Hello,
I am trying to convert calendar dates (Month, Day, Year) into Julian Days
Â
Product code Bureau of Meteorology station number Month Day Year Date_mdy Date
4102001 70014 1 1 1939 1/01/1939 1/01/1939
4102001 70014 1 2 1939 1/02/1939 2/01/1939
4102001 70014 1 3 1939 1/03/1939 3/01/1939
410
hello
I want a code which can correct the spelling mistakes as well as
grammatical mistakes in the sentences...that is if I am writing following
sentence:
I want too meet my frnd bt due to hectic shcedule I cant
I want output in following way:
I want to met my friend but due to hectic schedule
On Jun 14, 2012, at 2:17 PM, Alaska_Man wrote:
Dr. Winsemius,
Really quick, a BACI is a Before-After-Control-Impact approach. I
have a long time series of sea cucumber density estimates, which are
taken at the same location(s) through time. Some are in areas
Impacted by sea otters and
I am working on minimization of sum of squared errors for a problem that has
2 box-constrained parameters.
I got the solution for this problem using "L-BFGS-B" method in optim
function using an R code as
res<-optim(par=c(parInit), fn=myFunction, method = c("L-BFGS-B"), lower =
parMin, upper = par
Hi
I'm just stepping through the decompose() function, in "stats". Does
this contained line of code not work if you have a time series ending
"unevenly" (i.e., middle of the year), or am I missing something?
season <- na.omit(c(as.numeric(window(season, start(x) +
c(1, 0), end(x))), as.numer
Hi,
I'm having trouble understanding how trunc is operating on vectors of
POSIXlt objects. Why does dates[1:4] in the last line return a bunch of NAs
even though dates look like it has all the right elements? This worries me
that something is off with my use of trunc. Is trunc not suppose to be
ve
Hello,
Using the Correspondence Analysis package (ca), I wonder if there is a way
to further customize the plot beyond the options given in plot.ca provided
by the mentioned package.
The correspondence analysis I am doing consists of two datasets sharing
only the rows, so the plot draw the total
To whoever is looking for the same thing as I was, I found a solution, or
sort of.
Here is the code:
flavors<-c("vanilla", "chocolate", "strawberry")
w <- gwindow("checkbox example")
gp <- ggroup(container=w)
glabel("Favorite flavors:",cont=gp)
cbg <- gtable(flavors, cont=gp, multiple=T)
#
Hi,
I have been trying to get the sub function to work but can't for the life of me
figure it out. I tried looking at ?sub but I think there is something that I am
just not understanding.
Example failed attempts
> sub("\\","/","G:\Compensation Audits_Reports\Step Audit\Steps off Step.accdb")
E
Thanks for your reply. To my surprise I can find one more strange behavior of
my 15X15 matrix "A", that is if I call the function chol(A) in the terminal it
decompose the matrix fine without any errors or warnings.
But if I call the function chol() within a function, which I have written in
o
a) Avoid mixing Date objects and POSIXt objects. The timezones will mess you up
in the conversions. Just eliminate the as.Date conversion entirely.
b) By the time you reach the do.call function call, amortsByYears is a matrix.
While a data.frame is a special kind of list, a matrix is not. Hence,
Folks,
I call the function calcAmorts like so:
calcAmorts(prevAm, amort, myDates)
Note that I use the package lubridate.
The last line where do.call is called to first divide all the rows by the first
row and then rbind gives the following error:
Error in do.call("rbind", apply(amor
--
Michael Sumner
Hobart, Australia
e-mail: mdsum...@gmail.com
[[alternative HTML version deleted]]
__
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/post
Hi,
Try this:
date1<-"X01.11.2010"
date2<-strftime(strptime(date1,"X%d.%m.%Y"),"%d.%b.%y")
date2
[1] "01.Nov.10"
A.K.
- Original Message -
From: juliane0212
To: r-help@r-project.org
Cc:
Sent: Thursday, June 14, 2012 10:09 AM
Subject: [R] Transform date - style
Hello,
is there any
Hello arun,
Thinking about it, I believe this one is reasonably solid.
