Hello all,
Thanks for all your replies. I have studied on it some more in the meantime,
and found indeed out that what I was trying to do was not correct to begin
with. Sorry to have wasted your time, but thanks for the comments.
--
View this message in context:
http://r.789695.n4.nabble.com/H
Hi Lucas
The HTML page is formatted by using tables in each of the cells
of the top-most table. As a result, the simple table is much more
complex. readHTMLTable() is intended for quick and easy tables.
For tables such as this, you have to implement more customized processors.
doc =
htmlParse
I understand where you are coming from, but the issue is that some
exploration of the data through graphs and the like, showed that patterns
could be seen. However with only 7 means it is extremely difficult to get
any kind of statistical evidence and as some mean values are the same some
of the te
Hi
With package:gmp, is this an expected behavior?
> rep(1:3, rep(3, 3))
[1] 1 1 1 2 2 2 3 3 3
> rep(as.bigz(1:3), rep(3, 3))
Big Integer ('bigz') object of length 9:
[1] 1 2 3 1 2 3 1 2 3
This code is used inside `outer`, so more worse
> outer(1:3, 1:3, `*`)
[,1] [,2] [,3]
[1,]12
AFAIK you need to use the 32bit version of R for getting the data. Then you can
save it into a more versatile format and re-read it if you really need to run R
in 64 bit mode.
---
Jeff NewmillerThe
... perhaps also worth mentioning:
"The combination of some data and an aching desire for an answer does
not ensure that a reasonable answer can be extracted from a given body
of data. "
-- John Tukey
-- Bert
On Tue, Mar 27, 2012 at 7:55 PM, Dragonwalker
wrote:
> Hello all,
> If someone could t
Inline.
On Tue, Mar 27, 2012 at 5:28 PM, Luisin Galindo, PhD
wrote:
> Dear ReXperts,
>
> I have the below text file output. I need to extract the T, QC, QO, QO-QC
> and WT columns for
> the data between T = 10 and T=150.
>
> Any ideas?
Lots. They all begin with:
?"[.data.frame"
-- Bert
>
> Than
You've got to be kidding!
You are requesting extensive statistical consulting from the R-Help
list. That is not the purpose of this list, nor is it reasonable to
expect remote statisticians unfamiliar with your work or state of
understanding (which appears to be rather sketchy) to provide reliable
It's a bit of a hack, but I think you can try something like this:
x <- c(1,2,3,4,5,2,5)
duplicated(x) | duplicated(x, fromLast=T)
Michael
On Tue, Mar 27, 2012 at 7:02 PM, z2.0 wrote:
> I'm sure there's a better way to do this using plyr. I just can't nail the
> right series of commands.
>
> I
The Design package is obsolete and has been replaced by the rms package. But
both are the same in this regard. By starting at zero I assume you really
meant to say that you wanted to plot one minus cumulative probability of
survival, i.e., cumulative incidence. Do this with survplot(...,
fun=fun
? subset
subset(x, (T > 10) & (T < 150), c("T", "QC", "QO", "QO-QC")
Michael
On Tue, Mar 27, 2012 at 8:28 PM, Luisin Galindo, PhD
wrote:
> Dear ReXperts,
>
> I have the below text file output. I need to extract the T, QC, QO, QO-QC
> and WT columns for
> the data between T = 10 and T=150.
>
> A
See one more R-script (to prepare data for MASF Control Chart) in my other
blog post:
http://itrubin.blogspot.com/2012/03/r-script-to-aggregate-etl-to-mysql.html
R-Script to Aggregate (ETL to MySQL) Actual data with Base-line data for
IT-Control Charts
--
View this message in context:
http://r
The official way to find the entry point for C code is to look in
/src/main/names.c but its often easier just to grep for it.
Either way, you wind up here:
http://svn.r-project.org/R/trunk/src/library/stats/src/arima.c
Hope this helps,
Michael
On Tue, Mar 27, 2012 at 10:45 PM, Fretheim, Alexand
Just additional comment to my previous comment:
The most important part of A Control chart and especially MASF chart is data
that is suppose to BE pre-processed (ETLed) to a Data Cubical format before
chart is actually plotted. I wrote R-script using RODBC package to
illustrate how that works. See
Good Night
I made different test to check normality and multinormality in my dataset,
but I don´t know which test is better.
