imnew jubil...@live.com.sg writes:
Hi, I'm currently having some problem connect .mdb file into R.
I've installed the RODBC packages and I do the code this way:
channel - odbcConnectAccess(C:/Users/Documents/XYZ)
I have a total of 5 tables in the .mdb database. any one can help me with
On 10/07/2012 17:35, stanislas rebaudet wrote:
Hello,
I run [R] 2.14 on Mac OS 10.6 and I've been desperately trying to install rpy2
in order to compute spatial statistics on QGIS 1.7.3.
rpy2 is dependent upon the rgdal [R]package that I've been unable to install in
spite of up to date
On Tue, Jul 10, 2012 at 4:53 PM, Peter Ehlers ehl...@ucalgary.ca wrote:
Both this and Liviu's original solution destroy the
factor nature of 'Species' (which may not matter, of
course). How about
(.xb - iris[ iris$Species=='zz', ])
.xb - .xb[1, ] # this probably shouldn't work, but it
This is a fascinating example. It is an accident that it looks like it
comes out even.
Here are both sets displayed using hex format where we see exactly what
bits are used
in the internal representation of the numbers.
dphex - function(x) {sprintf(%+.13a, x)}
ones - c(-1.1,
+
Hello R community,
I am attempting to run multiple logistic regressions (multinomial, via
package 'nnet'), with Automated Model Selection (dredge, package 'MuMIn').
The aim is to reduce the number of predictor variables by assessing relative
performance of each variable, which can be done in a
Dear Jessica
thank you for the scale solution to my problem.
I tried to manually scale my data (scaling up and removing decimals),
however, this resulted in the same error message.
It remains vague to me what the precise meaning of...
the model matrix should be sensibly scaled with all columns
All,
How come i=1 in the first case, but i=2 in the second case. The second case
seems to work, but the first case does not.
David.
findruns - function(x,k) {
+ n - length(x)
+ runs - NULL
+ for (i in 1:(n-k+1)) {
+ if (all(x[i:(i+k-1)]==1)) runs - c(runs,i)
+ browser(i1)
+ }
+
there is a data frame £¬x
weekly.returns
2010-1-4-0.015933327
2010-1-11-0.015042868
2010-1-180.005350297
2010-1-25-0.049324703
2010-2-1-0.052674121
when i input
colnames(data)[1] - 'date'
it becomes
date
2010-1-4
Hi,
Let's say you original data.frame is called df1, you can do:
df2 - data.frame(date=row.names(df1), weekly.returns=df1[[1]])
Next time, think about giving us you sample data through the dput()
function. It makes it way easier.
HTH,
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR
I'm not sure either, the wordings starnge and english isn't my primary language
I would GUESS, that it means that all columns values should be between 0,1 or
1,2 or -0.5,0.5 or something like that. The scale function scales the columns
to be comparable between each other, by dividing each
On 07/11/2012 05:54 AM, Rich Shepard wrote:
Before reading water chemistry into a data frame I removed all missing
data. Yet when I try to run cenros() to summarize a specific chemical I get
an error that I do not understand:
with( subset(chem, param=='Ag'), cenros(quant,ceneq1) )
Error in
Hi,
On Wednesday, July 11, 2012, darnold wrote:
All,
How come i=1 in the first case, but i=2 in the second case. The second case
seems to work, but the first case does not.
It dos work: whne your code gets to browser() R calls that function and
passes it the argument given. If you expect
Hello,
Try
make.row - function(x, skip = 14, column){
dat - read.table(x, skip = skip - 1, header = TRUE,
stringsAsFactors = FALSE)
vpNum - dat$vpNum[1]
trial - length(dat[[ column ]])
correct - sum(dat$correct == 1)
result - c(vpNum, trial, correct, correct/trial)
You actually jsut need to say what the comment char and what the
na.strings are:
read.table(filename, sep=;, skip=3, header=TRUE, na.string=#N/A,
comment.char=)
Uwe Ligges
On 10.07.2012 19:30, arun wrote:
Hello Ravi,
I was not aware that your dataset have special character # before NA.
