Hi all,
Here is a description of my case. I am sorry if my question is also
statistic related but it is difficult to disentangle. I will however try
to make it only R applied.
My response is a growth constant k and my descriptor is prey biomass
NP and time series is of 21 years.
I applied
Hi Monnand,
Is this what you are looking for?
data[grep(prefix1,data$name),]
data[grep(prefix2,data$name),]
Jim
On Wed, Jan 28, 2015 at 6:51 PM, Monnand monn...@gmail.com wrote:
Hi all,
This really annoyed since I thought this would be easy with some higher
order function.
Here is what I
A new release 0.13.1 of matrixStats is now on CRAN
[http://cran.r-project.org/package=matrixStats]. The source code is
available on GitHub [https://github.com/HenrikBengtsson/matrixStats].
WHAT DOES IT DO?
The matrixStats package provides highly optimized functions for
computing common
Utiliza collapse en vez de sep dentro de la función paste().
Un ejemplo:
DT=data.table(ID=1:4,ERR1=c(1,1,NA,NA),ERR2=c(NA,2,2,NA),ERR3=c(3,3,3,NA))
DT
ID ERR1 ERR2 ERR3
1: 11 NA3
2: 2123
3: 3 NA23
4: 4 NA NA NA
Hola,
Si tienes su expresión algebráica, podrías ajustarla con un ajuste no
lineal nls().
nls() viene por defecto dentro del paquete stats y si no fuese posible
hacerla converger, puedes utilizar otros paquetes (nlstools, nls2) que
modifican nls() que utilizan otros algoritmos de convergencia.
Hola Hector, buenos d�as:Hay un m�todo generalista para maximizar funciones en
R que quiz�s te valga, prueba lo siguiente a ver qu� tal (por supuesto, si te
fijas en la definici�n de la funci�n de verosimilitud, ves que la he cuadrado
a mano al intervalo que trataslibrary(optimx)muestra - c(50,
Hola Hector,
No soy experto, pero en
http://r.789695.n4.nabble.com/Fitting-weibull-exponential-and-lognormal-distributions-to-left-truncated-data-td869977.html
http://www.r-bloggers.com/r-help-follow-up-truncated-exponential/
http://www.jstatsoft.org/v16/c02/paper
hay algunas ideas.Espero
Saludos cordiales.
Lamentablemente lo que dice Carlos no es correcto. Cuando la distribucin exponencial la truncamos en un intervalo la constante de la funcin de densidad (para que integre 1) tiene una dependencia (complicada) del parmetro que multiplica al exponente, con lo cual la ecuacin de
Hola Hector, buenos d�as:
Hay un m�todo generalista para maximizar funciones en R que quiz�s te valga,
prueba lo siguiente a ver qu� tal (por supuesto, si te fijas en la definici�n
de la funci�n de verosimilitud, ves que la he cuadrado a mano al intervalo
que tratas
library(optimx)
muestra -
Dear Xochitl,
Have a look at gls() from the nlme package. It allows you to fit auto
correlated errors.
gls(k ~ NPw, correlation = corAR1(form = ~ Time))
Best regards,
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek / Research Institute for Nature and
Forest
team Biometrie
Hi Kate,
Maybe you want:
seq(2,length(x),by=2)
Jim
On Thu, Jan 29, 2015 at 10:55 AM, Kate Ignatius kate.ignat...@gmail.com wrote:
I have genetic data as follows (simple example, actual data is much larger):
comb =
ID1 A A T G C T G C G T C G T A
ID2 G C T G C C T G C T G T T T
And I
The two datasets below are excerpts from much larger datasets. Note that
there are duplicate dates in both dat1 and dat2, e.g., 2009-10-14.
dat1 - read.table(textConnection(Date ConcAve
2009-07-08 7
2009-08-26 1
2009-08-26 2
2009-09-15 2
2009-10-14 2
2009-10-14
Hi,
You did the harder, it remains the easier
listMatrices - vector(list, 3)
doAll - function(A){
B - subset(A, (A[,1] %in% c(1,2) A[,2] %in% c(1,2)) |
(A[,1] %in% c(3) A[,2] %in% c(1) ) |
(A[,1] %in% c(4) A[,2] %in% c(1:4)) )
C - subset(A,
I have genetic data as follows (simple example, actual data is much larger):
comb =
ID1 A A T G C T G C G T C G T A
ID2 G C T G C C T G C T G T T T
And I wish to get an output like this:
ID1 AA TG CT GC GT CG TA
ID2 GC TG CC TG CT GT TT
That is, paste every two columns together.
I have
comment inline
David Winsemius wrote on 24.01.2015 21:08:
On Jan 23, 2015, at 5:54 PM, JohnDee wrote:
Heinz Tuechler wrote
At 07:40 21.06.2009, J Dougherty wrote:
[...]
There are other ways of regarding the FET. Since it is precisely
what it says
- an exact test - you can argue that you
Doh, can't believe I missed that. Sorry Bert.
On Tue, Jan 27, 2015 at 5:08 PM, Bert Gunter gunter.ber...@gene.com wrote:
Well, the OP already referred to segplot.
But, see, he shouldn't be doing this plot in the first place. Yes, I
know it's fairly standard in science, but it's a bad idea
with(dat1, ave(integer(length(Date)), Date, FUN=seq_along))
[1] 1 1 2 1 1 2 1 2 1 2 1
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Wed, Jan 28, 2015 at 4:54 PM, Morway, Eric emor...@usgs.gov wrote:
The two datasets below are excerpts from much larger datasets. Note that
there are
eek!
Chel Hee,anything that complicated should engender fear and trembling.
