On 25.04.2011 20:05, shuangyan wrote:
Hello, i got this package from the paper: Nonparametric Covariance Function
estimation for Functional and Longitudinal data.
http://stat.wharton.upenn.edu/~tcai/paper/html/Covariance-Function.html.
Thanks lot!
0. Please do read the posting guide!!! This
On Tue, 26 Apr 2011, Wincent wrote:
I believe this posting is placed, and I take the liberty to re-direct
to the r-help mailing list.
But this is *NOT* the support list for rJava: that is
http://rosuda.org/lists.shtml
In particular the author of rJava does not monitor R-help.
Regards,
Hi
d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
ncol=3, byrow=TRUE))
Change character value NA to missing value NA
d[d[,3]==NA,3]-NA
If you want drop any unused levels of a factor just use
factor(d[,3])
[1] xx yy NA
Levels: xx yy
Regards
Petr
r-help-boun...@r-project.org napsal
Dear users,
I'm trying to get a dot plot but always end up with a boxplot. Can
someone please tell me what I am doing wrong?
df - structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L,
+ 2L, 2L), .Label = c(base, tip), class = factor), Sq = c(274836,
+ 0.74182, 0.709205, 0.984552,
On 26 April 2011 12:08, Ivan Calandra ivan.calan...@uni-hamburg.de wrote:
Dear users,
I'm trying to get a dot plot but always end up with a boxplot. Can someone
please tell me what I am doing wrong?
df - structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L,
+ 2L, 2L), .Label =
On 2011-04-26 03:08, Ivan Calandra wrote:
Dear users,
I'm trying to get a dot plot but always end up with a boxplot. Can
someone please tell me what I am doing wrong?
df- structure(list(FACETTE = structure(c(1L, 1L, 1L, 1L, 2L, 2L,
+ 2L, 2L), .Label = c(base, tip), class = factor), Sq =
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more efficient
way to accomplish the same, or not?
set.seed(1)
a - matrix(rnorm(9), 3 ,3)
sec.large - as.vector(apply(a, 1, order, decreasing=T)[2,])
ans -
would this work (shorter)?
apply(a, 1, function(x) x[order(x)[2]])
On Tue, Apr 26, 2011 at 5:31 PM, Lars Bishop lars...@gmail.com wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more efficient
way to
one way is the following:
a - matrix(rnorm(9), 3 ,3)
aa - a[order(row(a), -a)]
matrix(aa, nrow(a), byrow = TRUE)[, 2]
I hope it helps.
Best,
Dimitris
On 4/26/2011 2:01 PM, Lars Bishop wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution,
On 2011-04-26 05:01, Lars Bishop wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more efficient
way to accomplish the same, or not?
set.seed(1)
a- matrix(rnorm(9), 3 ,3)
sec.large- as.vector(apply(a,
On Apr 26, 2011, at 8:01 AM, Lars Bishop wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more
efficient
way to accomplish the same, or not?
set.seed(1)
a - matrix(rnorm(9), 3 ,3)
sec.large -
Hello all-
What I'm going for here is a stack of four time series plots that use a common
X axis on the bottom plot. And setting up customized tick marks on each plot
that help illustrate the respective trends. Here's my code:
Start
fig_data - ts(read.csv(F:/mydata.csv),
Dear Bob,
Thanks for your answer! I didn't think factors would be a problem for plot()
Ivan
Le 4/26/2011 12:21, Bob O'Hara a écrit :
On 26 April 2011 12:08, Ivan Calandra ivan.calan...@uni-hamburg.de
mailto:ivan.calan...@uni-hamburg.de wrote:
Dear users,
I'm trying to get a dot
Hello,
Does anybody know how to make the hat correctly appears in the label of
this plot (with this cex.lab coefficient) :
plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5)
The hat does not completely appear on my graph, it is like cut on the left
side.
It tried to change the margin :
Dear R users,
I have a Fortran code that I would like to compile and call from R later.
I have never worked with Fortran before. Does anyone know the steps to create
Fortran DLLs for R on a Windows PC.
