Hi
If you want a formatted output then some of the packages may be of use
library(R2THML)
?HTML.ftable
library(xtable)
? xtable
there may be functions in the Hmisc /rms packages as well
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
Armidale NSW 2351
E
Have you really never encountered scientific notation?
http://en.m.wikipedia.org/wiki/Scientific_notation
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#.
This is the help-you-learn-R mailing list, not the do-my-work-for-me mailing
list. I don't know where the latter is.
To assist you in the former vein, I recommend that you read the posting guide
mentioned at the footer of every email on this list. It will tell you some
useful hints on interpret
Yes.
see:
?lm
?coef
?write.csv
also, if
fit <- lm(Y~X)
then:
slope_p-value <- summary(fit)[[4]][2,4]
Double check that last one for yourself.
Good luck.
hoguejm wrote
> I want to export the coefficient values and p-values from my regression
> output into a spreadsheet. Is there a way to
I am very very new with R.
Needless to say, your comment is a bit confusing. My regression name is
"mylogit"
Anyway to write a code that I can copy paste? Thanks
--
View this message in context:
http://r.789695.n4.nabble.com/Is-there-a-way-to-export-regression-output-to-an-excel-spreadshe
EstimateStd. Error z value
Pr(>|z|)
(Intercept) 2.338e+00
2.400e+03 0.0010.99922
fac
I want to export the coefficient values and p-values from my regression
output into a spreadsheet. Is there a way to do this?
--
View this message in context:
http://r.789695.n4.nabble.com/Is-there-a-way-to-export-regression-output-to-an-excel-spreadsheet-tp4649155.html
Sent from the R help mai
On 10/11/12 12:08, Martin J Reed wrote:
Rolf and Duncan
Many thanks. Your answers pointed me to a refinement that is closer to what I
want:
rm(list=intersect(ls(".GlobalEnv"),ls("package:reedgraph")),
pos=".GlobalEnv")
This only removes items that are "masked" by GlobalEnv from my pa
On Fri, Nov 9, 2012 at 9:10 PM, Tom Roche wrote:
>
> How best to implement a decorator pattern in R? What I mean, why I ask:
>
> A group of ncdf4 users is trying to make it easier (notably for ourselves :-)
> to write correctly IOAPI-formatted netCDF. Since IOAPI
>
> http://www.baronams.com/produ
Rolf and Duncan
Many thanks. Your answers pointed me to a refinement that is closer to what I
want:
rm(list=intersect(ls(".GlobalEnv"),ls("package:reedgraph")),
pos=".GlobalEnv")
This only removes items that are "masked" by GlobalEnv from my package.
As this is a bit long for some of th
Dear R-Users,
I am using pgmm in the plm package to estimate a dynamic models with panel
data. Besides the lagged dependent variable, I also have some other
endogenous variables. Does the pgmm have an argument that allows me to
specify these endogenous variables and their instruments? I didn't
On Fri, Nov 9, 2012 at 5:21 PM, tesssa wrote:
> How do you lm throw away excess data points. I am following the
> documentation with no success
Why would you want to throw away this data? I think that's generally
considered sub-optimal model-fitting technique, thought it might
explain your diffi
HI,
I found the reason why my first suggestion didn't work out:
mylist<-list(df1=data.frame(a=seq(1,10,1),
b=sample(1:15,10,replace=FALSE),prov=c(rep('QUE', 5), rep('BC', 5))),
df2=data.frame(prov=c(rep('Quebec', 5), rep('AB',
5)),a=seq(1,10,1),b=sample(1:25,10,replace=FALSE)))
lapply(myli
Note that the column-wise conversion I suggested might be better
done on the matrix R before conversion to a data.frame. E.g.
> R <- rbind(
list(Letter="a", Integer=1L, Complex=1+1i),
list(Letter="b", Integer=2L, Complex=2+2i))
> Rconverted <- lapply(structure(seq_len(ncol(R
Hello,
As you know R better it will take you less and less time to get it
right, and almost surely less and less lines of code to do the same
thing. Here's a one liner:
set.seed(1510)
s=numeric(length=10)
for(i in 1:10){
pop=(rbeta(n=20,shape1=2,shape2=1))
s[i]=sum(pop)
}
set.se
On Nov 9, 2012, at 11:21 AM, Sam Steingold wrote:
> The following code:
> --8<---cut here---start->8---
>> myfun <- function (x) list(x=x,y=x*x)
>> z <- as.data.frame(do.call(rbind,lapply(1:3,function(x)
Winsemius' Corollary:
Kurt Gödel's Incomprehension The
How best to implement a decorator pattern in R? What I mean, why I ask:
A group of ncdf4 users is trying to make it easier (notably for ourselves :-)
to write correctly IOAPI-formatted netCDF. Since IOAPI
http://www.baronams.com/products/ioapi/
decorates netCDF, one obvious way to do a package
sink.number() will give you an idea of how many 'sinks' you might have nested.
