Hello Group,
Is there an easy way to query a data.frame or data.table (this is
fast!) for multiple conditions?
I don't want to use a SQL kind of statement.
I am looking for something like a subset with multiple conditions.
Any tips of the like the binary search methodology used for data.table
Dear,
I want to analyze an outcome in an RCT using lme but I am not sure that I have
chosen the right way for the model.
We measured the outcome three times repeatedly in the same patient. One time
before intervention and two times after intervention. I wanted to adjust for
the correlated data
That does not remedy the situation in any case, take the following function
fun - function(x) dnorm(x, -500, 50) + dnorm(x, 500, 50)
that has a 'mode' of 0 again. Interestingly, if I transform it by 1/x, to
integrate I again have to reduce the error tolerance to at least 1e-10:
In https://stat.ethz.ch/pipermail/r-help/2008-March/156868.html I found what
linear separability means. But what can I do if I find such a situation in my
data? Field (2005) suggest to reduce the number of predictors or increase the
number of cases. But I am not sure whether I can, as an
KT_rfan wrote:
I'm attempting to create a data frame with correlations between every pair
of variables in a data frame, so that I can then sort by the value of the
correlation coefficient and see which pairs of variables are most strongly
correlated.
The sm2vec function in the corpcor
Sounds just like the subset function (?)
x - as.data.frame(matrix(sample(5, 100, rep=TRUE), ncol=10))
subset(x, V1 3 V2 5)
Michael
On 3 December 2010 19:05, Santosh Srinivas santosh.srini...@gmail.com wrote:
Hello Group,
Is there an easy way to query a data.frame or data.table (this is
.
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Or simpler:
df[, paste(A,C,sep=)]
df[[paste(A,C,sep=)]]
Or:
x - paste(A,C,sep=)
df[,x]
df[[x]]
Btw, you don't need to use cbind(), data.frame() does it already
Ivan
Le 12/3/2010 08:21, Santosh Srinivas a écrit :
try this ..
df[,colnames(df)==paste(A,C,sep=)]
On Fri, Dec 3, 2010 at 12:05
Steffen Fleischer wrote:
..
We measured the outcome three times repeatedly in the same patient. One
time before intervention and two times after intervention. I wanted to
adjust for the correlated data in the repeated measurement and baseline
differences in the variable in order to get
On 03-Dec-10 08:34:25, soeren.vo...@eawag.ch wrote:
In https://stat.ethz.ch/pipermail/r-help/2008-March/156868.html I found
what linear separability means. But what can I do if I find such a
situation in my data? Field (2005) suggest to reduce the number of
predictors or increase the number of
or even shorter
df[,paste(A,C,sep=)]
Santosh Srinivas santosh.srini...@gmail.com wrote in message
news:aanlktikcjy7bvyfbwuwmrq4dhg4pbdau+qh_7+k+b...@mail.gmail.com...
try this ..
df[,colnames(df)==paste(A,C,sep=)]
On Fri, Dec 3, 2010 at 12:05 PM, Yuan Jian jayuan2...@yahoo.com wrote:
Hello,
See below
Le 12/3/2010 06:54, Berwin A Turlach a écrit :
On Thu, 2 Dec 2010 23:34:02 -0500
David Winsemiusdwinsem...@comcast.net wrote:
[...] Erik is telling you that your use of ncol-4 got evaluated to
4 and that the name of the resulting object was ignored, howevert the
value of the
David Winsemius wrote:
On Dec 2, 2010, at 3:47 PM, Michael Friendly wrote:
I'm looking for an R method to produce latex versions of tables for
table/array objects of 3 or more dimensions,
Some time ago, I did a quick hack. Don't know if it helps, it was good
enough for my purpose.
Thanks!
Following Prof. Ripley's reply I inserted the following lines into
StructTS (makeBSM)
if (nf 3)
{ ind - 2L }
else
{ ind - 3:nf
T[cbind(ind + 1L, ind)] - 1
}
This worked and gave me the same variance estimates as the dlm package.
