Re: [R] Deparse substitute assign with list elements
Thanks, Bill, I should have googled more carefully:
http://stackoverflow.com/questions/9561053/assign-values-to-a-list-element-in-r
So, remove
assign(nm, tmp, parent.frame())
and add
txt - paste( nm, '-', tmp, sep='' )
eval( parse(text=txt), parent.frame() )
in `foo()` will do the trick.
Bests
Sören
On 15.05.2015, at 00:02, William Dunlap wdun...@tibco.com wrote:
You could use a 'replacement function' named 'bar-', whose last argument
is called 'value', and use bar(variable) - newValue where you currently
use foo(variable, newValue).
bar - function(x) {
x + 3
}
`bar-` - function(x, value) {
bar(value)
}
a - NA
bar(a) - 4
a
# [1] 7
b - list(NA, NA)
bar(b[[1]]) - 4
b
#[[1]]
#[1] 7
#
#[[2]]
#[1] NA
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 14, 2015 at 11:28 AM, soeren.vo...@posteo.ch wrote:
Hello,
When I use function `foo` with list elements (example 2), it defines a new
object named `b[[1]]`, which is not what I want. How can I change the
function code to show the desired behaviour for all data structures passed to
the function? Or is there a more appropriate way to sort of pass by
references in a function?
Thanks
Sören
src
bar - function(x) {
return( x + 3 )
}
foo - function(x, value) {
nm - deparse(substitute(x))
tmp - bar( value )
assign(nm, tmp, parent.frame())
}
# 1)
a - NA
foo(a, 4)
a # 7, fine :-)
# 2)
b - list(NA, NA)
foo(b[[1]], 4) # the first list item should be 7
b # wanted 7 but still list with two NAs :-(
/src
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[R] Deparse substitute assign with list elements
Hello,
When I use function `foo` with list elements (example 2), it defines a new
object named `b[[1]]`, which is not what I want. How can I change the function
code to show the desired behaviour for all data structures passed to the
function? Or is there a more appropriate way to sort of pass by references
in a function?
Thanks
Sören
src
bar - function(x) {
return( x + 3 )
}
foo - function(x, value) {
nm - deparse(substitute(x))
tmp - bar( value )
assign(nm, tmp, parent.frame())
}
# 1)
a - NA
foo(a, 4)
a # 7, fine :-)
# 2)
b - list(NA, NA)
foo(b[[1]], 4) # the first list item should be 7
b # wanted 7 but still list with two NAs :-(
/src
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[R] Get previous R digest volumes/issues
Hello, (how) can I download/re-retrieve/order previous R-** digest volumes/issues to my mailbox for local browsing? Thank you, Sören __ 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] Instances of a C++ class in R
On 28.04.2011, at 12:18, soeren.vo...@uzh.ch wrote: On 27.04.2011, at 11:59, soeren.vo...@uzh.ch wrote: We are working on a class in C++. The files compile fine (R CMD SHLIB ...) and run in R. A bzipped tar archive with source code can be downloaded from here: http://sovo.md-hh.com/files/GUTS3.tar.bz In R, dyn.load(GUTS.so) generates an instance of the GUTS class. (How) Is it possible to generate (within R) several instances of this class? (E.g., we would like to consider several experiments, each of which is represented by an instance of GUTS.) Hm, no comment so far. Is the question unclear? Did we miss something obvious? Is there other resources to post the issue to? Regards, Carlo/Sören For the files: the discussion continued here: https://stat.ethz.ch/pipermail/r-devel/2011-May/060922.html and is now here: http://lists.r-forge.r-project.org/pipermail/rcpp-devel/2011-May/002261.html Regards Sören -- soeren.vo...@uzh.ch, carlo.alb...@eawag.ch __ 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] Instances of a C++ class in R
On 27.04.2011, at 11:59, soeren.vo...@uzh.ch wrote: We are working on a class in C++. The files compile fine (R CMD SHLIB ...) and run in R. A bzipped tar archive with source code can be downloaded from here: http://sovo.md-hh.com/files/GUTS3.tar.bz In R, dyn.load(GUTS.so) generates an instance of the GUTS class. (How) Is it possible to generate (within R) several instances of this class? (E.g., we would like to consider several experiments, each of which is represented by an instance of GUTS.) Hm, no comment so far. Is the question unclear? Did we miss something obvious? Is there other resources to post the issue to? Regards, Carlo/Sören -- soeren.vo...@uzh.ch, carlo.alb...@eawag.ch __ 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] Instances of a C++ class in R
Hello We are working on a class in C++. The files compile fine (R CMD SHLIB ...) and run in R. A bzipped tar archive with source code can be downloaded from here: http://sovo.md-hh.com/files/GUTS3.tar.bz In R, dyn.load(GUTS.so) generates an instance of the GUTS class. (How) Is it possible to generate (within R) several instances of this class? (E.g., we would like to consider several experiments, each of which is represented by an instance of GUTS.) Thanks for help and advices, Sören -- soeren.vo...@uzh.ch, carlo.alb...@eawag.ch __ 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] Nagelkerke R square for Prediction data
Hello
I found some small postings dated to 22 Oct 2008 on the message subject.