Dear All,
The first issue of the fourth volume of The R Journal is now available
at http://journal.r-project.org/current.html
Thanks to everyone involved.
Martyn Plummer
Editor-in-Chief
---
This message and its attachments
On 10.07.2012 10:48, WATSON Mick wrote:
Dear All
According to the identify.hclust documentation the function cuts the tree at the
vertical position of the pointer and highlights the cluster containing the horizontal
position of the pointer.
When I carry out this, the tree isn't cut where I
On 10.07.2012 05:37, Manish Gupta wrote:
Hi,
I am working on stacked bar plot and want to add marker(arrow) in stacked
bar plot.
DF=data.frame(names=c(tomato, potato, cabbage, sukuma-wiki,
terere), freq=c(7,4,5,8,20))
barplot(as.matrix(DF[,2]), col=heat.colors(length(DF[,2])), legend=DF[,1],
On 10.07.2012 02:55, Angelr wrote:
I am calling quantiles as follows. I don't understand why sometimes the
columns (data values) above 95% are returned as NULL!! When I drop the
percentile down to 92%, I see colums appearing. Why would any quantile be
empty? I see sometimes that 95% percentile
Incidentally, the OP might want to take a look at ?rle.
Michael
On Jul 11, 2012, at 3:42 AM, Sarah Goslee sarah.gos...@gmail.com wrote:
Hi,
On Wednesday, July 11, 2012, darnold wrote:
All,
How come i=1 in the first case, but i=2 in the second case. The second case
seems to work, but
Packaging ist still too complicated for my level i guess, can't get it to work
properly, and it seems a lot of work.
However apparently putting a . in front of functions will hide them, saw that
in another script, don't know what its about, but it works, so who cares..
On 10.07.2012, at 17:02,
On Jul 10, 2012, at 2:05 PM, Rui Barradas wrote:
Hello,
Em 10-07-2012 18:59, Peter Ehlers escreveu:
On 2012-07-10 08:50, Brian Diggs wrote:
On 7/10/2012 7:53 AM, Peter Ehlers wrote:
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
On Tue, Jul 10, 2012 at 4:02 PM, Jessica Streicher
j.streic...@micromata.de wrote:
Forget about that, i'm stupid and can't use the tools available...
Here's something I wrote specially for all you wonderful stupid
people out there...
https://gist.github.com/3025606
What it tries to do is
On Tue, Jul 10, 2012 at 9:15 AM, Liviu Andronic landronim...@gmail.com wrote:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
Hi,
i need to get functions name from R file
example
test.R file having 2 functions like
add-function()
{
a+b
}
sub-function()
{
a-b
}
how to get function name from test.R file
i need output is functions are add,sub.
Thanks
B.Purushothaman
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On Wed, 11 Jul 2012, Jim Lemon wrote:
I don't have the NADA package, but I suspect that the cenros function is
doing something like dividing by zero. With that much data, it may be hard
to pinpoint where this is occurring. I would cut my data in half, run it,
cut the remainder in half, run it
Dear Rui,
thank you VERY much.
You helped me a lot!
I've just added the following:
rsort - ratios[order(ratios$vpNum),]
Now the test subjects are arranged according to their vpNum.
Thanks a lot again!
--
View this message in context:
Thanks Barry, that helped a lot, modified it a bit and it does pretty much what
i wanted.
On 11.07.2012, at 15:08, Barry Rowlingson wrote:
On Tue, Jul 10, 2012 at 4:02 PM, Jessica Streicher
j.streic...@micromata.de wrote:
Forget about that, i'm stupid and can't use the tools available...
Hi,
how about this:
sandbox-new.env()
sys.source(test.R,envir=sandbox)
#all objects defined in test.r
(tstf-ls(env=sandbox))
#just the functions
tstf[sapply(tstf, function(n) exists(n, envir = sandbox, mode =
function, inherits = FALSE))]
hth
Eik
Am 11.07.2012 15:22, schrieb purushothaman:
Why fails nls with singular gradient here?