It's only obvious when someone points it out :)
fubar is not created because, in the test x 3 returned FALSE, which
means the cat function doesn't get used, which means the y arg (fubar
- 6) is never required and therefore not evaluated.
Evil isn't it ?
Michael
On 3 December 2010 20:18, Ivan
Hi
r-help-boun...@r-project.org napsal dne 02.12.2010 17:26:20:
On Dec 2, 2010, at 11:10 AM, David Lyon wrote:
Thanks David
Do you have the url link that details the worked solution to my
problem:
that we have an extensive collection of documentation suitable for
beginners
Arf, yes it makes sense now!
So the idea here is: never use - in function argument...
Thanks for the explanation!
Regards,
Ivan
Le 12/3/2010 10:48, Michael Bedward a écrit :
It's only obvious when someone points it out :)
fubar is not created because, in the test x 3 returned FALSE, which
On Thu, 2010-12-02 at 11:19 -0600, Christine Dolph wrote:
Hi, Thanks very much for your response.
Thanks Christy,
Apologies if I sounded off-hand or dismissive yesterday. It was a busy
day, and as your mail lacked a reproducible example nor the code you
ran, I wanted to deal with the
Hi
Hm. Does str(poli) revealed said explicitly that poli is data frame? I
presume that it told you that it is list. In that case you need to use
appropriate indexing, which suits structure of your list.
Regards
r-help-boun...@r-project.org napsal dne 02.12.2010 22:06:15:
Simplified
G'day Ivan,
On Fri, 03 Dec 2010 10:54:58 +0100
Ivan Calandra ivan.calan...@uni-hamburg.de wrote:
Arf, yes it makes sense now!
Well, my original post said: R has lazy evaluation and the
assignment takes place when the function evaluates the argument :)
So the idea here is: never use - in
Hello,
All you need are explained at Help/Main/User guide/HTML from Tinn-R menu.
Basically:
1. Close R;
2. Menu R/Configure/ Pemanent (Rprofile.site): Tinn-R will generate a
small script inside of the file Rprofile.site (located on the folder
'etc' where R is installed);
3. If there are some
Le 12/3/2010 11:35, Berwin A Turlach a écrit :
G'day Ivan,
On Fri, 03 Dec 2010 10:54:58 +0100
Ivan Calandraivan.calan...@uni-hamburg.de wrote:
Arf, yes it makes sense now!
Well, my original post said: R has lazy evaluation and the
assignment takes place when the function evaluates the
mbedward wrote:
Sounds just like the subset function (?)
..
Or try sqldf if you feel at home in the SQL empire.
Dieter
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View this message in context:
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Sent from the R help mailing list archive
On 12/03/2010 06:54 AM, Berwin A Turlach wrote:
On Thu, 2 Dec 2010 23:34:02 -0500
David Winsemiusdwinsem...@comcast.net wrote:
[...] Erik is telling you that your use of ncol-4 got evaluated to
4 and that the name of the resulting object was ignored, howevert the
value of the operation was
On Fri, 2010-12-03 at 09:58 +, Gavin Simpson wrote:
On Thu, 2010-12-02 at 11:19 -0600, Christine Dolph wrote:
Hi, Thanks very much for your response.
Thanks Christy,
Apologies if I sounded off-hand or dismissive yesterday. It was a busy
day, and as your mail lacked a reproducible
On 02/12/2010 9:59 PM, Rolf Turner wrote:
On 3/12/2010, at 3:48 PM, David Scott wrote:
On 03/12/10 14:33, Duncan Murdoch wrote:
SNIP
I think the fill=TRUE option arrived about 10 years ago, in R 1.2.0.