Recently, I have been working with binary logistic regressions. I didn't use
the design package. Yet, I needed the fit indices. Therefore, I wrote a small
function to output the Nagelkerke's R, and the Cox--Snell R from a fitted
model. I am no professional programmer by far, yet, I hope, that the code is
okay and that it may be of some use to others -- or subject to useful
improvement. Please let me know, if you find errors.
Regards,
Sören
Rcsnagel - function(mod) {
llnull - mod$null.deviance
llmod - mod$deviance
n - length(mod$fitted.values)
Rcs - 1 - exp( (mod$deviance - mod$null.deviance) / n )
Rnagel - Rcs / (1 - exp(-llnull/n))
out - list('Rcs'=Rcs, 'Rnagel'=Rnagel)
class(out) - c(list, table)
return(out)
}
y - sample(c(T, F), 50, repl=T)
x - sample(1:7, 50, repl=T)
mod - glm(y ~ x, family=binomial(logit))
Rcsnagel(mod)
--
Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
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[R] Multinomial Analysis
I want to analyse data with an unordered, multi-level outcome variable, y. I am asking for the appropriate method (or R procedure) to use for this analysis. N - 500 set.seed(1234) data0 - data.frame(y = as.factor(sample(LETTERS[1:3], N, repl = T, + prob = c(10, 12, 14))), x1 = sample(1:7, N, repl = T, prob = c(8, + 8, 9, 15, 9, 9, 8)), x2 = sample(1:7, N, repl = T, prob = 7:1), + x3 = round(rnorm(N, 0, 1), 0) + 4) In [1], the use of mlogit is described (same package name). mlogit produces a nice output, but for some reason it does not accept the . operator in my formula. In addition, it requires me to transform my data frame before running the analysis: library(mlogit) data1 - mlogit.data(data0, varying = NULL, choice = y, shape = wide) summary(mod1 - mlogit(y ~ 1 | ., data = data1, reflevel = A)) Error in terms.formula(object) : '.' in formula and no 'data' argument form2 - paste(y, ~ 1 | , paste(names(data0)[-1], sep = , + collapse = + ), sep = , collapse = ) summary(mod2 - mlogit(as.formula(form2), data = data1, reflevel = A)) Call: mlogit(formula = as.formula(form2), data = data1, reflevel = A, method = nr, print.level = 0) Frequencies of alternatives: A B C 0.282 0.338 0.380 nr method 3 iterations, 0h:0m:0s g'(-H)^-1g = 8.8E-07 gradient close to zero Coefficients : Estimate Std. Error t-value Pr(|t|) altB 0.269645 0.569943 0.4731 0.6361 altC 0.642654 0.551250 1.1658 0.2437 altB:x1 0.018484 0.061061 0.3027 0.7621 altC:x1 0.075489 0.059837 1.2616 0.2071 altB:x2 -0.102349 0.068194 -1.5008 0.1334 altC:x2 -0.031934 0.065491 -0.4876 0.6258 altB:x3 0.034394 0.115189 0.2986 0.7653 altC:x3 -0.137739 0.112124 -1.2284 0.2193 Log-Likelihood: -542.05 McFadden R^2: 0.0065888 Likelihood ratio test : chisq = 7.1903 (p.value=0.30361) This is no big problem if the data frame holds only few variables, however, it can easily become confusing if the number variables in the data frame increases. So, I can use the function multinom (package nnet), as described in [2] and [3]: library(nnet) summary(mod3 - multinom(y ~ ., data = data0, na.action = na.omit, + Hess = T, model = T)) # weights: 15 (8 variable) initial value 549.306144 iter 10 value 542.075434 final value 542.046320 converged Call: multinom(formula = y ~ ., data = data0, na.action = na.omit, Hess = T, model = T) Coefficients: (Intercept) x1 x2 x3 B 0.2696466 0.01848389 -0.10234898 0.03439345 C 0.6426552 0.07548893 -0.03193395 -0.13773888 Std. Errors: (Intercept) x1 x2x3 B 0.5699428 0.06106070 0.06819403 0.1151887 C 0.5512500 0.05983652 0.06549051 0.1121242 Residual Deviance: 1084.093 AIC: 1100.093 The estimates produced by this function are the same like in the analysis above (mlogit), and in principle, I should be able to calculate the z and fit statistics by hand. However, this seem more long-winded than using the mlogit function. Alternatively, I may describe my results in the following way: Given that a person knows B and C (which is the case for all my cases), what drives some to use B, and what drives some to use C? Basically, I am wondering whether this analysis can be done using a set of two, parallel binary logistic regressions, one describing the step from A to B, and another one describing the step from A to C, while keeping predictors the same in both regressions. More generally, I am not sure whether each multinomial logistic model could not just as well be analysed using consecutive binary logistic regressions. This has two advantages in my opinion, one, the model statistics and fit measures are well-described in binary models compared to multinomial ones (see [1] at the bottom), and two, they are much easier to communicate. But I am wondering, analogously to multiple mean comparisons, whether I run into any traps, and I have no idea what these traps look like. I would kindly appreciate any advice. Thanks, Sören [1] http://www.ats.ucla.edu/stat/r/dae/mlogit.htm [2] Venables, W. N. and Ripley, B. D. (2002). Modern applied statistics with S (4th ed.). New York: Springer. [3] http://www.stat.washington.edu/quinn/classes/536/S/multinomexample.html __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Linear separation
