Displaying 20 results from an estimated 10000 matches similar to: "create block diagonal with each rows"
2013 Feb 04
1
How to obtain the model/equation at each level automatically in a regression model with a few factors
I am wondering how to obtain the model/equation at each level automatically
in a regression model with a few factors
without looking at summary of the lm model. For example, consider
lm.factors <- lm(y ~ x1 + factor(x2)*factor(x3)+x4*factor(x5))
The coefficients of lm.factors in summary(lm.factors) might be complicated.
I would like to have the equation at each level from lm.factor.
Could you
2011 Jun 14
2
Off-topic: (Simple?) Random Sampling when n is a random variable
Hi everyone,
I'm involved in a discussion with a colleague. He suggested a sample
design for a finite-sized process that (to all intents and purposes)
involves tossing a coin and examining the unit if the coin shows
Heads.
I should emphasize that we're both approaching the problem from a
design-based sampling theory point of view. So I have no argument
about the appropriateness of the
2013 Jan 16
3
matrix manipulation with its rows
Dear R users,
I have a question about matrix manipulation with its rows.
Plz see the simple example below
sample <- list(matrix(1:6, nr=2,nc=3), matrix(7:12, nr=2,nc=3),
matrix(13:18,nr=2,nc=3))
> sample
[[1]]
[,1] [,2] [,3]
[1,] 1 3 5
[2,] 2 4 6
[[2]]
[,1] [,2] [,3]
[1,] 7 9 11
[2,] 8 10 12
[[3]]
[,1] [,2] [,3]
[1,] 13 15 17
[2,]
2013 Jan 17
2
Explore patterns with GAM
Dear all,
I new to r and I would like your help.
I want to explore the patterns (unimodal, monotonically increased/decreased)
of species richness~altitude using GAM in R. Although I run the gam function
in mgcv package I do not know how to manually define knots and degrees of
freedom.
Any help would be greatly appreciated.
Spyros
--
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2013 Jan 17
1
plotting from dataframes
thanks to your guys help I am closer to solving my problem but I have some
small problem. So let's say I start with
>data
number day hour
1 17 10
2 17 11
3 17 6
4 18 4
5 18 10
6 19 8
7 19 8
I want to split to odd days, which I am able to do, I call this object
frames, which looks like:
> frames
$`1`
c1 day1 hour1
1 1 17 10
2 2 17 11
3 3 17 6
$`2`
c1 day1
2011 Aug 11
5
generate two sets of random numbers that are correlated
Dear R users
I'd like to generate two sets of random numbers with a fixed correlation
coefficient, say .4, using R.
Any suggestion will be greatly appreciated.
Regards,
Kathryn Lord
--
View this message in context: http://r.789695.n4.nabble.com/generate-two-sets-of-random-numbers-that-are-correlated-tp3735695p3735695.html
Sent from the R help mailing list archive at Nabble.com.
2013 Jan 15
1
Random Forest Error for Factor to Character column
Hi,
Can someone please offer me some guidance?
I imported some data. One of the columns called "JOBTITLE" when imported was imported as a factor column with 416 levels.
I subset the data in such a way that only 4 levels have data in "JOBTITLE" and tried running randomForest but it complained about "JOBTITLE" having more than 32 categories. I know that is the limit
2010 May 24
2
Table to matrix
Dear R users,
I am trying to make this (3 by 10) matrix A
--A----------------------------------------------------
0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0
0 0.5 0.5 0 0 0 0 0 0 0
-------------------------------------------------------
from "mass.func"
--mass.func-------------------------------------------
> mass.func
$`00`
prop
5
1
$`10`
2009 Jul 28
3
character vector -> numeric matrix ??
Dear R users...
I'd like to change this character vector, "zz",
zz <- c("12","56","89")
to the following numeric matrix.
[,1] [,2]
[1,] 1 2
[2,] 5 6
[3,] 8 9
Actually, "zz" vector has a long length.
Any comments will be greatly appreciated.
Kathryn Lord
--
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2010 Apr 16
3
VERY SIMPLE QUESTION
Dear R users,
I am looking for more efficient way to compute the followings
--------------------------------------------------------------------------
a <- matrix(c(1,1,1,1,2,2,2,2),4,2)
b <- matrix(c(1,2,3,4),4,1)
Eventually, I want to get this matrix, `c`.
c <- matrix(c(1/1,1/2,1/3,1/4,2/1,2/2,2/3,2/4),4,2)
--------------------------------------------------------------------------
2008 Oct 08
3
Re move repeated values
Dear R users,
I'd like to make this data
rem.y = c(-1,0,2,4,5)
from
y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5).