To verify univariate normality I checked: shapiro.test, cvm.test, ad.test,
lillie.test, sf.test or jaque.bera.test and
To verify multivariate normal distribution I use mar
Hello,
I'm new to R and am trying to access data from a Microsoft Access 2010
database. I've read through the RODBC package vignette and seem to
understand the instructions of the commands but can't see to connect to the
database. I have installed the Access Database Engine from the Microsoft
webs
try anova if they have same observe points.
conditions<-rep(c(1,2),each=1600)
response<-sample (1:20,1600*2, replace= TRUE)
points<-rep(rep(1:400, each=4),2)
replicates<-rep(1:4,800)
obs<-data.frame(response,conditions,points,replicates)
fit <- lm(response~conditions+points,data=obs)
anova(fit)
Dear ReXperts,
I have the below text file output. I need to extract the T, QC, QO, QO-QC
and WT columns for
the data between T = 10 and T=150.
Any ideas?
Thanks in advance.
1 D C ---CAT-- T
Hello all,
If someone could take a little time to help me then I would be very
grateful.
I studied piping plovers last summer. I watched each chick within a brood
for 5 minutes and recorded behaviour, habitat use and foraging rate.
There were two Sites, the first with 4 broods and the second with 3
I'm sure there's a better way to do this using plyr. I just can't nail the
right series of commands.
I've got a vector of strings. I want to remove all where the string's count
within the vector is > 1.
Right now I'm using:
Where x2 is the original data.frame and my character strings live i
I want to plot many points and want to use circles. The filling color
depends on variable a. if a=1, then not fill
if a=2 then fill with red, if a=3 then fill with blue, if a=4, fill
half with red and half with blue. Can anyone tell me how to plot the
case "a=4"? Thanks a lot
_
Dear People
I can't figure out how to fix this problem: rgl won't run under R
2.14.2 (it was working for me before under 2.14.0). The error message
is:
> library(rgl)
Error : .onLoad failed in loadNamespace() for 'rgl', details:
call: dyn.load(file, DLLpath = DLLpath, ...)
error: una
Could you please post a small example of your data and code which gives you
this error. Your assumed error distribution sounds reasonable. I am
interested as to why you have zeros... you have sites with species richness
==0 ??
Lívia Dorneles Audino wrote
>
> I'm trying to make a glmm to i
Hello,
>
> my idea is to get results like this:
> user, sector, source, destine, count, average
> 7 1 22 22 4 186.25 #
> (109+100+214+322)
> 7 2 161 97 1 68
> 7 2 97 97
Bert,
Try posting on the R-sig-ME list for help with mixed models.
Cheer,
Neil
On Wed, Mar 28, 2012 at 1:16 AM, Bert Harris wrote:
> HI all,
>
> I am planning to get Zuur et al.'s new book when it comes out, but until
> then I was wondering if anyone could suggest examples of zero inflated o
Dear R,
Thanks for helping me locate the source for the StructTS method from
stats, but I've run in to a roadblock in reverse engineering it to locate a
formula for its forecasting because it calls some compiled C code, a function
called KalmanLike. I've looked through that R library that
I wasn't thinking straight.
> old.data= 11:20
> recalc.please= (old.data%%2==0)
> old.data[recalc.please]
[1] 12 14 16 18 20
> new.data[recalc.please]= old.data[recalc.please]^2
Error in new.data[recalc.please] = old.data[recalc.please]^2 :
object 'new.data' not found
# this is where I had give
Dear R-list,
I'm queering a M$ Access database with the sqlQuery function from the RODBC
library. As I cannot make a working example with a database here is an
illustrative example,
library(RODBC)
mdbConnect<-odbcConnectAccess("S:/data/ ... /databse.mdb")
data <- sqlQuery(mdbConnect, "select id
It is (at least for me) really unclear what the problem is, or how
it's related to mclapply.
You say
" this works fine, except that what I want to get NA's in the return
positions that were not recalculated. then, I can write
>
> newdata$y <- ifelse ( is.na(olddata$y), mc.byselectrows( olddata,
There is a mailing list R-Sig-Geo which is more appropriate for
questions about the rgdal and related packages.