In https://stat.ethz.ch/pipermail/r-help/2008-March/156868.html I found what linear separability means. But what can I do if I find such a situation in my data? Field (2005) suggest to reduce the number of predictors or increase the number of cases. But I am not sure whether I can, as an alternative, take the findings from my analysis and report them. And if so, how can I find the linear combination of the predictors that separates the zeros from the ones? Below a small example to illustrate the situation. set.seed(123) df - data.frame( 'y'=c(rep(FALSE, 6), rep(TRUE, 14)), 'x1'=c(sample(1:2, 6, repl=T), sample(3:5, 14, repl=T)), 'x2'=c(sample(4:7, 6, repl=T), sample(1:3, 14, repl=T)), 'x3'=round(rnorm(20, 4, 2), 0) ) df[17:18, c(2, 3)] - df[17:18, c(3, 2)] glm(y ~ ., data=df[, -3], family=binomial(logit)) glm(y ~ ., data=df, family=binomial(logit)) Thanks, Sören -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] slicing list with matrices
A list contains several matrices. Over all matrices (list elements) I'd like to access one matrix cell: m - matrix(1:9, nrow=3, dimnames=list(LETTERS[1:3], letters[1:3])) l - list(m1=m, m2=m*2, m3=m*3) l[[3]] # works l[[3]][1:2, ] # works l[[1:3]][1, 1] # does not work How can I slice all C-c combinations in the list? Sören -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM +41 76 233 3637, http://www.eawag.ch, http://sozmod.eawag.ch __ 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] S: appropriate significance tests
Hello (1) How can I compare two Pearson correlation coefficients for significant differences without the use of the raw data? (2) How can I compare two linear regression coefficients for significant differences without the use of the raw data? (3) How can I compare two multiple correlation coefficients (as produced in a linear regression) for significant differences, again, if the raw data should not be used? src type=perhaps useful da - data.frame( y1=sample(1:5, 40, repl=T), y2=sample(1:5, 40, repl=T), x1=sample(1:5, 40, repl=T), x2=sample(1:5, 40, repl=T) ) cro1 - cor(da$y1, da$x1) cro2 - cor(da$y2, da$x1) lmo1 - lm(y1 ~ x1 + x2, data=da) lmo2 - lm(y2 ~ x1 + x2, data=da) /src Thanks, Sören __ 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] S: appropriate significance tests
Thanks Greg for the additional remarks. Basically I have two questions, let me try to specify them as follows: (1) Height and intelligence may correlate at, say, X, but speed and finger length may correlate at Y. Despite any sense of such a statement, is X significantly larger than Y? How can I perform this significance test (of two correlation coefficients) if I don't have the original data, but only the correlation coefficients, the degrees of freedom and the standard errors? (2) Let's assume I've studied whether height (as independent) has an influence on (a, regression 1) intelligence and (b, regression 2) speed. Suppose I found out that my linear regression between intelligence and height, (1), reveals a standardised estimate of .6 for height, while for the regression between speed and height, (2), reveals an estimate of .65 for height. Is the estimate for height in regression 1 significantly higher than that for height in regression 2? I have asked a couple of persons how much they like music by Robbie Williams or Nora Jones. Additionally I recorded their hair colour and their attitude towards pop music (ATP). After my analysis, I'd like to state that for RW and NJ neither hair colour nor ATP is of *different* importance, whereas the hair colour is much more important for liking RW. I thus thought that I should compare two regression estimates for significant differences since saying that the one is lower/higher than the other doesn't satisfy my editor. On 20.10.2010, at 20:48, Greg Snow wrote: It is not completely clear what question you are trying to answer or what you are trying to accomplish. But here are some additional questions that may help: What tests would you use if you could use the original data? What assumptions are you willing to make about the data and/or statistics? What are you trying to accomplish? -- 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 soeren.vo...@eawag.ch Sent: Wednesday, October 20, 2010 7:33 AM To: r-help@r-project.org Subject: [R] S: appropriate significance tests Hello (1) How can I compare two Pearson correlation coefficients for significant differences without the use of the raw data? (2) How can I compare two linear regression coefficients for significant differences without the use of the raw data? (3) How can I compare two multiple correlation coefficients (as produced in a linear regression) for significant differences, again, if the raw data should not be used? src type=perhaps useful da - data.frame( y1=sample(1:5, 40, repl=T), y2=sample(1:5, 40, repl=T), x1=sample(1:5, 40, repl=T), x2=sample(1:5, 40, repl=T) ) cro1 - cor(da$y1, da$x1) cro2 - cor(da$y2, da$x1) lmo1 - lm(y1 ~ x1 + x2, data=da) lmo2 - lm(y2 ~ x1 + x2, data=da) /src Thanks, Sören __ 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] Sweave line breaks
Hello, when I print x in Sweave, the lines do not wrap. However, I want them to wrap (perhaps at a specified width). How? Thanks, *S* keep.source=TRUE= x - Lorem ipsum dolor sit amet, consectetur adipisicing elit, sed do eiusmod tempor incididunt ut labore et dolore magna aliqua. Ut enim ad minim veniam, quis nostrud exercitation ullamco laboris nisi ut aliquip ex ea commodo consequat. Duis aute irure dolor in reprehenderit in voluptate velit esse cillum dolore eu fugiat nulla pariatur. Excepteur sint occaecat cupidatat non proident, sunt in culpa qui officia deserunt mollit anim id est laborum. print(x) @ __ 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] ROpenOffice (which requires Rcompression)
