Actually it just the parameterization that is causing trouble near k=0
let u = (x-z)/a
then the problematic part of your function is
(1- k*u)^(1/k)
take the log to get
log(1-k*u)/k
= -(k*u +k^2*u^2/2 + ...)/k
= -(u +k u^2/2 + ..)
so your function is exp(-u - ku^2/2 - ...)
and
Hi there,
What I'm trying to do is to calculate the line coefficients of a local
linear regression model by hand. I'm aware that there are many great
packages out there that calculate the local expectation E(y|x) with
local linear regression, but that's not what I need. I need the
coefficients of
Hello,
Looking at the bfast help page is says that the output component of
the returned object is a list where the elements correspond to results
for each iteration of the fitting algorithm.
Michael
On 13 October 2010 15:58, CALEF ALEJANDRO RODRIGUEZ CUEVAS
alejandro.rodriguez.cue...@gmail.com
Yes, that is what I what...
Thanks.
Feng
On Wed, Oct 13, 2010 at 6:38 AM, Michael Bedward
michael.bedw...@gmail.comwrote:
Hello Feng,
I think you just want this...
lapply(A, function(x) apply(x[,,-c(1,2)], c(1,2), mean))
Michael
On 13 October 2010 04:00, Feng Li feng...@stat.su.se
benchmark(
+ all_eq = {isTRUE(all.equal.numeric(x,y))},
+ dfrm = {compare-data.frame(id-seq(1,43e3,1),x,y);
+ compare$id[compare$x!=compare$y]},
+ int = {(compare$x!=compare$y)},
+ slf = {differences-compare$id[compare$x!=compare$y]},
+ replications=1000)
test replications
Hello!
I am using box plot and one of my boxes has only one whisker. How can I
change this?
Code:
bp-c(2.7, 3.1, 3.5, 8.95)
Methode1 - quantile(bp,type = 7)
Methode2 - quantile(bp,type = 2)
d-data.frame(Methode1,Methode2)
boxplot(d,ylab = Beispiel 1,range = 1.5)
The subject says it all :).
So I have a table and I want it do be ploted like a table so to say and cant
find any function/package that dose it for me.
Anyone know of one?
//Joel
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Sent
Try with more data points?!
You have only five points, the last one being considered as outlier.
Note that boxplot() requires a numeric vector for specifying data from
which the boxplots are to be produced!
HTH
Ivan
Le 10/13/2010 09:50, tom a écrit :
Hello!
I am using box plot and one of
Seems like I must be missing something but ?plot specifically ?plot.table
Is that not ploted like a table? If it is not, do you know of any
examples we could look at?
Cheers,
Josh
On Wed, Oct 13, 2010 at 1:00 AM, Joel joda2...@student.uu.se wrote:
The subject says it all :).
So I have a
Dear everyone,
I would like to create a kite chart in which I plot densities (width of the
vertical kites) in relation to sediment depth (on reversed y-axis) for 6
different locations (Distances from seep site, on x-axis on top of the
plot). The dataset I would like to use is:
It should look something like this (not at all relevant except the look)
http://r.789695.n4.nabble.com/file/n2993297/tableR19.jpg
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Oh, what about latex() in package Hmisc? Creating a table seems like
more of a job for a typesetting program, conveniently R has great
interfaces to many. You might also look at R2HTML and the ilk.
On Wed, Oct 13, 2010 at 1:14 AM, Joel joda2...@student.uu.se wrote:
It should look something
Hi,
grid.table in gridExtra might give you some inspiration.
HTH,
baptiste
On 13 October 2010 10:14, Joel joda2...@student.uu.se wrote:
It should look something like this (not at all relevant except the look)
http://r.789695.n4.nabble.com/file/n2993297/tableR19.jpg
--
View this message
R community,
I am trying to write a code that fills in data gaps in a time series. I
have no R or statistics background at all but the use of R is proving to be
a large portion of my PhD research.