I've added a 'txt0' just in case it wouldn't like shorter.
txt0 <- "my name name is micky"
txt1 <- "my name name name is micky"
txt2 <- "my name name name name is micky"
pat <- "(\\w+\\s)\\1+"
gsub(pat, "\\1", txt0)
gsub(p
Dr. Winsemius,
Really quick, a BACI is a Before-After-Control-Impact approach. I have a long
time series of sea cucumber density estimates, which are taken at the same
location(s) through time. Some are in areas Impacted by sea otters and some
are in areas Not Impacted by sea otters (two leve
Using the animation package, I am trying to show fish movement over a jpeg of
our study area. I want to show fish detection at each of 5 sites on each day
they are detected. Each of the five fish species will have their own color
(not coded yet). I am able to create the animation, though the data i
Hi All,
This question may not belong here, but I asked this on the
R-SIG-FINANCE list and so far have not got any reply, so was hoping
someone here may help.
I have a basic query about TSRV and was hoping you all can shed some
light on the issue.
I have 22500 records for each day. So if I take t
Thank you!
This works much better.
Best regards,
Luigi
-Original Message-
From: David L Carlson [mailto:dcarl...@tamu.edu]
Sent: 14 June 2012 16:44
To: 'Luigi'
Cc: r-help@r-project.org
Subject: RE: [R] Median line with stripchart
This automates things a bit once we re-organize your data
Hi
I use the flexmix function for clustering.
when i give the command plot(flexmixobjext) , it gives the rootogram plot
can anyone help me to understand the plot
-
Thanks in Advance
Arun
--
View this message in context:
http://r.789695.n4.nabble.com/Rootogram-for-flexmix-function-t
problem solved. if one adjusts in r the time span the way i did it, r really
only has data available for this time span. to the opposite in eviews. if
one sets in eviews a time span via adjusting the sample size, it still uses
the observations outside the sample size i.e. in cointegrationn tests wh
I just tried to figure out why R does not calculate the same johansen test
statistics as eviews does:
I imported data:
sy=read.csv("sy.csv",sep=";",header=TRUE)
si=read.csv("si.csv",sep=";",header=TRUE)
merged them.
swe=merge.zoo(si,sy)
when i test swe for cointegration with the johansen test, i
Hi,
If you need the difference between two dates from the dataset,
Try this:
>dat1<-data.frame(DATETIME=c("1/1/2010 0:10","1/1/2010 0:20","1/1/2010
0:30"),HEADER1=c(197.19,203.88,206.56),HEADER2=c(100.08,100.10,100.04))
>dat1$DATETIME<-strptime(dat1$DATETIME, "%d/%m/%Y %H:%M")
>difftime(dat
On Thu, Jun 14, 2012 at 02:24:20PM -0400, cowboy wrote:
> thank you, Petr.
> This is exactly what I'm looking for in my post.
> An related question can be how to get an arbitrary weight, say if row1
> and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
> row 2 have 2 common value
The second page (mmo-champion.com) doesn't contain a
node.
To scrape the data from the page, you will have to explore its
HTML structure.
D.
On 6/14/12 9:31 AM, Moon Eunyoung wrote:
> Hi R experts,
>
> I have been playing with library(XML) recently and found out that
> readHTMLTable workls
thank you, Petr.
This is exactly what I'm looking for in my post.
An related question can be how to get an arbitrary weight, say if row1
and row 2 have 1 common value 1, then assign a weight 10, if row 1 and
row 2 have 2 common value 1, then assign a weight 12. I'm not so sure
how to expand your me
I do think this is more of a Bioconductor question -- but no worries,
they're all much nicer there than we are here and won't flame you if
you double post :-)
Best,
Michael
On Thu, Jun 14, 2012 at 12:37 PM, Hans Thompson
wrote:
> I know that there are quite a few packages out that there for clus
thanks a alot !!! I got it :)
On Thu, Jun 14, 2012 at 8:07 PM, Rui Barradas [via R] <
ml-node+s789695n463338...@n4.nabble.com> wrote:
> Hello,
>
> Try
>
> sink("sentiment.txt")
> classify_polarity(...etc...)