Hello, for my data preparation and administration (data, labels, etc.) I use OpenOffice.org ODS spreadsheet files with several sheets in one file. However, I find it inconvenient to export every single sheet to a csv file whenever I apply changes to the labelling or so, and I haven't found a plugin or an application which does this batch in OOorg. Is there new development on a direct import package for ODS files? I've googled around one afternoon, found ROpenOffice (http://www.omegahat.org/ROpenOffice/) which requires Rcompression (http://www.mail-archive.com/r-help@r-project.org/msg55480.html), which in turn is quite outdated. Is the whole idea dead or did I miss something important when googling? *S* (R version 2.11.1 (2010-05-31), x86_64-apple-darwin9.8.0) -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] Default For list.len Argument (list output truncated)
Hello How can I set a default for the 'list.len' argument in 'str'? x - as.data.frame(matrix(1:1000, ncol=1000)) str(x) formals(str) - alist(object=, ...=, list.len=1000) args(str); formals(str) str(x) Does not display errors, but does not work either. Thanks for help Sören __ 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] S: Book for time series analysis with R
Hello Could you recommend a printed/electronic book that teaches time series analysis (using R) for students? I am searching for something similar to the MASS book, with a level lower but close, for TSA. Easy examples would be fine to understand deeper statistical procedures. Functions, results, and their practical interpretation should be explained next to the R code. Thanks for help and/or suggestions! Sören -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM +41 76 233 3637, http://www.eawag.ch, http://sozmod.eawag.ch __ 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] Adding NAs to data.frame
Hello, after the creation of a data.frame I like to add NAs as follows:
n - 743;
x - runif(n, 1, 7);
Y - runif(n, 1, 7);
Ag6 - runif(n, 1, 7);
df - data.frame(x, Y, Ag6);
# a list with positions:
v - apply(df, 2, function(x) sample(n, sample(1:ceiling(5*n/100), 1), repl=F));
# a loop too much?
for (i in 1:length(df)){
df[unlist(v[i]), i] - NA;
}
summary(df);
This works fine but it appears to me that there is a more elegant and simple
way -- which one?
Thanks
Sören
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[R] apply fun to df returning a matrix
Hello, a data.frame, df, holds the numerics, x, y, and z. A function,
fun, should return some arbitrary statistics about the arguments, e.g.
the sum or anything else. What I want to do is to apply this function
to every pair of variables in df, and the return should be a matrix as
found with cov. How can I achieve that? Thanks, Sören
df - data.frame(x=1:10, y=11:20, z=21:30);
fun - function(x){
return(sum(x));
}
# and now???
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Re: [R] apply fun to df returning a matrix
Hi Mohamed, thanks for your answer. Anyway, the how to is exactly my
problem, since ...
fun2 - function(x){
please_use_aggregate_and_apply_in_some_way_and_return_the_output_of_my_example_as_requested
(fun(x));
}
fun2(df);
... unfortunately returns an error ;-). Could you please give a simple
example? Thanks, Sören
On 30.04.2010, at 12:59, Mohamed Lajnef wrote:
Hi Soeren
Apply or aggregate functions
best regards
M
soeren.vo...@eawag.ch a écrit :
Hello, a data.frame, df, holds the numerics, x, y, and z. A
function, fun, should return some arbitrary statistics about the
arguments, e.g. the sum or anything else. What I want to do is to
apply this function to every pair of variables in df, and the
return should be a matrix as found with cov. How can I achieve
that? Thanks, Sören
df - data.frame(x=1:10, y=11:20, z=21:30);
fun - function(x){
return(sum(x));
}
# and now???
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Re: [R] apply fun to df returning a matrix
Thanks for the code, that was exactly what I was looking for. Regards,
Sören
On 30.04.2010, at 14:04, David Winsemius wrote:
On Apr 30, 2010, at 6:59 AM, Mohamed Lajnef wrote:
Hi Soeren
Apply or aggregate functions
Probably needs combn as well. Could do it all with numeric indices,
but this effort with character vectors seems acceptable:
fun - function(x){ cnms - colnames(x)
return(apply(combn(cnms,2), 2, function(y) sum(x[,y])))
}
fun(df)
#[1] 210 310 410
I do have a question about returning a matrix though. Did you mena
that you wanted the pairs of sums rather than the sums of pairs. In
that case:
fun2 - function(x){cnms - colnames(x)
return(apply(combn(cnms,2), c(1,2), function(y) sum(x[,y])))
}
fun2(df)
# [,1] [,2] [,3]
#[1,] 55 55 155
#[2,] 155 255 255
--
David.
best regards
M
soeren.vo...@eawag.ch a écrit :
Hello, a data.frame, df, holds the numerics, x, y, and z. A
function, fun, should return some arbitrary statistics about the
arguments, e.g. the sum or anything else. What I want to do is to
apply this function to every pair of variables in df, and the
return should be a matrix as found with cov. How can I achieve
that? Thanks, Sören
df - data.frame(x=1:10, y=11:20, z=21:30);
fun - function(x){
return(sum(x));
}
# and now???
--
Mohamed Lajnef,IE INSERM U955 eq 15
Pôle de Psychiatrie
Hôpital CHENEVIER
40, rue Mesly
94010 CRETEIL Cedex FRANCE
mohamed.laj...@inserm.fr
tel : 01 49 81 31 31 (poste 18470)
Sec : 01 49 81 32 90
fax : 01 49 81 30 99
David Winsemius, MD
West Hartford, CT
--
Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
__
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and provide commented, minimal, self-contained, reproducible code.
