Hi,thanks a lot for the answer. I think that my problem is before packages. I
have a large amount of different datasets and I had a random look on their
histograms. I know that some of them look like the combination I have explained
(exponential+pareto, exponential+gamma) but I am not sure still
yes of course, and the answer is latex() in the Hmisc package.
Why were you excluding it?
Details follow
Rich
The current release of the Hmisc package has this capability on
Macintosh and Linux.
For Windows, you need the next release 3.14-7 which is available now at github.
## windows needs the
Perhaps
as.data.frame(VarCorr(pcrpred))
?
Best,
Ista
On Mon, Dec 8, 2014 at 4:15 PM, GMAIL wrote:
> I have a mer object that has fixed and random effects (lmer).
>
> How do I extract the variance or standard deviation estimates for the random
> and fixed effects? Here is a simplified version
I have a mer object that has fixed and random effects (lmer).
How do I extract the variance or standard deviation estimates for the random
and fixed effects? Here is a simplified version of my question:
pcrpred <- lmer(PCR ~ (1|TIME) + (1|ID), data = mydataPCRlong)
pcrpred
This gives a long
2 ideas (haven't tried them):
1. if your data is in a data frame, did you try using the by function?
Seems it would do the grouping for you.
2. Since you mention the cpu cores, you could use libraries like foreach
and %dopar% or mcapply.
I would try 1. and see if it provides a sufficient speed-u
Hello,
I use the following code to draw a map of Europe. The colour of the
countries depends on a value called "EPI2.0" in a data frame called
epidata.
epimap=joinCountryData2Map(epidata,joinCode="NAME",nameCountryColumn="Country",nameJoinColumn="Country")
mapCountryData(epimap, nameColumnTo
Thank you all. That was very helpful. Farnoosh
On Saturday, December 6, 2014 7:41 PM, Chel Hee Lee
wrote:
It seems that you would like to make a spaghetti plot (in a longitudinal
data analysis). You can use the function 'interaction.plot()'.
> with(my.df, interaction.plot(TIME
Hi,
I have a simple question. I know there are plenty of packages out
there that can provide code to generate a table in latex. But I was
wondering whether there was one out there where I can generate a table
from my data (which ever way I please) then allow me to save it as a
pdf?
Thanks
K.
Rolf Turner auckland.ac.nz> writes:
>
>
> I know nothing about the bbmle package and its mle2() function, but it
> is a general truth that if you need to constrain a parameter to be
> positive in an optimisation procedure a simple and effective approach is
> to reparameterize using exp().
Hi all,I am having some heavy tailed data and I am trying to think of the more
appropriate package for the fitting.The canonical try should be something like
exponential and pareto or exponential + gamma (or gamma + gamma with different
shape parameters). I am trying to have one distribution tha
Have you considered "distr" and related packages?
If this does not solve your problem, have you considered
searching with "findFn" in the "sos" package? If that still does not
produce sufficient enlightenment, please try this list again with
"commented, minimal, self-contained, r
On 8 Dec 2014, at 21:21, apeshifter wrote:
> The last relic of the afore-mentioned for-loop that goes through all the
> word pairs and tries to calculate some statistics on them is the following
> line of code:
>> typefreq.after1[i]<-length(unique(word2[which(word1==word1[i])]))
> (where word1 a
Have you looked at the merge() function?
Here is an example. I don't know if it resembles your problem.
> M1 <- data.frame(V1=letters[1:3], V2=LETTERS[26:24], N1=101:103)
> M2 <- data.frame(V1=letters[c(3,1,2,3,2)],
V2=LETTERS[c(23,26,22,24,24)], N2=c(1003,1001,1002,1003,1002))
> merge(M
The data.table package might be of use to you, but lacking a reproducible
example [1] I think I will leave figuring out just how to you.
Being on Nabble you may not be able to see the footer appended to every
message on this MAILING LIST. For your benefit, here it is:
* R-help@r-project.org m
Below...
On Mon, 8 Dec 2014, David Lambert wrote:
I have 2 data frames, M1[n,20] and M2[m,30].
What does this mean? It might be intended to convey matrix dimensions, but
these are not matrices and that is not R syntax.
If V1 and V2 are the same in both M1 and M2, then append V3-V30 from M
Actually, the zero-length look-ahead expression is enough to get the job
done:
> strsplit(c(":sad", "happy:", "happy:sad", ":happy:sad:subdued:"),
split="(?=:)", perl=TRUE)
[[1]]
[1] ":" "sad"
[[2]]
[1] "happy" ":"
[[3]]
[1] "happy" ":" "sad"
[[4]]
[1] ":" "happy" ":" "sad"
strsplit(split=":") does almost what you want, but it omits the colons from
the output. You can use perl zero-length look-ahead and look-behind
operators in the split argument to get the colons as well:
> strsplit(c(":sad", "happy:", "happy:sad"), split="(?<=:)|(?=:)",
perl=TRUE)
[[1]]
[1] ":"
On 08/12/2014 6:25 AM, Frederic Ntirenganya wrote:
Hi All,
i would like to write a srcipt which ruturn the random numbers which are
binary numbers.
