a) Please post in plain text.
b) Although you mention Excel, it does not seem to be relevant. (This is
a good thing, but mentioning it is a red herring.)
c) Your definition of X is not executable R code. If you have already
imported data from elsewhere, you can use dput to make it easy for
On Nov 25, 2012, at 10:50 PM, Jack Tanner wrote:
I have some large-ish files that are the output of save() from R
2.15.1, which
that version can load() just fine. After upgrading to 2.15.2, load()
no longer
works on these files. Is this a known issue?
No longer works?
--
David
Thank you Uwe Ligges-3
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__
R-help@r-project.org mailing list
Hi all,
I'd like to fit different models using a loop together with the jags
package.
to do this, I load the following packages runjags, R2jags and coda
and use the function jags to fit my models.
The problem is that the following a windowx with the following message
appear between each model
Hello
I'm using some R code in order to use the model BG-NBD implementation
I'm using Java and I call R by using RCaller.
I must admit that i'm really really new to R so maybe I'm doing something
wrong.
I use the code on this URL: http://code.google.com/p/clv-master-thesis/
Hi
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of David Winsemius
Sent: Monday, November 26, 2012 9:11 AM
To: Jack Tanner
Cc: r-h...@stat.math.ethz.ch
Subject: Re: [R] compatibility of load() in R 2.15.2
On Nov 25,
Hello,
Why mail a question just to me? Post to the list and the odds of getting
more answers (and better) are bigger.
As for your question, the problem is in the call to glm, you don't need
the prefix 'train$' in the formula, the argument 'data' solves that and
when predicting R will look for
I cannot figure out how to tune the minor tic marks on the date axis of a
zoo plot.
I read hundreds of CSV files from a zip archive transparently. The
time/date strings I convert to POSIXct format, order them and then make a
zoo object as there may be cases which have unequal time stamping. As
On Mon, Nov 26, 2012 at 6:36 AM, Alex van der Spek do...@xs4all.nl wrote:
I cannot figure out how to tune the minor tic marks on the date axis of a
zoo plot.
I read hundreds of CSV files from a zip archive transparently. The
time/date strings I convert to POSIXct format, order them and then
Dear all,
I am using the book Generalized Linera Models and Extension by Hardin and
Hilbe (second edition, 2007) at the moment. The authors suggest that
instead of OLS models, the log link is generally used for response data
that take only positive values on the continuous scale. Of course they
Webinar signup:
Advances in Gradient Boosting: the Power of Post-Processing
December 14, 10-11 a.m., PST
Webinar Registration:
http://2.salford-systems.com/gradientboosting-and-post-processing/
Course Outline:
*Gradient Boosting and Post-Processing:
o What is missing from
Not unless we have more information. Please read the Posting Guide to see how
to make it easier for people to answer your question.
Best,
Andy
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Oritteropus
Sent: Thursday, November
Dear all,
I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase). I
keep getting an error about the channel when using sqlQuery(). However,
sqlTables() does not complain about the channel and lists all tables in the
database. If I try sqlFetch(), then R crashes.
I'm happy
John Kane
Kingston ON Canada
-Original Message-
From: landronim...@gmail.com
Sent: Sun, 25 Nov 2012 15:02:16 +0100
To: jrkrid...@inbox.com
Subject: Re: [R] RExcel, ROOo and LibreOffice inquiry
On Sun, Nov 25, 2012 at 2:38 PM, John Kane jrkrid...@inbox.com wrote:
Can you supply
First a statistical issue: The survfit routine will produce predicted survival curves for
any requested combination of the covariates in the original model. This is not the same
thing as an adjusted survival curve. Confusion on this is prevalent, however. True
adjustment requires a
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r - as.data.frame(sapply(s,difftime2string),stringsAsFactors=FALSE)
names(r) - c(string)
On Nov 26, 2012, at 7:37 AM, ONKELINX, Thierry thierry.onkel...@inbo.be
wrote:
Dear all,
I'm trying to connect to an MSAccess database (ArcGIS personal geodatabase).