[R] Return a variable name
Hello,
how can I return the name of a variable, say a$b, from a function?
fun - function(x){
return(substitute(x));
}
a - data.frame(b=1:10);
fun(a$b)
... returns a$b, but this is a type language, thus I can't use it as a
character string, can I? How?
Thanks for help,
Sören
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and provide commented, minimal, self-contained, reproducible code.
[R] cbind, row names
Hello, I read the help as well as the examples, but I can not figure out why the following code does not produce the *given* row names, x and y: x - 1:20 y - 21:40 rbind( x=cbind(N=length(x), M=mean(x), SD=sd(x)), y=cbind(N=length(y), M=mean(y), SD=sd(y)) ) Could you please help? Thank you Sören __ 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] Exact test for count data
Hello!
Bortz, Lienert, Boehnke (2008, pp. 140--142) suggest an exact
polynomial test for low frequency tables. I used it recently, and
thus, created the code attached. Maybe someone would use (and likely
modify) it or incorporate it into their package.
Sören
References:
Bortz, J., Lienert, G. A., Boehnke, K. (2008). Verteilungsfreie
Methoden in der Biostatistik. Berlin: Springer
Code:
library(forensim)
myfun - function(cur, exp, p.obs, force=F){
if(length(cur) != length(exp)){
stop(wrong length of dimensions, try transpose the matrix\n)
}
# keep users from hot laptops
if(!force){
if(Cmn(sum(obs), length(obs)) (2 * (10^6))){
stop(Use option \force=T\ to compute more than 2 million
combinations.\n)
}
}
p.cur - factorial(sum(cur)) / prod(sapply(cur, function(x)
factorial(x))) * prod(exp ^ cur)
if(p.cur = p.obs){
return(p.cur)
}
}
# Trial 1, Book example
d.sta - Sys.time()
exp - c(.28, .08, .04, .42, .12, .06)
exp - exp/sum(exp)
obs - c(0, 3, 0, 0, 0, 0)
Cmn(sum(obs), length(obs)) # 56
p0 - factorial(sum(obs)) / prod(sapply(obs, function(x)
factorial(x))) * prod(exp ^ obs)
mat - comb(sum(obs), length(obs))
sum(unlist(apply(mat, 1, function(x) myfun(x, exp, p0
d.sto - Sys.time()
difftime(d.sto, d.sta) # Time difference of 0.1103680 secs
# Trial 2
d.sta - Sys.time()
exp - c(3, 5, 7, 4, 5, 4, 2, 9, 1)
exp - exp/sum(exp)
obs - c(3, 0, 2, 0, 1, 1, 0, 5, 3) # now with a longer vector
Cmn(sum(obs), length(obs)) # 490314
p0 - factorial(sum(obs)) / prod(sapply(obs, function(x)
factorial(x))) * prod(exp ^ obs)
mat - comb(sum(obs), length(obs))
sum(unlist(apply(mat, 1, function(x) myfun(x, exp, p0
d.sto - Sys.time()
difftime(d.sto, d.sta) # Time difference of 1.050846 mins
# Trial 3
d.sta - Sys.time()
exp - c(3, 5, 7, 4, 5, 4, 2, 9, 1)
exp - exp/sum(exp)
obs - c(3, 2, 2, 0, 1, 1, 0, 5, 3) # changed 0 to 2 on position 2
Cmn(sum(obs), length(obs)) # 1081575
p0 - factorial(sum(obs)) / prod(sapply(obs, function(x)
factorial(x))) * prod(exp ^ obs)
mat - comb(sum(obs), length(obs))
sum(unlist(apply(mat, 1, function(x) myfun(x, exp, p0
d.sto - Sys.time()
difftime(d.sto, d.sta) # Time difference of 2.425888 mins
# Trial 4
d.sta - Sys.time()
exp - c(3, 5, 7, 4, 5, 4, 2, 9, 1)
exp - exp/sum(exp)
obs - c(3, 2, 2, 0, 2, 1, 0, 5, 3) # changed 1 to 2 on position 5
Cmn(sum(obs), length(obs)) # 1562275
p0 - factorial(sum(obs)) / prod(sapply(obs, function(x)
factorial(x))) * prod(exp ^ obs)
mat - comb(sum(obs), length(obs))
sum(unlist(apply(mat, 1, function(x) myfun(x, exp, p0
d.sto - Sys.time()
difftime(d.sto, d.sta) # Time difference of 3.658092 mins
# Trial 5
d.sta - Sys.time()
exp - c(3, 5, 7, 4, 5, 4, 2, 9, 1)
exp - exp/sum(exp)
obs - c(3, 3, 2, 0, 2, 1, 0, 5, 3) # changed 1 to 2 on position 5
Cmn(sum(obs), length(obs)) # 2220075
p0 - factorial(sum(obs)) / prod(sapply(obs, function(x)
factorial(x))) * prod(exp ^ obs)
mat - comb(sum(obs), length(obs))
sum(unlist(apply(mat, 1, function(x) myfun(x, exp, p0
d.sto - Sys.time()
difftime(d.sto, d.sta) # Time difference of 3.658092 mins
--
Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
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and provide commented, minimal, self-contained, reproducible code.