> sink()
>
>
> If this doesn't do it see
>
> ?capture.output
>
> Hope this helps,
>
> Rui
Though not exactly what you asked for, I find the output of dput() to be useful
dput(moransl, file = "moransl.txt")
results in
list("Moran.I.First", structure(list(observed = 0.06988288, expected =
-0.03225806,
sd = 0.02513276, p.value = 4.822722e-05), .Names = c("observed",
"expected", "s
Below are two equivalent solutions.
study_df <- data.frame(course = c(rep('Mathematics', 80 + 15),
rep('Physics', 32 + 24),
rep('Biology', 18 + 29)),
A = c(rep(1, 80), rep(0, 15),
Hi Carlos,
Thanks for your suggestions. I saw Rui's reply about the same problem using
rle. It looks very solid. I was trying replicate the same thing with "gsub",
but it was not working in that way.
For example,
txt1<-"my name name name is micky"
gsub("\\b(\\w+)\\b(\\s+)\\1\\2","",txt
Dear Rxperts,
I am back to favoRite! I would need your favoR, please!
Is there a way to use read "format" arguments for "POSIXct" or "strptime"
from the xml file without specifying format="..." in these functions? The
format of data-time values are present in the xml file as shown below.
Please
Thanks and with
datlines <- as.data.frame(inp[( grep("", inp)[1]+5 ):(grep("",
inp)[1]-1)]);
I get the data as needed.
Thanks again
H.
- Original Message -
On Jun 14, 2012, at 10:23 AM, Halldór Björnsson wrote:
> Hi,
>
> I am trying to read in weather balloon data, where each file
thanks for reply..I got it...:)
On Thu, Jun 14, 2012 at 8:41 PM, Rui Barradas [via R] <
ml-node+s789695n4633393...@n4.nabble.com> wrote:
> Hello,
>
> I don't understand, do you want to output:
>
> 1st line: the original input line
> 2nd line: " I want output in following format " <-- this line
>
Thanks,
even better
- Original Message -
On Jun 14, 2012, at 12:18 PM, Halldór Björnsson wrote:
> Thanks and with
>
> datlines <- as.data.frame(inp[( grep("", inp)[1]+5 ):(grep(" PRE>", inp)[1]-1)]);
>
I suggest this instead.
> read.fwf(textConnection(datlines), widths=rep(7,11))
V1
Hi,
Try this:
sub("(I)\\((.*?)\\)","\\2",t1)
[1] "(Ithis) test"
A.K.
- Original Message -
From: nalluri pratap
To: r-help@r-project.org
Cc:
Sent: Thursday, June 14, 2012 7:03 AM
Subject: [R] gsub
Hi,
I have a string t1="(Ithis) I(test)". I need to get t2="(Ithis) test".
Can som
Hi R experts,
I have been playing with library(XML) recently and found out that
readHTMLTable workls flawlessly for some website, but it does give me an
error like below
... Error in function (classes, fdef, mtable) :
unable to find an inherited method for function "readHTMLTable", for
signat
I know that there are quite a few packages out that there for cluster
analysis. The problem that I am facing is finding a package that will not
incorporate all my samples into clusters but just the samples that fit a
threshold (that I have not set yet and may need help finding the right
level) for
On 14.06.2012 01:48, Sarah Henderson wrote:
Hello to all --
I'm hoping that someone more knowledgeable than me can shed some light
on a problem I have been having. In point form:
- I am running XP on a 64-bit processor
- I run 5 automated R tasks every morning using .bat files and a
little ut
On Thu, Jun 14, 2012 at 01:11:45PM +, G. Dai wrote:
> Dear Rlisters,
> I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
>
> To elaborate, let's consider an example. Assume we have the following
> 3 by 4 matrix A with elements either 0 or 1,
> 0 1 1 0
> 1 0 1
On Jun 14, 2012, at 12:18 PM, Halldór Björnsson wrote:
Thanks and with
datlines <- as.data.frame(inp[( grep("", inp)[1]+5 ):(grep("PRE>", inp)[1]-1)]);
Er, ... are you sure? I got a factorized mess when I did that.
> str(datlines)
'data.frame': 98 obs. of 1 variable:
$ inp[(grep("", i
On Jun 13, 2012, at 7:36 PM, Alaska_Man wrote:
Hello,
I am performing a BACI analysis with ANOVA using the following glm:
I admit I had no idea what a "BACI analysis" might be. Looking it up
it appears to be a cross-over design and my statistical betters have
sternly warned me about this
First thing is to supply data in a usable form See ?dput for one easy way of
doing it.