Much simpler and more efficient (if I understand correctly)
i - seq.int(1L,length(ID1),by = 2L)
paste0(ID1[i],ID1[i+1])
That gives a vector of paired letters. If you want a single character
string, just collapse with a
Hi Bert! yes, you are VERY correct!!! Why am I making this simple thing
so complicated??? ;) Thank you so much for your nice lesson!
Chel Hee Lee
On 01/28/2015 09:59 PM, Bert Gunter wrote:
eek!
Chel Hee,anything that complicated should engender fear and trembling.
Much simpler and more
I like the way presented by William Dunlap in the previous post. You
may also try this:
dat1$item - Reduce(c,lapply(table(dat1$Date), seq_len))
dat2$item - Reduce(c,lapply(table(dat2$Date), seq_len))
dat1
Date ConcAve item
1 2009-07-08 71
2 2009-08-26 11
3
I am using just the first row of your data (i.e. ID1).
ID1 - c(A, A, T, G, C, T, G, C, G, T, C, G,
T, A)
do.call(c,lapply(tapply(ID1, gl(7,2), c), paste, collapse=))
1234567
AA TG CT GC GT CG TA
Is this what you are looking for? I hope this helps.
Chel Hee Lee
Kate, here's a solution that uses regular expressions, rather than vector
manipulation:
mystr = ID1 A A T G C T G C G T C G T A
gsub( ([ACGT]) ([ACGT]), \\1\\2, mystr)
[1] ID1 AA TG CT GC GT CG TA
-John
-Original Message-
From: R-help [mailto:r-help-boun...@r-project.org] On
On 01/27/2015 02:54 AM, Bert Gunter wrote:
Huh??
ifelse(TRUE, a - 2L, a - 3L)
[1] 2
a
[1] 2
Please clarify.
In Bioconductor ifelse() is a generic function (with methods for Rle
objects) so all its arguments are evaluated before dispatch can
happen. You can reproduce with:
Hi Kathryn,
If you construct a list of your logical conditions, you can pass that
to a function that evaluates them one by one and returns a list of the
resulting subsets.
subsets-list(B=(A[,1] %in% c(1,2) A[,2] %in% c(1,2)) | (A[,1] %in%
c(3) A[,2] %in% c(1)) | (A[,1] %in% c(4) A[,2] %in%
I need a function that reverse calibrates radiocarbon dates (similar to the
R_Simulate command in Oxcal). In essence, the user inputs a calendar date
and the function outputs a possible radiocarbon date based on the
probability distribution at that point on the radiocarbon calibration curve.
I
Hi:
Don't know about performance, but this is fairly simple for operating
on atomic vectors:
x - c(A, A, G, T, C, G)
apply(embed(x, 2), 1, paste0, collapse = )
[1] AA GA TG CT GC
Check the help page of embed() for details.
Dennis
On Wed, Jan 28, 2015 at 3:55 PM, Kate Ignatius
Dear R-help,
I have df.1001 as a data frame with rows columns of values.
I also have other data frames named similarly, i.e., df.*.
I used DFName from:
DFName - ls(pattern = glob2rx(df.*))[1]
would like to pass on DFName to another function, like:
length(DFName[, 1])
however, when I run:
Hi,
I think what you did is correct. The row name 'comp168081_c2_seq1' is not
in the filter data frame. Neither is the row as the following small example
shows. Actually the dim(datawithoutVF) 171417 12
is different from dim(data) 171471 12. Or 171417 is not the same as
171471
Hi,
I define a function named starStop with three inputs: start, stop and
increment.
startStop - function(start, stop, increment) {
+ for (i in seq(start, stop, increment+1)) cat(i,-,i+increment,\n)}
startStop(12, 35, 5)
12 - 17
18 - 23
24 - 29
30 - 35
startStop(42, 83, 5)
42 - 47
48
I need a function that reverse calibrates radiocarbon dates (similar to
the R_Simulate command in Oxcal). In essence, the user inputs a calendar
date and the function outputs a possible radiocarbon date based on the
probability distribution at that point on the radiocarbon calibration
curve.
I
Hi,
Here is my implementation:
combine - function(x){
+ odd - x[1:length(x) %% 2 == 1]
+ even - x[1:length(x) %%2 == 0]
+ paste0(odd,even)}
temp - letters[1:24]
temp
[1] a b c d e f g h i j k l m n o p q r
s t u v w x
combine(temp)
[1] ab cd ef gh ij kl mn op qr st uv wx
--
View this
Hi,
I think you need quotation around I like the following:
status
2010 2011 2012
1 AAA
2 AII
3 AAA
4 UUU
5 AAA
6 III
7 UII
8 AUA
9 IAU
10III
apply(start,2,function(x)
This approach is fraught with dangers.
I recommend that you put all of those data frames into a list and have your
function accept the list and the name and use the list indexing operator
mylist[[DFName]] to refer to it. Having functions that go fishing around in the
global environment will be
Hello,
In July/August/September 2015 we will be running again a series of
statistics courses in Australia.
Confirmed courses:
1. Darwin: Data exploration, regression, GLM and GAM with introduction to R
2. Sydney: Introduction to mixed modelling, GLMM and MCMC with R
3. Canberra:
Dear R experts,
Suppose I have a matrix A below.
a - rep(1:4, each=5)
b - rep(1:5, 4)
c - rnorm(20)
A - cbind(a,b,c)
A
a bc
[1,] 1 1 0.761806718
[2,] 1 2 0.239734573
[3,] 1 3 -0.728339238
[4,] 1 4 -0.121946174
[5,] 1 5 -0.131909077
[6,] 2 1 -0.069790098
[7,] 2 2
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