I post a minimal example on the bottom and would be very
happy if someone could help me.
Kind regards,
###
# define some constants
smallc - 0.0001
t - seq(0,1,0.001)
t0 - 0.5
tau1 - 0.02
# generate yy(t)
yy - 1/2 * ( 1- tanh((t - t0)/smallc)
On 11/07/2012 11:04 AM, Jonas Stein wrote:
Why fails nls with singular gradient here?
I post a minimal example on the bottom and would be very
happy if someone could help me.
Take a look at the predicted values at your starting fit: there's a
discontinuity at 0.4, which sure makes it look as
Hi all,
I just try to get familiar with levelplot() for generating heatmaps. I
have x, y, z-data stored in a file: e.g.:
file.csv:
0 0 0.1
0 1 0.5
0 2 0.4
1 0 0.3
1 1 0.4
1 2 0.6
...
I can use scan() for generating the matrix for R:
inp - scan(file.csv, list(0, 0, 0))
How can I feed inp into
I might suggest you use read.csv() instead of scan if your data really
is a csv (what you provided isn't).
Once you have it in R, I'd recommend you use a formula interface to
levelplot(). Taking an example from the help page [which you can
access by typing ?levelplot and run the examples by
# One more question, Joshua: let instead of merging tickers
# I would like to put prices from an OHLC object
# in weekly format, then selecting just the close prices.
# What would be a code to do it?
# I guess:
data = new.env()
ticker.list - c('SPY', 'TLT', 'GLD')
getSymbols(ticker.list, env =
Hi Don,
Sorry that this may be too late for what you were working on, but wanted to
reply just in case you have to tackle this in the future.. :)
library(SAScii)
x - read.SAScii( d:/fh4.txt , d:/sas input script.sas )
Anthony
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Read your data using read.table with as.is=TRUE
then you could reformat the date column to make R understand that it is a
date, depending on your raw date format.
Thanks.
--
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Sent from the R help
Hi,
Try this:
dat1-read.table(text=
weekly.returns
2010-1-4 -0.015933327
2010-1-11 -0.015042868
2010-1-18 0.005350297
2010-1-25 -0.049324703
2010-2-1 -0.052674121
,sep=,header=TRUE)
dat2-data.frame(date=row.names(dat1),weekly.returns=dat1[1])
read the part on power fitting:
http://www.itc.nl/~rossiter/teach/R/R_CurveFit.pdf
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Sent from the R help mailing list archive at Nabble.com.
__
Thanks a lot for the guidance. I have another text file with a time stamp and
an empty column as given below:
First line: Skip this line
Second line: skip this line
Third line: skip this line
variable1
Your error comes from the source file makebin
it's because your sequence identifiers are not in ascending order:
// check if the sequence and event/time
// identifiers are in ascending order and
// determine the number of sequences.
ns = l = h = 0;
for (i = 0; i LENGTH(sx);
Hmm, I guess that's not exactly what I needed. It's my fault; you gave me
what I asked for, I was just asking for the wrong thing.
What I'm hoping to do is something similar to:
aggregate(dat$SunScore, by=list(h), median)
except I'd like to do it by half hour instead of hour. I guess what I
Hi,
Try this:
.xa-iris[1,][rep(NA,length(iris),1),]
.xa
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#NA NA NA NA NA NA
#or
.xb-iris[1,][rep(NA,ncol(iris),1),]
.xb
# Sepal.Length Sepal.Width Petal.Length Petal.Width Species
#NA
Hi,
Here i have an matrix like this,
ABCPQRXYZ MNO
-- --- --
367 15
2 122415
20 5 1 2
25 50 15 35
i need to get the
You can try this:
for (i in 1:length(ds)) {
dummy-ds[[i]];
dummy[is.nan(dummy)]-0
ds[[i]]-dummy
}
if is.nan doesn't work, replace with is.na
-
Yasir Kaheil
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Thank you chamilka.