In any case, assuming those dates and times are character values something like
this should work but not tested on your data.
Assuming the data frame is called dtime
dtime[,1] <- strptime(dtime[,1], "%d/%
Hi Noah,
I did ask basically the same question about a year ago and there
wasn't anything around
(http://tolstoy.newcastle.edu.au/R/e14/help/11/06/3651.html)
Although I agree that R would be very suitable for this kind of
calculations exist. I guess one reason is that a decision tree is not
really
And I'd like to add, just for the purpose of learning about R ... even if
wishes to use the loop version, there appears to be a misunderstanding of
R syntax.
The expression
1:225*100
does not produce 22500 numbers to put into the matrix, as apparently
expected.
Compare:
> 1:3*5
[1] 5 10
Just for make the archives more complete and simplifing the life of the
following readers.
I think to have solved my problem using the caret packages.
In this package there is a function named createData Partition that after
defining a column of interest in a data.frame allows to split a dataset in
What makes you think that with 12 elements the dimensions should be 3x4? It
seems that 4x3 would be equally valid.
In every use of matrices that I have encountered, the dimensions have been
relatable to some known quantity from the problem context, and matrix
dimensioning has been an exercise i
Can you explain why n=12 should result in 3x4 instead of 2x6 or 6x2 or
4x3 or 1x12 ?
On Thu, Jun 14, 2012 at 8:51 AM, karthicklakshman
wrote:
> Dear R experts,
>
> I am interested in getting the dimensions for the matrix dynamically, based
> on the the number of elements in a matrix for example.
You have been asked before (by me!) to give reproducible examples
using dput()...
You might need the diff() function.
Michael
On Thu, Jun 14, 2012 at 5:08 AM, Rantony wrote:
> Hi,
>
>
> Here, i have a matrix like this
>
> MyMatrix <-
>
> *DATETIME HEADER1 HEADER2*
> 1/1/2010 0
On Jun 14, 2012, at 12:18 PM, Halldór Björnsson wrote:
Thanks and with
datlines <- as.data.frame(inp[( grep("", inp)[1]+5 ):(grep("PRE>", inp)[1]-1)]);
I suggest this instead.
> read.fwf(textConnection(datlines), widths=rep(7,11))
V1V2V3V4 V5 V6 V7 V8V9 V10
If the number of elements is 12, the dimensions could be 1x12, 12x1, 2x6,
6x2, 3x4, or 4x3. How did you decide on 3x4?
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Original Message-
If you have your data in x, you can try something like
matrix(x,length(x)/4,4)
hth
kd
2012.06.14. 14:51 keltezéssel, karthicklakshman írta:
Dear R experts,
I am interested in getting the dimensions for the matrix dynamically, based
on the the number of elements in a matrix for example. if th
Why would those figues (dim= 3X4, dim=5X4) be the dimensions? Could not they
equally be 4X3 an 4X5 or am I completely misunderstanding the question?
John Kane
Kingston ON Canada
> -Original Message-
> From: karthick.laksh...@gmail.com
> Sent: Thu, 14 Jun 2012 05:51:24 -0700 (PDT)
> To:
On Jun 14, 2012, at 10:23 AM, Halldór Björnsson wrote:
Hi,
I am trying to read in weather balloon data, where each file has a
header of fixed length and
a trailing section of a fixed length. The data section (the table)
is of variable length.
An example of the data is on:
http://weather
This automates things a bit once we re-organize your data.
my.newdata <- stack(data.frame(my.data), select=c(Control, Case))
stripchart(values~ind, my.newdata, method = "stack", offset=1/3,
vertical = TRUE, pch=19)
medians <- tapply(my.newdata$values, my.newdata$ind, median)
points(c(1, 2), me
But the meaning of a 3x4 table is rather different than the meaning of
a 1x12 table.
Regardless, you probably want to start with integer factorization, and
can read more
about implementations in R here (and elsewhere):
http://tolstoy.newcastle.edu.au/R/help/05/01/10007.html
Sarah
On Thu, Jun 14,
Would something like ggplot's transparacy option help?
