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
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
Hello, (how) can I download/re-retrieve/order previous R-** digest
volumes/issues to my mailbox for local browsing? Thank you, Sören
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PLEASE do read the posting guide
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:
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
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
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
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 =
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
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
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
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
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
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
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
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,
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?
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
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
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,
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
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
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.,
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
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
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
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
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
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;
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
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
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)
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
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
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)
###
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
### 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
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
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
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
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
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
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
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
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