On Fri, Nov 9, 2012 at 2:07 PM, Dan Baron wrote:
> Oftentimes I want to make outputs to be displayed on the R console.
> However, after I execute a program with a sink command in it the R console
> becomes unresponsive
This is a case where you start by breaking down you complex one-liner
into separate statements and examine what the results are at each
point. This is what I would have to do with the script you posted.
I think as Bill pointed out, one of your function calls is probably
creating a result that you
HI,
If you don't want to list the column names explicitly,
you can try this:
z1<-as.data.frame(do.call(rbind,lapply(1:3,function(x)
c(a=paste("a",x,sep=""),unlist(do.call(c,list(b=myfun(x),c=myfun(x*x*x
z2<-z1[,-1]
z2[]<-sapply(z2,function(x) as.numeric(as.character(x)))
data.frame(a=z1
On Nov 9, 2012, at 12:52 PM, Jacob B. wrote:
> Hello,
>
> I'm trying to fit a logistic curve to data but I'm having a hard time
> discovering how. Every tutorial I've come across either assumes the
> logistic curve has 0
> I simply have two vectors, a and b, of equal length with no missing
> d
There are still a lot of things that you did not specify, but maybe this
will get you started:
n.groups <- sample(3:6, 1)
n.species <- sample(2:10, n.groups, replace=TRUE)
n.animals <- sample( 10:25, sum(n.species), replace=TRUE )
mu.g <- rnorm(n.groups, 100, 50)
mu.species <- rnorm( sum(n.species
Actually it was not tessa but the original document that was at fault.
I suspect you may be right that the 0 should not be there but it was in the
document.
In any case look at these
conc = c(10, 20, 30, 40, 50)
signal = c (4, 22, 44, 60, 82)
lm.r1 <- lm( signal ~ conc)
s
On Fri, Nov 9, 2012 at 3:12 PM, Andrea Goijman
wrote:
> Thanks Sarah,
>
> Ok, I'll start over... forget the functions that I wrote before and lets
> start from scratch.
> Suppose I want to simulate data; for example n species that belong to g
> groups. Lets say 10 species, where 4 belong to group
Your call to rbind() creates matrix of mode "list". Thus every element
can be of a different type, although you "know" that there is a pattern
to the types. E.g.,
> R <- rbind(
list(Letter="a", Integer=1L, Complex=1+1i),
list(Letter="b", Integer=2L, Complex=2+2i))
> str(R)
Hi,
I would start by installing the packages from CRAN instead of omegahat:
install.packages(c("XML", "RCurl"))
If you continue to have problems, try posting on the R-SIG-MAC list
(https://stat.ethz.ch/mailman/listinfo/r-sig-mac)
Best,
Ista
On Fri, Nov 9, 2012 at 12:53 PM, tmbassassin wrote:
> -Original Message-
> From: sarah.gos...@gmail.com
> Sent: Fri, 9 Nov 2012 14:34:57 -0500
> To: jrkrid...@inbox.com
> Subject: Re: [R] lm function - strange error
>
> On Fri, Nov 9, 2012 at 2:20 PM, John Kane wrote:
>>
>>> -Original Message-
>>> From: dwinsem...@comcast.net
>>>
Thanks Sarah,
Ok, I'll start over... forget the functions that I wrote before and lets
start from scratch.
Suppose I want to simulate data; for example n species that belong to g
groups. Lets say 10 species, where 4 belong to group 1, 3 to group 2, and
the rest to group 3.
I want to simulate those
http://www.statmethods.net/stats/descriptives.html
the psych package seems to do what you wan
set.seed(235)
mydata <- data.frame(aa = rnorm(20), bb = rnorm(20) , cc = rnorm(20))
library(psych)
describe(mydata)
John Kane
Kingston ON Canada
> -Original Message-
> From: powella...@g
On Nov 9, 2012, at 9:52 AM, Ricky Corp wrote:
> I am new to R and learned to program 10 years ago in C++. I am currently
> working a project that looks at the distribution of randomly generated beta
> values. I take 20 random beta values find their sum, repeat 10 times.
