pNames is a vector of character c(R,P[m],...) and vparams is a
vector of characters representing the values with the # of digits limited
to
4. What I'm struggling with is to get the pNames recognized as objects
similar to those in the expression above, but using variables rather than
explicitly.
In addition, is is to all go into a function to build the expression for
varying numbers of elements (e.g. maybe just R = or P[m] =, both, or even
more. The return from this function would be the expr that goes into a
text(... command or,as Eik used, a main= plot parameter.
http://n4.nabble.com/file/n1460088/help.png
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Re: [R] merging data frames gives all NAs
On 2/1/2010 5:51 PM, David Winsemius wrote:
I figured this out finally. I really believe that the R help write-ups
are sorely lacking. As soon as I looked at
http://www.statmethods.net/management/merging.html, it was obvious:
Adding Columns
To merge two dataframes (datasets) horizontally, use the *merge*
function. In most cases, you join two dataframes by one or more common
key variables (i.e., an inner join).
|# merge two dataframes by ID
total - merge(dataframeA,dataframeB,by=ID)|
|# merge two dataframes by ID and Country
total - merge(dataframeA,dataframeB,by=c(ID,Country)) |
Adding Rows
To join two dataframes (datasets) vertically, use the* rbind* function.
The two dataframes *must* have the same variables, but they do not have
to be in the same order.
|total - rbind(dataframeA, dataframeB) |
I needed to add rows, and had to use rbind. If the help for merge said
To merge two dataframes (datasets) horizontally I would have known
right away that it was the wrong function to use.
Thanks for the help,
Jim Rome
On Feb 1, 2010, at 5:30 PM, David Winsemius wrote:
On Feb 1, 2010, at 5:16 PM, James Rome wrote:
Dear kind R helpers,
I have a vector of runway names in rwy (31R, 31L,... the number
is user selectable)
arrgnd is a data frame with data for all flights and all runways,
with a Runway column.
I am trying to subset arrgnd into a dat frame for each selected
runway, and then combine them back together using the following code:
for (j in 1:nr) {# nr = number of user-selected runways
Safer would be:
for (j in seq_along(rwy) {
ar4rw = arrgnd[arrgnd$Runway==rwy[j],]
Clearer would be :
ar4rw - subset(arrgnd, Runway= j) # and I think the NA line's
will also disappear.
^ == ^
if (j == 1) {
arrw = ar4rw
}
else {
arrw = merge(arrw, ar4rw)
}
}
You really should give us something like:
dput(rwy)
dput( head(arrgnd, 10) )
but, the merge step gives me a data frame with all NAs. In addition,
ar4rw always gets a row with NAs at the start, which I do not
understand. There are no rows with all NAs in the arrgnd data frame.
ar4rw[1:2,] # first time through for 31R
DateTime Date month hour minute quarter weekday IATA ICAO
Flight
NA NA NANA NA NA NA NA NA NA NA
529 1/1/09 21:46 2009-01-01 1 21 46 87 5 TA
TAI TAI570
AircraftType Tail Arrived STA Runway FromTo Delay
NA NA NA NA NA NA NANA
529 A320 N496TA 21:46:58 22:3031R MSLP /KJFK 0
Operatordq gw
NA NA NA NA
529 TACA INTERNATIONAL AIRLINES 2009-01-01 87 1
ar4rw[1:2,] # second time through for 31L
DateTime Date month hour minute quarter weekday IATA ICAO
Flight
NA NA NANA NA NA NA NA NA NA NA
552 1/1/09 23:03 2009-01-01 1 23 3 92 5 AA
AAL AAL22
AircraftType Tail Arrived STA RunwayFromTo Delay
Operator
NA NA NA NA NA NA NANA NA
552 B762 N329AA 23:03:35 23:1031L LAX /JFK 0
AMERICAN AIRLINES
dq gw
NA NA NA
But after the merge, I get all NAs. What am I doing wrong?
The data layout gets mangled and I cannot tell what rows are being
matched to what. Use dput to convey an unambiguous, and easily
replicated example.
Thanks,
Jim Rome
552 2009-01-01 92 1
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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.