I keep getting an error about the channel when using sqlQuery(). However,
sqlTables() does not complain about the channel
(from R-help Digest, Vol 117, Issue 26)
Date: Sun, 25 Nov 2012 07:49:05 -0800 (PST)
From: billycorgcandi...@gmail.com
To:r-help@r-project.org
Subject: [R] creation of an high frequency series
Message-ID:1353858545067-4650744.p...@n4.nabble.com
Content-Type: text/plain; charset=us-ascii
Hi R
Hi,
Here im not able to connect with MS-SQLSERVER database with *.R-File.
Previously i was able to do in R in different machine and configuration was
R Ver-2.11.1
Package Installed : RODBC_1.3-2
And now where im doinig now there the same code im trying to execute but its
not connecting.
You should read your error messages more carefully, especially: could not
find function!
take a look at loadworkbook and loadWorkbook.
Bart
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Hello,
Suppose that i have a dataframe
a - read.dta(banca_impresa.dta)
i have a column with 17900 obs like
1
2
3
1
6
7
8
3
4
4
and i want to know the number of the different values so in this case it
would be 7
How can i do?
Thank you
--
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I am trying to call a r file named es.r which have lotes of R functions.
These R functions are internally calling java functions by using .jnew()
and .jcall(). I have added
necessary jar's to the Classpath and I am able to run es.r from command
prompt .
But when I tried to call it
Goodmorning,
I'moneafazrtrbhoquhasa
variablefactorcomtwoNivesCandH.Queoaatravésdlinear
regressionrelationshipetrehaSaerifthe
variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat
showsthe two curvesHandCLIRand
Selecionar tudo
Thank you
Ana C. Rocha Rua
Hello members,
I have this data frame with 3 columns,
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
Here, the rows 4 , 5 has type B and
Hi
I have set of records seperated by a separator say $$$ i want to get the
values in a dataframe.
eq
qwer$$12$$qwre
ewrtr$7789$ewwe
I want the output as\
V1 V2V3
qwer 12 qwre
ewrtr 7789 ewwwe
Please help me
-
Thanks in Advance
Arun
--
View this message in
Hey again!
I finally, after some work done before, had time to apply the code.
The sorting of the table did not work well or maybe something was
misunderstood.
I have a table with 973 rows and 1329 col (ascii/text file). I want to sort
the table that all columns are one under each other so that
this is wrong because with the command unique it counts the only values
that are unique ..
in my column, for instance, 1 and 4 are not unique so the formula doesn't
work in my case
--
View this message in context:
length(unique(a))
Hard Core wrote
Hello,
Suppose that i have a dataframe
a - read.dta(banca_impresa.dta)
i have a column with 17900 obs like
1
2
3
1
6
7
8
3
4
4
and i want to know the number of the different values so in this case it
would be 7
How can i do?
Thank
HI Marcel,
Sorry, that was a mistake.
I guess this should be the one:
df$condition2-ifelse(df$condition1=1,runif(length(df$condition1),0,1),df$condition1)
df$condition2
#[1] 0.5207776 0.5227388 0.5196520 0.6552820 2.000 2.000 2.000
#[8] 2.000 3.000 3.000 3.000
That did it!
Thanks so much as always.
I emailed the question to you because I think you are an R expert based
on all the suggestions, feedback and codes I have received from you in
the past.
Yes, I do look for answers in the open
forum but when it comes to a question for which the
It seems that your sqlTables also give no results.
So there must be something wrong with the ODBC connect.
I didn't use odbcConnectAccess, but made an ODBC connection, and then used
ODBCconnect to connect to our database.
Maybe you can define a user DSN and try it this way?