>
> Here is my code t
Andrea,
I don't understand what you want. Why don't you start over, and
explain what the inputs are, and what your desired outputs are.
Using your previous code:
# pre-specified number of groups
groups <- 3
# what is species? It needs to be defined before
# we can assign a value to a component o
> * arun [2012-11-09 11:33:43 -0800]:
>
> z2<-within(z1,{b.x<-as.numeric(as.character(b.x));b.y<-as.numeric(as.character(b.y));c.x<-as.numeric(as.character(c.x));c.y<-as.numeric(as.character(c.y))})
1. I don't want to have to list all the column names explicitly
2. I find the num->char->num conv
Yes, I tried your suggestion, but it didn't help. It just creates a
tri-dimentional array for p... and that is not what I want p per species
(within groups) to be an unbalanced array...
I need to have a single value of p per species, and the total number of
elements in p would be whatever number r
Need help!
I know how to get the mean and kurtosis for a single variable but I am
reading in an excel file that has several columns representing variables. I
need a way to find descriptive statistics across ALL the variables?
[[alternative HTML version deleted]]
__
We can't really diagnose the problem without a repeatable example
(that is, what's in your R file?).
But my suspicion is that you've called sink(filename) more than once.
Simply type
sink()
at the command prompt until you get the warning:
In sink() : no sink to remove
And then all should be well
Andrea,
I simply meant that I couldn't run your code assigning a value to
species$groups
because the code didn't include any information about creating the R
object species. Thus, I changed the name of that R object to
speciesgroups and altered your code so that it runs by removing the
erroneous c
HI,
It is better to have a reproducible example to work with. I hope this helps
you.
library(reshape)
library(zoo)
set.seed(132)
list1<-lapply(1:13,function(x) lapply(1:12,function(x)
data.frame(col1=sample(1:50,5,replace=TRUE),col2=sample(1:150,5,replace=TRUE
names(list1)<-2000:2012
na
Hi Sarah,
Maybe I expressed myself wrong, but so far, I don't have the species; I'm
just wanting to generate simulated data. For example, creating an
unbalanced (and random) number of species per group, and then run the
for-loops
for example:
Group1: 3 species
Group2: 5 species
Group3: 8 species
Hi Tessa, I agree with John. I think you've made a typo, but looking at
your data I think the zero concentration should not be there. Try plotting
it.
conc = c(10, 20, 30, 40, 50)
signal = c (4, 22, 44, 60, 82)
plot(signal~conc)
abline(lm(signal~conc))
2012/11/9 John Kane
> Tessa,
>
> Would y
Hello,
I'm trying to fit a logistic curve to data but I'm having a hard time
discovering how. Every tutorial I've come across either assumes the
logistic curve has 0http://www.apsnet.org/EDCENTER/ADVANCED/TOPICS/ECOLOGYANDEPIDEMIOLOGYINR/DISEASEPROGRESS/Pages/NonlinearRegression.aspx,
which assum
On Fri, Nov 9, 2012 at 2:20 PM, John Kane wrote:
>
>> -Original Message-
>> From: dwinsem...@comcast.net
>> Sent: Fri, 9 Nov 2012 10:56:48 -0800
>> To: tesara...@gmail.com
>> Subject: Re: [R] lm function - strange error
>>
>>
>> On Nov 9, 2012, at 9:16 AM, tesssa wrote:
>>
>>> Sorry it is
Hello,
I already tried and looked at the bfast package (very nice package by the
way!) as I am working on VI time series as well.
However, my model is definitely not linear, so in worst case scenario my
idea was to use the bfast package to find the breakpoints (with the
harmonic fit) and then to
Oftentimes I want to make outputs to be displayed on the R console.
