Birth order is a factor. You can't have a fractional birth order - e.g.
ORDER = 1.5 makes no sense. You are either first, second, third, etc.
"Order" in your data can only take two unique values, and thus anything
plotted against birth order is only going to vary along a vertical line
above perm
Greg,
I really like that TeachingDemos::SnowsPenultimateNormalityTest()… even the
tortuous way to always return a p-value == 0:
# the following function works for current implementations of R
# to my knowledge, eventually it may need to be expanded
is.rational <- function(x){
rep( TRUE, leng
You can use setattr() in the data.table package. It can be used too on
data.frames or other objects.
Best,
Philippe Grosjean
On 22 Feb 2014, at 03:13, Smart Guy wrote:
> Hi All
>
> I am having problem running the 'attributes' command to set a attribute on
> each column of a large dataset. Da
Hi All
I am having problem running the 'attributes' command to set a attribute on
each column of a large dataset. Dataset has 80 columns and 312407 rows. Its
taking more than 60 seconds to set simple attributes like split=TRUE,
usermissing=FALSE.
Here is the source code, assuming Dataset1 is the
I am using a gam for a model predictor and need to forecast into the future
based off data collected earlier. I would like to predict fish weight on
day 10 from measures taken on a fish from day 2. This seems simple, but
using predict, I have only been able to predict weight for a given day. A
s
Second!!
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom."
H. Gilbert Welch
On Fri, Feb 21, 2014 at 3:44 PM, Rolf Turner wrote:
> On 22/02/14 11:53, Greg Snow wrote:
>
>
>
>
Hi Ioanna,
If you need to paste the colnames if there are multiple 1's per row:
You could try:
A<-data.frame(A=c(10,100,1000,30,50,60,300,3,4,2,20,35,45),B=c(0,1,1,1,0,0,0,0,0,1,0,0,1),C=c(0,0,0,0,1,1,0,0,0,0,1,1,1),D=c(1,0,0,0,0,0,1,0,0,1,NA,1,1))
apply(A[,-1],1,function(x) {x1 <-paste(colnames(A[
**Appologies for cross-posting with Stack Overflow**
An example of the problem:
load(url('http://andrewcd.berkeley.edu/sdat'))
head(sdat)
library(plm)
fem =
plm(y~T+G:t,data=sdat,effect="twoways",model="within",index=c("ID","t"))
summary(fem)
lsdvm = lm(y~ID+T+G:t,data=
Thanks. This helped greatly. I’m sorry about sending the html message, and
hope this one is plain-text. Dan.
On Feb 21, 2014, at 7:00 AM, Prof Brian Ripley wrote:
> On 21/02/2014 10:40, Daniel Kelley wrote:
>> I’m wondering whether anyone can help me with a translation exercise. I
>> have
On 22/02/14 11:53, Greg Snow wrote:
Why are you testing your data for normality? For large sample sizes
the normality tests often give a meaningful answer to a meaningless
question (for small samples they give a meaningless answer to a
meaningful question).
Fortune!!!
cheers,
Rolf
___
I would like to add some boxes with special extent to my plot.
Example
gh <- raster()
gh[] <- 1:ncell(gh)
SP <- spsample(Spatial(bbox=bbox(gh)), 10, type="random")
Then plot them
levelplot(gh, col.regions = rev(terrain.colors(255)), cuts=254,
margin=FALSE) +
layer(sp.points(SP
On 22/02/14 11:04, Rui Barradas wrote:
Hello,
Not answering directly to your question, if the sample size is a
documented problem with shapiro.test and you want a normality test, why
don't you use ?ks.test?
m <- mean(HP_TrinityK25$V2)
s <- sd(HP_TrinityK25$V2)
ks.test(HP_TrinityK25$V2, "pnorm"
Rui,
Note this quote from the last paragraph of the Details section of ?ks.test:
"If a single-sample test is used, the parameters specified in '...'
must be pre-specified and not estimated from the data."
Which is the exact opposite of your example.
Gonzalo,
Why are you testing your dat
Hello,
Not answering directly to your question, if the sample size is a
documented problem with shapiro.test and you want a normality test, why
don't you use ?ks.test?
m <- mean(HP_TrinityK25$V2)
s <- sd(HP_TrinityK25$V2)
ks.test(HP_TrinityK25$V2, "pnorm", m, s)
Hope this helps,
Rui Barra
Hello,
The answer is yes, it is possible. I don't know how to plot a curved
arrow but the rest should be possible to do using
?plot.default
?lines
?text
And please, post to R-Help, the odds of you getting more and better
answers are greater.
Hope this helps,
Rui Barradas
Em 21-02-2014 20
On Feb 21, 2014, at 11:00 AM, Luigi Marongiu wrote:
> Dear all,
>
> I would like to draw a Venn plot for data represented by 6 variables. I
> know how to do this using the package venneuler (which requires rJava).
> However this package does not report the numbers of elements within each
> class
On Feb 19, 2014, at 11:19 AM, PQuery wrote:
> Dear all,
>
> I have a data frame with a status column and some condition columns. (a dput
> of part of it is listed below).