The comment in the NEWS file suggests it was in response to some strange
csv
I'm trying to read intraday zoo but running into issues (again) ...
what am I missing here? (the date doesn't seem to read in correctly)
head(dat)
TrdDate TrdTime impliedVol
1 20090102 09:55:03 0.3610715
2 20090102 09:55:04 0.3637943
3 20090102 09:55:05 0.3752375
4 20090102 09:55:05
A common point made in discussion of contrasts, type I, II, III SS etc
is that for sensible comparisons one should use contrasts that are
'orthogonal in the row-basis of the model matrix' (to quote from
http://finzi.psych.upenn.edu/R/Rhelp02/archive/111550.html)
Question: How would one check,
On Dec 3, 2010, at 3:44 AM, Michael Bedward wrote:
Sounds just like the subset function (?)
x - as.data.frame(matrix(sample(5, 100, rep=TRUE), ncol=10))
subset(x, )
With data,table:
require(data.table)
xd - as.data.table(x)
xd[ V1 3 V2 5 , ]
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10
[1,] 4
Date: Fri, 3 Dec 2010 08:07:15 +0100
From: rom...@r-enthusiasts.com
To: r-help@r-project.org
CC: diklev...@gmail.com
Subject: Re: [R] Strange problems with compiling dll
Hello,
Your question is more appropriate on the R-devel mailing list.
Hello
How could I include in the model a value for lambda in a glm family=poisson
model?
Or alpha and beta in a glm family=gamma(link=inverse) model?
Thanks
Rosario
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R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
On Dec 3, 2010, at 4:14 AM, Keith Jewell wrote:
or even shorter
df[,paste(A,C,sep=)]
Other grepping methods that generalize better to partial matches:
df[ , grep(^AC$, colnames(df))]
df[ grep(^AC$, colnames(df)) ]
--
David.
Santosh Srinivas santosh.srini...@gmail.com wrote in message
On Fri, Dec 3, 2010 at 7:37 AM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
I'm trying to read intraday zoo but running into issues (again) ...
what am I missing here? (the date doesn't seem to read in correctly)
head(dat)
TrdDate TrdTime impliedVol
1 20090102 09:55:03 0.3610715
Your function does NOT have a mode at zero. It is bimodal with 2 modes:
-500 and 500. So, my approach still works. Zero is a stationary point
where the gradient is zero, but it is not a mode (the second derivative at
zero is not negative but it is zero).
Ravi.
Dear R help list,
I'm fitting a 'variable coefficient model' in the MGCV package and I want to
plot the different smoothers I get for each factor level in one graph.
So, I do something like this to fit the gam:
Mtest - gam(outcome ~ s(age, by=as.numeric(gender==0)) +
On 12/3/2010 4:19 AM, Dieter Menne wrote:
Some time ago, I did a quick hack. Don't know if it helps, it was good
enough for my purpose.
## RNW file
[snip]
Yes, thanks -- that's a good start. Ideally, I'd like to have something
that could also take a file= argument and write to a file
Hi Steve,
The short answer is that there is typically no reason to check (beyond
looking to see what contrasts have been defined for the factors in the
model; see ?contrasts) because the rules are pretty simple.
1) the design matrix is orthogonal on the row-basis (i.e., the columns
sum to zero)
There's a pretty good section in the R book by Crawley on contrast
statements in R, including some discussion of the contrasts being
orthogonal.
I would say you should just make your own table and sort it out there- if
you have equal sample sizes, then the contrast coefficients along the row
Hello,
I have a dataframe assigning various scores on around 20 variables to a list of
countries. The scores are rated on a scale of (D, C, B, A) and there are also
some not rated ones (NR) and others are left blank (NA). I now wanted to
transfer the scores into numeric values (such as NR=0,
Dear R-users,
I'm studing a DB, structured like this (just a little part of my dataset):
_
Site
Latitude
Longitude
Year
Tot-Prod
Total_Density
dmp
Dendoudi-1
Dear all,
I am currently looking for a book about support vector machines for
regression and classification and am a bit lost since they are plenty of
books dealing with this subject. I am not totally new to the field and
would like to get more information on that subject for later use with
think this might help you for start.
http://www.kernel-machines.org/frequently-asked-questions
:)
--
View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
Hi
If you had your variables as factors you can change levels of each factor.
levels(some.factor) - c(0,25,50,27,100)
fac-factor(sample(LETTERS[1:5],20, replace=TRUE) )
fac
[1] A A D A D A D A C B A A D D D B D E C B
Levels: A B C D E
levels(fac)-1:5
fac
[1] 1 1 4 1 4 1 4 1 3 2 1 1 4 4 4 2
I solved the 1° problem with this command:
matrix2plot2[,sensor_data]-as.numeric(as.character((matrix2plot2[,sensor_data])))
In fact before the previous command I if write:
str(matrix2plot2[,sensor_data])
I get:
Factor w/ 10 levels 131.22,148.532,..: 4 6 4 6 9 8 9 8 3 1 ...