Bart
--
View this
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
'x' and 'y' lengths differ). With more than one value I don't get
I am following instructions online for cluster analysis using the mclust
package, and keep getting errors.
http://www.statmethods.net/advstats/cluster.html
These are the instructions (there is no sample dataset unfortunately):
# Model Based Clustering
library(mclust)
fit - Mclust(mydata)
Hey
The code need some corrections and I would kindly ask for help.
Say:
I have a table with two columns:
col1=LST and col2=NDVI
i would like to sort all data by NDVI.
in reality the NDVI ranges between 0 and 1 (although some values might be
minus also).
I want to sort by NDVI values and then
My apologies.
I still do not understand the difference; good luck.
Hard Core wrote
this is wrong because with the command unique it counts the only values
that are unique ..
in my column, for instance, 1 and 4 are not unique so the formula doesn't
work in my case
--
View this message
Hi,
maybe somebody would be kind enough to help a bloody and unprofessional
beginner like me (and I hope I did not overlook the answer to my question on
the website). I've imported a csv data frame into R, but I can't run a
regression because R interprets 4 out of 5 variables as factors (rather
You state that you want $$$ as the separator, but your example has
$$ and $ so I'm assuming an indeterminate number of consecutive
$.
x - c(qwer$$12$$qwre, ewrtr$7789$ewwe)
x - strsplit(x, \\$+)
x - do.call(rbind, x)
x - data.frame(x, stringsAsFactors = FALSE)
x$X2 - as.numeric(x$X2)
I've also
x - c(1, 2, 3, 1, 6, 7, 8, 3, 4, 4)
length(unique(x))
[1] 7
On Mon, Nov 26, 2012 at 5:20 AM, Hard Core gi...@hotmail.it wrote:
Hello,
Suppose that i have a dataframe
a - read.dta(banca_impresa.dta)
i have a column with 17900 obs like
1
2
3
1
6
7
8
3
4
4
and i want to know the
Most likely there's something amiss in your csv file: R won't convert
numeric data to factors unless there are non-numeric characters
included.
First check your csv file for errors. If that doesn't solve your
problem, please provide a reproducible example.
Hi,
Imagine the column is named XX.
Type:
nrow(table(banca_impresa$XX))
and you'll get how many different categories there are in that column.
(If you type
table(banca_impresa$XX)
you'll get the frequencies).
José
José Iparraguirre
Chief Economist
Age UK
T 020 303 31482
E
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
s - summary(as.numeric(v), ...)
r -
On Nov 26, 2012, at 3:53 AM, R_Antony antony.akk...@ge.com wrote:
Hi,
Here im not able to connect with MS-SQLSERVER database with *.R-File.
Previously i was able to do in R in different machine and configuration was
R Ver-2.11.1
Package Installed : RODBC_1.3-2
And now where im doinig
Hello,
You're wrong, unique doesn't count unique returns a vector without
repetitions.
unique(a) # includes 1 and 4
[1] 1 2 3 6 7 8 4
length(unique(a))
[1] 7
Hope this helps,
Rui Barradas
Em 26-11-2012 14:55, Hard Core escreveu:
this is wrong because with the command unique it counts the
Robert -- it may be that some of your factor variables cannot be
coerced to numeric because they have non-numeric elements.
This is why we ask for a reproducible example when posting. I
suspect just a few lines from the data frame would tell the story.
On 26-Nov-12, at 7:59 AM, EcoFranc
Lorenzo,
I'd suggest posting such questions to the R-sig-geo list, which seems
more suitable.
The book Applied Spatial Data Analysis with R written by Roger
Bivand, Edzer Pebesma and Virgilio Gómez-Rubio has exactly what you want.
Also when reading a CSV file, if you do not want characters columns
converted to factors, use 'as.is = TRUE' as one of the parameters. If
you have a column that is a factor, to convert it to numeric you have
to do the following:
as.numeric(as.character(factorColumn))
notice that you have to
EcoFranc,
di you remember to use the argument
header=TRUE
?
If not, the variable name is interpreted as data and will flip you into
factor rather than numeric.