Is anyone aware of a manual (or does anyone know how to) that explains:
- What tools
require(quantmod)
getSymbols(GLD)
library(manipulate)
manipulate(
plot(Cl(GLD), xlim=c(0,x.max)),
x.max=slider(10,100))
Why the graph is not plotting?
thanks
veepsirtt
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Hello,
How to plot the time series values of YHOO symbol?,
getSymbols(YHOO,src=google)
library(manipulate)
manipulate(
plot(YHOO, xlim=c(0,x.max)),
x.max=slider(10,250))
while running this code I am getting this message
Warning message:
In plot.xts(YHOO, xlim = c(0, x.max)) :
only the
Hello,
Does anybody know how to make the hat correctly appears in the label of
this plot (with this cex.lab coefficient) :
plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5)
The hat does not completely appear on my graph, it is like cut on the left
side.
It tried to change the margin :
Hi,
Does anyone know how to extract the VIP (Variable influence on Projection)
from a Partial Least Squares model?
I'm using Package PLS.
Thanks in Advance
Tommy
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plot(1:10, 1:10, ylab=)
mtext(side=2, expression(hat(h)),cex = 1.5, line=2.5)
Rich
On Tue, Apr 26, 2011 at 7:45 AM, BMichel michel.bertrand@gmail.comwrote:
Hello,
Does anybody know how to make the hat correctly appears in the label of
this plot (with this cex.lab coefficient) :
Dennis, this is really great, thanks a lot!
Do you know how to prevent the result from omitting the first 2
values. I mean - it starts (within each group) with the 3rd row but
omits the first 2...
Dimitri
On Mon, Apr 25, 2011 at 5:31 PM, Dennis Murphy djmu...@gmail.com wrote:
Hi:
I think the
On 26/04/2011 8:01 AM, BMichel wrote:
Hello,
Does anybody know how to make the hat correctly appears in the label of
this plot (with this cex.lab coefficient) :
plot(1:10, 1:10,ylab = expression(hat(h)),cex.lab = 1.5)
The hat does not completely appear on my graph, it is like cut on the left
On 26/04/2011 8:19 AM, A.Noufaily wrote:
Dear R users,
I have a Fortran code that I would like to compile and call from R later.
I have never worked with Fortran before. Does anyone know the steps to create
Fortran DLLs for R on a Windows PC.
Is anyone aware of a manual (or does anyone know
On 2011-04-26 05:26, Dimitris Rizopoulos wrote:
one way is the following:
a- matrix(rnorm(9), 3 ,3)
aa- a[order(row(a), -a)]
matrix(aa, nrow(a), byrow = TRUE)[, 2]
That's clever, Dmitris. And very fast!
Lars: my apologies; I didn't read your request
carefully. You asked for more _efficient_
Thanks a lot to all of you!
On Tue, Apr 26, 2011 at 10:34 AM, Peter Ehlers ehl...@ucalgary.ca wrote:
On 2011-04-26 05:26, Dimitris Rizopoulos wrote:
one way is the following:
a- matrix(rnorm(9), 3 ,3)
aa- a[order(row(a), -a)]
matrix(aa, nrow(a), byrow = TRUE)[, 2]
That's clever,
I would probably still have to manually specify the values for row 1
and row 2 - and then loop through groups. Something like:
mydata[!is.na(mydata$myvalue),new.value]-mydata[!is.na(mydata$myvalue),myvalue]*0.5
# this calculates the values for row 1
h - function(x) embed(x, 3) %*% c(0.5, 0.35,
On Apr 26, 2011, at 14:36 , David Winsemius wrote:
On Apr 26, 2011, at 8:01 AM, Lars Bishop wrote:
Hi,
I need to extract the second largest element from each row of a
matrix. Below is my solution, but I think there should be a more efficient
way to accomplish the same, or not?
Dear Community,
I have the following two variables, which I have split according to a
factor:
*y1*
[1]
1
2
3 [2]
2
3
4
and
*y2*
A B [1]
1 4
2 5
3 6 [2]
2 5
3 6
4 7
Now I need the following Vector Autoregressive Models:
VAR(cbind(y1[1],y2[1]$A)), VAR(cbind(y1[1],y2[1]$B)),
Dear community,
As it's explained I've tried the following:
model- lm(log(v1)~log(v2)+v3, data=dat)
newax- expand.grid(
v2 = seq(min(log(dat$v2)), max(log(dat$v2)), length=100),
v3= seq(min(dat$v3), max(dat$v3), length=100)
)
fit - predict(model,newax)
graph -
Thank you so much!