David Winsemius, MD
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PLEASE do read the posting guide
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Heritage Laboratories
West Hartford, CT
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Re: [R] List of object properties
On Feb 2, 2010, at 1:08 PM, Philipp Rappold wrote: Dear all, I have a simple question: How can I retrieve a list with all properties of an object? Example: If I fit a regression with model - coxph(Surv()~...) I know that I can access the coefficients with model $coefficients[...] afterwards, but how do I know which other variables are available and what are their names? ?str Thanks and all the best Philipp -- David Winsemius, MD Heritage Laboratories 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] Deleting many columns of a data frame with the same name in a row
On Feb 2, 2010, at 12:04 PM, Steve Lianoglou wrote: Hi, On Tue, Feb 2, 2010 at 11:47 AM, anna lippelann...@hotmail.com wrote: This is what I just found now but I guess there is a simpler way: datas[which(names(datas)==A)]- list(rep(NULL,length(which(names(datas)==A but it worked For what it's worth, you could also have done: clean - datas[,-which(names(datas)==A)] (note that indexing with a negative vector removes those rows/columns from your object (instead of picking them)). But only with numeric vectors. -steve -- Steve Lianoglou Graduate Student: Computational Systems Biology | Memorial Sloan-Kettering Cancer Center | Weill Medical College of Cornell University Contact Info: http://cbio.mskcc.org/~lianos/contact __ 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. David Winsemius, MD Heritage Laboratories 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] Subset and point plot
I tried your code and a funny thing happened: I got an error object 'Foodin' not found Hmmm; could you try to post something _reproducible_ (and minimal, please)? But here are some suggestions: 1. investigate '%in%' 2. put a comma after Wgt in your plot call 3. see what you get from c(red,blue,purple,pink)[a4] 4. assuming that 'Tanks' is a factor, see what you get from c(red,blue,purple,pink)[Tanks] -Peter Ehlers Marlin Keith Cox wrote: OK, I need help plotting. I have column headings of Day, Wgt, Foodin, Rep, Grp and Tanks. Rep=c(1,2,3) and Tanks=c(a1,a2,a3,a4,a5,a6, c1,c2,c3,c4,c5,c6, h1,h2,h3,h4,h5,h6). I created a subset where I only would like Rep=2, and Tanks=c(a4,c4,h4) and would like to graph (points) of Wgt and Day. I would think that I only need 3 colors, but when I run with only 3, only 2 lines show up. When I add a 4th color (pink in this case), I get the 3rd line. If I subset for c(a1,a2,a3), it works using three colors. Should be simple, but I dont see it. rm(list=ls()) daily-subset(raw, subset=Foodin1 Rep==2 Grp==fed Tanks==a4| Foodin1 Rep==2 Grp==fed Tanks==c4|Foodin1 Rep==2 Grp==fed Tanks==h4) attach(daily) x11() par(cex=1.4) plot(Day, Wgt col=c(red,blue,purple,pink)[Tanks]) detach(daily) -- Peter Ehlers University of Calgary __ 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 frames gives all NAs
James Rome wrote: On 2/1/2010 5:51 PM, David Winsemius wrote: I figured this out finally. I really believe that the R help write-ups are sorely lacking. The help docs are probably not the best way to learn R, but they are great for users of the functions. I have found that after going through an introduction book on R or online tutorial (plus experience), that the help system in R is really, really good at *documenting the behavior of the functions*, which is the point of them. As another general hint from someone who has learned R slowly over time, when something happens that you don't understand on real data, construct a minimal example data.frame and try out your code on that. Also, learning how to use browser() or the debug package has been very useful. __ 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] List of object properties
Philipp, Check ?str which displays the structure of R objects. And do use extractor functions when available: coef(yourModel) instead of yourModel$coef -Peter Ehlers Philipp Rappold wrote: Dear all, I have a simple question: How can I retrieve a list with all properties of an object? Example: If I fit a regression with model - coxph(Surv()~...) I know that I can access the coefficients with model$coefficients[...] afterwards, but how do I know which other variables are available and what are their names? Thanks and all the best Philipp __ 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 Ehlers University of Calgary __ 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] population variance and sample variance