Example: The first group : 1
second : 0
third : 1010111100
in such away that i can make it
I want to do the following: if a string does not contain a colon (:),
no change is needed; if it contains one or more colons, break the
string into multiple strings using the colon as a separator. For
example, "happy:" becomes
"happy" ":"
":sad" turns to
":" "sad"
and "happy:sad" changes to
"h
Hi All,
i would like to write a srcipt which ruturn the random numbers which are
binary numbers.
Example: The first group : 1
second : 0
third : 1010111100
in such away that i can make iterartions.
Frederic Ntirenganya
Maseno University,
Af
(I am sorry if you have received this email twice but it does not look sent on
my client)
Hi all,I am having some heavy tailed data and I am trying to think of the more
appropriate package for the fitting.The canonical try should be something like
exponential and pareto or exponential + ga
Dear all,
for the past two weeks, I've been working on a script to retrieve word pairs
and calculate some of their statistics using R. Everything seemed to work
fine until I switched from a small test dataset to the 'real thing' and
noticed what a runtime monster I had devised!
I could reduce p
Hello!
I am performing a sentiment analysis of 2.000 negative and positive reviews.
I think my code needs improvement because I am getting accuracy 68 % and
the running duration of the code is 20 minutes!! Please find below a part
of the code.
"
# Read data from their directories
pos <- Corpus(Di
I have 2 data frames, M1[n,20] and M2[m,30].
If V1 and V2 are the same in both M1 and M2, then append V3-V30 from M2 onto M1.
Otherwise, continue searching for a match.
M1 is complete for all V1 and V2. M2 is missing observations for V1 or V2, or
both.
I can't figure this one out, except
You are still posting in HTML, and it is continuing to impede this
conversation. Learn how to post in plain text before posting again. Gmail
does have this option.
You are not using dput, as previously asked, either. Read the web page I
referenced to learn how to send R data unambiguously.
Y
On 08/12/14 21:18, Ragia Ibrahim wrote:
Hi,
Kindly I had a data frame looks like this
x y
1 3
2 2
3 1
4 3
and I want to add column z that sum cumulativly like this
x y z
1 3 3
2 2 5
3 1 6
4 3 9
how to do this?
(1) Learn to use R. This is very basic; read some introductory
material. Start wi
my.data$z <- cumsum(my.data$y)
Yes, the function you need is even in your message subject.
> On Dec 8, 2014, at 12:18 AM, Ragia Ibrahim wrote:
>
> Hi,
> Kindly I had a data frame looks like this
> x y
> 1 3
> 2 2
> 3 1
> 4 3
> and I want to add column z that sum cumulativly like this
Hello,
If your dataset is named 'dat', try
dat$z <- cumsum(dat$y)
Hope this helps,
Rui Barradas
Em 08-12-2014 08:18, Ragia Ibrahim escreveu:
Hi,
Kindly I had a data frame looks like this
x y
1 3
2 2
3 1
4 3
and I want to add column z that sum cumulativly like this
x y z
1 3 3
2 2 5
3 1 6
4
I know nothing about the bbmle package and its mle2() function, but it
is a general truth that if you need to constrain a parameter to be
positive in an optimisation procedure a simple and effective approach is
to reparameterize using exp().
I.e. represent xmin as exp(lxmin) (say) and use l
Hi
Is this what you want?
by(data$time, list(data$size), function(x) t.test(x))
Petr
> -Original Message-
> From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Paul
> Johnston
> Sent: Tuesday, December 02, 2014 10:17 AM
> To: r-h...@lists.r-project.org
> Subject: [R] Passing
Please don't post in HTML... you may not recognize it, but the receiving
end does not necessarily (and in this case did not) look like the sending
end, and the cleanup can impede answers you are hoping to get.
In many cases, loops can be vectorized. However, near as I can tell this
is an exam
Great, many thanks for your help Jeff.
Apologies for the HTML format, I'll be more careful next time.
Arnaud
On 08/12/2014 08:25, Jeff Newmiller wrote:
Please don't post in HTML... you may not recognize it, but the
receiving end does not necessarily (and in this case did not) look
like the send
Hi,
Kindly I had a data frame looks like this
x y
1 3
2 2
3 1
4 3
and I want to add column z that sum cumulativly like this
x y z
1 3 3
2 2 5
3 1 6
4 3 9
how to do this?
Regards
Ragia
[[alternative HTML version deleted]]
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Dear all,
I am fitting models to data with mle2 function of the bbmle package.In
specific, I want to fit a power-law distribution model, as defined here
(http://arxiv.org/pdf/cond-mat/0412004v3.pdf), to data.
However, one of the parameters - xmin -, must be necessarily greater than zero.
What ca
There is a system wide installation for the university computer of r and
Rcmdr (R-Commander)
There a a few computer with the following message the user tries to open
an excel sheet with R.Commander:
(I assume the german message is from the german operating system win 7)
> library(RODBC, pos=
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