If by "read the information GDType" you mean to get that "Int16"
description you can get it by delving into the attributes of the
GDALinfo return value, for example:
f <- system.file("p
On 2012-03-27 15:11, Ben Bolker wrote:
Lívia Dorneles Audino gmail.com> writes:
I'm trying to make a glmm to identify the relationship between insect
species richness with fragment size, isolation and time (different years).
I already tried to analyse it using poisson distribution error, but
I solved some of my problems, but the one that remains is
that reading the two-dimensional arrays into R transposes the matrix.
The arrays I want to read are unequal interval time multi series with
the first column being the times which are converted in java from
calendar CnYrMoDaHrMnScDCMQ or C
I believe it was 2008.
Hadley
On Mon, Mar 26, 2012 at 11:46 AM, Marina Doucerain
wrote:
> Hello,
>
> I'm wondering what was the year (or year range) of collection for the data
> included in the 'diamonds' dataset in ggplot2.
> This information would be very helpful in interpreting the 'price' var
Lívia Dorneles Audino gmail.com> writes:
>
> I'm trying to make a glmm to identify the relationship between insect
> species richness with fragment size, isolation and time (different years).
> I already tried to analyse it using poisson distribution error, but I
> always face with the following
hi, I'm a beginner of tcltk packages.
I'm making some gui for some function and want to change the background
color that is grey in default.
anybody who knows the way that changes the color of it plz teach me how to
do that.
Forthemore is there a nice manual for tclck?
Thanks.
--
View this mess
Dear R-helpers
I am wondering if there is an option to the survplot function in the design
package that allows for drawing Kaplan-Meier plots starting from 0 instead of
1, similar like fun = 'event' in the standard plotting function used on a
survfit object.
I apologize in advance for having mi
On 27/03/2012 19:30, Justin Haynes wrote:
There may very well be a better solution, but this works.
format(strptime(dayofyear, format="%j"), format="%m-%d")
The answer depends on the year (think leap years), so I think you need
strptime(paste("2008", dayofyear), format="%Y %j")
Probably a be
Hello Dr. Winsemius,
Not sure how or if the use of NAs you describe applies to my case. I'll go back
to this again when the ggplot2 book arrives. It may be that this will provide a
helpful insight then.
Thanks,
Paul
--- On Fri, 3/23/12, David Winsemius wrote:
> From: David Winsemius
> Subj
I wrote in my previous message the following Octave code:
[Octave code:]
octave:1> x=1;
octave:2> save -ascii testdata.mat x
Forget the "-ascii". It should be "-text" or nothing ("-text" is the
default).
By the way, read.octave() does not really "fail" (it does return a
value), but the result is
Hello,
I am modelling positive continuous data (including zeros) using the ZAGA
distribution in GAMLSS and want to use the model for predictions. My final
model includes smoothers (pb()) for the mu and nu parameter.
First, I "blindly" used the default options for predictions but noticed that I
R tries hard to keep you from committing scientific abuse.
As stated, your problem seems to me akin to
1. Given that a man's age can be modelled as a function
of the grayness of his hair,
2. predict a man's age from the temperature in Barcelona.
Your calibration relates 'abs' and 'conc'. No
> test2=list(numeric(0),c(10,20));
> test1=list(c(1),c(1,2,3,4));
> for (i in 1:2){
> tryCatch(wilcox.test(test1[[i]],test2[[i]]),error = function(e) NULL);
> }
>
> I cannot get the p-value of the test for i=2.
You didn't store the results of wilcox.test anywhere.
First make it work for data th
FORTUNE!!!
-- Bert
On Tue, Mar 27, 2012 at 11:44 AM, Peter Ehlers wrote:
>
> R tries hard to keep you from committing scientific abuse.
> As stated, your problem seems to me akin to
>
> 1. Given that a man's age can be modelled as a function
> of the grayness of his hair,
> 2. predict a man's a
Hi,
On Tue, Mar 27, 2012 at 6:05 AM, Alekseiy Beloshitskiy
wrote:
> Hi All,
>
> Here is the case. I want to build classification model (SVM). Some of
> variables for this model are categorical attributes which represent words
> (usually 3-10 words - query for search in google). For example:
>
R tries hard to keep you from committing scientific abuse.