So far my code identifies where and the number of new entries required but I
do not know how to
Hi,
I would like to estimate a panel model (small N large T, fixed effects),
but would need robust standard errors for that. In particular, I am
worried about potential serial correlation for a given individual (not so
much about correlation in the cross section).
From the documentation, it
Ivan Calandra wrote:
Try with more data points?!
You have only five points, the last one being considered as outlier.
Note that boxplot() requires a numeric vector for specifying data from
which the boxplots are to be produced!
But why is only one of the boxplots missing his whisker?
Thx for the ideas will try them out.
Have a wonderful day
//Joel
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__
Hello Ellen,
First up I think you can use reshape to get your data into a form that
kiteChart will work with...
# assuming your matrix or data.frame is called X
Xwide - reshape(X, timevar=depth, idvar=distance, direction=wide)
# replace NAs with 0 (don't think kiteChart likes NA)
Xwide[
Oops, sorry, I left out a step in that last post
After replace NAs with 0 in Xwide...
# use distance col as row names
rownames( Xwide ) - Xwide[ , 1 ]
Xwide - Xwide[ , -1 ]
kiteChart( Xwide )
On 13 October 2010 19:49, Michael Bedward michael.bedw...@gmail.com wrote:
Hello Ellen,
First up I
Well, you don't use the same data for both.
Type 2 and 7 give different values for the 25 and 75%, which correspond
more or less to the box hinges.
If you take a look at how the whiskers are defined (look at ?boxplot.stats):
|coef| this determines how far the plot ‘whiskers’ extend out from
On 10/12/2010 01:09 AM, Panos Hadjinicolaou wrote:
Dear R colleagues,
I am trying to plot some geophysical data as a filled contour on a continent
map and so far the guidance from the R-help archives has been invaluable. The
only bit that still eludes me is the colour key (legend) coming with
On Wed, 13 Oct 2010, Max Brown wrote:
Hi,
I would like to estimate a panel model (small N large T, fixed effects),
but would need robust standard errors for that. In particular, I am
worried about potential serial correlation for a given individual (not so
much about correlation in the cross
On 10/12/2010 08:58 PM, Łukasz Ręcławowicz wrote:
Hi,
I don't know how to sample such data, it can't be done by row sampling
as default method on matrix in boot.
Function takes matrix and returns single coefficient.
#There is a macro but I want use R :)
Hi Laura,
If you want ANOVA output, ask for it! A general strategy that almost
always works in R is to fit 2 models, one without the term(s) you want
to test, and one with. Then use the anova() function to test them.
(models must be nested, and in the lmer() case you need to use REML =
FALSE).
Hello,
I've attached the result I'v'e got by applying your code (thanks for this!),
but I seem to have horizontal kites instead of vertical kites. I need to
rotate the entire graph, so to speak..
I've tried using viewpoint (found it on some forum), but this only seems to
work with lattice
I've tried hard to find a way to exponentiate each element of a whole matrix
such that if I start with A
A = [ 2 3
2 4]
I can get back B
B = [ 7.38 20.08
7.38 54.60]
I've tried
B - exp(A) but no luck.
Thanks
J
===
Dr. Jim Maas
University of
On Wed, Oct 13, 2010 at 11:51 AM, Maas James Dr (MED) j.m...@uea.ac.uk wrote:
I've tried hard to find a way to exponentiate each element of a whole matrix
such that if I start with A
A = [ 2 3
2 4]
I can get back B
B = [ 7.38 20.08
7.38 54.60]
I've tried
B - exp(A)
On Wed, 13 Oct 2010 11:51:39 +0100
Maas James Dr (MED) j.m...@uea.ac.uk wrote:
I've tried hard to find a way to exponentiate each element of a whole
matrix such that if I start with A
A = [ 2 3
2 4]
I can get back B
B = [ 7.38 20.08
7.38 54.60]
I've tried
B
Dear R user fellows,
I would like to ask you about the package 'rgenoud' which is a genetic
optimization tool.