Lisa
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Hey,
i have a question about how to reorganize a data frame in the easiest way.
my example: what would be the easiest way to bring a data.frame such like this:
nr height age Seed
1 1 4.51 3 301
2 15 10.89 5 301
3 29 28.72 10 301
4 43 41.74 15 301
5 57 52.70 20
A different approach is to use order() to sort
first by row number and then break the ties by
value. It is quick when there are lots of short
rows.
f1 - function (x)
+apply(x, 1, function(row) sort(row, decreasing = TRUE)[2])
f2 - function (x)
+ -apply(-x, 1, function(row)
And I hit the send button before adding the timings for
when there were lots of columns and few rows. f3 changes
from the best to the worst in this case. There is rarely
one most efficient function for all datasets.
x - t(x)
benchmark(r1 - f1(x), r2 - f2(x), r3 - f3(x), r4 - f4(x),
try this:
x
nr height age Seed
1 1 4.51 3 301
2 15 10.89 5 301
3 29 28.72 10 301
4 43 41.74 15 301
5 57 52.70 20 301
6 71 60.92 25 301
7 2 4.55 3 303
8 16 10.92 5 303
9 30 29.07 10 303
10 44 42.83 15 303
11 58 53.88 20 303
12 72 63.39 25
On 2011-04-26 09:11, William Dunlap wrote:
A different approach is to use order() to sort
first by row number and then break the ties by
value. It is quick when there are lots of short
rows.
Bill,
That's what Dmitris did, too; his solution falls between
f3 and f4. Unless speed were
On Tue, Apr 26, 2011 at 11:18 AM, agent dunham crossp...@hotmail.comwrote:
Dear community,
As it's explained I've tried the following:
model- lm(log(v1)~log(v2)+v3, data=dat)
newax- expand.grid(
v2 = seq(min(log(dat$v2)), max(log(dat$v2)), length=100),
v3= seq(min(dat$v3),
On Tue, Apr 26, 2011 at 10:51:33AM +0200, Petr PIKAL wrote:
Hi
d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
ncol=3, byrow=TRUE))
Change character value NA to missing value NA
d[d[,3]==NA,3]-NA
If you want drop any unused levels of a factor just use
factor(d[,3])
[1] xx
Hi,
I am looking into a way to hook into the R coercion framework to allow
me to convert table-like data stored within a COM object into a
data.frame.
Some of our COM objects have their own table-like data storage, and from
R's point of view it's an object (EXTPTRSXP) decoarated with a sepcial
Hello,
given that a self made function produces multiple outputs, is there a
possibility to separate latter by stars or simply by blank lines? Are there
funtions for the arrangement of the output in general?
Thank you.
Best Regards
[[alternative HTML version deleted]]
On Apr 26, 2011, at 18:52 , Petr Savicky wrote:
On Tue, Apr 26, 2011 at 10:51:33AM +0200, Petr PIKAL wrote:
Hi
d-data.frame(matrix(c(ww,ww,xx,yy,ww,yy,xx,yy,NA),
ncol=3, byrow=TRUE))
Change character value NA to missing value NA
d[d[,3]==NA,3]-NA
If you want drop any unused
How do I extract the standard error of the parameter estimates?
Also, if I would like to add two parameters together (x+y), can I use this
equation to calculate the new standard error?:
x = parameter 1
y = parameter 2
xSE = SE parameter 1
ySE = SE parameter 2
Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp - c(aa, aA, ab)
How do I determine the first one have the two same “a”, and the second and
third have the different characters? Thanks in advance.
Lisa
--
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Hello again all,
I am responding to my own earlier post about a non-conformable arguments
error with the predict() function of the pls package (
http://cran.r-project.org/web/packages/pls/) in R 2.13.0 (running in Ubuntu
10.10).