However, after I execute a program with a sink command in it the R console
becomes unresponsive. Meaning that the following occurs in R console:
> source("Program_containing_sink.R")
> a<-1
> a
>
>sink()
>a
>
R help says that sin
Hi,
May be this helps:
z1<-as.data.frame(do.call(rbind,lapply(1:3,function(x)
c(a=paste("a",x,sep=""),unlist(do.call(c,list(b=myfun(x),c=myfun(x*x*x
z2<-within(z1,{b.x<-as.numeric(as.character(b.x));b.y<-as.numeric(as.character(b.y));c.x<-as.numeric(as.character(c.x));c.y<-as.numeric(as.ch
Hi Andrea,
Without knowing what species is, I can't run your code as is, but try this:
groups<-3
speciesgroups <- as.integer(runif(groups,1,10))
# I'd use sample(1:10, groups, replace=TRUE)
p<-array(NA,dim=speciesgroups)
Mostly you're trying to use c() on something that's already a vector.
Sa
> -Original Message-
> From: dwinsem...@comcast.net
> Sent: Fri, 9 Nov 2012 10:56:48 -0800
> To: tesara...@gmail.com
> Subject: Re: [R] lm function - strange error
>
>
> On Nov 9, 2012, at 9:16 AM, tesssa wrote:
>
>> Sorry it is not a comma but a tilda. The R.help message editor Pl
On Nov 9, 2012, at 9:52 AM, Haszun wrote:
> Hi, can someone tell me what is wrong in this function:
>
> ilecyfr=function(x){
> if (x=0){
> return(1)
> } else {
> k=0
> while (abs(x)/10^k >0) {
> k=k+1
> } return(k)
> }}
>
So if x == 0 (that's your first error and possibly a fatal one, but let
On Fri, 9 Nov 2012, Thomas Coquet wrote:
Hello,
I already tried and looked at the bfast package (very nice package by the
way!) as I am working on VI time series as well.
Good! :-)
However, my model is definitely not linear,
Not even after taking logs or some other transformation?
In prin
Tessa,
Would you please leave in the context --that is your early remarks and mine?
Most of us on the R-help list do no use nabble and it is really annoying to
have to go there to see what was discussed. I suspect that a lot of readers
just don't bother and nabble users lose a lot of possible
Hi,
If you move the read.csv statement outside the function and instead
pass the data as an argument, then you can provide a reproducible
example using dput(head(drugs, 20)) or a similar fake dataset. It's
really hard to debug code without knowing what kind of data it is
supposed to act upon.
In
The following code:
--8<---cut here---start->8---
> myfun <- function (x) list(x=x,y=x*x)
> z <- as.data.frame(do.call(rbind,lapply(1:3,function(x)
> c(a=paste("a",x,sep=""),as.list(unlist(list(b=myfun(x),c=myfun(x*x*x
> z
a b.x b.y c.x c.y
1 a1 1
And your problem is?
Please note that the test for equality is == while the preferred
assignment operator is <-
Beyond that, some statement of what you're trying to accomplish and an
actual executable example would be useful.
Sarah
On Fri, Nov 9, 2012 at 12:53 PM, Haszun wrote:
> *
> ilecyfr=f
I know this seems like a very easy question (and maye it is) but I've been
trying to simulate nested data and been unsucessful so far..
I want to simulate a varying number of species within a group; and then
create an array to store the results of my for-loop. For example:
groups<-3
species$group
On Fri, 9 Nov 2012, thomas88 wrote:
Hello,
I have done some research about breakpoints (I am not a statistician) and I
found out about the breakpoint, strucchange and segmented packages in R
allowing to find breakpoints assuming linear model.
However, I would like to fit a periodic time series
On Nov 9, 2012, at 9:16 AM, tesssa wrote:
> Sorry it is not a comma but a tilda. The R.help message editor Please
> replace the dash with a tilda as lm function requires
It might be interesting to have such a supernatural entity zooming around the
Rhelp Universe answering prayers from newbi
On Nov 9, 2012, at 9:07 AM, rlcorp wrote:
> I also would like to sort the values that are greater than 10 in one vector
> and less than or equal to 10 in another vector.
>
x <- 1:20
x[x>10]
x[x<=10]
--
David Winsemius, MD
Alameda, CA, USA
__
R-he
Hi everybody!
In a case-control study I have already sampled cases stratifying for sex
(0,1) and age (<62y, >=62y). I need to sample a group of controls with the
same characteristics (which I can easily do) plus one more: the level of
relatedness. Therefore controls should be matched to cases for s
Hi,
I am using the Random Forest package to classify observations into one of two
classes. My data is unbalanced with the minority class accounting for 7% of
total data set.