From: chamilka [via R] [mailto:ml-node+s789695n4636142...@n4.nabble.com]
Sent: Wednesday, July 11, 2012 8:13 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: MODE , VARIANCE , NTH PERCENTAILE
Din't you try sapply function?
I tried it for you.
Just convert your
I'm stuck on a seemingly simple problem. I'm trying to subset the data by
several numbers and it cuts out half of the rows. Here is the sample code:
test - as.matrix(c(1,1,1,1,3,3,7,7,7,7))
Count - tapply(test[,1], test[,1], length) # count for each value
spp - unique(test[,1])
Count1 -
I see that this is your input: c(1,1,1,1,3,3,7,7,7,7)
what do you want the output to be?
what could the multiple values of AllMax be?
-
Yasir Kaheil
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Sent from the R help
An my easy but not very useful answer is that this particular subset
probably violates some assumption of the cenros() model. I myself would
start with simple inspections of the data, such as
with( subset(chem, param=='Ag'), table(ceneq1) )
with( subset(chem, param=='Ag'), qqnorm(quant) )
Hello useRs,
I'm having trouble trying to produce axis labels running parallel to the
axes in 3d space. Is their a way to do this when producing 3d graphs using
the rgl package?
Thanks for your help,
Patrick
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Why does one want to replace a zero-row data.frame
with a one-row data.frame of NA's? Unless this is for
an external program that cannot handle zero-row inputs,
this suggests that there is an unnecessary limitation (i.e.,
a bug) in the R code that uses this data.frame.
Bill Dunlap
Spotfire,
On Wed, Jul 11, 2012 at 1:49 PM, Cren oscar.soppe...@bancaakros.it wrote:
# One more question, Joshua: let instead of merging tickers
# I would like to put prices from an OHLC object
# in weekly format, then selecting just the close prices.
# What would be a code to do it?
# I guess:
data =
On Wed, Jul 11, 2012 at 3:08 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
On Wed, Jul 11, 2012 at 3:07 PM, R. Michael Weylandt
michael.weyla...@gmail.com wrote:
On Wed, Jul 11, 2012 at 1:49 PM, Cren oscar.soppe...@bancaakros.it wrote:
# One more question, Joshua: let instead of
Hi. Maybe this:
ct - table(test)
as.numeric(names(ct[ct==max(ct)]))
test[test[,1]%in%as.numeric(names(ct[ct==max(ct)])),,drop=FALSE]
?
Andrija
On Wed, Jul 11, 2012 at 8:33 PM, Amanduh320 aadam...@uwo.ca wrote:
I'm stuck on a seemingly simple problem. I'm trying to subset the data by
several
On Wed, 11 Jul 2012, MacQueen, Don wrote:
An my easy but not very useful answer is that this particular subset
probably violates some assumption of the cenros() model. I myself would
start with simple inspections of the data, such as
with( subset(chem, param=='Ag'), table(ceneq1) )
with(
Hi everyone!
I already posted
http://r.789695.n4.nabble.com/Declaring-a-density-function-with-for-loop-td4635699.html
a question on finding density values of a new Binomial like distribution
which has the following pmf:
http://r.789695.n4.nabble.com/file/n4636134/kb.png
Thank fully
Hi Bill
Many thanks for your help.
Cheers
R
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: 10 July 2012 17:22
To: Raghuraman Ramachandran; r-help@r-project.org
Subject: RE: Help with vectors and rollapply
It looks like you already have the zoo package loaded
Hi Dude!!
It seems that your column names are not sequentially arranged, I guess,
If so u just reorder them and do it your way...it will work..
for example,
colnames(data)[2]–weekly.returns ## sometimes u can use like this as
well, if u want to assign at last most column
-- Forwarded message --
From: umesh khatri khatriumes...@gmail.com
Date: Wed, 11 Jul 2012 16:15:11 +0530
Subject: Poisson ridge regression
To: r-h...@rproject.org
I'm student of Masters in Statistics (Actuarial) from Central
University of Rajasthan, India. I am doing a major
Thank you! It is fixed now.