Example bascially from http://had.co.nz/ggplot2/geom_point.html
# Plot large data set.
d <- ggplot(diamonds, aes(carat, price)) + geom_point()
d
# plot large data set with transparancy set to 1/10 th
p = ggplot(diamonds, aes(carat, price))
I think you're right -- prob probably isn't quite what you need (at
least, directly): constrained sampling like this is a little trickier
-- I'll leave this to someone who knows more than me.
Michael
On Thu, Jun 14, 2012 at 9:07 AM, Guido Leoni wrote:
> Sorry I'm not sure that prob is suitable f
But there are multiple shapes for matrices with 12 elements -- how did
you get 3x4 ? You also could have had 1x12, 2x6, 3x4, 4x3, 6x2,12x1
If you have an R object, you can use dim() on it. [Or perhaps slightly
more robustly, NCOL() and NROW()]
Michael
On Thu, Jun 14, 2012 at 7:51 AM, karthicklak
Try this. It will turn off plotting the points, with 186 values, the points
are covering up the line.
x <- runif(186)*10
plot(ecdf(x), do.points=FALSE)
?plot.ecdf # to get help on the plot method used to plot ecdf
?plot.stepfun # to get help on the plot method used to plot stepfun
Dear Rlisters,
I'm writing to ask how to manipulate a matrix or dataframe in a specific way.
To elaborate, let's consider an example. Assume we have the following
3 by 4 matrix A with elements either 0 or 1,
0 1 1 0
1 0 1 1
0 0 0 1
>From the original matrix A, I'd like to generate a new
Hello,
Hello,
If the input matrix is symmetric, positive definite then the Cholesky
decomposition algorithm is stable. That's why it is so used in
statistics, where many times the matrices meet those conditions.
Therefore, the matrix isn't symmetric, positive definite to begin with.
Rui Barr
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Luigi
> Sent: 14 June 2012 00:44
> To: dcarl...@tamu.edu
> Cc: r-help@r-project.org
> Subject: Re: [R] Median line with stripchart
>
> Sorry for the inconvenience,
> I have ad
Hello,
I don't understand, do you want to output:
1st line: the original input line
2nd line: " I want output in following format " <-- this line
3rd line and following: the input line broken by the patterns
Then just
f <- function(x, pattern) unlist(strsplit(x, pattern))
x <- "I love to wa
Hi,
I am trying to read in weather balloon data, where each file has a header of
fixed length and
a trailing section of a fixed length. The data section (the table) is of
variable length.
An example of the data is on:
http://weather.uwyo.edu/cgi-bin/sounding?region=naconf&TYPE=TEXT%3ALIST&YEAR
Sorry I'm not sure that prob is suitable for my purposes(but i'm quite
newbie with R).
If I correctly understand prob allows to set a weight for each row in the
original dataset in order to include the rows on the basis of their
weights). ... I'm not sure to correctly understanding ;-)
In my case a
Try this:
old.date <- "X01.11.2010"
new.date <- format(as.Date(old.date,format="X%d.%m.%Y"),"%d.%b.%y")
[1] "01.Nov.10"
--
View this message in context:
http://r.789695.n4.nabble.com/Transform-date-style-tp4633377p4633381.html
Sent from the R help mailing list archive at Nabble.com.
_
Hi,
Here, i have a matrix like this
MyMatrix <-
*DATETIMEHEADER1HEADER2*
1/1/2010 0:10 197.1947 100.0859
1/1/2010 0:20 203.8811 100.1013
1/1/2010 0:30 206.564 100.0433
1/1/2010 0:40 207.9563
Hello,
Thanks, I wasn't really liking it very much but it more of a diffuse
feeling than of a founded thought. It's good to see an example.
Rui Barradas
Em 14-06-2012 14:22, David Winsemius escreveu:
On Jun 14, 2012, at 3:20 AM, Rui Barradas wrote:
Hello,
Now the output of str() says 'da
> -Original Message-
> First, i use the BOXPLOT() and SUBSET() to produce the box
> plot of all the 5 funds performance individually:
> ...
> *So the FIRST QUESTION is how to eliminate the other 5
> strategies on the Y-aix? *
either use
boxplot(NetReturn~factor(Strategy),data=filtere
Hello,
Try
t1 <- "(Ithis) I(test)"
t2 <- "(Ithis) test"
t3 <- sub("I\\((.+)\\)", "\\1", t1)
t2 == t3 #[1] TRUE
If there are more than one occurences, use 'gsub'.