As stated, your problem seems to me akin to
1. Given that a man's age can be modelled as a function
of the grayness of his hair,
2. predict a man's age from the temperature in Barcelona.
Your calibration relates 'abs' and 'conc'. Now
On Mar 27, 2012, at 2:36 PM, C Lin wrote:
I'm sorry. I do appreciate you are trying to help. However, what I
am trying to do is not exactly the same as in FAQ.
If I do the following:
test2=list(numeric(0),c(10,20));
test1=list(c(1),c(1,2,3,4));
for (i in 1:2){
tryCatch(wilcox.test(test1[[i
I'm sorry. I do appreciate you are trying to help. However, what I am trying to
do is not exactly the same as in FAQ.
If I do the following:
test2=list(numeric(0),c(10,20));
test1=list(c(1),c(1,2,3,4));
for (i in 1:2){
tryCatch(wilcox.test(test1[[i]],test2[[i]]),error = function(e) NULL);
}
Hi,
On Tue, Mar 27, 2012 at 1:03 PM, Heba S wrote:
>
> Hello,I am trying to install a newer version of R (R 2.14.2) from this
> linkhttp://cran.r-project.org/bin/macosx/
> However I am getting an error that it can not be installed on my computer. My
> Mac is version 10.6.8. Can you please advi
There may very well be a better solution, but this works.
format(strptime(dayofyear, format="%j"), format="%m-%d")
On Tue, Mar 27, 2012 at 11:12 AM, Sam Albers wrote:
> Hello,
>
> I am having trouble figuring out how to convert a Day of Year integer
> back into a Date format. For example I have
On Mar 27, 2012, at 2:18 PM, C Lin wrote:
> As a matter of fact, I did read the FAQ. However, in the FAQ coef()
> is used to return the coefficients of lm() if it succeeded.
> I cannot find similar function for pvalue.
So your question has nothing to do with the subject line? If you are
tryi
On 27-03-2012, at 19:24, Nederjaard wrote:
> Hello,
>
> I'm new here, but will try to be as specific and complete as possible. I'm
> trying to use “lm“ to first estimate parameter values from a set of
> calibration measurements, and then later to use those estimates to calculate
> another set o
As a matter of fact, I did read the FAQ. However, in the FAQ coef() is used to
return the coefficients of lm() if it succeeded.
I cannot find similar function for pvalue.
> CC: r-help@r-project.org
> From: dwinsem...@comcast.net
> To: bac...@hotmail.com
> Subject: Re: [R] ignore error getting
Hello,
I am having trouble figuring out how to convert a Day of Year integer
back into a Date format. For example I have the following:
date <-
c('2008-01-01','2008-01-02','2008-01-03','2008-01-04','2008-01-05','2008-01-06','2008-01-07',
'2008-01-08','2008-01-09','2008-01-10','2008-01-11','2008-
Hello to everyone.
I´m using this function to download some information from a website.
This is the URL:
http://164.77.222.61/climatologia/php/vientoMaximo8.php?IdEstacion=330007&FechaIni=01-1-1980
If you go to that website you´ll find a table with meteorological
information. One column is called "
Hi,
On Tue, Mar 27, 2012 at 10:35 AM, yx78 wrote:
> In the package lasso2, there is a Prostate Data. To find coefficients in the
> prostate cancer example we could impose L1 constraint on the parameters.
>
> code is:
> data(Prostate)
> p.mean <- apply(Prostate, 5,mean)
> pros <- sweep(Prostate,
Hello,
I'm new here, but will try to be as specific and complete as possible. I'm
trying to use âlmâ to first estimate parameter values from a set of
calibration measurements, and then later to use those estimates to calculate
another set of values with âpredict.lmâ.
First I have a calib
On Mar 27, 2012, at 12:56 PM, C Lin wrote:
Dear All,
How do I ignore an error and still getting result of next iteration.
I am trying to do wilcox.test on a loop, when the test fail, I would
like to continue doing the next iteration and getting the p-value.