That is, I need to remove repeated values.
Here is my code, but I don't think it is efficient. How could I improve
this?
#------------------------------------------------------------------------
y = c(-1,-1,0,2,2,2,2,4,4,5,5,5,5,5)
n=length(y)
for (i in 1:n) #
2009 Jul 27
2
Splitting matrix into several small matrices
Dear R users...
I need to split this matrix(or dataframe), for example,
z <- matrix(c(13,1,1,1,1,12,0,0,0,0,8,1,0,1,1,8,0,1,0,0,
10,1,1,1,1,3,0,1,0,0,3,1,0,1,1,6,1,1,1,1),8,5,byrow = T)
> z
[,1] [,2] [,3] [,4] [,5]
[1,] 13 1 1 1 1
[2,] 12 0 0 0 0
[3,] 8 1 0 1 1
[4,] 9 0 1 0 0
[5,] 10 1 1 1 1
2011 Apr 28
1
Undefined columns selected
This is part of my program. I am getting an error, that I cannot figure out, any help would very much appreciated, thanks.
# subset variables
arc <- arc[,c("SNAP", "code", "ncode", "var", "n_total")]
Error in `[.data.frame`(arc, , c("SNAP", "code", "ncode", :
undefined columns selected
arc$N_eff <-
2011 May 02
1
Optimization - n dimension matrix
Dear all,
I am facing the following problem in optimization:
w = (d, o1, ..., op, m1, ..., mq) is a 1 + p + q vector
I want to determine:
w = argmin (a - d(w))' A (a - d(w))
where a is a 1xK marix, A is the covariance matrix of vector a, d(w) is a
1xK vector which parameters are functions of parameters d, o1 .. op, m1 ..
mq.
Is there some function to solve this problem easily? I know
2011 May 02
1
UNIX-like "cut" command in R
The R "cut" command is entirely different from the UNIX "cut" command.
The latter retains selected fields in a line of text. I can do that kind
of manipulation using sub() or gsub(), but it is tedious. I assume there
is an R function that will do this, but I don't know its name. Can you
tell me?
I'm also guessing that there is a web page somewhere that will tell
2011 May 04
1
problem with package "adapt" for R in Mac
Hi,
How i can install the package "adapt" in some version of R for mac?
i try in 2.13, 2.9,2.7 and other previous versions... and nothing happens.
and another question: There are some packages that do the same but that it
is implemented for mac? (calculate integrals in 2 or more dimmensions).
help me please, it's for an important work.
greetings.
--
Matías Hernán Ramírez
2011 May 05
1
functions pandit and treebase in the package apTreeshape
Hello.
I'm trying to use the functions pandit and treebase. They are in the package apTreeshape. Once I've loaded the package, R responses:
- no function pandit/treebase.
Somebody knows why or what is the reason?
Thanks,
Arnau.
------------------------------------------------------------
Arnau Mir Torres
Edifici A. Turmeda
Campus UIB
Ctra. Valldemossa, km. 7,5
07122 Palma de Mca.
2011 May 04
1
bivariate linear interpolation
Hi,
I have three matrices (X,Y,P) with the same dimension. The X,Y grid is
regular and I want to
perform linear interpolation to pick out certain points. In matlab
appropriate call is
something like
Pout=interp2(X,Y,P,Xout,Yout, method="linear")
where Xout and Yout are the locations where I want the Pout data
(typically a different grid).
(Scipy has this routine in
2011 May 04
1
two-way group mean prediction in survreg with three factors
I'm fitting a regression model for censored data with three categorical
predictors, say A, B, C. My final model based on the survreg function is
Surv(..) ~ A*(B+C).
I know the three-way group mean estimates can be computed using the predict
function. But is there any way to obtain two-way group mean estimates, say
estimated group mean for (A1, B1)-group? The sample group means don't
2011 May 03
1
delete excel id automatically generated
Dear community,
I uploaded an excel with read.xls. My xls file actually have a column which
is an id, ("plot" is the id) :
plot height area
34 7.6 5.4
85 3.2 4.1
89 5.4 8.4
121 6.7 6.2
...
1325 2.1 1.5
However R uses another id, this way:
r id plot height area
1 34 7.6 5.4
2 85 3.2 4.1
3 89 5.4 8.4
4 121