I have the following:
a #note that the 28 is a row.name
GHP GP T Tn
28 2.2194 2.6561 2.9007 3.2988
min(as.numeric(a))
2.9007
min(as.numeric(as.character(a)))
2.9007
as.numeric(as.character(a)) #What's going on here???
[1] 33. 29. 2.9007 3.2988
We can't say because we don't know how a was created.
Please email the output from
str(a)
and
dput(a)
Yours sincerely / Med venlig hilsen
Frede Aakmann Tøgersen
Specialist, M.Sc., Ph.D.
Plant Performance Modeling
Technology Service Solutions
T +45 9730 5135
M +45 2547 6050
a is ofcourse a subset of data.frame, a ROW of the original table specifically.
str(a)
'data.frame': 1 obs. of 4 variables:
$ GHP: Factor w/ 51 levels 0.0944,0.1446,..: 33
$ GP : Factor w/ 51 levels 0.1755,0.3582,..: 29
$ T : num 2.9
$ Tn : num 3.3
dput(a)
structure(list(GHP =
On 09-04-2014, at 00:56, David Winsemius dwinsem...@comcast.net wrote:
On Apr 8, 2014, at 2:41 PM, Bert Gunter wrote:
Just for the fun of it, I searched for R package moses extreme
reaction test on google. The 8th hit was the package DescTools, but
that included the 3 earlier hits and
Now we are able to help you. What you see is an artefact of an object in T of
class 'factor'
## So please, see ?factor
## first 2 columns subset of a factor
str(a)
'data.frame': 1 obs. of 4 variables:
$ GHP: Factor w/ 51 levels 0.0944,0.1446,..: 33
$ GP : Factor w/ 51 levels
Hi Dila,
If 'dat' is the dataset:
dat$C[dat$B==0] - 0
A.K.
On Wednesday, April 9, 2014 1:30 AM, dila radi dilarad...@gmail.com wrote:
Dear arun
Yes indeed..if the value in B is O..next value in the same row also 0..
other value than 0 remains unchanged.
On Apr 9, 2014 12:48 PM, arun
Hi,
Try:
min(sapply(lapply(a,as.character),as.numeric))
#[1] 2.2194
A.K.
On Wednesday, April 9, 2014 2:27 AM, Sachinthaka Abeywardana
sachin.abeyward...@gmail.com wrote:
a is ofcourse a subset of data.frame, a ROW of the original table specifically.
str(a)
'data.frame': 1 obs. of 4
On 04/09/2014 01:12 PM, Hurr wrote:
This still puts tick marks at data points:
h=c(1,2,3,4,5,6,7,8,9)
v=c(9,8,7,6,5,4,3,2,1)
plot(h,v,xaxt='n')
x-c(1.6,2.6,6.6,9.6,12.9)
axis_labels-1/pretty(x)
axis(1,at=pretty(x),labels=axis_labels)
How do I get the axis to approx 13 ?
Thanks Jim, Labels for the following are at 0,2,4,6,8,10,12, not data, good.
and omitted where lack of room.
h=c(1.2,2.4,3.1,4.7,5.3,6.2,7.6,8.8,9.7)
v=c(9,8,7,6,5,4,3,2,1)
plot(h,v,xaxt='n',xlim=c(-1,13))
x-c(1.6,2.6,6.6,9.6,12.9)
axis_labels-1/pretty(x)
axis(1,at=pretty(x),labels=axis_labels)
On 04/09/2014 06:17 PM, Hurr wrote:
Thanks Jim, Labels for the following are at 0,2,4,6,8,10,12, not data, good.
and omitted where lack of room.
h=c(1.2,2.4,3.1,4.7,5.3,6.2,7.6,8.8,9.7)
v=c(9,8,7,6,5,4,3,2,1)
plot(h,v,xaxt='n',xlim=c(-1,13))
x-c(1.6,2.6,6.6,9.6,12.9)
axis_labels-1/pretty(x)
On Wednesday, April 9, 2014 1:26 AM, arun smartpink...@yahoo.com wrote:
Hi Atem,
If you change colnames(mat1)[-(2:3)] - ' ' in 'lst1New'
lst1New - lapply(lst1,function(x) {lst2 - setNames(lapply(x,function(y) {dat
- read.table(y,sep= ,header=TRUE, stringsAsFactors=FALSE); dat[,1:104]} ),
Hello all,
I wish to add to the
installrhttp://cran.r-project.org/web/packages/installr/package the
ability to check for new versions of R once every X units of
time (maybe once every two weeks).