[R] searching code for combination of vector
For a given numeric vector v of length n and sum s, is there a ready- to-run code that returns every combination of v in n summing up to s? Example for n=3 and s=2: v - c(2, 0, 0) # find some coding here that returns [1] 2 0 0 [2] 1 1 0 [3] 1 0 1 [4] 0 2 0 [5] 0 1 1 [6] 0 0 2 Thanks Sören -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] Define return values of a function
I have created a function to do something:
i - factor(sample(c(A, B, C, NA), 793, rep=T, prob=c(8, 7, 5,
1)))
k - factor(sample(c(X, Y, Z, NA), 793, rep=T, prob=c(12, 7, 9,
1)))
mytable - function(x){
xtb - x
btx - x
# do more with x, not relevant here
cat(The table has been created, see here:\n)
print(xtb)
list(table=xtb, elbat=btx)
}
tbl - table(i, k)
mytable(tbl) # (1)
z - mytable(tbl) # (2)
str(z) # (3)
(1) Wanted: outputs the string and the table properly. *Unwanted*:
outputs the list elements.
(2) and (3) Wanted: outputs the string properly. Wanted: assigns the
list properly.
How can I get rid of the *unwanted* part? That is, how do I define
what the functions prints and -- on the other hand -- what it returns
without printing?
Thanks
Sören
--
Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] shrink list by mathed entries
On 14.11.2009, at 03:58, David Winsemius wrote:
On Nov 13, 2009, at 11:19 AM, soeren.vo...@eawag.ch wrote:
a - c(Mama, Papa, Papa; Mama, , Sammy; Mama; Papa)
a - strsplit(a, ; )
mama - rep(F, length(a))
mama[sapply(a, function(x) { sum(x==Mama) }, simplify=T) 0] - T
[...]
... produces the variables mama and papa correctly. But how do
I remove all Mama list entries
[...]
Maybe you should explain what you were trying to do? Perhaps:
a[!mama]
[...]
I would sidestep that confusing sequence of logical assignments and
just do this:
a[ -grep(Mama, a) ]
[...]
Explanation of what I want to do: This code is PHP, maybe rather crude
but it works the way I want it and explains my goal:
#!/usr/bin/php
?php
error_reporting(E_ALL);
ini_set('display_errors', true);
ini_set('html_errors', false);
$strings = array(Mama, Papa, Papa; Mama, , Sammy; Mama;
Papa, Josh, Mama);
$vars = array(Mama, Papa, Sammy);
$i=0;
foreach($strings as $line){
$line = explode(; , $line);
$matches = array_intersect($line, $vars);
$diffs = array_diff($line, $vars);
foreach($matches as $match){
eval(\$$match.[.$i.] = 1;); // no easier way
}
foreach($diffs as $diff){
$others[$i] = $diff;
}
$i++;
}
print_r($Mama); // array with elements 0, 2, 4, and 6 set to 1
print_r($Papa); // array with elements 1, 2, and 4, set to 1
print_r($Sammy); // array with element 4 set to 1
print_r($others); // array with elements 3 set to , and 5 set to
Josh
?
Sören
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[R] Craddock-Flood Test in R?
Hello The Craddock-Flood Test is recommended for large tables with small degrees of freedom and low-frequency cells. Is there an R procedure and/or package which does the test? Thank you for your help! Sören Vogel -- Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] shrink list by mathed entries
Hello
a - c(Mama, Papa, Papa; Mama, , Sammy; Mama; Papa)
a - strsplit(a, ; )
mama - rep(F, length(a))
mama[sapply(a, function(x) { sum(x==Mama) }, simplify=T) 0] - T
papa - rep(F, length(a))
papa[sapply(a, function(x) { sum(x==Papa) }, simplify=T) 0] - T
# ... more variables
... produces the variables mama and papa correctly. But how do I
remove all Mama list entries in a in the same run, that is, shrink
the list by what was already matched?
Thank you for your help!
Sören Vogel
--
Sören Vogel, Dipl.-Psych. (Univ.), PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
__
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and provide commented, minimal, self-contained, reproducible code.