Hope this helps,
Rui Barradas
Em 14-06-2012 12:03, nalluri pratap escreveu:
Hi,
I have a string t1="(Ithis) I(test)". I need
Hello,
is there any posibility to automatically transform this date notation
"X01.11.2010" into "01.Nov.10" ?
I did not see it before starting my simulation and now I have to deal with
it.
It is no problem for my calculations but for my graphical outputs I would
like
> sub("I\\((.*?)\\)", "\\1", t1) # nongreedy
[1] "(Ithis) test"
On Thu, Jun 14, 2012 at 7:03 AM, nalluri pratap wrote:
>
> Hi,
>
> I have a string t1="(Ithis) I(test)". I need to get t2="(Ithis) test".
>
> Can someone look into this. ?I have tried using gsub("I[^)]","",t1) , but
> didn't get th
Hello,
Try
sink("sentiment.txt")
classify_polarity(...etc...)
sink()
If this doesn't do it see
?capture.output
Hope this helps,
Rui Barradas
Em 14-06-2012 09:13, raishilpa escreveu:
hello,
I am using following command
classify_polarity(documents,algorithm="bayes",verbose=TRUE)
output i
Dear R experts,
I am interested in getting the dimensions for the matrix dynamically, based
on the the number of elements in a matrix for example. if the number is 12,
I should get dim= 3X4, if it is 20, dim=5X4.
please help me do this.
Thank you
Regards
karthick
--
View this message in conte
On Jun 14, 2012, at 13:03 , nalluri pratap wrote:
>
> Hi,
>
> I have a string t1="(Ithis) I(test)". I need to get t2="(Ithis) test".
>
> Can someone look into this. ?
Well, you can, it is your problem
You need to look into the hairier parts of regular expression syntax. Looks
like th
Hi All,
I have used glm to model my data, I have two factors and a covariate as
described in the example code below (mod.1).
I have been able to "force" glht to perform multiple comparisons by creating
a combined variable for the factors, accepting that there will be a loss of
statistical power
Hi David,
summary(res)$coefficients[,"Pr(>|z|)"] or summary(res)$coefficients[,4]
M
Regrads
Le 14/06/12 12:44, David Studer a écrit :
> Hi everyone!
>
> Can anyone tell me, how to obtain p.values from a linear model?
>
> Example:
> mod1<-lm(dV~iV1+iV2)
>
> Now, I can get the coefficients with
On Jun 14, 2012, at 3:20 AM, Rui Barradas wrote:
Hello,
Now the output of str() says 'dat' is a list not a data.frame.
That's why R is complaining about dimensions (lack of, in this case).
Try
dat2 <- data.frame(do.call(cbind, dat), stringsAsFactors=FALSE)
The construction data.frame(cb
sample() takes a prob = argument which lets you supply weights, which
need not sum to one so, if I understand you, you could just pass TRUEs
and FALSEs for those rows you want. If I'm wrong about that last bit,
I'm still pretty confident sample(prob = ) is the way to go.
Best,
Michael
On Thu, Jun
Your matrix is not symmetric, positive definite. If you don't know
what this means, you shouldn't be using chol()
This may be because it isn't to begin with, or due to numerical error,
it doesn't behave as one in the decomposition. My relative ignorance
of numeric methods for linear algebra preven
Hello,
I try to use xtable under odfWeave.
With
<>=
library(xtable)
xtable(iris)
@
this results in an endless list of "xmlParseEntityRef: no name"
On internet I find that results should be tex:
<>=
However, now the result is:
'arg' shoul
Hi,
I'am using the sentient analysis available in "R2.15".
Now the challenge that I'am facing is that the Naïve Baye's has an inbuilt list
of 6500 words in which it has been trained.
So my question that can I increase the number of words on which this algorithm
is trained?
In case i can increas
Have you manually searched the CRAN Task Views? Sometimes you'll find
something there that searches won't pick up.