I would like to keep a time stamp somewhere, that would stay persistent
across R sessions (i.e.:
I would suggest that unlist(lapply(... should always be preferable to
sapply(lapply... if you want to convert data in a data frame to a
vector. I can't see any reason to run the same loop twice. But check
timings -- maybe I'm overly sensitive to unimportant aesthetics.
Cheers,
Bert
Bert Gunter
Sagnik raises the question as to why the psych package does not offer the
‘equamax’ rotation.
It is because all rotations are handled through the GPArotation package which
does not offer equamax.
Sagnik also points out that if the requested rotation is not available, fa
defaults to
Hi,
Try:
dat - structure(list(Custom = c(Judi, Judi, Ben, Tom, Tom,
Bill, Lindy, Shary, Judu, Judu, Billy, Tommy, Tommy,
Benjum, Linda, Shiry), Gender = c(Female, Female, Male,
Male, Male, Male, Female, Female, Female, Female,
Male, Male, Male, Male, Female, Female), Product = c(A,
B, A, A,
I don't understand this error or how to resolve it.
Error in get(name, envir = asNamespace(pkg), inherits = FALSE) :
object '.Source' not found
Using R version 3.1.0 beta (2014-03-28 r65330) -- Spring Dance
# output from console ##
library(quantmod)
Loading required package:
I am unable to install flowDensity package from bioconductor in R
version 3.0 or 3.1.
did anyone have the same problems with this.
Thanks,
Raghu
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PLEASE do read the
Thanks all, I'll try some of these suggestions out but it seems like a
raw string ability could come in helpful -- there aren't any packages
out there that have this capability?
--j
On Tue, Apr 8, 2014 at 1:23 PM, Jeff Newmiller jdnew...@dcn.davis.ca.us wrote:
What is wrong with
winpath -
I posted this on SO, got no response. Sorry about cross-posting,
but atleast I would like to know if I am understading something totally
wrong
I am using a `glmulti` wrapper for glmer (binomial) and the `summary` is:
`This is glmulti 1.0.7, Apr. 2013.
Length Class Mode
0 NULL NULL`
[Sounds like a question for R-devel]
On Wed, Apr 9, 2014 at 5:02 AM, Tal Galili tal.gal...@gmail.com wrote:
Hello all,
I wish to add to the
installrhttp://cran.r-project.org/web/packages/installr/package the
ability to check for new versions of R once every X units of
time (maybe once
On 04/09/2014 07:23 AM, Raghu wrote:
I am unable to install flowDensity package from bioconductor in R version 3.0 or
3.1.
did anyone have the same problems with this.
Please ask questions about Bioconductor packages on the Bioconductor mailing
list
At 09:31 08/04/2014, Julien Riou wrote:
Dear R community,
Comments in line below
I'm working on a meta-analysis of prevalence data. The aim is to get
estimates of prevalence at the country level. The main issue is that the
disease is highly correlated with age, and the sample ages of
Due in large part to my nonexistent statistics vocabulary, I'm not sure
what the name is for the type of analysis I'm trying to do. I have,
however, started to cobble something together (below) that is undoubtedly
reinventing the wheel (i.e., someone has probably already written something
to do
Dear mates,
I am bit of stuck with a coding, all I have to do is to clean and tranform
the missing data -97 of the attribute ej into a normal number value
in a range from 0 to 22. My steps are :
1) first return the subsets, as check, they do sum to original
2) manipulate the attribute ej
Have you read An Introduction to R (or a web tutorial) to learn R?
This looks straightforward, if I understand you correctly.
?sample
and indexing operations might be all you need.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information
Hi,
Try:
datNew - read.csv(customer_samples.csv,stringsAsFactors=FALSE)
#I could reproduce similar error message with:
dat[] - lapply(dat,as.factor)
dat1 - within(dat, Categ - ave(Product, Custom, FUN= function(x)
if(length(x)1) A and B else x))
#Warning messages:
1: In
I'm looking for a method to analyse behavior data that can be assigned to
multiple categories
as a proportion with all the categories adding to 1 and some explanatory
variables and repeated measures on some individuals.
Data would look something like this
AnimalID Behavior1 Behavior2 Behavior3
I am testing a call to a C function from R, using .C interface. The test
consists in passing a numeric vector to the C function with no entries,
dynamically allocates n positions, makes attributions and return the vector
to R.
I'm using Calloc from R.h. The prototype of the function is
type*
Hi, how do you convert the illumina ProbeID to entrezID? I've used LIMMA to
identify top genes but i'm not sure how to apply illuminaHumanv4.db to get the
EntrezID.