Probably a simple typo, but just to keep things straight: you want to divide by n when describing the standard deviation of a sample, and divide by n-1 when estimating a population standard deviation (your initial description had it backwards I think). On Tue, Feb 2, 2010 at 5:25 PM, Peng Yu pengyu...@gmail.com wrote: On Mon, Oct 19, 2009 at 12:53 PM, Kingsford Jones kingsfordjo...@gmail.com wrote: sum((x-mean(x))^2)/(n) [1] 0.4894708 ((n-1)/n) * var(x) [1] 0.4894708 But this is not a built-in function in R to do so, right? hth, Kingsford On Mon, Oct 19, 2009 at 9:30 AM, Peng Yu pengyu...@gmail.com wrote: It seems that var() computes sample variance. It is straight forward to compute population variance from sample variance. However, I feel that it is still convenient to have a function that can compute population variance. Is there a population variance function available in R? $ Rscript var.R set.seed(0) n = 4 x = rnorm(n) var(x) [1] 0.6526278 sum((x-mean(x))^2)/(n-1) [1] 0.6526278 __ 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. -- Ista Zahn Graduate student University of Rochester Department of Clinical and Social Psychology http://yourpsyche.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] factorial map?
hello! i'm dealing with the following: i've collected a factor covariable at irregularly placed sampling points along a line with spatial informations, i.e.: dataset-c(x-coordinates, y-coordinates, level-of-factor) the factor describes the density of vegetation between 0 (no ground cover) and 5 (almost complete cover). id like to produce a map similar to the ones the akima package enables me to (function in akima is called: interp(x,y,z-value...), then use image(...) on the resulting data). but: as the akima library works suitable with continuous z-values (topography, air pressure, whatever is measured can be described a number), my resulting map looks terribly patchy (of course, since the levels of the factor cannot be interpolated) and contains no valuable information. thus my question: i imagine something like a growing-cells mechanism, which lets every sample grow until its borders touch the borders of an adjacent sample, resulting in something like a honeycomb-structure. that would give me a cellular pattern of my factor spread across the map! __ 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 frames gives all NAs
I agree. I have a foot of books on R now, for example the R Book by Michael Crowly. But so far, Googling the archives of this list has been the most help. Nonetheless, if I cannot understand the documentation of a function, then the documentation needs to be updated. For example, there needs to be a Returns section at the top of every function, so one can see what type of thing the function returns. Merge() needs to start with To merge two dataframes (datasets) horizontally, use the *merge* function. rather than Merge two data frames by common columns or row names, or do other versions of database /join/ operations which does not at all say that it does a horizontal merge if one does not know SQL. I do know SQL, and it is still not clear to me. And the/ merge/ documentation should then refer users to/ rbind/ for vertical merges. I hope that someone on the list can take actually change this file for the benefit of others. Thanks, Jim On 2/2/2010 2:00 PM, Erik Iverson wrote: James Rome wrote: On 2/1/2010 5:51 PM, David Winsemius wrote: I figured this out finally. I really believe that the R help write-ups are sorely lacking. The help docs are probably not the best way to learn R, but they are great for users of the functions. I have found that after going through an introduction book on R or online tutorial (plus experience), that the help system in R is really, really good at *documenting the behavior of the functions*, which is the point of them. As another general hint from someone who has learned R slowly over time, when something happens that you don't understand on real data, construct a minimal example data.frame and try out your code on that. Also, learning how to use browser() or the debug package has been very useful. [[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] merging data frames gives all NAs
Yeah, sometimes the vocabulary we bring to a task does not match up
(or merge properly) with the vocabulary that the developers use. In
this case the merge operation is one that has a precise meaning in
database lingo, which apparently you do not have background in. My
experience in trying to append objects ran into similar frustrations
early in my R endeavors. For the life of me, I could not find any
instances of append in the index of the references I was using.