I ran the function 'genoud' with two variables to be minimized by the
following command.
result-genoud(fn,nvars=2,starting.values=c(0.5,0),
pop.size=1000, max.generations=10,
On 10/13/2010 07:11 PM, elpape wrote:
Dear everyone,
I would like to create a kite chart in which I plot densities (width of the
vertical kites) in relation to sediment depth (on reversed y-axis) for 6
different locations (Distances from seep site, on x-axis on top of the
plot). The dataset I
Super ! An option for vertical plotting would be very nice.
Michael
On 13 October 2010 22:19, Jim Lemon j...@bitwrit.com.au wrote:
On 10/13/2010 07:11 PM, elpape wrote:
Dear everyone,
I would like to create a kite chart in which I plot densities (width of
the
vertical kites) in relation
Hi:
David was on the right track...
library(reshape) # for the melt() function below
rnorm(100,5,3) - A
rnorm(100,7,3) - B
rnorm(100,4,1) - C
df - melt(data.frame(A, B, C))
names(df)[1] - 'gp'
histogram(~ value | gp, data=df, layout=c(3,1), nint=50,
panel=function(x, ..., groups){
Hi:
This recent thread revealed that a package on R-forge for calculating earth
movers distance is available:
http://r.789695.n4.nabble.com/Measure-Difference-Between-Two-Distributions-td2712281.html#a2713505
HTH,
Dennis
On Tue, Oct 12, 2010 at 7:39 PM, Michael Bedward
Hi:
Perhaps
?append
for simple insertions...
HTH,
Dennis
On Wed, Oct 13, 2010 at 1:24 AM, dpender d.pen...@civil.gla.ac.uk wrote:
R community,
I am trying to write a code that fills in data gaps in a time series. I
have no R or statistics background at all but the use of R is proving
I am trying to convert an array from numeric values back to date and time
format. The code I have used is as follows;
for (i in 0:(length(DateTime3)-1)) {
DateTime3[i] - (strptime(start, %m/%d/%Y %H:%M)+ i*interval)
where start - [1] 1/1/1981 00:00
However the created array
Ah, that's interesting. I'll have a look because it's bound to be
better than my effort.
Many thanks Dennis.
Michael
On 13 October 2010 22:36, Dennis Murphy djmu...@gmail.com wrote:
Hi:
This recent thread revealed that a package on R-forge for calculating earth
movers distance is available:
Thanks all, works perfect.
Alejo
2010/10/13 Dennis Murphy djmu...@gmail.com
Hi:
David was on the right track...
library(reshape) # for the melt() function below
rnorm(100,5,3) - A
rnorm(100,7,3) - B
rnorm(100,4,1) - C
df - melt(data.frame(A, B, C))
names(df)[1] - 'gp'
HI useRs,
Is it required to remove mean before using ARIMA models
thanks
nuncio
--
Nuncio.M
Research Scientist
National Center for Antarctic and Ocean research
Head land Sada
Vasco da Gamma
Goa-403804
[[alternative HTML version deleted]]
Wow! Thanks!
On 13 October 2010 13:23, mbedward [via R]
ml-node+2993501-2047744113-199...@n4.nabble.comml-node%2b2993501-2047744113-199...@n4.nabble.com
wrote:
Super ! An option for vertical plotting would be very nice.
Michael
On 13 October 2010 22:19, Jim Lemon [hidden
Hi:
On Tue, Oct 12, 2010 at 8:59 PM, Laura Halderman lk...@pitt.edu wrote:
Hello. I am new to R and new to linear mixed effects modeling. I am
trying to model some data which has two factors. Each factor has three
levels rather than continuous data. Specifically, we measured speech at
Dennis,
Thanks for that. The problem now is that I am trying to use it in a for
loop. Based on the example before, 2 entries are required after H[3] as
specified by O. The problem is that when inserting values the length of the
array changes so I don't know how to tell the loop that. This is
Hi R-helpers , i am learning nnet package now.I have seen that the nnet can
be used for regression and classification.Can you give me more insights on
the other data mining and predictive techniques that the nnet package can be
used for? If possible, can you send me links for sample datasets with
I have to make correction in my error message which I introduced in my
original message. Sorry for my mistake.