And after the
Katharina -
I think something like this may be helpful:
z = data.frame(matrix(sample(c(LETTERS[1:4],'NR',NA),100,replace=TRUE),20,5))
codes = c(A=100,B=27,C=50,D=25,NR=0)
newz = sapply(z,function(x)codes[x])
- Phil Spector
Dear [R] Users,
I have implemented a linear model with this syntax:
model- lm (var_dependent ~ var_indipendent + factor + var_indipendent :
factor, dataframe)
anova (model)
Response: var_dependent
Df Sum Sq Mean Sq F valuePr(F)
On 02/12/10 17:49:37, Andrew Agrimson wrote:
I've been comparing results from kmeans() in R to PROC FASTCLUS in SAS
and I'm getting drastically different results with a real life data set.
[...] Has anybody looked into the differences in the implementations or
have any thoughts on the matter?
David,
Thanks for the comments.
I think, though, that I have found the answer to my own post.
Question: How would one check, in R, that [contrasts .. are
'orthogonal in the row-basis of the model matrix'] for a particular
fitted linear model object?
?lm illustrates the use of crossprod() for
I am using the MICE package to do multiple imputation. Once I generated my
imputed datasets, I wanted to make certain variables binary and certain
variables factors. I figured out how to convert my mids file to a dataframe
and edit it using the complete function, but evidently, the pool
function
On 03/12/10 16:23:33, manuel.martin wrote:
I am currently looking for a book about support vector machines for
regression and classification and am a bit lost since they are plenty of
books dealing with this subject. I am not totally new to the field and
would like to get more information on
Francesco:
1. You need to seek local statistical help.
2. The answer to your question is: it depends in how you define
influence significantly. If you define it as the interaction term
is significant then, by definition the answer is yes. If you want to
understand what is going on and make
HI,
I have a dataframe like this:
nametype
A t1
B t2
C t1
D t1
E t3
Ft2
how can I have a sub dataframe based with the column type like this:
(for type = t1)
nametype
A t1
C t1
D t1
(for type = t2)
Hello everyone,
I am not too sure if any of you are familiar with simulating RNA-seq data. I
am interested in simulating data to perform Fisher's Exact Test. Can you
help?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
Hi,
On Fri, Dec 3, 2010 at 11:26 AM, alcesgabbo alcesga...@hotmail.com wrote:
HI,
I have a dataframe like this:
name type
A t1
B t2
C t1
D t1
E t3
F t2
how can I have a sub dataframe based with the column type like this:
On Dec 3, 2010, at 11:26 AM, alcesgabbo wrote:
HI,
I have a dataframe like this:
nametype
A t1
B t2
C t1
D t1
E t3
Ft2
how can I have a sub dataframe based with the column type like
this:
?subset
(for type = t1)
name
Hi all,
I have a dataset called ,dataSet1'. The time column is given in a numeric
code beginning with the year and ending with the minutes. Frist I tried the
strptime() function to solve the problem. It gave me just the date back (and
not the date and time). There is also the ISOdatetime function
Since I posted last night I've been exploring the possibilities of how the
two implementations could be different. The underlying algorithm appears to
be slightly but I think the main difference between the two is how the
initial seeds are chosen. In FASTCLUS I believe it's some sort of random
On 03.12.2010 16:31, Alexander Salim wrote:
Hi all,
I have a dataset called ,dataSet1'. The time column is given in a numeric
code beginning with the year and ending with the minutes. Frist I tried the
strptime() function to solve the problem. It gave me just the date back (and
not the date
On Dec 3, 2010, at 11:30 AM, Jim Silverton wrote:
Hello everyone,
I am not too sure if any of you are familiar with simulating RNA-seq
data. I
am interested in simulating data to perform Fisher's Exact Test. Can
you
help?