[R] when use which()
Hello: # some code to assign with and without which q - 1:20; q[c(9, 12, 14)] - NA r - 1:20; r[c(8:9, 12:15)] - NA s - 1:20; s[c(8:9, 12:15)] - NA r[q 16] - 0 s[which(q 16)] - 0 r;s # both: 0 0 0 0 0 0 0 0 NA 0 0 NA 0 NA 0 16 17 18 19 20 r - 1:20; r[c(8:9, 12:15)] - NA s - 1:20; s[c(8:9, 12:15)] - NA r[is.na(q)] - 30 s[which(is.na(q))] - 30 r;s # both: 1 2 3 4 5 6 7 NA 30 10 11 30 NA 30 NA 16 17 18 19 20 So it appears to me that a[b] - c delivers the same results as a[which(b)] - c. Is there any situation where the assignment with/ out which indeed makes a difference? Thanks for help Regards Sören __ 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] Formatted contingency tables with (%)
Quite often, I need those tables: x - sample(c(a, b, c), 40, rep=T) y - sample(c(X, Y), 40, rep=T) (tbl - table(x, y)) (z - as.factor(paste(as.vector(tbl), (, round(prop.table(as.vector(tbl)) * 100, 1), %), sep=))) matrix(as.factor(z), nrow=3, dimnames=dimnames(tbl)) But the result looks ugly and is not copypaste-able for LaTeX verbatim or table environment, moreover, the \ is not what I want in the printout. How to achieve: y x X Y a 3 (7.5%) 7 (17.5%) b 9 (22.5%) 5 (12.5%) c 6 (15.0%) 10 (25.0%) Thank you for help or hints. Sören __ 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] pca vs. pfa: dimension reduction
Can't make sense of calculated results and hope I'll find help here. I've collected answers from about 600 persons concerning three variables. I hypothesise those three variables to be components (or indicators) of one latent factor. In order to reduce data (vars), I had the following idea: Calculate the factor underlying these three vars. Use the loadings and the original var values to construct an new (artificial) var: (B1 * X1) + (B2 * X2) + (B3 * X3) = ArtVar (brackets for readability). Use ArtVar for further analysis of the data, that is, as predictor etc. In my (I realise, elementary) psychological statistics readings I was taught to use pca for these problems. Referring to Venables Ripley (2002, chapter 11), I applied princomp to my vars. But the outcome shows 4 components -- which is obviously not what I want. Reading further I found factanal, which produces loadings on the one specified factor very fine. But since this is a contradiction to theoretical introductions in so many texts I'm completely confused whether I'm right with these calculations. (1) Is there an easy example, which explains the differences between pca and pfa? (2) Which R procedure should I use to get what I want? Thank you for your help Sören Refs.: Venables, W. N., and Ripley, B. D. (2002). Modern applied statistics with S (4th edition). New York: Springer. __ 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] sunflowerplot error
A sunflowerplot crossing two categorial variables with NAs fails: ### sample: start ### set.seed(20) a - c(letters[1:4]) z - c(letters[23:26]) fa - factor(sample(rep.int(a, 1000), 100, replace=T), levels=a, ordered=T) fz - factor(sample(rep.int(z, 1000), 100, replace=T), levels=z, ordered=T) sunflowerplot(fa, fz) # okay, but: r - xyTable(fa, fz) length(r$x)==length(r$y) length(r$x)==length(r$number) # TRUE, TRUE is.na(fa) - sort(sample(1:100, 3)) sunflowerplot(fa, fz) # Error in rep.int(i.multi, number[number 1]) : invalid 'times' value s - xyTable(fa, fz) length(s$x)==length(s$y) length(s$x)==length(s$number) # TRUE, TRUE ### sample: end ### Seems to fail due to NAs, but (1) why and (2) how to get by? Thanks, *Sören* __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] chisq.test: decreasing p-value
A Likert scale may have produced counts of answers per category.
According to theory I may expect equality over the categories. A
statistical test shall reveal the actual equality in my sample.
When applying a chi square test with increasing number of repetitions
(simulate.p.value) over a fixed sample, the p-value decreases
dramatically (looks as if converge to zero).
(1) Why?
(2) (If this test is wrong), then which test can check what I want to
check, that is: are the two distributions of frequencies (observed and
expected) in principle the same?
(3) By the way, how to deal with low frequency cells?