Finally, I had the following error message after running the function '
genoud'.
Error in optim(foo.vals, fn = fn1, gr = gr1, method = optim.method, control
= control) :
non-finite
Dear R-users,
is anybody aware of some package or routine to implement nonparametric
Multivariate Analysis of Covariance (MANCOVA) using matrices instead of
single variable names? I found something for parametric MANCOVA which still
requires single variables to be used (ffmanova,vegan), but since
I have to make correction in my error message which I introduced in my
original message. Sorry for my mistake.
Finally, I had the following error message after running the function
'genoud'.
Error in optim(foo.vals, fn = fn1, gr = gr1, method = optim.method, control
= control) :
non-finite
Dear R-users,
I am working with R version 2.10.1 and package RODBC Version: 1.3-2 under
windows.
Say I have a table testtable (in an Access data base), which has many
different columns, among them a character column X with integer-like data
as 0012345.
Using sqlFetch, I'd like to assure that
Try this:
sapply(0:(length(DateTime3)-1), function(i)as.character(strptime(start,
%m/%d/%Y %H:%M) + i * interval))
On Wed, Oct 13, 2010 at 8:51 AM, dpender d.pen...@civil.gla.ac.uk wrote:
I am trying to convert an array from numeric values back to date and time
format. The code I have
Hi there,
I'm working on a biological dataset that contains up to 40 species and 100
sites each. The response variable is index (trend), dependent is year - each
combination has an index for each year between 1994 and 2008 inclusive. So
my input file is along the following lines: species, site,
Dear all R users, I was trying to create a polymonial using the polynom() of
PolynomF package. My problem is, if I pass coefficients as simple
numerical values, it is working, but for matrix coefficient it is not. Here
is my try:
library(PolynomF)
z - polynom()
## following is working
p1 - 1-
Thanks Henrique,
Do you have any idea why the first entry doesn't have the time as the start
specified is 1/1/1981 00:00?
Is it something to do with being at midnight?
Doug
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Dear useRs:
I use pROC package to compute the bootstrap C.I. of AUC.
The command was as follows:
roc1-roc(all$D,all$pre,ci=TRUE,boot.n=200)
However, the result was:
Area under the curve: 0.5903
95% CI: 0.479-0.7016 (DeLong)
Why the C.I. was computed by the Delong Method?
Yao Zhu
Department
Try this:
sapply(0::(length(DateTime3)-1), function(i)format(strptime(start,
%m/%d/%Y %H:%M) + i * interval, %Y-%m-%d %H:%M:%S))
On Wed, Oct 13, 2010 at 10:16 AM, dpender d.pen...@civil.gla.ac.uk wrote:
Thanks Henrique,
Do you have any idea why the first entry doesn't have the time as the
On Oct 13, 2010, at 5:06 AM, Ivan Calandra wrote:
Well, you don't use the same data for both.
Type 2 and 7 give different values for the 25 and 75%, which
correspond more or less to the box hinges.
If you take a look at how the whiskers are defined (look at ?
boxplot.stats):
|coef| this
Perfect.
Thanks
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R-help@r-project.org mailing list
Hello,
I am aware this may be an obscure problem difficult to advice about, but
just in case... I am calling a C function from R on an iMac (almost shining:
bought by my institution this year) and gives a memory not mapped error.
Nevertheless, exactly the same code runs without problem in a
Hello.