On Mon, Nov 26, 2012 at 12:15 PM, jim holtman jholt...@gmail.com wrote:
Also when reading a CSV file, if you do not want characters
Hello,
Try the following.
x - read.table(text=
C1 C2 TYPE
10 20 A
33 44 B
66 80 A
111 140 B
144 220 B
300 340 A
380 449 A
455 500 B
510 520 A
540 580 B
, header = TRUE)
x
fun - function(x){
mn - which.min(x$C1)
mx - which.max(x$C2)
c(C1 = x$C1[mn], C2 = x$C2[mx], TYPE =
try this:
x - read.table(text = C1 C2 TYPE
+ 10 20 A
+ 33 44 B
+ 66 80 A
+ 111 140 B
+ 144 220 B
+ 300 340 A
+ 380 449 A
+ 455 500 B
+ 510 520 A
+ 540 580 B, header = TRUE, as.is = TRUE)
# mark
-- Forwarded message --
From: Catarina Maia catarinaramosm...@gmail.com
Date: 2012/11/22
Subject: prediction problem
To: r-help@r-project.org
Hello,
I am using the mda package and in particular the fda routine to
classify/predict in terms of color to a set of 20 samples for
try this: (provide sample data next time)
# create some test data
x - data.frame(LST = runif(1000), NDVI = runif(1000, 0, 1))
head(x,10) # show some data
LST NDVI
1 0.86542839 0.95129647
2 0.88910058 0.75971649
3 0.44086718 0.86532140
4 0.99879370 0.05511501
5 0.02401490
Hi,
You could do this:
Lines-qwer$$12$$qwre
ewrtr$7789$ewwe
res-unlist(strsplit(Lines,split=\\$|\n))
as.data.frame(matrix(res[res!=],nrow=2,byrow=TRUE),stringsAsFactors=FALSE)
V1 V2 V3
#1 qwer 12 qwre
#2 ewrtr 7789 ewwe
A.K.
- Original Message -
From:
Hello Petr Savicky, hello all,
I have a situation similar to the previous one,
I need to group a data.frame in a specific way,
col1 col2 score
2873 3192 319
4268 4451 183
5389 5534 145
6622 10622 4000
12631 17853 5222
20408 20615 207
21595 21838 243
23121 2313918
the out
In the example below, I don't see any $$$ separators. Are you sure this is
supposed to be the separate or just a single dollar sign? If this is the case,
you don't specify what is to happen when multiple separators appear next to
each other. From your example, it appears that two separators
I noticed the following problem with cacheSweave.
If I want to print the result of a list object, with cache=TRUE option, if I
just use summary(x), the output would not appear in the tex file. If I use
print(summary(x)) instead, the output would appear.
With cache=FALSE option, however,
does anybody have a suggestion as to how to use R to fit some date to a
cosine function and then have some output statistics that will evaluate the
fit?
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Sent from the R help mailing list archive at
The message is referring to the format of your input data. The package needs
a matrix or vector format.
I suggest using this:
xm- as.matrix(x)
return.modwt-modwt(xm, filter=la8, n.levels=5, boundary=periodic,
fast=TRUE)
--
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To whom It May Concern,
I am working on a dimensional reduction problem using Smoothly Clipped
Asolute Deviation (SCAD) Penalty according to Variable Selection via
Nonconcave Penalized Likelihood and its Oracle Properties, J.Fan and R. Li,
JASA, Dec 2001. I found an R package named *ncvreg* which
Yes, I'm sorry ... i was checking for another column so i made a mistake.
Thank you people ;)
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Sent from the R help mailing list archive at
Dear All:
I would any appreciate any help with this plot I am struggling with.
I have 4 estimates (95% CIs) I want to plot. I want the CI lines to be
horizontal on each plotted point. I was trying to tweak some old codes (was
for a vertical CI lines) into horizontal but not much dice.