I am glad that you found that material helpful, but I think its use of
the terms join or merge are incorrect in a database framework as
well, so I do not think it could be used as an unambiguous guide. Your
use of combine was likewise ambiguous. In composing questions to R-
help, it is advised that you post a small example and illustrate what
you want to see as a result.
--
David.
On Feb 2, 2010, at 1:47 PM, James Rome wrote:
On 2/1/2010 5:51 PM, David Winsemius wrote:
I figured this out finally. I really believe that the R help write-
ups are sorely lacking.
You should ponder whether you actually know enough to criticize the
help page when it describes the merge function as performing database
join operations. My guess is that you don't. The help page are not to
be designed to teach basic computer programming concepts.
As soon as I looked at http://www.statmethods.net/management/merging.html
, it was obvious:
Adding Columns
To merge two dataframes (datasets) horizontally, use the merge
function. In most cases, you join two dataframes by one or more
common key variables (i.e., an inner join).
# merge two dataframes by ID
total - merge(dataframeA,dataframeB,by=ID)
# merge two dataframes by ID and Country
total - merge(dataframeA,dataframeB,by=c(ID,Country))
Adding Rows
To join two dataframes (datasets) vertically, use the rbind
function. The two dataframes must have the same variables, but they
do not have to be in the same order.
total - rbind(dataframeA, dataframeB)
I needed to add rows, and had to use rbind. If the help for merge
said To merge two dataframes (datasets) horizontally I would have
known right away that it was the wrong function to use.
Thanks for the help,
Jim Rome
On Feb 1, 2010, at 5:30 PM, David Winsemius wrote:
On Feb 1, 2010, at 5:16 PM, James Rome wrote:
Dear kind R helpers,
I have a vector of runway names in rwy (31R, 31L,... the
number is user selectable)
arrgnd is a data frame with data for all flights and all runways,
with a Runway column.
I am trying to subset arrgnd into a dat frame for each selected
runway, and then combine them back together using the following
code:
for (j in 1:nr) {# nr = number of user-selected runways
Safer would be:
for (j in seq_along(rwy) {
ar4rw = arrgnd[arrgnd$Runway==rwy[j],]
Clearer would be :
ar4rw - subset(arrgnd, Runway= j) # and I think the NA
line's will also disappear.
^ == ^
if (j == 1) {
arrw = ar4rw
}
else {
arrw = merge(arrw, ar4rw)
}
}
You really should give us something like:
dput(rwy)
dput( head(arrgnd, 10) )
but, the merge step gives me a data frame with all NAs. In
addition, ar4rw always gets a row with NAs at the start, which I
do not understand. There are no rows with all NAs in the arrgnd
data frame.
ar4rw[1:2,] # first time through for 31R
DateTime Date month hour minute quarter weekday IATA
ICAO Flight
NA NA NANA NA NA NA NA NA NA NA
529 1/1/09 21:46 2009-01-01 1 21 46 87 5
TA TAI TAI570
AircraftType Tail Arrived STA Runway FromTo Delay
NA NA NA NA NA NA NANA
529 A320 N496TA 21:46:58 22:3031R MSLP /KJFK 0
Operatordq gw
NA NA NA NA
529 TACA INTERNATIONAL AIRLINES 2009-01-01 87 1
ar4rw[1:2,] # second time through for 31L
DateTime Date month hour minute quarter weekday IATA
ICAO Flight
NA NA NANA NA NA NA NA NA NA NA
552 1/1/09 23:03 2009-01-01 1 23 3 92 5
AA AAL AAL22
AircraftType Tail Arrived STA RunwayFromTo
Delay Operator
NA NA NA NA NA NA NANA NA
552 B762 N329AA 23:03:35 23:1031L LAX /JFK 0
AMERICAN AIRLINES
dq gw
NA NA NA
But after the merge, I get all NAs. What am I doing wrong?
The data layout gets mangled and I cannot tell what rows are being
matched to what. Use dput to convey an unambiguous, and easily
replicated example.
Thanks,
Jim Rome
552 2009-01-01 92 1
__
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.