> I would like to know if:
>
> 1) There are more chances to have a "status" of "1" when more than one
> conditions have the
I may have been able to answer my own questions by reading the e1071
source. It looks like the features are just converted to doubles with
as.double(x). And, I haven't found where in the code yet, but it looks
like it's not scaling the factors which explains why I'm missing rows
in the scale file.
Dear R users,
Please help with with this maybe basic question. I am trying to see if my
data is normal but is a large file and the test does not work.
I keep getting the message : "Error in shapiro.test(x = HP_TrinityK25$V2)
: sample size must be between 3 and 5000"
thanks!
shapiro.test(x=HP_Tri
I have a trained SVM that I want to export with write.svm and
eventually use in libSVM. Some of my features are factors. Standard
libSVM only works with features that are doubles, so I need to figure
out how my features should be represented and used.
How does e1071 treat factors in an SVM? For fe
Dear all,
I would like to draw a Venn plot for data represented by 6 variables. I
know how to do this using the package venneuler (which requires rJava).
However this package does not report the numbers of elements within each
class.
Do you know how to display the number of elements using either
Dear all,
I would like to draw a Venn plot for data represented by 6 variables. I
know how to do this using the package venneuler (which requires rJava).
However this package does not report the numbers of elements within each
class.
Do you know how to display the number of elements using either
Hi,
I perfectly agree that ggplot package is the perfect solution, but if
you prefer the old-fashion plot, you can find a plot_errorbar function
in the package phenology:
plot_errbar(1:100, rnorm(100, 1, 2),
xlab="axe x", ylab="axe y", bty="n", xlim=c(1,100),
errbar.x=2, errbar.
This merely translates the matrix multiplication I used into explicit
arithmetic!
Nor does it generalize without extra manipulation to get the correct
arithmetic expression.
-- Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
"Data is not information. Information is not knowl
The problem is resolved already. Please don't include this question in future
mailing list
Umair Durrani
email: umairdurr...@outlook.com
> Subject: Re: [R] Detecting Vehicle locations using R
> From: jdnew...@dcn.davis.ca.us
> Date: Thu, 20 Feb 2014 20:06:28 -0800
> To: umairdurr...@outlook.co
Hi IOanna,
Do you have rows with multiple '1's? If not, you could also try:
A$Variable <- c("none",names(A)[-1])[1+with(A,B+2*C+3*D)]
A.K.
On Friday, February 21, 2014 4:44 AM, ioanna ioannou wrote:
Thank you very much. One further question.
Assuming that for some points there is no classi
This is easy to do in the approach I showed.
Instead of:
> names(A)[-1][as.matrix(A[,-1])%*%(seq_len(ncol(A)-1))]
modify it to:
> c("none",names(A)[-1])[as.matrix(A[,-1])%*%seq_len(ncol(A)-1)+1]
[1] "D""B""B""B""C""C""D""none"
Cheers,
Bert
Bert Gunter
Genentech
On 21/02/2014 10:40, Daniel Kelley wrote:
I’m wondering whether anyone can help me with a translation exercise. I have a
package named “oce”, which does oceanographic processing, and I’d like to make
it produce graphs with labels that work in different languages. For example,
in English I wr
Prof Brian Ripley writes:
> On 21/02/2014 10:07, Rainer M Krug wrote:
>> Hi
>>
>> I want to assign the function & and | to a variable, because I want to
>> specify as a function argument if inside the function & or | should be
>> used.
>>
>>link <- &
>>
>> does not work, and
>>
>>link <-
On 21/02/2014 10:07, Rainer M Krug wrote:
Hi
I want to assign the function & and | to a variable, because I want to
specify as a function argument if inside the function & or | should be
used.
link <- &
does not work, and
link <- "&"
results in the string "&" being assigned to link.
Im wondering whether anyone can help me with a translation exercise. I have a
package named oce, which does oceanographic processing, and Id like to make
it produce graphs with labels that work in different languages. For example,
in English I write Depth and in Spanish Id like to write
Hi
I want to assign the function & and | to a variable, because I want to
specify as a function argument if inside the function & or | should be
used.
link <- &
does not work, and
link <- "&"
results in the string "&" being assigned to link.
So how can I assign the logical function to th
Thank you very much. One further question.
Assuming that for some points there is no classification for example:
A<-data.frame(A=c(10,100,1000,30,50,60,300,3),
B=c(0,1,1,1,0,0,0,0),
C=c(0,0,0,0,1,1,0,0),
D=c(1,0,0,0,0,0,1,0))
Hi
I have fit a model, and estimated the risk difference(RD) for the
interaction term,but I don't know how to estimate the standard error and
confidence interval for RD. The code are as follows
table(dat$A3)
# 0 1
#167762
table(dat$A17)
# 0 1 2 3
#129982 23429
Hi,
Thanks for the reply, I will wait a couple of days and eventually post
elsewhere unless I find the solution myself.
Best.
--
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