However, now when I'm trying with hist(skatter) i get this message: Error
in hist.default(skatter) : 'x' must be numeric
I don't know what I'm doing wrong but it worked perfectly on windows some
weeks ago.
My data skater looks like this:
skatter
I can't seem to determine how to get the name of a list member to
substitute:
ll - list(a1 = a,a2 = b)
t1[t==ll[0], v] - 99
why doesn't this substitute to:
t1[t==a, v] - 99
Thank you!
--
Charles Stangor
Professor
[[alternative HTML version deleted]]
Thank you! It is fixed now.
However, now when I'm trying with hist(skatter) i get this message: Error
in hist.default(skatter) : 'x' must be numeric
I don't know what I'm doing wrong but it worked perfectly on windows some
weeks ago.
My data skatter looks like this:
skatter
I have a data assimilation problem that might be amenable to the use of GAMS,
but I am not sure how feasible it is to implement. I was told the R mailing
list was a great resource.
My observations are spatiotemporal salinity in the San Francisco Bay at a
number of instruments over a few days.
Hello everyone,
I'm relatively new to the RCurl package, and I'm having some issues with
reading in my data the way I want to. I have binary data that I've only been
able to read in correctly using getBinaryURL(), but I also want to read the
data in chunks (it's a huge timeseries dataset).
Hello,
I've been using: tmp.df = read.xls(filename, stringsAsFactors = FALSE) to
read in my files. Even though I get the There were 50 or more warnings thing,
for the most part most of the data is read in correctly. However, there are a
few select rows where there are values but they are
Hi Everyone,
I could use help developing a for loop. I have a dataset with tallies for
a number of species within 7 different size classes. I need to uncollate the
data into the rawest form (ie: each row a different individual) retrain
the metadata associated with each row (Date, Recorder,
I can't seem to determine how to get the name of a list member to
substitute:
ll - list(a1 = a,a2 = b)
t1[t==ll[0], v] - 99
why doesn't this substitute to:
t1[t==a, v] - 99
Thank you!
[[alternative HTML version deleted]]
__
Hi,
I have a list called ds which has the following attributes:
attributes(ds)
$names
[1] adj.r.squared fstatisticintercept slope
[5] std.error tstatistic
I want to replace all the NaN is ds with 0, and after searching past posts I
found I can hardcode it like this:
Hi,
I have two dataframes:
The first, df1, contains some missing data:
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
The second, df2, contains the following:
cola colb colc cold cole
1 1.4 0.8
Dear everyone,
I'm student of Masters in Statistics (Actuarial) from Central
University of Rajasthan, India. I am doing a major project work as a
part of the degree. My major project deals with fitting a glm model
for the data of car insurance. I'm facing the problem of
multicollinearity for this
I want the output to be 1,1,1,1,7,7,7,7
The multiple values of AllMax are 1 and 7.
I think I've figured it out though, I added a loop at the end:
test - as.matrix(c(1,1,1,1,3,3,7,7,7,7))
Count - tapply(test[,1], test[,1], length) # count for each value
spp - unique(test[,1])
Count1 -
Dear Simon,
Thanks for the quick reply.
Unfortunately I don't have access to Pinheiro and Bates. I tried googling
the pdSymm and lme but I still cannot get the syntax right.
In my model, I only have 1 random factor with repetitions (groups) (e.g. 2
records per each level)
I am pasting bellow a
yes just use %in% instead of == AllMax.. but also use table for count
-
Yasir Kaheil
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Thanks. That worked.
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R-help@r-project.org mailing list
Hello,
I wanted to know if there is a simple way of getting the inverse cdf for a
KDE estimate of a density (using the ks or KernSmooth packages) in R ?
The method I'm using now is to perform a numerical integration of the pdf
to get the cdf and then doing a search for the desired probablity
Hi,
I've installed the igraph package and have been otherwise using it
successfully, but when I try to use graph.bfs I get the error:
could not find function graph.bfs
Moreover, I don't seem to have the documentation installed either. (per
?graph.bfs and ??graph.bfs).