David Winsemius, MD
Heritage Laboratories
West Hartford, CT
[R] mysterious extra spaces appearing in expression paste
Hi, i'm trying to put a legend on some figures and they're coming out a bit wonky. here's an example: a - c(1:10) par(mfrow=c(2,1)) plot(a,type=s,lwd=3) leg - c(expression(paste(data1 (,rho,=1))), expression(paste(data2 (,rho,=0.0 legend(bottomright,legend=leg,col=c(1,2),lwd=3) plot(a,type=s,lwd=3) leg - c(expression(paste(data1 \n(,rho,=1))), expression(paste(data2 \n(,rho,=0.0 legend(bottomright,legend=leg,col=c(1,2),lwd=3, y.intersp = 2.0,adj=c(0,1.5)) the problem is that lots of extra space appears between the open parenthesis and the rho. what can be done? (i'm running version 2.10.1 (2009-12-14) on fedora 12) Thanks! George Locke __ 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] StructTS standard errors
I just did a search on StructTS on the same topic and ended up here. Did this issue ever get resolved here or elsewhere? Regards, Myron -- View this message in context: http://n4.nabble.com/StructTS-standard-errors-tp966079p1460216.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] Import fixed-format ascii file with mixed record types
Thanks Jim, As I know nothing about Perl (is it really necessary?), I think that I will try with David's approach. What I want to do is to for a pseudo-panel with my survey data (that is collected periodically), and process it with the survey package. The resulting data set will be big (around 3.2 million records and 400 MBytes). Hug --- On Mon, 2/1/10, Jim Holtman jholt...@gmail.com wrote: From: Jim Holtman jholt...@gmail.com Subject: Re: [R] Import fixed-format ascii file with mixed record types To: trece por ciento el13porcie...@yahoo.com Cc: David Winsemius dwinsem...@comcast.net, r-help@r-project.org r-help@r-project.org Date: Monday, February 1, 2010, 4:43 PM you probably need to split your file into several files according to the format. you can use something like perl or do it within R by doing readline and then separating the lines. you can use grep, split and textConnection to do most of the work. What is the problem you are trying to solve? Sent from my iPhone. On Feb 1, 2010, at 14:33, trece por ciento el13porcie...@yahoo.com wrote: Thanks David, but can read.fwf cope with different record types? For example, if recordtype is the 4th character, I could have: 011125678 --- This is record Type 1 011136779 --- This is record Type 1 011124943 --- This is record Type 1 011286711 --- This is record Type 2 011234872 --- This is record Type 2 011135628 --- This is record Type 1 So, how can I tell read.fwf to take the correct type into account? Thanks again, Hug --- On Mon, 2/1/10, David Winsemius dwinsem...@comcast.net wrote: From: David Winsemius dwinsem...@comcast.net Subject: Re: [R] Import fixed-format ascii file with mixed record types To: trece por ciento el13porcie...@yahoo.com Cc: r-help@r-project.org Date: Monday, February 1, 2010, 12:01 PM On Feb 1, 2010, at 11:40 AM, trece por ciento wrote: I need to import several ascii files in fixed format with two different record types. The data comes from European Labor Force Surveys, wich is a household survey. The first record type is for people over 16 years, and the second much sorter is for people aged 15 or less (this record has a filler with several blanks to get the same record length). The files tipically have 16 records, with 176 characters per record, the data is numeric, corresponding to 102 variables, mostly integers (seven variables have two decimals). My opertating system is Windows XP. My questions: 1. Wich do you think is the best way to import the files into R? ?read.fwf 2. Could you give me any references or examples? There are examples in the help page. Thanking you in advance, Hug [[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. David Winsemius, MD Heritage Laboratories 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-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] Import fixed-format ascii file with mixed record types