I tried to do tryCatch or try but
Hello,I am trying to install a newer version of R (R 2.14.2) from this
linkhttp://cran.r-project.org/bin/macosx/
However I am getting an error that it can not be installed on my computer. My
Mac is version 10.6.8. Can you please advise me what the problem. I need the
newer version to install t
>
> Hi,
>
> I used
> GDALinfo("MOD13Q1.A2001049.h13v11.005.2007002215512.250m_16_days_EVI.tif")
> and
> got the results:
>
> rows10
> columns 11
> bands 1
> origin.x150701.4
> origin.y7744897
> res.x 250
> res.y 250
> ysign -1
> oblique.x 0
>
Sorry last message was not completed before sending
Please below
On Tue, Mar 27, 2012 at 5:36 PM, HJ YAN wrote:
> Thank you very much Gerrit, for the nice hints!
>
> Just done some more googling and reaserches on this and trying to
> answering it myself...
>
> Below is the code that work
Inline:
On Tue, Mar 27, 2012 at 10:00 AM, Weidong Gu wrote:
> Hi,
>
> your code has errors: apply function only has 1 or 2 as margin.
FALSE. Please re-read the Help files. It works as expected with
arbitrary higher dim arrays.
-- Bert
>
> bound is used as turning parameter for summation of a
Hi,
I'm afraid that the function read.octave from package "foreign" has
some problems with the ASCII data format exported by new versions of
Octave (later than 3.2.X). It fails even for a simple case as:
[Octave code:]
octave:1> x=1;
octave:2> save -ascii testdata.mat x
[Now in R:]
> octavedata
Hi,
your code has errors: apply function only has 1 or 2 as margin.
bound is used as turning parameter for summation of absolute
coefficients. lasso runs on a grid of the turning parameter for
varying strength of shrinkage. so each turning value may yield
different sets of coefficients and values
Dear All,
How do I ignore an error and still getting result of next iteration.
I am trying to do wilcox.test on a loop, when the test fail, I would like to
continue doing the next iteration and getting the p-value.
I tried to do tryCatch or try but I cannot retrieve the p-value if the test is
I'm trying to make a glmm to identify the relationship between insect
species richness with fragment size, isolation and time (different years).
I already tried to analyse it using poisson distribution error, but I
always face with the following warning:
*glm.fit: fitted probabilities numerically 0
It's really not suggested etiquette to thread-jack, but generally, the
more you can tell to read.table (particularly the colClasses, nrows,
as.is, and stringsAsFactors arguments) the faster it will be able to
read things by skipping various necessary checks.
Michael
On Tue, Mar 27, 2012 at 12:07
Ah, thanks. I am new to R and was unaware of the from/to parameters for
the plot function. I thought xlim and ylim served that purpose. Thanks
again!
-Chad
On Tue, Mar 27, 2012 at 3:31 AM, Matthieu Dubois wrote:
> Dear Chad,
>
> your problem is linked to (1) the function returning NaNs from
yet another way:
> city<-data.frame(city="Barcelona",cod=1)
> city<-rbind(city,data.frame(city="Madrid",cod=2))
> city<-rbind(city,data.frame(city="Lisbon",cod=3))
> city<-rbind(city,data.frame(city="Milan",cod=4))
> city<-rbind(city,data.frame(city="London",cod=5))
>
> travel<-data.frame(pos=1,So
Hello,
this code, works perfectly
temp <- merge(travel, city, by.x="Source", by.y="cod")
result <- merge(temp, city, by.x="Destine", by.y="cod")
The problem was the construction of the data frame, had a parenthesis in
city<-rbind(city,data.frame(city="Lisbon",cod=3))),
I tried to dele
Note that you can actually drop the line defining the big list "x". I
thought it would be needed, but it turns out to be unnecessary after
cleaning up the second half: cutting off that allocation might save
you even more time.
Best,
Michael
On Tue, Mar 27, 2012 at 11:14 AM, Kurinji Pandiyan
wrot
I realised that I removed the link to the question but forgot to remove the
text regarding it. Sorry. I am not sure if I am supposed to link to other
forums, but I can add the links as needed (as the format is clearer).
I actually have one more question though in regards to which data to use.