I have heard the 10-Fold Cross validation can help me with improving
classification. But being new at most of this it's
Hi, can someone tell me what is wrong in this function:
ilecyfr=function(x){
if (x=0){
return(1)
} else {
k=0
while (abs(x)/10^k >0) {
k=k+1
} return(k)
}}
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-tp4649082.html
Sent from the R help mailing list archive at N
so...
i'm trying to integrate google analytics with R (statistics)-- i know it's
possible, there's a lot of documentation and some walkthroughs on how to do
it, but i can't for the LIFE of me get it to work or even know where to
begin.
for starters, the directions i'm following are from this web
I am new to R and learned to program 10 years ago in C++. I am currently
working a project that looks at the distribution of randomly generated beta
values. I take 20 random beta values find their sum, repeat 10 times.
Here is my code that it took me 4 hours to get
s=numeric(length=10)
fo
I am new to R and learned to program 10 years ago in C++. I am currently
working a project that looks at the distribution of randomly generated beta
values. I take 20 random beta values find their sum, repeat 10 times.
Here is my code that it took me 4 hours to get
s=numeric(length=10)
fo
Here is the document
http://www.montefiore.ulg.ac.be/~kvansteen/GBIO0009-1/ac20092010/Class8/Using%20R%20for%20linear%20regression.pdf
--
View this message in context:
http://r.789695.n4.nabble.com/lm-function-strange-error-tp4649065p4649080.html
Sent from the R help mailing list archive at Na
*
ilecyfr=function(x){
if (x=0){
return(1)
} else {
k=0
while (abs(x)/10^k >1) {
k=k+1
} return(k)
}}
--
View this message in context:
http://r.789695.n4.nabble.com/Problem-with-tp4649082p4649084.html
Sent from the R help mailing list archive at Nabble.com.
___
Hello fellow R-ers,
I have spent some time on this and it is driving me NUTS! I am sure there is a
solution, so please help.
I am trying to create a function that will plot different lines for subsets of
a dataset. For example, I am trying to look at different drug groups (drug2),
let's say 1,2
How do you lm throw away excess data points. I am following the
documentation with no success
--
View this message in context:
http://r.789695.n4.nabble.com/lm-function-na-action-tp4649075.html
Sent from the R help mailing list archive at Nabble.com.
___
Hi,
I'm using kinship2 to calculate heritabilty, but I would like calculate in
GenABEL too.
I trying the code:
> require(kinship2)
> require(GenABEL)
> pedig = with(Dados, pedigree(id=IID, dadid=PAT, momid=MAT, sex=SEX,
famid=FID, missid=0))
> kmat = kinship(pedig)
> (mod1
Sorry it is not a comma but a tilda. The R.help message editor Please
replace the dash with a tilda as lm function requires
--
View this message in context:
http://r.789695.n4.nabble.com/lm-function-strange-error-tp4649065p4649074.html
Sent from the R help mailing list archive at Nabbl
Hello,
I have done some research about breakpoints (I am not a statistician) and I
found out about the breakpoint, strucchange and segmented packages in R
allowing to find breakpoints assuming linear model.
However, I would like to fit a periodic time series with a non linear
(periodic) model, an
I also would like to sort the values that are greater than 10 in one vector
and less than or equal to 10 in another vector.
--
View this message in context:
http://r.789695.n4.nabble.com/Running-a-loop-of-random-generated-sums-tp4649013p4649070.html
Sent from the R help mailing list archive at
On Nov 9, 2012, at 11:23 AM, Bert Gunter wrote:
> Marc et. al:
>
> On Fri, Nov 9, 2012 at 9:05 AM, Marc Schwartz wrote:
>> On Nov 9, 2012, at 10:50 AM, Eiko Fried wrote:
>>
>>> A colleague wrote the following syntax for me:
>>>
>>> D = read.csv("x.csv")
>>>
>>> ## Convert -999 to NA
>>> fo
Thank you both!! Great to know. I've posted my question to
r-sig-mixed-models.
Thanks again!
Annie
On 11/9/12 12:53 PM, "R. Michael Weylandt"
wrote:
>On Fri, Nov 9, 2012 at 5:27 PM, Bert Gunter
>wrote:
>> Well, you've posted to the wrong list!
>>
>> First off, you're almost always better off po
On Fri, Nov 9, 2012 at 5:27 PM, Bert Gunter wrote:
> Well, you've posted to the wrong list!