r - c(10, 100, 500, 1000, 2000, 5000)
v - c(35, 40, 45, 45, 40, 35)
sapply(list(r), function (x) { chisq.test(v, p=c(rep.int(40, 6)),
rescale.p=T, simulate.p.value=T, B=x)$p.value })
Thank you, Sören
--
Sören Vogel, PhD-Student, Eawag, Dept. SIAM
http://www.eawag.ch, http://sozmod.eawag.ch
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] chisq.test: decreasing p-value
Thanks to Peter, David, and Michael! After having corrected the coding error, the p values converge to particular value, not necessarily zero. The whole story is, 634 respondents in 6 different areas marked their answer on a 7-step Likert scale (very bad, bad, ..., very good -- later recoded to 5 scale levels). The statistical question now is, do the answer's distributions (amount of goods, bads etc.) in either area differ from the mean answer-distribution calculated with summing up all goods, bads, etc. Anyway an omnibus chi square would not answer my question, and due to spurious significances I'd rather go back to my chi square book ;-) (for the interested, see http://sozmod.eawag.ch/files/file.Robj for the entire table). Thanks for your help Sören __ 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] Summary of data.frame according to colnames and grouping factor
A dataframe holds 3 vars, each checked true or false (1, 0). Another var holds the grouping, r and s: ### start:example set.seed(20) d - data.frame(sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T), sample(c(0, 1), 20, replace=T)) names(d) - c(A, B, C) e - rep(c(r, s), 10) ### end:example How do I get the count of 1's (or any other function) applied over each var according to the grouping? That is: Desired output table: A B C r count count count s......... or likewise transposed. I'd like to use the table for textual display and/or barplot creation. Thx, Sören -- Sören Vogel, PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] Recode factor into binary factor-level vars
How to I recode a factor into a binary data frame according to the factor levels: ### example:start set.seed(20) l - sample(rep.int(c(locA, locB, locC, locD), 100), 10, replace=T) # [1] locD locD locD locD locB locA locA locA locD locA ### example:end What I want in the end is the following: m$locA: 0, 0, 0, 0, 0, 1, 1, 1, 0, 1 m$locB: 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 m$locC: 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 m$locD: 1, 1, 1, 1, 0, 0, 0, 0, 1, 0 Instead of 0, NA's would also be fine. Thanks, Sören -- Sören Vogel, PhD-Student, Eawag, Dept. SIAM http://www.eawag.ch, http://sozmod.eawag.ch __ 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] Summary grouped by factor
### example:start v - sample(rnorm(200), 100, replace=T) k - rep.int(c(locA, locB, locC, locD), 25) tapply(v, k, summary) ### example:end ... (hopefully) produces 4 summaries of v according to k group membership. How can I transform the output into a nice table with the croups as columns and the interesting statistics as lines? Thx, Sören __ 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] Summary grouped by factor
On 06.03.2009, at 16:48, soeren.vo...@eawag.ch wrote: ### example:start v - sample(rnorm(200), 100, replace=T) k - rep.int(c(locA, locB, locC, locD), 25) tapply(v, k, summary) ### example:end ... (hopefully) produces 4 summaries of v according to k group membership. How can I transform the output into a nice table with the croups as columns and the interesting statistics as lines? ### Right??? and good??? solution: sapply(by(v, list(area=k), function(x)x, simplify=F), summary) Sören (again) __ 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] Grouped Boxplot
Pls forgive me heavy-handed data generation -- newby ;-) ### start ### # example data g - rep.int(c(A, B, C, D), 125) t - rnorm(5000) a - sample(t, 500, replace=TRUE) b - sample(t, 500, replace=TRUE) # what I actually want to have: boxplot(a | b ~ g) # but that does obviously not produce what I want, instead i - data.frame(g, a, rep(one, length(g))) j - data.frame(g, b, rep(two, length(g))) names(i) - c(Group, Number, Word) names(j) - c(Group, Number, Word) k - as.data.frame(rbind(i, j)) boxplot(k$Number ~ k$Word * k$Group) ### end ### Questions: (1) Is there an easier way? (2) How can I get additional space between the A:D groups? Thank you Sören __ 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] add absolute value to bars in barplot
Hello, r-h...@r-project.orgbarplot(twcons.area, beside=T, col=c(green4, blue, red3, gray), xlab=estate, ylab=number of persons, ylim=c(0, 110), legend.text=c(treated, mix, untreated, NA)) produces a barplot very fine. In addition, I'd like to get the bars' absolute values on the top of the bars. How can I produce this in an easy way? Thanks Sören __ 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] cross tabulation: convert frequencies to percentages
Hello, might be rather easy for R pros, but I've been searching to the dead end to ... twsource.area - table(twsource, area, useNA=ifany) gives me a nice cross tabulation of frequencies of two factors, but now I want to convert to pecentages of those absolute values. In addition I'd like an extra column and an extra row with absolute sums. I know, Excel or the likes will produce it more easily, but how would the procedure look like in R? Thanks, Sören __ 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 grouped histograms
r11 -- r16 are variables showing a reason for usage of a product in 6 different situations. Each variable is a factor with 4 levels imported from a SPSS sav file with labels ranging from not important to very important, and NA's for a sample of N = 276. (1) I need a chi square test of independence showing that the reason does not differ depending on the situation. (2) I need a single coloured histogram plot. The x axis should be grouped by the 6 situation with small space between the groups, each group should show different bars for each factor value (not important, little..., ..., NA's), but NA's is not necessary. I've been googling the whole day, searching the mailing list and handbooks, and struggled through the somewhat R programmer specific documentation. Beeing a newby in R, I'm now afraid that I have to go back to SPSS and Excel where my tasks would be a work of an hour. But I was told euphoric that R may solve many of the problems I have (and don't like) with SPSS, or having to separate calculation (SPSS, Excel by hand) and plotting (GNUplot). So my two questions are: How can I easily solve my 2 tasks? Secondly, is R really recommended for R newbys in daily work? Thank you for any help Sören __ 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] read.spss: variable.labels
Hi,
how can I attach variable labels originally read by read.spss() to the
resulting variables?
pre
X - read.spss('data.sav', use.value.labels = TRUE, to.data.frame =
TRUE, trim.factor.names = TRUE, trim_values = TRUE, reencode = UTF-8)
names(X) - tolower(names(X))
attach(X)
/pre
Thank you
Sören
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Regression with nominal data
Hi, y is nominal (3 categories), x1 to 3 is scale. What I want is a regression, showing the probability to fall in one of the three categories of y according to the x. How can I perform such a regression in R? Thanks for your help Sören __ 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.