-- Bert
On Thu, Jun 14, 2012 at 4:12 AM, ya wrote:
> Hi all,
>
> I have a question, is there any R package dealing with latent transition
> analysis with both categorical and conti
Hi,
I've kind of a tricky question, which I don't know how to solve yet:
I get multiple dataframes loaded (readRDS) in a loop function. Each loaded
dataframe contains two columns one with a var-name and one with a value. The
rownumber (order) is very important as it is a value of the rank (1:x
On 06/14/2012 07:08 PM, field.cady wrote:
I'm working with a large dataset - large enough that when I do a scatter plot
the points all blur together, so I want to plot their density by color - a
heat map or something like that. I've used smoothScatter for tasks like
this, but the problem is that
On Jun 14, 2012, at 12:14 , syrvn wrote:
> Hello!
>
> I found out that it is possible to open the windows explorer with a
> predefined path via the cmd.exe program using the following command:
>
> explorer PATH
>
> Back in R using the following command opens up the windows explorer:
>
> syste
Dear friends,
When I do Cholesky decomposition for a 15x15 matrix using the function chol(),
I get the following error for which I do not understand the meaning of the error
Error in chol.default(M_cov) :
the leading minor of order 10 is not positive definite
When I searched online for simila
Hi all,
I have a question, is there any R package dealing with latent transition
analysis with both categorical and continuous indicators? So far what I found
from GOOGLE are only packages dealing with latent class analysis. So what about
the longitudinal situation? Any way we could look at the
Dear David,
Try
summary(mod1)$coef[,4]
Best
Ozgur
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R-help@r-project.org mailing list
ht
Dear list I wish to extract from a population genotypized for 10 SNP a
subsample of the same population of size n with similar allele frequencies.
Essentially i have a matrix of 200 rows (df) like this
Name,Condition,rs1385699_X,rs6625163_X,rs962458_X,Rs4658627_1,
sample01,Case,1,1,1,-1
sample02,Co
Hi,
I have a string t1="(Ithis) I(test)". I need to get t2="(Ithis) test".
Can someone look into this. ?I have tried using gsub("I[^)]","",t1) , but
didn't get the required result.
Thanks,
Pratap
[[alternative HTML version deleted]]
__
R-
Hello,
What do you want to do with these p-values?
Best Regards
Le 12/06/14 19:44, David Studer a écrit :
Hi everyone!
Can anyone tell me, how to obtain p.values from a linear model?
Example:
mod1<-lm(dV~iV1+iV2)
Now, I can get the coefficients with mod1$coef
But how can I get p-values? ($
Hi everyone!
Can anyone tell me, how to obtain p.values from a linear model?
Example:
mod1<-lm(dV~iV1+iV2)
Now, I can get the coefficients with mod1$coef
But how can I get p-values? ($p.values seems to work with cor.test() only)
Thank you!
[[alternative HTML version deleted]]
Thank you so much Sarah and Jim, both solutions worked perfectly!!
I still have so much to learn about R, and this fantastic team motivates me
a lot!
I'd like to congratulate to all people that makes this work so well, it's
so useful for those of us that are just beginning with R... and surely to
On Thu, Jun 14, 2012 at 01:08:53AM -0700, mogwai84 wrote:
> Hi all,
>
> I've got a very long numeric string. I want to find the lowest number that
> isn't in that string.
>
> E.G
>
> String = 123456
> Therefore 7 is the lowest number not in that string
>
> E.G.2
>
> String = 1234567891011
> Th
Hello!
I found out that it is possible to open the windows explorer with a
predefined path via the cmd.exe program using the following command:
explorer PATH
Back in R using the following command opens up the windows explorer:
system("explorer", intern=TRUE)
However, when I specify a path R re
I'm working with a large dataset - large enough that when I do a scatter plot
the points all blur together, so I want to plot their density by color - a
heat map or something like that. I've used smoothScatter for tasks like
this, but the problem is that my current dataset really only looks good o
Good Afternoon,
I'm trying to create a cdf plot, with the following code.It works well,
but I have little doubt, if you can help solve.When I create the plot,
like the graph line would still not appear with point
#cdf
x<-table(Dataset$Apcode)
View(s)
hist(s)
*plot(ecdf(x))*
x<-1
On Thu, Jun 14, 2012 at 01:08:53AM -0700, mogwai84 wrote:
> Hi all,
>
> I've got a very long numeric string. I want to find the lowest number that
> isn't in that string.
>
> E.G
>
> String = 123456
> Therefore 7 is the lowest number not in that string
>
> E.G.2
>
> String = 1234567891011
> Th
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