Thanks in advance,
.kripa
[[alternative HTML version deleted]]
You might be thinking of bootstrap analysis. Note that measures are often
correlated, so beware of just introducing randomness one measure at a time.
Discussions about background theory are not really R-specific, so further
chatter about this would not be on topic here, but the term bootstrap
I am trying to generate a binary matrix where every row in the matrix is
guaranteed to have at least one 1. Ideally, I would like most rowSums to be
equal to 2 or 3 with some 1s and some 4s. But, rowSums cannot be equal to 0.
I can tinker with the vector of probability weights, but in doing so
Doran, Harold HDoran at air.org writes:
I am trying to generate a binary matrix where
row
in the matrix is guaranteed to have at
least one 1.
Ideally, I would like most rowSums to be equal
2 or 3
with some 1s and some 4s. But,
rowSums cannot be equal
to 0.
I can tinker with the
Thanks very much to both Bill Dunlap and Hadley Wickham.
Per Bill's second suggestion, match.call(eval(bo[[1]]), bo)
returned the desired plot(x = 0, y = 1) [with bo -
body(function()plot(0, 1)].
Hadley's standardise_call(bo) [from library(pryr)] is much
simpler.
A bit kludgey but how about:
dimMat - matrix(0, 1000, 4)
for(i in 1:1000){
while(sum(dimMat[i, ] - sample(c(0,1), 4, replace = TRUE, prob = c(.3, .7)))==0)
dimMat[i, ] - sample(c(0,1), 4, replace = TRUE, prob = c(.3, .7))
}
table(rowSums(dimMat))
Clint BowmanINTERNET:
You could randomly assign 1 to a single column in each row and
then use binomial draws on the remaining 0's:
set.seed(42)
dimMat - matrix(0, 1000, 4)
dimMat[cbind(1:1000, sample.int(4, 1000, replace=TRUE))] - 1
dimMat[dimMat1] - sample(0:1, 3000, replace=TRUE, prob=c(.6,
.4))
Thank you very much for the reply Henrik, you gave great pointers for me to
look at.
I am now curious about R.cache (although I am not sure it will help in my
case, I'll need to look at it further).
Also, I was debating if to use R-help or R-devel for this, since I was not
sure from the
On Wed, Apr 9, 2014 at 11:27 AM, Cassiano dos Santos crn...@gmail.com wrote:
I am testing a call to a C function from R, using .C interface. The test
consists in passing a numeric vector to the C function with no entries,
dynamically allocates n positions, makes attributions and return the
Cassiano dos Santos crns13 at gmail.com writes:
I am testing a call to a C function from R, using .C interface. The test
consists in passing a numeric vector to the C function with no entries,
dynamically allocates n positions, makes attributions and return the
vector to R.
Asking on
What we've covered so far is of great value.
For a neater plot,
the next step will be to learn how to put
numbers with units at each tick mark.
I suppose I can form the number-unit string myself in
separate code and put the tickmark in a place that
I calculate in separate code.
But I need to
hi
i want to simulate multivariate dichotomous data matrix with categories
(0,1) and n=1000 and p=10.
thanks alot in advance
[[alternative HTML version deleted]]
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Hi,
Suppose your data is similar to below:
dat - structure(list(Custom = c(Judi, Judi, Ben, Tom, Tom,
Bill, Lindy, Shary, Judu, Judu, Billy, Tommy, Tommy,
Benjum, Linda, Shiry, Shiry, Shiry, Judu, Billy,
Tommy, Lindy), Gender = c(Female, Female, Male, Male,
Male, Male, Female, Female, Female,
On Apr 9, 2014, at 6:17 PM, Hurr wrote:
What we've covered so far is of great value.
For a neater plot,
the next step will be to learn how to put
numbers with units at each tick mark.
I suppose I can form the number-unit string myself in
separate code and put the tickmark in a place that
On Apr 9, 2014, at 10:25 AM, Brian Battaile wrote:
I'm looking for a method to analyse behavior data that can be assigned to
multiple categories
as a proportion with all the categories adding to 1 and some explanatory
variables and repeated measures on some individuals.
Maybe expanding your
Hello,
I'm been playing around trying to get some maps working in ggplot ggmap.
I have a series of maps I'm trying to make and I want to fix the bubble
area size of banana catches to the specified breaks. I.e so banana catches
of 2 will always be a specified size so when I make the next map,
On 10/04/14 09:28, thanoon younis wrote:
hi
i want to simulate multivariate dichotomous data matrix with categories
(0,1) and n=1000 and p=10.
Nobody's stopping you! :-)
cheers,
Rolf Turner
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