If i
Guys, let me add my 5 coins into your interesting discussion.
I have ~10Gb txt file with train data for my model. It has about 150 millions
rows for 12 variables.
When I load it into memory (just run only one row!):
train<-read.table(file="/training.txt")
while loading it takes ~28Gb of RAM (It
The title() function also has parameter 'line' where you can specify the
margin line in which the text should be displayed.
How many lines of margin should be around the figure region of the plot
can be specified before plotting by par(mar=c(bottom,left,top,right)),
in text lines. margin lines are
You should use mixed effects modeling to analyze data of this sort.
This is not a topic that has generally been covered by introductory
classes, so you should consult with a professional statistician on
your problem, or educate yourself well beyond the novice level (this
takes more than just readin
On Mar 27, 2012, at 9:39 AM, Gerrit Eichner wrote:
Hi, HJ,
see
?plotmath
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
On Tue, 27 Mar 2012, HJ YAN wrote:
Dear R-help,
I am tryin
In the package lasso2, there is a Prostate Data. To find coefficients in the
prostate cancer example we could impose L1 constraint on the parameters.
code is:
data(Prostate)
p.mean <- apply(Prostate, 5,mean)
pros <- sweep(Prostate, 5, p.mean, "-")
p.std <- apply(pros, 5, var)
pros <- swe
Hi,
I have similarity value between string pairs in a mysql database.
I need to construct the distance matrix which hclust can take and cluster
the strings. Most of the examples I came across show how to construct the
distance matrix using dist function.
How can I code to construct distance matri
Good Afternoon,
I believe that my to the problem, the R has a more effective solution.
in place the use the loop
I have the following set of data, and needs to extract some sections.
user poscommunications source v_destine
7 1 109 2222
7 2 100 222
On Mar 27, 2012, at 3:37 AM, peter dalgaard wrote:
On Mar 26, 2012, at 17:33 , David Winsemius wrote:
The usual approach to that problem is to use sapply:
x <- list()
x <- sapply(1:10, function(z) x[[z]] <- 1:z )
Yikes!
If that works, it is only by coincidence (The pre-assignment to
Hello,
I encountered a situation similar as the one described by Tal above :
I use the RODBC library to export multiple dataframes into different sheets
of an Excel file.
My dataframes contain Character, Date and Numeric columns.
library("RODBC")
channel <- odbcConnectExcel(xls.file = myXlsFile
*I'm still a R noob, just had a couple of lectures about it in our research
master.
There is a Deal or no deal experiment where I have to write some code for.
Someone wrote a website to gather the data and write it in a .xlsx file.
These are seperate files for seperate participants so first I have
Hello Mike,
I don't think I did, but I fixed the issue by loading each package before
use. The second issue was solved by removing a variable that was used to
create two other categorical variables. I think it must have been
recognising this.
Thanks for the help.
--
View this message in context:
Thank you for the modified script! I have now tried on different datasets
and it works very well and is dramatically faster than my original script!
I really appreciate the help.
Kurinji
On Fri, Mar 23, 2012 at 1:33 PM, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> Taking a look at
HI all,
I am planning to get Zuur et al.'s new book when it comes out, but until
then I was wondering if anyone could suggest examples of zero inflated or
hurdle GAMMs. I have count data with many zeros, non-linear relationships,
and site as a random effect.
Thank you!
Bert Harris, University of
Hello,
I have been attempting to set up a lme and have looked at numerous posts
including 'R's lmer cheat-sheet' as well as reading a number of papers and
other resources including R help, but I am still a little confused on how to
write my model (I thought I had it).
I have asked a number of ques
Hello,
I am new at using R.
I would like to use the following functions of the "clim.pact" package:
ncdfcont and retrieve.nc
I have installed the package "clim.pact" in Rstudio.
I have downloaded the "ncdf pack" from unicar (including ncdump and ncgen).
The ncdf file I'm working on is called "es
No idea what a "mean median histogram" is but you may wish to check
out ?tapply or library(plyr), both of which are designed for this
split-apply-combine paradigm.
Michael
On Tue, Mar 27, 2012 at 12:51 AM, arunkumar wrote:
> Hi
>
> I have records like like this
>
> X1 X2 State
> 34
:)
yes! I agree!