In principle Achim is right, by default vcovHC.plm does things the Arellano
way, clustering by group and therefore giving SEs which are robust to general
heteroskedasticity and serial correlation. The problem with your data, though,
is that this estimator is N-consistent, so it is
Hi:
On Wed, Oct 13, 2010 at 5:31 AM, dpender d.pen...@civil.gla.ac.uk wrote:
Dennis,
Thanks for that. The problem now is that I am trying to use it in a for
loop. Based on the example before, 2 entries are required after H[3] as
specified by O. The problem is that when inserting values
Dear R-community,
Using bwplot, how can I put the whiskers at percentile 5 and percentile 95,
in place of the default position coef=1.5??
Using panel=panel.bwstrip, whiskerpos=0.05, from the package agsemisc gives
satisfaction, but changes the appearance of my boxplot and works with an old
Dear All,
I was trying to use the lmList function to get the lmList graphic but have a
problem creating the graphic. My data has missing values and an error
occurred
as stated below.
###
library(nlme)
a
schoolid spring score
Please take a look at the documentation for lmList regarding
the na.action= argument. It should be a *function*, not TRUE
or FALSE. For example, try
lmList(score ~ childid | spring, data=a, na.action=na.omit)
- Phil Spector
Greetings
I'm having difficulty witht the strptime function. I can't seem to
figure a way to strip month (name) and year and create separate columns
from a column with MM/DD/ formatted dates. Can anyone help?
Cheers
Kurt
***
On Tue, Oct 12, 2010 at 8:54 PM, Eric Hu eric...@gilead.com wrote:
Hi,
I am trying to see if I can use R to perform more rigorous regression
analysis. I wonder if the fingerprint package is able to handle pipeline
pilot fingerprints (ECFC6 etc) now.
Currently no - does Pipeline Pilot out put
Thanks Giovanni and Achim!
I will try out some of the things you suggested. Let's see how far I get :-)
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PLEASE do read the posting guide
On Oct 13, 2010, at 10:05 AM, Christophe Bouffioux wrote:
Dear R-community,
Using bwplot, how can I put the whiskers at percentile 5 and
percentile 95,
in place of the default position coef=1.5??
Using panel=panel.bwstrip, whiskerpos=0.05, from the package
agsemisc gives
satisfaction,
hello list,
i'd very much appreciate help with setting up the
contrast for a 2-factorial crossed design.
here is a toy example:
library(multcomp)
dat-data.frame(fac1=gl(4,8,labels=LETTERS[1:4]),
fac2=rep(c(I,II),16),y=rnorm(32,1,1))
mod-lm(y~fac1*fac2,data=dat)
## the
newmat - oldmat[ c(rep(FALSE,19),TRUE), ]
Or
newmat - oldmat[ seq(20, nrow(oldmat), 20), ]
--
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
Dear All,
I have 4 or 5 contour plots that I need to overlay. Currently they are maps
showing hot and cold areas for specific elements.
I would like to combine these plots into one map, where each color will
correspond to a different element and you would be able to see the areas that
each
I don't know of any tools that would do this generally (RODBC could do it if
there were specific column names in the 1st row and a given column with
information to identify the rows, but this seems unlikely from your
description). A couple of possibilities:
In Excel you can highlight the
Dear R community,
I am struggling a bit with a probably fairly simple task. I need to use some
already existing functions as argument for a new function that I am going to
create. 'dataset' is an argument, and it comprises objects named
'mean_test', 'sd_test', 'kurt_test' and so on. 'arg1' tells
Also look at textplot in the gplots package and addtable2plot in the plotrix
package.
--
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-
On Oct 13, 2010, at 11:44 AM, Jeremy Olson wrote:
Dear All,
I have 4 or 5 contour plots that I need to overlay. Currently they
are maps showing hot and cold areas for specific elements.
Providing paste-able examples is the standard way to present such
problems.
I would like to combine
If I understand you correctly, just arg2(arg1) will work fine...One of
the nice features of R.
Example:
f - function(dat,fun=mean,...)