You are likely to get an audience with a greater density of
Hi,
I am implementing a function which generates about 10 .pdf plots in the
current directory.
I need the graphic to fit into a LaTeX-Presentation-slide, so the outer
margin should be removed (this is the way to do it, right?):
I am having trouble finding out where to put the par command(s).
Dear R-users,
Why variables that appear correlated with dependent variable in a scatterplot,
results not correlated in the summary of linear model, and vice versa?
I mean, variable Longitude (see the example below) is correlated (***) with
dependent variable in the linear model. But
In formatR 0.1-5, the dependency on the gWidgets and animation
packages are removed. The GUI by gWidgets is optional now.
Meanwhile, the function tidy.source() has been moved from the
animation package to this package.
library(formatR)
tidy.source(textConnection('
# rotation of the word
Hey everyone,
I know that I can call 'R' from other scripts, and that I can make
command calls from 'R' (e.g., using system() ). But how can I get 'R' to
RETURN values to the script that called it. E.g., I would like to be able
to do something like the following (as a simpler example) from
Yes, Albyn. I do not think that this is a dangerous behavior of the tool
(`integrate'). It is certainly a dangerous use of the tool. One will be
hard-pressed to find a numerical algorithm/software that is fool-proof in
the sense that it always gives you either the correct results or warns you
Dear all, given a daily time series data, I am able to calculate monthly
average, quarterly average like:
library(zoo)
dat - zooreg(rnorm(500), start=as.Date(2000-01-01), frequency=1)
mo.ave - aggregate(dat, as.yearmon(index(dat)), mean)
head(dat)
head(mo.ave)
However is there any direct way
Francesco,
My guess would be collinearity of the predictors. The linear model
gives you the best fit to all of the predictors at once; unless the
predictors are orthogonal (which in a case like this is certainly not
the case), there is no guarantee that the parameter estimates which
give the best
On 03/12/2010 12:22 PM, Alexx Hardt wrote:
Hi,
I am implementing a function which generates about 10 .pdf plots in the
current directory.
I need the graphic to fit into a LaTeX-Presentation-slide, so the outer
margin should be removed (this is the way to do it, right?):
I am having trouble
Dear Steven,
I am so happy, that you answered me! I tried what you said to put all the
variable in one dataframe. The Pretreatment is not really necessary, because it
didn't show any significance. I didn't copy it into the help, because I tried
to concentrate on the essential things.
Here is
Data:
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Am 03.12.2010 20:31, schrieb Duncan Murdoch:
On 03/12/2010 12:22 PM, Alexx Hardt wrote:
Hi,
I am implementing a function which generates about 10 .pdf plots in the
current directory.
I need the graphic to fit into a LaTeX-Presentation-slide, so the outer
margin should be removed (this is the
Lilith,
No the big mystery is the Tukey test. I just can't find the mistake, it
keeps telling me, that
there are less than two groups
...
### Tukey test ##
summary(glht(PAM.lme, linfct = mcp(Provenancef = Tukey)))
Error message:
Fehler in glht.matrix(model = list(modelStruct =
Dear All,
When I have a numeric vector, I want to change it to one character value.
For example,
When I have
test - c(4, 5, 3, 2)
I want to change it to 4 5 3 2(one character value)
How can I get it?
Thanks in advance,
Soyeon
__
R-help@r-project.org
t - c(4, 5, 3, 2)
paste(test,collapse=' ')
[1] 4 5 3 2
- Phil Spector
Statistical Computing Facility
Department of Statistics
UC
Hi Mike,
Is this what you want?