I'm using RStudio
Did you use read.table(filename, ..., dec=,) when importing
the data (so 30,3 is read as the number 30 and 3 tenths instead
of as the character string 30,3)?
Whenever importing data follow up by using str() or summary()
on its output, before doing any further analysis. E.g.,
bad -
Din't you try sapply function?
I tried it for you.
Just convert your matrix into a data frame using as.data.frame and then
* rantony*
ABC PQR XYZ MNO
[1,] 3 6 7 15
[2,] 2 12 24 15
[3,] 20 5 1 2
[4,] 25 50 15 35
* rantony=as.data.frame(rantony)*
*
Dear all,
This is what I'd like to do (I have an implementation using for loops, which I
designed before I realised just how slow R is at executing them - this process
currently takes days to run).
I have a large dataframe containing corporate bond data, columns are:
BondID
Date (goes back
Hey guys,
So I'm working with a project where I manage a database within R, and I'm
developing a script/function that will automatically run my queries in R
depending on the date parameters passed in.
The problem is that when I create variables for the dates, and use those
variables in my sqldf
Thanks Peter.
We did manage to solve the problem using the approach bellow.
Hab - cbind(seq(1,21),habitat)
hab - matrix(NA,nrow=3,ncol=7)
for (i in 1:3) {hab[i,] - Hab[habitat==i,1]}
The problem was that the hab matrix was not a multiple of habitat.
--
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I can't seem to determine how to get the name of a list member to
substitute as a variable name:
ll - list(a1 = a,a2 = b)
t1[t==ll[1], v] - 99
why doesn't this substitute to:
t1[t==a, v] - 99
Thank you!
--
Charles Stangor
Professor
--
Charles Stangor
Professor
[[alternative
Hello,
If I understanding it, what you want are the values below.
half.hours - function(x){
s - strsplit(as.character(x), :)
H - as.integer(sapply(s, `[`, 1))
M - as.integer(sapply(s, `[`, 2))
h - (H*60 + M)/60
floor(h/0.5)*0.5
}
half.hours(dat$SunTime)
thanks!
2012/7/11 Yasir kah...@gmail.com
Your error comes from the source file makebin
it's because your sequence identifiers are not in ascending order:
// check if the sequence and event/time
// identifiers are in ascending order and
// determine the number of sequences.
ns
On Wed, Jul 11, 2012 at 9:56 PM, William Dunlap wdun...@tibco.com wrote:
Why does one want to replace a zero-row data.frame
with a one-row data.frame of NA's? Unless this is for
an external program that cannot handle zero-row inputs,
this suggests that there is an unnecessary limitation
Hi, Prof. Ripley, Bert:
Thanks for the comments.
Might there be a function similar to c but retains more
attributes than just names?
Before I write such, I felt a need to ask if anyone knows where
it may already have been done -- and if people have suggestions for how
Hello,
Please don't post datasets like this, it's unusable by us. Use dput().
?dput
dput(head(myData, 20)) # post the output of this.
Paste the output of that command in a post. It starts with 'structure'.
Don't worry if it looks awkward, it is, on the contrary, very usefull.
All we need
It should. Reproducible example of it not?
Michael
On Jul 11, 2012, at 3:06 PM, Charles Stangor cstan...@gmail.com wrote:
I can't seem to determine how to get the name of a list member to
substitute as a variable name:
ll - list(a1 = a,a2 = b)
t1[t==ll[1], v] - 99
why doesn't this
Hello,
A one-liner could be
df1 - read.table(text=
cola colb colc cold cole
1NA59 NA 17
2NA6 NA 14 NA
3 3NA 11 15 19
4 48 12 NA 20
, header=TRUE)
df2 - read.table(text=
cola colb colc cold cole
1 1.4 0.8 0.02 1.6 0.6
, header=TRUE)
On Wed, Jul 11, 2012 at 11:16 AM, scstrein scstr...@ncsu.edu wrote:
Hey guys,
So I'm working with a project where I manage a database within R, and I'm
developing a script/function that will automatically run my queries in R
depending on the date parameters passed in.