{
fun(dat, ...) ## ... allows extra arguments to fun
}
f( rnorm(10), trim=.05) ## trimmed mean
x - 1:10
f(x, fun = max,na.rm=TRUE)
Cheers,
Bert
On Wed,
Thaks for your quick reply Bert, although I doubt it works.
The reason is that the names of the objects of the dataset, all end with the
sufix '_test' and therefore I need to attach/paste/glue this suffix to the
'arg2' of the function. Any other idea?
--
View this message in context:
Got it , thanks!
From: David Winsemius [via R]
[mailto:ml-node+2992845-2070125287-199...@n4.nabble.com]
Sent: Tuesday, October 12, 2010 3:28 PM
To: Steve Swope
Subject: Re: Plotting Y axis labels within a loop
On Oct 12, 2010, at 5:54 PM, Steve Swope wrote:
When I plot y axis
From what I read, you want something like this:
myfunction-function(dataset,arg1,arg2)
{
func = match.fun(arg2)
argument = dataset[, match(paste(arg1,_test, sep=), names(dataset))]
result=func(argument)
return(result)
}
On Wed, Oct 13, 2010 at 9:28 AM, Manta mantin...@libero.it wrote:
Is this the sort of thing you are looking for?
f - function(testName, dataset, ...) {
+ testFunc - match.fun(paste(testName, _test, sep=))
+ testFunc(dataset, ...)
+ }
m_test - function(x) the m test
p_test - function(x) the p test
f(m, 17)
[1] the m test
Bill Dunlap
This often happens when your C code uses memory that
it did not allocate, particularly when it reads or
writes just a little beyond the end of a memory block.
On some platforms or if you are lucky there is unused
memory between blocks of allocated memory and you don't
see a problem. Other
If I have a character array:
list = c(A, B, C)
how do I access the third element without doing list[3]. Can't I find the
index of C using a particular function?
--
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Sent from
which(list==C)
[1] 3
See ?which
On 10/13/2010 11:56 AM, lord12 wrote:
If I have a character array:
list = c(A, B, C)
how do I access the third element without doing list[3]. Can't I find the
index of C using a particular function?
--
Christian Raschke
Department of Economics
and
Hello all,
I had a very strange looking problem that turned out to be due to unexpected
(by me at least) format changes to one of my data files. We have a small lab
study in which each run is represented by a row in a tab-delimited file; each
row identifies a repetition of the experiment and
Try this:
library(RDCOMClient)
xl - COMCreate(Excel.Application)
wk - xl$Workbooks()$Open(Book1.xlsx)
do.call(cbind, wk$Sheets(1)$Range(B50:C55)$Value())
On Tue, Oct 12, 2010 at 9:17 PM, Jeevan Duggempudi jdug...@yahoo.comwrote:
Hello all,
I have a business user who generates monthly
Try:
ci.auc(all$D,all$pre,m=b)
2010/10/13 zhu yao mailzhu...@gmail.com:
Dear useRs:
I use pROC package to compute the bootstrap C.I. of AUC.
The command was as follows:
roc1-roc(all$D,all$pre,ci=TRUE,boot.n=200)
--
Miłego dnia
__
How about emacs?
albyn
On Wed, Oct 13, 2010 at 01:13:03PM -0400, Schwab,Wilhelm K wrote:
Hello all,
.
Have any of you found a nice (or at least predictable) way to use OO Calc to
edit files like this? If it insists on thinking for me, I wish it would
think in 24 hour time and 4 digit
On Wed, Oct 13, 2010 at 10:13 AM, Schwab,Wilhelm K
bsch...@anest.ufl.edu wrote:
Hello all,
I had a very strange looking problem that turned out to be due to unexpected
(by me at least) format changes to one of my data files. We have a small lab
study in which each run is represented by a
On Oct 13, 2010, at 12:13 PM, Schwab,Wilhelm K wrote:
Hello all,
I had a very strange looking problem that turned out to be due to unexpected
(by me at least) format changes to one of my data files. We have a small lab
study in which each run is represented by a row in a tab-delimited
On Wed, Oct 13, 2010 at 10:13 AM, Schwab,Wilhelm K
bsch...@anest.ufl.edu wrote:
Hello all,
I had a very strange looking problem that turned out to be due to unexpected
(by me at least) format changes to one of my data files. We have a small lab
study in which each run is represented by a
Albyn,
I'll look into it. In fact, I have a small book on it that I bought in my very
early days of using Linux. I quickly found TeX Maker (for the obvious),
Code::Blocks for C/C++ and I would not have started the move without a working
Smalltalk (http://pharo-project.org/home).