I believe I have narrowed down the cause of the error. My new
Hi Lisa,
Is this what you have in mind?
temp - c(aa, aA, ab)
temp == temp[1]
[1] TRUE FALSE FALSE
HTH,
Jorge
On Tue, Apr 26, 2011 at 2:09 PM, Lisa wrote:
Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp - c(aa, aA,
Hi Lisa,
On Tue, Apr 26, 2011 at 2:09 PM, Lisa lisa...@gmail.com wrote:
Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp - c(aa, aA, ab)
How do I determine the first one have the two same “a”, and the second and
third have
This will handle varying length strings if you want to test for the
same character:
temp - c(aa, aA, ab)
x - strsplit(temp, '')
x
[[1]]
[1] a a
[[2]]
[1] a A
[[3]]
[1] a b
sapply(x, function(z) all(z[1] == z))
[1] TRUE FALSE FALSE
On Tue, Apr 26, 2011 at 2:09 PM, Lisa
Try this:
sapply(apply(sapply(temp, charToRaw), 2, unique), length) == 1
On Tue, Apr 26, 2011 at 3:09 PM, Lisa lisa...@gmail.com wrote:
Dear all,
I just want to determine if the characters in a character string are the
same or not. For example,
temp - c(aa, aA, ab)
How do I determine the
Thanks Icn for pointing that out, but I don't understand it.
My use of .jcall to return a type double scalar or String worked.
My use of .jfield to get a one dimensional static array worked.
My use of .jfield to get a static scalar constant did not work.
My use of .jfield to get a two dimensional
On Tue, 2011-04-26 at 19:46 +0200, ivan wrote:
Hello,
given that a self made function produces multiple outputs, is there a
possibility to separate latter by stars or simply by blank lines? Are there
funtions for the arrangement of the output in general?
You can create a class for your
statconnDCOM, availagble from rcom.univie.ac.at
is a (D)COM server accessing R.
rcom (available from CRAN) is an R package
which can turn R into a COM server or into a COM client.
These tools have their own mailing list which can
be subscribed at rcom.univie.ac.at.
On 4/26/2011 6:20 PM, Von
Dear Luke !
Thanh you for the lovely packages. I have used it, it woks fine for Linux, XP,
but concerning about windows 7 - 64 bits.
I have problem with function makeCluster(). with R 13.0 version.
I wonder it may caused a update problem or ???
Do you have any hint ???
Best
Truc
I am trying a heatmap.
I have 30 individuals and 1,000 genes.
Some components I want for this heatmap plot: (1) the 'color-key'; (2) a
banner across the top to emphasize how the hierarchical clustering has
correctly split the data into the two groups;
I use the 'heatmap.2' function in the
Hi All,
I am trying to use the segmented package to determine the break point
in a simple quadratic regression. I am receiving the error (Some)
estimated psi out of range
I searched the message board and found others that have had the same
issue, but no clear solutions have been
Hi,
Rserve allows one to render plots into jpeg with R code such as:
jpeg(filename, ...)
plot(...)
dev.off()
However, substituting png, tiff, or bmp device redirection fails on Rserve
with eval error on OSX.
png(filename, ...)
plot(...)
dev.off()
The above in an eval
Thanks a lot.
Lisa
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I have a large amount of data which I break down into a collection of
vectors of 100-125 values each. I would like to test the normality of the
vectors and compare them. In the interactive mode I can test any one vector
using the Shapiro-Wilk test or the Kolmogorov-Smirnov test. My problem is
Yes. That is what I want. Thank you very much.
Lisa
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On 2011-04-26 13:15, Bruce Kindseth wrote:
I have a large amount of data which I break down into a collection of
vectors of 100-125 values each. I would like to test the normality of the
vectors and compare them. In the interactive mode I can test any one vector
using the Shapiro-Wilk test or
On Tue, 2011-04-26 at 16:15 -0400, Bruce Kindseth wrote:
I have a large amount of data which I break down into a collection of
vectors of 100-125 values each. I would like to test the normality of the
vectors and compare them. In the interactive mode I can test any one vector
using the
Hi Bruce,
One way is via apply()
# some data
set.seed(123)
X - matrix(rnorm(100), ncol = 5)
X
# tests
t(apply(X, 2, function(x){
sw - shapiro.test(x)
c(sw$statistic, P = sw$p.value)
}))
See ?apply and ?str and ?shapiro.test for more information.