>
> First off, you're almost always better off posting to an R SIG list
> when one exists in your area of concern,as it does here:
> r-sig-mixed-models.
>
> Second, this appears to be primarily a statistics
Well, you've posted to the wrong list!
First off, you're almost always better off posting to an R SIG list
when one exists in your area of concern,as it does here:
r-sig-mixed-models.
Second, this appears to be primarily a statistics question, and R-help
is not a statistics help list (though, I a
Can you provide a reference or link to the document? I think that you have a
misprint or a typo in it.
the na.omit is a reference to missing data that is accounted for in the vector,
which would appear as NA in the data.
For example this works just fine for me
conc = c(0, 10, 20, 30, 40, 50)
s
Marc et. al:
On Fri, Nov 9, 2012 at 9:05 AM, Marc Schwartz wrote:
> On Nov 9, 2012, at 10:50 AM, Eiko Fried wrote:
>
>> A colleague wrote the following syntax for me:
>>
>> D = read.csv("x.csv")
>>
>> ## Convert -999 to NA
>> for (k in 1:dim(D)[2]) {
>>I = which(D[,k]==-999)
>>if (length
This is a bug in the unmentioned VGAM package and you have to report it
to its maintainer (hence CCing Thomas Yee).
Uwe Ligges
On 09.11.2012 10:08, Alena888 wrote:
I would like to ude the function "qdagum", to compite quantiles of dagum
distribution, but if I use the form:
>qdagum(0.1, 4, 7
Hi,
This is my first time posting to the list so please forgive any breeches
of etiquette!
I am new to mixed-effects modeling.
This is my dataset:
subject treatment day replicate outcome
1 1 1 1 0.0
1 1 4 1 0.0
1 1 8
On Nov 9, 2012, at 10:50 AM, Eiko Fried wrote:
> A colleague wrote the following syntax for me:
>
> D = read.csv("x.csv")
>
> ## Convert -999 to NA
> for (k in 1:dim(D)[2]) {
>I = which(D[,k]==-999)
>if (length(I) > 0) {
>D[I,k] = NA
>}
> }
>
> The dataset has many missing
Many thanks to Prof. Frank Harrell for the comments and for the relevant
considerations.
In the meantime, I find an R code to estimate the net reclassification
improvement (NRI) in the context of censored survival outcomes published
by Benjamin French, Paramita Saha-Chaudhuri, Bonnie Ky, Thomas P
On Fri, Nov 9, 2012 at 4:37 PM, tesssa wrote:
> I am following a document teaching how to use regression and right at the
> onset I get an R error. I understand that variables "conc" and "signal"
> should have the same length but I am using the what R manual suggests to
> drop the error and it i
I am following a document teaching how to use regression and right at the
onset I get an R error. I understand that variables "conc" and "signal"
should have the same length but I am using the what R manual suggests to
drop the error and it is not cooperating. What gives? The manual says
that
A colleague wrote the following syntax for me:
D = read.csv("x.csv")
## Convert -999 to NA
for (k in 1:dim(D)[2]) {
I = which(D[,k]==-999)
if (length(I) > 0) {
D[I,k] = NA
}
}
The dataset has many missing values. I am running several regressions on
this dataset, and want to e
Hi,
> * Thibault Helleputte [2012-11-09
> 09:22:11 +0100]:
>
> The next release of LiblineaR should offer the possibility of using
> sparse matrices. However, the next release date is not fixed yet...
thanks.
> On Thu, Nov 8, 2012 at 10:07 PM, Sam Steingold wrote:
>> >> It would also be nice
On Nov 9, 2012, at 5:10 AM, Tagmarie wrote:
> I have a data frame somewhat like this:
>
> myframe <- data.frame (ID=c(2,3,4,5), Hunger =c(415,452,550,318 ))
> myframe
>
> Now I would like to add a column to the right which summarizes the values
> for Hunger somewhat to reduce the number of val
Werner,
Yes, it exists. See the likert function in the HH package.
Run the examples in the help file ?likert and the demo to see all the
capabilities of the
likert function, particularly multiple groupings within the sample
population.