On Mon, Mar 26, 2012 at 10:51:17AM -0700, Bert Gunter wrote:
> Fortunes candidate?!
> -- Bert
>
> On Mon, Mar 26, 2012 at 10:24 AM, Sarah Goslee wrote:
> < The OP wrote>
> "The problem is that it gives the result that I want."
>
> : That's a new sort of problem.
>
>
>
>
Benilton,
*
*
*Thank you you are quite right!!*
*
*
*Regards,*
*Tom
*
On Tue, Mar 27, 2012 at 9:35 AM, Benilton Carvalho <
beniltoncarva...@gmail.com> wrote:
> You probably want:
>
> sql<-"UPDATE testtable SET vals=21 WHERE countries='NewZealand'"
> dbGetQuery(con, sql)
>
> instead...
>
> b
>
>
Hi, HJ,
see
?plotmath
Hth -- Gerrit
-
Dr. Gerrit Eichner Mathematical Institute, Room 212
gerrit.eich...@math.uni-giessen.de Justus-Liebig-University Giessen
Tel: +49-(0)641-99-32104 Arndtstr. 2
You probably want:
sql<-"UPDATE testtable SET vals=21 WHERE countries='NewZealand'"
dbGetQuery(con, sql)
instead...
b
On 27 March 2012 14:18, Thomas Adams wrote:
> All:
>
> I am using RSqlite and want to be able to update individual values in a
> record, such as with this simple example:
>
>
Thanks guys for all the replies.
"It is an urban myth that using 'apply' functions will deliver better
performance than 'for' loops. It may even worsen performance or create
obstacles when it is improperly used with dataframes. Most of the
benefits come from improving readability and main
Hi Jim!
Thank you so much for the very helpful hints!!
I am learning 'split' now and it seems very useful..
HJ
On Tue, Mar 27, 2012 at 12:58 PM, jim holtman wrote:
> Why not use 'split' and get all the groups at once:
>
> result <- split(Calandra, list(Calandra$Day, Calandra$Season, drop = TRU
Dear R-help,
I am trying to express myself as best as I can here. If you also use Latex
to edit math reports or other languages with similar editing method,
you'll see what I'm talking about. My sincere appologies if my question is
not clear enough to some extend, as also I'm not able to provide
Hm.. so what you need is either
- one new feature for each activity that has a binary value
e.g.:
cust_id , cycling, swimming, cooking
1001 , 1 , 0, 1
- one new feature that has a value corresponding to a certain combination of
activities
so if you had just the three
All:
I am using RSqlite and want to be able to update individual values in a
record, such as with this simple example:
library(RSQLite)
drv<-dbDriver("SQLite")
con<-dbConnect(drv,"test.db")
my.data<-data.frame(countries=c("US","UK","Canada","Australia","NewZealand"),vals=c(52,36,74,10,98))
dbWrit
Right,
I was also thinking about it, but since I have few thousands of unique words I
'm not quite sure how it will work
I just posted my question with more detailed description here:
http://stats.stackexchange.com/questions/25355/multi-value-categorical-attributes-how-r
Really interesting case
On 27/03/12 01:09, Benilton Carvalho wrote:
> I need to read in csv files, created by 3rd party, with fields
> containing single quotes (as shown below).
>
> "header1","header2","header3","header4"
> "field1r1","field2r1","field3r1","field4r1"
> "field1r2","field2r2","field3r2PartA), field3r2PartB
Thanks Henrique...
giving it a try now, but it'll take a good while, given the file size.
Cheers,
b
On 27 March 2012 02:35, Henrique Dallazuanna wrote:
> Benilton,
>
> Try this:
>
> read.table(textConnection(gsub('","', "','", gsub('^\"|\"$', "'",
> readLines('../teste.csv', sep = ',', quo
Dear R-help,
I am using R 2.14.1 on Windows 7 with the 'gfcure' package (cure rate model).
I have included the treatment variable in the cure part of the model as shown
below:
Ø ref_treat <-
gfcure(Surv(rem.Remtime,rem.Rcens)~1,~1+strata(drpa)+factor(treat(delcure)),data=delcure,dist="loglogi
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