HTH,
Jorge
Here, I have following generic function:
Fn1 - function(x) {
return(x) # assume x is calculated in previous steps
.
return(y) # assume y is calculated in previous steps
..
As soon as you execute the 'return' , the value is returned. What you
did not show is did the code have if-then-else to go down separate
paths.
On Tue, Apr 26, 2011 at 5:29 PM, Bogaso Christofer
bogaso.christo...@gmail.com wrote:
Here, I have following generic function:
Fn1 - function(x) {
On Mon, Apr 25, 2011 at 12:57:46AM +0200, peter dalgaard wrote:
I have a set of data of the form (x, y1, y2) where x is the
independent variable and (y1, y2) is the response pair. The model is
some messy non-linear function:
(y1, y2) = f(x; param1, param2, ..., paramk) + (y1error,
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
*0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want to set zero the entire
row.
The desired output looks like the following. In this case row 3 and row 5
are set
On Sun, Apr 24, 2011 at 07:02:48PM -0400, Ravi Varadhan wrote:
Julian,
You have not specified your problem fully. What is the nature of f? Is f a
scalar function or is it a vector function (2-dim)?
It's something like this (only a bit worse):
given x, work out alpha from
cos(alpha) =
On Tue, Apr 26, 2011 at 2:28 PM, Val valkr...@gmail.com wrote:
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
* 0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want to set zero the entire
row.
The desired
Try this:
replace(m, m[,ncol(m)] 0, 0)
On Tue, Apr 26, 2011 at 6:28 PM, Val valkr...@gmail.com wrote:
Hi all,
Assume I have a matrix
xv= [1 0 0 0 0 12,
0 1 0 0 0 10,
* 0 0 1 0 0 -9,*
0 0 0 1 0 20,
* 0 0 0 0 1 -5]*
if the last column of xv less than 0 then I want
Hi everyone,
I am running the 'gls' command (least squares method) for a number of data
out of which many are zeros. I strongly believe that the output is wrong and
I think that this is due to the large number of zero values included in my
dataset.
I would like to ask if there is a command that
Hi Vana,
I am not sure what package gls() is in off hand, but many model
fitting functions have a subset argument. If not, supposing your data
is in dat, and the variable with the zeros in it that are concerning
you is X, then something like:
newdat - dat[dat[, X] != 0, ]
and now fit gls() on
Hi,
One way would be:
summary(nls.object)[[coefficients]][, Std. Error]
If you have a hankering to do it yourself rather than go through the
summary formula, the code here will get you there:
getAnywhere(summary.nls)
If you are going to be doing it a lot, creating a little function might be
On 2011-04-26 15:11, Joshua Wiley wrote:
Hi,
One way would be:
summary(nls.object)[[coefficients]][, Std. Error]
If you have a hankering to do it yourself rather than go through the
summary formula, the code here will get you there:
getAnywhere(summary.nls)
If you are going to be doing it a
Hi, All,
I want to have the 1%, 2%, 3%, ... contours for Dirichlet distribution. I
need the exact contour circles (mathematically) instead of contour plots.
Can anyone help me with this?
Many thanks,
Cindy
[[alternative HTML version deleted]]
Hi:
One approach is to remove the top two observations from each group.
Here's one way:
ddply(mydata, .(group), function(d) tail(d, -2))
Now apply the previous procedure to this data subset.
HTH,
Dennis
On Tue, Apr 26, 2011 at 7:18 AM, Dimitri Liakhovitski
dimitri.liakhovit...@gmail.com
Hello. I am trying to save an object which I created using assign as
following:
assign(paste(NombreAlgoritmo,_Portafolio,sep=),PortafolioInicial)
save(get(paste(Algoritmo,_Portafolio,sep=)),file=paste(camino,\\Libreria\\Portafolio\\Port_,NombreAlgoritmo,\\,NombreAlgoritmo,_Portafolio.Rdata,sep=))
Dear Rxperts
Below is a small vector of values of zeros and non-zeros... was wondering if
there is an efficient way to get the block sizes of submatrices of a big
matrix similar to the one shown below? diagonal elements can be zero too.