#!/bin/bash
myTest=$(Rscript testing.R.r)
echo myTest contains
echo $myTest
Note that testing.R.r will have to cat() or print() myResult (e.g, my
testing.R.r contains
myResult - paste(Hello World)
cat(myResult)
Best,
Ista
On Fri, Dec 3, 2010 at 1:23 PM, Mike
1) You have redefined the command list which creates lists - not a great
idea.
2) See lapply; for example. Try something like:
list.of.df - lapply(list.of.filenames,read.csv)
list.of.results - lapply(list.of.df,your.application.function)
Regards,
David
--
View this message in context:
I am iterating through multiple files and storing dataframes in a list. I
want to perform calculations on these dataframes. How would I go about doing
this?
I do:
num_files- list.files()
list = c()
for(x in num_files)
{
frame = read.table(x)
list = c(list, assign(paste(frame,x),frame))
}
For
Dear all,
I have a dataset that looks like
id var1 var2 var4 var7 var8
10.0 0.10.30.90.0
20.4 0.60.00.00.2
30.0 0.00.00.80.7
Some columns are missed, for example, here the fourth (var3), sixth(var5)
and seventh (var6)
Lisa -
I think something like this will do what you want:
mydf =
data.frame(id=1:3,var1=c(0,.4,0),var2=c(.1,.6,0),var4=c(.3,0,0),var7=c(.9,0,.8),var8=c(0,.2,.7))
allvars = c('id',paste('var',1:8,sep=''))
mydf[,allvars[!allvars %in% names(mydf)]] = 0
mydf = mydf[,sort(names(mydf))]
mydf
On 12/04/2010 01:25 AM, Katharina Noussi wrote:
Hello,
I have a dataframe assigning various scores on around 20 variables to a list of
countries. The scores are rated on a scale of (D, C, B, A) and there are also
some not rated ones (NR) and others are left blank (NA). I now wanted to
transfer
On 03/12/2010 3:00 PM, Alexx Hardt wrote:
Am 03.12.2010 20:31, schrieb Duncan Murdoch:
On 03/12/2010 12:22 PM, Alexx Hardt wrote:
Hi,
I am implementing a function which generates about 10 .pdf plots in the
current directory.
I need the graphic to fit into a LaTeX-Presentation-slide, so
Just to close this thread, Lilith provided the data which was in a .csv text
file
and had multiple lines of blank data at the end
species;code;treatment;pretreatment;provenance;greenhouse;individual;leaf;Date;DataPAM
Ae;c-ae-1-1-3;C;C;1;1;3;1;25.05.10 14:00; 0.665
.
.
.
Hello Everyone,
I'm trying to use SAS to replicate some results obtained in R. I was wondering
if anyone call tell me the SAS equivalent of the code that appears below.
fm.glm.x - glm(resp ~ . - 1, data = as.data.frame(mm.x),
na.action = na.exclude, family = binomial(link = probit))
ummm...
This is an R list. Shouldn't you be posting this query on a SAS list?
-- Bert
(and why would you want to do this anyway?! ... but we won't go there.)
On Fri, Dec 3, 2010 at 1:42 PM, Paul Miller pjmiller...@yahoo.com wrote:
Hello Everyone,
I'm trying to use SAS to replicate some
see below.
On Fri, Dec 3, 2010 at 3:23 PM, Mike Williamson this.is@gmail.com wrote:
Hey everyone,
I know that I can call 'R' from other scripts, and that I can make
command calls from 'R' (e.g., using system() ). But how can I get 'R' to
RETURN values to the script that called it.
On Dec 3, 2010, at 22:42 , Paul Miller wrote:
Hello Everyone,
I'm trying to use SAS to replicate some results obtained in R. I was
wondering if anyone call tell me the SAS equivalent of the code that appears
below.
fm.glm.x - glm(resp ~ . - 1, data = as.data.frame(mm.x),
Hi Bert and others,
I've also tried posting my question to a SAS list. I'm doing the analyses for
my study in SAS and so would like to be able to keep everything in that
program. Still not familiar enough with R to use it to complete the statistical
analysis and reporting for a study. Working
Dear R experts:
I searched cran (and r-help) for nash equilibrium and game but
nothing stuck out. has someone written a numerical nash optimizer for
two players?
player a has choices x1,x2,x3,... and cares about (maximizes)
pa(x1,x2,x3,...,y1,y2,y3)
player b has choices y1,y2,y3,..., and cares
I think Christophe Dutang is writing a package for generalized Nash
Equilibria models called GNE.