For editing
Peter,
vi is *really* primitive =:0 R is a little late because I tend to do shape
changes prior to invoking R. However, I could load tweak and re-save and then
bring R back into it later. I never would have thought of it. Thanks!
Bill
From:
Hi Rajarshi,
Here is a post I found from Pipeline pilot community help pages:
https://community.accelrys.com/message/3466
Eric
-Original Message-
From: Rajarshi Guha [mailto:rajarshi.g...@gmail.com]
Sent: Wednesday, October 13, 2010 7:52 AM
To: Eric Hu
Cc: r-help@r-project.org
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
1 104
296
3101
So I do:
d -data.frame(exp_num=c(1,2,3)); /* Just 3
Thank you very much for your kind reply.
I used gdb, and it returns a reason KERN_INVALID_ADDRESS on a very simple
operation (obtaining the time step for a numerical integration). Please see
bellow.
But, um, I solved it by changing the function's arguments in the C code from
unsigned long int to
Dear all,
I am trying to run a loop in my codes, but the software returns an error:
subscript out of bounds
I dont understand exactly why this is happenning. My codes are the following:
rm(list=ls()) #remove almost everything in the memory
set.seed(180185)
nsim - 10
mresultx -
Hi,
this should be an easy one, but I can't figure it out.
I have a vector of tests, with their units between brackets (if they have
units).
eg tests - c(pH, Assay (%), Impurity A(%), content (mg/ml))
Now I would like to hava a function where I use a test as input, and which
returns the units
try this:
n - 3
data.frame(Exp = seq_len(n),
Heads = rbinom(n, 200, 0.5))
I hope it helps.
Best,
Dimitris
On 10/13/2010 7:28 PM, Shiv wrote:
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column names in a data frame, I ran into the
Shiv wrote:
I am trying a simple toin coss simulation, of say 200 coin tosses. The idea
is to have a data frame like so:
Experiment#Number_Of_Heads
1 104
296
3101
So I do:
d -data.frame(exp_num=c(1,2,3)); /*
On 13/10/2010 2:19 PM, Jon Zadra wrote:
Hi,
I've looked all over for a solution to this, but haven't had much look
in specifying what I want to do with appropriate search terms. Thus I'm
turning to R-help.
In the process of trying to write a simple function to rename individual
column
I don't know much about the iMac. R's .C() passes
R-language integers as pointers to 32-bit ints. If
your iMac is 64-bit and sizeof(long)==8 in your compiler
(pretty common for 64-bit compilers but not so for
Microsoft Windows compilers) then Long[0] will
use the next 64 bits to make an
It will get a good look, as will gnumeric - thanks to all!
Bill
From: Albyn Jones [jo...@reed.edu]
Sent: Wednesday, October 13, 2010 2:14 PM
To: Schwab,Wilhelm K
Subject: Re: [R] [OT] (slightly) - OpenOffice Calc and text files
emacs shows you exactly
Try this:
replace(gsub(.*\\((.*)\\)$, \\1, tests), !grepl(\\(.*\\), tests), )
On Wed, Oct 13, 2010 at 3:16 PM, Bart Joosen bartjoo...@hotmail.com wrote:
Hi,
this should be an easy one, but I can't figure it out.
I have a vector of tests, with their units between brackets (if they have
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