Rows with only a diagonal element may be considered as a
Yes, if there are no control statements such as if that prevent that return
function from being activated.
---
Jeff Newmiller The . . Go Live...
DCN:jdnew...@dcn.davis.ca.us Basics: ##.#. ##.#. Live Go...
Live: OO#..
Hi all,
I tried to plot the density curve using the data from simulation. I am
sure that the data should be exponentially distributed, but the plot
of density curve always starts from (0,0) which is not the case for
exponential distribution. Is there any way around this, to keep the
In barplot for degree distribution x-axis is not seen.
See the example below
g = barabasi.game(500, 0.4)
dd1 = degree.distribution(g)
plot(dd1, xlab=degree, ylab = frequency)
whereas barplot doesnot have any x-axis
barplot(dd1, xlab = degree, ylab = frequency)
Please see the figures
Greetings from Rio de Janeiro, Brazil.
I am looking for advice / references on binary logistic regression
with weighted least squares (using lrm weights), on the following
context:
1) unbalanced sample (n0=1, n1=700);
2) sampling weights used to rebalance the sample (w0=1, w1=14.29); e
3)
Hey Everyone!
I´m a quite new R user .. I found a problem that I'd like to share with you
and help me find a solution.
I have a large txt. file which I opened with read.table command, and what I
understood from many R manuals is that I have a kind of matrix readed with
read.table,
I've used
I would generally use the coef() extractor function if
it's available (and it is for nls()). ?nls has an example:
coef(summary(fm1DNase1))
I knew about coef() on model objects, but I did not know it had
methods for their summaries. What wonderful information!
Josh
which is a matrix from
Is this what you were looking for as output. You did not show what
the output would look like:
x
var1 var2 X. varN
1 122 nnn1 …1
2 213 nnn2 …2
3 422 nnn4 …2
4 432… …3
5 441… …4
6 500… …4
7 550… …4
str(x)
'data.frame': 7 obs. of 4
On Tue, Apr 26, 2011 at 2:21 PM, Schatzi adele_thomp...@cargill.com wrote:
How do I extract the standard error of the parameter estimates?
Also, if I would like to add two parameters together (x+y), can I use this
equation to calculate the new standard error?:
x = parameter 1
y = parameter 2
Hi:
Try this (and note the use of vectorization rather than a loop):
rate - 3
dta - -log(1 - runif(1000))/rate
hist(dta, nclass = 30, probability = TRUE)
x - c(0.001, seq(0, 3, by = 0.01))
lines(x, dexp(x, rate = 3))
This is the difference in timings between the vectorized and iterative
methods
Tena koe Santosh
It is not clear to me precisely what your blocking rules are (e.g., where
should matrix[4,4] go, is matrix[5:7,5:7] to be considered as a different block
to matrix[8:10,8:10]). Also, what happens if there is an isolated 1 (e.g., in
location matrix[9,6])? However, I imagine
Hi:
Firstly, you should have mentioned that you were using the igraph
package; with over 2700 R packages available on CRAN, it's
unreasonable to expect folks to know to which package a particular
function resides, but you may not have been aware of that fact. It
turns out that dd1 is a numeric
On Apr 26, 2011, at 7:37 PM, Luis Felipe Parra wrote:
Hello. I am trying to save an object which I created using assign as
following:
assign(paste(NombreAlgoritmo,_Portafolio,sep=),PortafolioInicial)
save(get(paste(Algoritmo,_Portafolio,sep=)),file=paste(camino,\
Sorry David, I understand what you mean but could you help with how it would
be done more specifically.
Thanks
On Wed, Apr 27, 2011 at 11:27 AM, David Winsemius dwinsem...@comcast.netwrote:
On Apr 26, 2011, at 7:37 PM, Luis Felipe Parra wrote:
Hello. I am trying to save an object which I
Hi:
Maybe this can help get you started. Reading your data into a matrix m,
m - structure(c(1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0,
0, 1, 1, 0, 0, 0, 0, 0, 0,
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