Many
Hi everyone,
I have tried to produce a table with reclassification (continuous NRI, p
value) using the reclassification() function in the PredictABEL package. My
problem is that I can only get the results printed on the screen, rather
than stored in an object (as when doing a regular
Sarah:
You may not agree, but the following avoids the IMHO ugly rbind() and
separate conversion to numeric by using scan():
## first, as before..
x - c(qwer$$12$$qwre, ewrtr$7789$ewwe)
x - strsplit(x, \\$+)
## Then, instead ...
x.convert -
I have a timeseries with some missing data points that I need smoothed
using a Savitzky-Golay. Right now I am using the sgolayfilt function
in the signal package. It fails if I have NA values in the timeseries.
If I replace NA with NULL sgolayfilt does not fail but it drops the
NULL values which
Try this code that uses segments to draw in the bars
# put data into a nicer form
y=c(1.73,1.30,2.30, 1.83,1.36,2.45,1.46,1.07,2.00,1.58,1.15,2.17)
ym - matrix(y
, ncol = 3
, byrow = TRUE
, dimnames = list(NULL, c(Val, Lower, Upper))
)
ym - cbind(ym, x = 1:4) # add the x-coord
Hello,
You had a typo in the lines() instruction, the parenthesis didn't close
after c(y...).
Anyway, I'm not sure I understand but to have horizontal lines, just
reverse the roles of x and y. (And change pch = _ to pch = |).
plot(y, x, type=n,axes=F, xlab=PR(95% CI),ylab= )
Another approach is to use gsub() followed by read.table():
x - c(qwer$$12$$qwre, ewrtr$7789$ewwe)
x - gsub(\\$+, \\$, x)
x - read.table(text=x, sep=$, stringsAsFactors=FALSE)
x
V1 V2 V3
1 qwer 12 qwre
2 ewrtr 7789 ewwe
--
David L
Thank you Marc,
I will study the material you sent and follow up on this at R-SIG-DB - should I
still have problems.
I'm using a Mac OS X 10.6.8
Thanks.
Raff.
Subject: Re: [R] remote connection to an Oracle database - using RODBC -
RMySQL..?
From: marc_schwa...@me.com
Date: Wed, 21
Hi,
Just a modification:
x1-gsub([$], ,x)
read.table(text=x1,sep=,header=FALSE,stringsAsFactors=FALSE)
# V1 V2 V3
#1 qwer 12 qwre
#2 ewrtr 7789 ewwe
A.K.
- Original Message -
From: David L Carlson dcarl...@tamu.edu
To: 'Bert Gunter' gunter.ber...@gene.com; 'Sarah Goslee'
On Nov 26, 2012, at 1:47 PM, Raffaello Vardavas r_varda...@hotmail.com wrote:
Thank you Marc,
I will study the material you sent and follow up on this at R-SIG-DB - should
I still have problems.
I'm using a Mac OS X 10.6.8
Thanks.
Raff.