The problem is that
On Wed, Jul 11, 2012 at 10:05 AM, Russell Bowdrey
russell.bowd...@justretirement.com wrote:
Dear all,
This is what I'd like to do (I have an implementation using for loops, which
I designed before I realised just how slow R is at executing them - this
process currently takes days to run).
Hi All,
I need to calculate VIF (variance inflation factor) for my linear regression
model. I found there was a function named vif in 'HH' package. I have two
questions:
1) I was able to install that package in my R under windows. But while
trying to install that package in UNIX, I
Hello,
That seems easy.
dat$variable1 - with(dat, paste(variable1, variable2))
dat$variable2 - dat$variable3
dat$variable3 -
Then convert variable1 to date/time using as.POSIXct or strptime
See ?strptime.
Hope this helps,
Rui Barradas
Em 11-07-2012 13:30, vioravis escreveu:
Thanks a lot
See the examples at
# install.pacages('car')
require(car)
?vif
HTH,
Jorge.-
On Wed, Jul 11, 2012 at 6:10 PM, Hui Du wrote:
Hi All,
I need to calculate VIF (variance inflation factor) for my linear
regression model. I found there was a function named vif in 'HH' package.
I have two
Thanks. But in UNIX side, I got the same error
In getDependencies(pkgs, dependencies, available, lib) :
package ââ¬Ëcarââ¬â¢ is not available
HXD
From: Jorge I Velez [mailto:jorgeivanve...@gmail.com]
Sent: Wednesday, July 11, 2012 3:19 PM
To: Hui Du
Cc: R-help
Subject: Re: [R] Variance
Dear R-ers,
I feel I am close, but can't get it quite right.
Thanks a lot for your help!
Dimitri
# I have 2 data frames:
x-data.frame(a=c(aa,aa,ab,ab,ba,ba,bb,bb),b=c(1:2,1:2,1:2,1:2),d=c(10,20,30,40,50,60,70,80))
Could you please include your sessionInfo() ?
Thank you,
Jorge.-
On Wed, Jul 11, 2012 at 6:27 PM, Hui Du hui...@dataventures.com wrote:
Thanks. But in UNIX side, I got the same error
** **
In getDependencies(pkgs, dependencies, available, lib) :
package âcarâ is not
Thank you for your consideration.
sessionInfo()
R version 2.11.1 (2010-05-31)
i686-pc-linux-gnu
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_US.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=C LC_MESSAGES=en_US.UTF-8
[7] LC_PAPER=en_US.UTF-8
Hi Dimitri,
Try creating a key for x and y and then merging the result by that
variable:
x$key - with(x, paste(a, b, sep = /))
y$key - with(y, paste(a2, b, sep = /))
merge(x, y, by = 'key')[, c(2:4, 8:9)]
HTH,
Jorge.-
On Wed, Jul 11, 2012 at 6:28 PM, Dimitri Liakhovitski wrote:
Dear R-ers,
In that case, I think that using a subscript of NA is the
best way to go. It works for both matrices and data.frames
(unlike an integer larger than nrow(data)) and its meaning
is pretty clear.
Also, you will probably get better results if the function
in your call to apply() returns the index
Hui Du,
There is probably something wrong with either your setup or your choice of
CRAN
mirror. Both HH and car are present in CRAN for all platforms.
---Your session info just arrived. R version 2.11.1 (2010-05-31) is over
two years old.
The current version is R version 2.15.1 (2012-06-22).
Why did you use the 'lower.tri' syntax?
Does this work for you?
lme(Y~Random, data = DATA,
random = list(Random = pdSymm(CovM,~Random)))
Kevin
On Wed, Jul 11, 2012 at 9:27 AM, Marcio mrese...@ufl.edu wrote:
Dear Simon,
Thanks for the quick reply.
Unfortunately I don't have access to
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