Discussion and example of multiple groupings are included in
Hi Ivo,
The only problem is that uniroot() actually uses Brent's algorithm, and is
based on the C code from netlib (there is also Fortran code available -
zeroin.f). Brent's algorithm is more sophisticated than a simple bisection
approach that you have vectorized. It combines bisection and se
Hey there,
I've got a list of lists with matrices:
A list with 13 entries (representing years), each of them another list with
12 matrices (representing one month). In each matrix there are as many rows
as there are hours in the different months and 2 columns, since two
meteorological parameters a
Sorry, just to clarify as I have put .S files in the later response but .csv
in the earlier, I have tried with both, changing to read.csv or read.table
as appropriate but get the same play of events.
H
--
View this message in context:
http://r.789695.n4.nabble.com/CCA-with-Vegan-Plot-problem-tp
Hi,
Try this:
myframe$newcol<-ifelse(myframe[,2]>=300 &
myframe[,2]<400,350,ifelse(myframe[,2]>=400
&myframe[,2]<500,450,ifelse(myframe[,2]>=500 & myframe[,2]<600,550,NA)))
myframe
# ID Hunger newcol
#1 2 415 450
#2 3 452 450
#3 4 550 550
#4 5 318 350
A.K.
-
HI,
Sorry, I didn't know that "prov" exists in different columns.
Try this:
#changed your example dataset to include one more column and reordered the
columns in list elements
mylist<-list(df1=data.frame(a=seq(1,10,1),
b=sample(1:15,10,replace=FALSE),prov=c(rep('QUE', 5), rep('BC', 5))),
df2=
On Fri, Nov 9, 2012 at 8:35 AM, sheen maria wrote:
> Sir,
> But in this price returns only the closing value (Cl(getsymbols("IBM"))
> i already refer the ?TreynorRatio
>
> But it not specify how to calculate the benchmark return.
> In this case what value i put as benchmark return series.
>
Ple
Thank you for the hint.
In batch mode, I got:
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
In interactive mode, I got:
R version 2.15.1 (2012-06-22)
Platform: x86_64-pc-mingw32/x64 (64-bit)
And I have found the cause. In my path, I had still included
C:\R-2.15.1\bin
Hello,
Try
myframe$new <- NA
myframe$new[300 <= myframe$Hunger & myframe$Hunger < 400] <- 350
myframe$new[400 <= myframe$Hunger & myframe$Hunger < 500] <- 450
myframe$new[500 <= myframe$Hunger & myframe$Hunger < 600] <- 550
myframe
Why reduce the number of values?
Hope this helps,
Rui Barrada
You told us neither what you did, nor what was not as you expected.
Nonetheless, try this example:
x <- c(9, 5, 7)
mybarplot <- barplot(x, ylim=c(0, 10))
text(mybarplot, x+.08, letters[1:3], pos=3)
Sarah
On Fri, Nov 9, 2012 at 6:33 AM, Geophagus wrote:
> Hi @ all,
> I try to set a labeling on
On Fri, 2012-11-09 at 05:16 -0800, rumble14 wrote:
> Thanks for replying. Apologies for not putting the code, I have not deviated
> from that used on the website I linked to.
>
> >Can you show the exact code you used please?
>
> I am using:
> #below is Species_1 etc
> org <- read.table ("Organis
Thanks,
I feel like I was so close!!!
KW
--
On Nov 9, 2012, at 4:48 AM, Rui Barradas wrote:
> Hello,
>
> Something like this?
>
>
> fun <- function(years, periods){
>res <- lapply(splitIt(1:12, periods), function(x) groupingStrings(years,
> x))
>names(res) <- as.integer(names(re
Le vendredi 09 novembre 2012 à 10:43 +, Franckx Laurent a écrit :
> Dear all
>
> I use a 64 bit Windows7 system with 16 GB memory.
> I have run an R script in batch mode with the following command:
> R CMD BATCH masterassignment_2012_11_09.r --save
> The execution was terminated with th
HI,
May be this:
res<- data.frame(apply(df,2,function(x)
ifelse(grepl("\\d+",x),LETTERS[x-9],NA)))
res
# y z
#1 A S
#2 B T
#3 C U
#4 D V
#or
apply(df,2,function(x) LETTERS[x-9])
A.K.
- Original Message -
From: Fabricius Domingos
To: r-help@r-project.org
Cc:
Sent: Friday, November
Thanks for replying. Apologies for not putting the code, I have not deviated
from that used on the website I linked to.
>Can you show the exact code you used please?
I am using:
#below is Species_1 etc
org <- read.table ("Organisms.S", header=TRUE, row.names=1)
#below is environmental variables
1 - 100 of 125 matches
Mail list logo