I am cc'ing him here.
I don't know if there are other packages out there. Christophe would know.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant
Hello Group,
I need a modification in the data.table example to get my intended
result shown below ... is there a more simple way!
dt - data.table(A = rep(1:3, each=4), B = rep(1:4, each=3), C = rep(1:2, 6))
dt[, transform(.SD,D=mean(A)), by=B]
The result I want is below ... which is probably
Might something like Nash-Sutcliffe Efficiency be relevant?
I found this as follows:
library(sos)
(n - ???nash)
# found 22 links in 11 packages
Hope this helps.
Spencer
On 12/3/2010 3:25 PM, Ravi Varadhan wrote:
I think Christophe Dutang is writing a package for generalized
Guys,
I am new to R so please excuse if I am not very clear.
My problem is: I have a 'for' loop in which I am defining a Dataframe df
with a SQL query.
First iteration gives a df with 31 rows(that's correct), however next
iterations also gives me max rows as 31. It's kinda stuck at that value.
No, Spencer. Nash-Sutcliff efficiency is due to John E. Nash. It is
unrelated to game theory.
The well-known Nash equilibrium in game theory is due to John Forbes Nash,
Jr.
Ravi.
---
Ravi Varadhan, Ph.D.
Assistant Professor,
Division of
Hello,
Please post a sample of your code so people here can understand what
you are trying to do.
Michael
On 4 December 2010 11:00, rushabhbm rushab...@gmail.com wrote:
Guys,
I am new to R so please excuse if I am not very clear.
My problem is: I have a 'for' loop in which I am defining a
for(i in 1:lengthBD)
{
bdid1 = basicdata[i,]
bdid = as.character(bdid1)
dq = paste(',bdid,',sep = )
dataquery = paste(select *,From Main_Data AS m JOIN Basic_Data AS b ON
m.BD_ID = b.BD_ID JOIN Point_System_Name AS p ON b.psn_id = p.psn_id JOIN
Trend_Location as tl ON b.tl_id = tl.tl_id, where
Hi there,
I am doing a test to see the the residual is distributed in the form of
t-distribution and trying to plot the residuals and the t-distribution to
compare in the graph.
Cheers,
Bill
- Original Message -
From: r-help-requ...@r-project.org
Date: Wednesday, December 1,
Hi,
I would like to know how R assigns the numeric code to a set of factors
in a vector. For example, I have a vector of 5 different factors in a
random order, and I want a color-coded plot by factors:
rfactor=as.factor(sample(letters[1:5], 50, replace=T))
rfactor
[1] c c c d b a b d d a a
On Fri, Dec 3, 2010 at 6:38 PM, Santosh Srinivas
santosh.srini...@gmail.com wrote:
Hello Group,
I need a modification in the data.table example to get my intended
result shown below ... is there a more simple way!
dt - data.table(A = rep(1:3, each=4), B = rep(1:4, each=3), C = rep(1:2, 6))
Hello Bill,
Have a look at the example at the bottom of the help page for ?qqplot
Michael
On 4 December 2010 11:19, 5...@queensu.ca wrote:
Hi there,
I am doing a test to see the the residual is distributed in the form of
t-distribution and trying to plot the residuals and the
It seems to be dependent upon the the character (assigned alphabetically)
which I found out by manually changing the order of appearance of the
characters in rfactor; the colour would stick to the character a, whether
this appears first in rfactor or not.
I as able to control the colours being
Hi all,
I'm curious whether the standard correlation calculation implemented
in stats::cor uses the underlying BLAS (at least in the default case
use = all.obs).
On a 32-bit linux system using R-2.11.1 compiled with GotoBLAS,
stats::cor is as fast as matrix multiplication, so it would appear
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