In that case, if you use RODBC, you will
Hi,
* Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03
-0800]:
For instance, if you want the min and max of `delay` within each group
defined by `share.id`, and let's assume `infl` is a data.frame, you
can do something like so:
R as.data.table(infl)
R setkey(infl,
Dear All,
I could use a bit of help here, this function is hard to figure out (for me at
least) I have the following so far:
PKindex-data.frame(Subject=c(1),time=c(1,2,3,4,6,10,12),conc=c(32,28,25,22,18,14,11))
Dose-200
Tinf -0.5
defun- function(time, y, parms) {
dCpdt - -parms[kel] *
Hi Sam,
On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold s...@gnu.org wrote:
Hi,
* Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03
-0800]:
For instance, if you want the min and max of `delay` within each group
defined by `share.id`, and let's assume `infl` is a
On Mon, Nov 26, 2012 at 4:46 PM, David Winsemius dwinsem...@comcast.net wrote:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
this overcomes the summary generation, but not printing:
--8---cut here---start-8---
summary.difftime - function (v, ...) {
On Mon, Nov 26, 2012 at 10:38 AM, alanaro...@sapo.pt wrote:
Goodmorning,
I'moneafazrtrbhoquhasa variablefactorcomtwoNivesCandH.Queoaatravésdlinear
regressionrelationshipetrehaSaerifthe
variablespthiedoseogmerparawanttheHtothatanddouCrespctinteRaficthat showsthe
two curvesHandCLIRand
On Mon, Nov 26, 2012 at 2:41 PM, Ludo Pagie l.pa...@nki.nl wrote:
Hi all,
there appears to be something strange with the plotting of tables of 1
dimension; if I attempt to make a plot of a table of characters with only
1 value I get an error (Error in xy.coords(x, y, xlabel, ylabel, log) :
hi Steve,
* Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 16:08:59
-0500]:
On Mon, Nov 26, 2012 at 3:13 PM, Sam Steingold s...@gnu.org wrote:
* Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-19 13:30:03
-0800]:
For instance, if you want the min and max of `delay`
* David Winsemius qjvafrz...@pbzpnfg.arg [2012-11-26 08:46:35 -0800]:
On Nov 26, 2012, at 7:14 AM, Sam Steingold wrote:
summary(infl), where infl$delay is a difftime vector, prints
...
delay
string:c(492.00 ms, 18.08 min, 1.77 hrs, 8.20 hrs, 8.13 hrs,
6.98 days)
secs :c( 0.5,
Hi,
On Mon, Nov 26, 2012 at 4:57 PM, Sam Steingold s...@gnu.org wrote:
[snip]
Could you please copy paste the output of `(head(infl, 20))` as
well as an approximation of what the result is that you want.
Don't know how dput got clipped in your reply from the quoted text I
wrote, but I actually
Hi,
* Steve Lianoglou znvyvatyvfg.ubarl...@tznvy.pbz [2012-11-26 17:32:21
-0500]:
--8---cut here---start-8---
f - data.frame(id=rep(1:3,4),country=rep(6:8,4),delay=1:12)
f
id country delay
1 1 6 1
2 2 7 2
3 3 8
It looks like summary.data.frame(d) calls format(d[[i]]) for i in
seq_len(ncol(d))
and pastes the results together into a table object for printing. Hence,
write
a format.summary.difftime if you want objects of class summary.difftime (which
I assume summary.difftime produces) to be formatted as
Hello,
Also gives an error in R 2.15.2 on Windows 7. I'd change when the
dimnames of the table are not integers to not numeric as that's what
the code for plot.table tests. And it seems to come from seq.int, since
with table value of 2, seq.int produces a vector of length 2 but the
table
Thanks a lot - almost there!
--8---cut here---start-8---
format.summary.difftime - function(sd, ...) {
t - matrix(sd$string)
rownames(t) - rownames(sd)
print(t)
format(as.table(t))
}
print.summary.difftime - function (sd, ...) {
print(format(sd),
Thanks, Arun! This pretty much does what I was looking for. Looks like I
should get more familiar with the apply functions, it seems like they can
solve a lot of these problems!
Best,
Steve
On Sun, Nov 25, 2012 at 8:09 PM, arun smartpink...@yahoo.com wrote:
HI Steve,
You could try this:
why do I see NULLs?!
because
... format.difftime does a reasonable job (except that it does not copy
the input names to its output).
Replace your call of the form
format(difftimeObject)
with
structure(format(difftimeObject), names=names(difftimeObject))
to work around this.
Bill
On Monday, November 26, 2012, Sam Steingold wrote:
[snip]
there is precisely one country for each id.
i.e., unique(country) is the same as country[1].
thanks a lot for the suggestion!