Thanks again, David I think that this could work. Final questions: 1. I have read that read.fwt could be slow for big tables (my tables have aprox. 16 records, with 176 characters of recordlenght, almost 28MBytes). Could that be a problem? 2. If using read.fwt is not a problem, wouldn't be better to read all the records by read.fwt into a dataframe with the Type 1 structure, and then process the Type 2 records in the dataframe adding new fields for these records (NULL valued for Type 1)? Hug --- On Mon, 2/1/10, David Winsemius dwinsem...@comcast.net wrote: From: David Winsemius dwinsem...@comcast.net Subject: Re: [R] Import fixed-format ascii file with mixed record types To: trece por ciento el13porcie...@yahoo.com Cc: r-help@r-project.org Date: Monday, February 1, 2010, 2:23 PM On Feb 1, 2010, at 2:33 PM, trece por ciento wrote: Thanks David, but can read.fwf cope with different record types? For example, if recordtype is the 4th character, I could have: 011125678 --- This is record Type 1 011136779 --- This is record Type 1 011124943 --- This is record Type 1 011286711 --- This is record Type 2 011234872 --- This is record Type 2 011135628 --- This is record Type 1 So, how can I tell read.fwf to take the correct type into account? You may need to separate the line-types first. If the numbers of lines are not too large then this would exemplify a strategy: txt - 011125678 + 011136779 + 011124943 + 011286711 + 011234872 + 011135628 substr(readLines(textConnection(txt)), 4,4) [1] 1 1 1 2 2 1 file1 - readLines(textConnection(txt))[substr(readLines(textConnection(txt)), 4,4) == 1] file2 - readLines(textConnection(txt))[substr(readLines(textConnection(txt)), 4,4) == 2] file1 [1] 011125678 011136779 011124943 011135628 file2 [1] 011286711 011234872 Then these text objects could be processed with read.fwf(textConnection(file1) and the same for file2. --David. Thanks again, Hug --- On Mon, 2/1/10, David Winsemius dwinsem...@comcast.net wrote: From: David Winsemius dwinsem...@comcast.net Subject: Re: [R] Import fixed-format ascii file with mixed record types To: trece por ciento el13porcie...@yahoo.com Cc: r-help@r-project.org Date: Monday, February 1, 2010, 12:01 PM On Feb 1, 2010, at 11:40 AM, trece por ciento wrote: I need to import several ascii files in fixed format with two different record types. The data comes from European Labor Force Surveys, wich is a household survey. The first record type is for people over 16 years, and the second much sorter is for people aged 15 or less (this record has a filler with several blanks to get the same record length). The files tipically have 16 records, with 176 characters per record, the data is numeric, corresponding to 102 variables, mostly integers (seven variables have two decimals). My opertating system is Windows XP. My questions: 1. Wich do you think is the best way to import the files into R? ?read.fwf 2. Could you give me any references or examples? There are examples in the help page. Thanking you in advance, Hug [[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. David Winsemius, MD Heritage Laboratories West Hartford, CT David Winsemius, MD Heritage Laboratories 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] Error with toString
isn't there a way to specify from which library I am taking the method from...like rjava.toString or base.toString.. - Anna Lippel -- View this message in context: http://n4.nabble.com/Error-with-toString-tp1290327p1460262.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] mysterious extra spaces appearing in expression paste
try putting sep= in your paste method - Anna Lippel -- View this message in context: http://n4.nabble.com/mysterious-extra-spaces-appearing-in-expression-paste-tp1460261p1460264.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] mysterious extra spaces appearing in expression paste
On Feb 2, 2010, at 2:13 PM, George Locke wrote: Hi, i'm trying to put a legend on some figures and they're coming out a bit wonky. here's an example: a - c(1:10) par(mfrow=c(2,1)) plot(a,type=s,lwd=3) leg - c(expression(paste(data1 (,rho,=1))), expression(paste(data2 (,rho,=0.0 legend(bottomright,legend=leg,col=c(1,2),lwd=3) plot(a,type=s,lwd=3) leg - c(expression(paste(data1 \n(,rho,=1))), expression(paste(data2 \n(,rho,=0.0 legend(bottomright,legend=leg,col=c(1,2),lwd=3, y.intersp = 2.0,adj=c(0,1.5)) the problem is that lots of extra space appears between the open parenthesis and the rho. You need to use sep= what can be done? (i'm running version 2.10.1 (2009-12-14) on fedora 12) Thanks! George Locke __ 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. David Winsemius, MD Heritage Laboratories 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.