R result - f[, list(min=min(delay), max=max(delay),
count=.N,country=country[1L]), by=share.id]
And is
Dear R help
I have conducted a fa() analysis, and I want to use fa.diagram to assess the
extent to which the 11 latent factors predict the 37 items in a psychological
battery. However, the display on the screen has very large font size for the
coefficients of the relationship between the 11
Dear Bill
Wow! Thank you so much for your rapid reply - you are such a kind person,
thank you!
I'll try fa.rgraph - thanks
Thanks
Best wishes
Brent
-Original Message-
From: William R Revelle [mailto:reve...@northwestern.edu]
Sent: Tuesday, 27 November 2012 5:54 p.m.
To: Brent Caldwell
On Mon, Nov 26, 2012 at 5:33 AM, Florian Weiler fweile...@jhubc.it wrote:
Dear all,
I am using the book Generalized Linera Models and Extension by Hardin and
Hilbe (second edition, 2007) at the moment. The authors suggest that
instead of OLS models, the log link is generally used for
The question says that there is an experiement to investigate the effect on
breathing rate when doing different types of exercise with wearing more
clothes or less clothes (factor A, coded 1,2).
The duration of exercise was 10min, 20min, 30min (factor B, coded 1,2,3).
Could you give m anymore
Thank you Paul,
It works.
Now I will study for repeating the same code for different days.
Regards,
Vincent
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HI,
Just a modification of Rui's solution:
x1-within(x,TYPE-as.character(TYPE))
group-cumsum(c(TRUE,x1$TYPE[-1]!=x1$TYPE[-length(x1$TYPE)]))
res-as.data.frame(do.call(rbind,lapply(split(x1,group),function(x)
c(C1=min(x[,1]),C2=max(x[,2]),TYPE=x[,3][1]))),stringsAsFactors=FALSE)
Any idea on the reason why the instruction
p1x-c(2, 1, 0, 0, 0, 7, 1, 0, 2, 0, 5, 0, 0, 0, 0, 0, 10, 1, 3, 0, 2, 0, 0,
0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 12, 0, 0,
1, 0, 0, 0, 0, 0, 0, 2, 1, 0, 1, 0, 4, 0, 1, 0, 0, 0, 2, 0, 0, 3, 0, 0, 1,
0, 1, 0, 0, 0, 3, 0, 0, 1,
Hi all,
I have this microarray large microarray data set (ALL) from which I would like
to subset or extract a set of data based on a factor ($mol.biol).I looked
up some example of subsetting in, picked up two commands and tried both but I
got error messages as follows
testset -
My problem is relatively straight forward, but I cannot seem to find a way to
make it work.
I have a RCBD with repeated measurements over time. I have created a fit
using the gamm4 package. My model is:
fit4a - gamm4(Rate ~ s(Time,by=trt,bs=cr)+trt,data=qual.11.dat,
Thanks Jim and Rui.
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__
R-help@r-project.org mailing list
Yeah, I meant 3 instead of 5.
This was just an example, it's not what Im really doing.
I am using a 'arules' package for data mining, and I have to pass and
'arg[]' element and use it as the new column name of a data.frame.
It's a bit complicated, so I used this example, and I would like to use
In this case, instead of naming the column 'c', it names it 'args[1]' as a
string, not a variable.
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Hello,
When I run the following glm model:
modelresult=glm(CID~WS+SS+DV+DS, data=kimu, family=binomial)
I get the following warning messages:
1: glm.fit: algorithm did not converge
2: glm.fit: fitted probabilities numerically 0 or 1 occurred
What I am trying to do is model my response
I am trying to call a r file named es.r which have lotes of R functions.
These R functions are internally calling java functions by using .jnew()
and .jcall(). I have added
necessary jar's to the Classpath and I am able to run es.r from command
prompt .
But when I tried to call it
Brent,
No, cex doesn't work (as you have discovered). That is a bug.
I will work on it.
In the meantime, try Rgraphviz called from fa.rgaph or
use the output from fa.graph which produces a dot file for processing with any
graphic package (including graphviz) which handles the dot language.
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