see:
?sample
library(vegan)
?rrarefy
not knowing your data structure, its hard to say.
camilleislande wrote
Hello
I have several populations where I have morphology and diet for each
individual. I am interested in the correlation between diet and
morphological distances. However the
Hi, Ista,
1.
/df.m - transform(df.m, Period = reorder(Period, -1*value))
ggplot(df.m, aes(x = Period, y = value/1e+06, fill = Region)) +
geom_bar(stat = identity, position = stack) /
I have followed your instruction but I found the sequence did not change.
2. I am looking for the stack
I am relatively new to R and scatterplot3d function. I need to draw a 3d
scatterplot and then add three different planes (each parallel to xy, yz and
zx with varying intercepts) to a 3d scatterplot. I got the plot running and
i was also able to produce one plane parallel to xy surface by using the
Hi everyone,
I'm working on a script trying to use foreach %dopar% but without success,
so I manage to run the code with foreach %do% and looks like this:
The code is part of a MCMC model for projects valuation, returning the most
important results (VPN, TIR, EVA, etc.) of the simulation.
On Sat, Feb 18, 2012 at 01:51:01AM +, Benilton Carvalho wrote:
Hi everyone,
For reasons beyond the scope of this message, I'd like to append a
NULL element to the end of a list.
tmp0 - list(a=1, b=NULL, c=3)
append(tmp0, c(d=4)) ## works as expected
append(tmp0, c(d=NULL)) ## list
On Fri, 17 Feb 2012, Joseph Wang wrote:
I was able to narrow down to a column that has different types using
featurefields-c(7, 17, 19, 20, 22, 33, 35, 36, 38, 44:132)
flag - rep(0, ncol(data))
for (i in featurefields) {
flag[i] - class(data[,i]) != class(validdata[,i])
}
A
Dear R-group,
I have run into a problem in estimating confidence intervals for the median
difference.
I want to establish a confidence interval at (1- alpha) level for the
difference between the
medians of two indipendent samples (size n and m), by using the Wilcoxon
distribution
or
I don't think so, but it's easy enough to rerun your previous commands,
especially if you've saved then in a file.
Sarah
On Feb 17, 2012 11:06 PM, YN Kim y2sile...@gmail.com wrote:
Hi all,
Is there any command or function to withdraw
a command performed already in R? For instance, after
It seems to me that your two approaches are calculating CIs for
different quantities. The bootstrap methods are calculating a CI for
the difference in medians, while the Wilcoxon approach is calculating a
CI for the median of the differences. If this were the mean, those
would be the same,
Hello
orderulcount
Group.1 Group.2 Group.3 xV5
7C L 0.0 30 C / L
19 C L 0.2 27 C / L
31 C L 0.4 15 C / L
43 C L 0.6 7 C / L
54 C L 0.8 2 C / L
10 C S 0.0 27 C / S
22 C S
Thank you very much for those suggestions, it really helped me a lot.
I just got another problem, i want to run flexmix GLM binomial model with
huge dataset (say 5 million rows), but the problem is i cant run flexmix
with this dataset as a whole in R ( it will crash). So is it fair to run the
with simpler expressions:
if I have a dataframe
A B C
is it possible de transform it to a matrix where nrow(A) colomns, nrow(B)
row and for each value of C, we put the corresponding value in the matrix.
Regards
Le 18 février 2012 13:57, Adel ESSAFI adeless...@gmail.com a écrit :
Hello
I am trying to use matrix algebra to get the beta coefficients from a simple
bivariate linear regression, y=f(x).
The coefficients should be computable using the following matrix algebra: t(X)Y
/ t(x)X
I have pasted the code I wrote below. I clearly odes not work both because it
returns a
Adel,
I don't fully understand your question. Please try to explain what you
want to do again.
In general to change a data frame to a matrix, all one needs to do is
use the as.matrix function.
# Create dataframe
mydataframe1 - data.frame(x=1:3, y=10:12)
mydataframe1
class(mydataframe1)
#
On Feb 18, 2012, at 8:35 AM, Adel ESSAFI wrote:
with simpler expressions:
if I have a dataframe
A B C
is it possible de transform it to a matrix where nrow(A) colomns,
nrow(B)
row and for each value of C, we put the corresponding value in the
matrix.
Perhaps:
xtabs(C ~ A + B,
Hi John: I don't understand what you're doing ( not saying that it's wrong.
I just don't
follow it ). Below is code for computing the coefficients using the matrix
way I follow.
Others may understand what you're doing and be able to fix it so I wouldn't
just
use below immediately.
xprimex -
Mark
Thank you!
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please
On 18-02-2012, at 14:36, John Sorkin wrote:
I am trying to use matrix algebra to get the beta coefficients from a simple
bivariate linear regression, y=f(x).
The coefficients should be computable using the following matrix algebra:
t(X)Y / t(x)X
I have pasted the code I wrote below. I
Thank you,
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call
The Wilcoxon test is not related to the difference in medians but rather to
the Hodges-Lehmann estimator, which is the median of all possible
differences of observations between sample 1 and sample 2. So what's needed
is a confidence interval for this estimate, obtainable from inverting the
Hi,
Here is what I get before reordering:
http://izahn.homedns.org/tmp/barplotOriginal.png
and after reordering:
http://izahn.homedns.org/tmp/barplotReordered.png
Are you saying you get something different? Or that this is not what you were
trying to do?
Best,
Ista
On Friday, February 17,
Esteemed useRs and Devs,
Attempts to update package:rJava to the latest version have failed. See
my code and output below.
Notably, as suggested here
http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing
$ sudo apt-get install r-cran-rjava
ran successfully without
On Fri, Feb 17, 2012 at 7:51 PM, Benilton Carvalho
beniltoncarva...@gmail.com wrote:
Hi everyone,
For reasons beyond the scope of this message, I'd like to append a
NULL element to the end of a list.
tmp0 - list(a=1, b=NULL, c=3)
append(tmp0, c(d=4)) ## works as expected
append(tmp0,
Thanks guys... I'm already embarrassed given how simple the solutions are. b
On Saturday, 18 February 2012, Hadley Wickham wrote:
On Fri, Feb 17, 2012 at 7:51 PM, Benilton Carvalho
beniltoncarva...@gmail.com javascript:; wrote:
Hi everyone,
For reasons beyond the scope of this message,
Hi again,
I studied with influence measures of logistic regression of some data by
using R and SPSS and I get different values of DFFITS and Cook's Distance
from SPSS and R outpus, but the coefficients, model formulas and leverage
values are the same.
Can anyone help me about ; Which is the
On Feb 18, 2012, at 10:44 AM, Karl Brand wrote:
Esteemed useRs and Devs,
Attempts to update package:rJava to the latest version have failed. See my
code and output below.
Notably, as suggested here
http://stackoverflow.com/questions/3311940/r-rjava-package-install-failing
$ sudo
The validate function in the rms package can do cross validation of
ols objects (ols is similar to lm, but with additional information),
the default is to do bootstrap validation, but you can specify
crossvalidation instead.
On Thu, Feb 16, 2012 at 10:44 AM, samuel-rosa
Marcos,
Untested because you didn't provide a reproducible example but my
guess, the problem is in the local assignment of MCPVMP*. The %do%
worked just because it operates in the same (local) environment for
all threads. I'll try to clarify with a a self contained example of
sourcing a script
maris478 wrote
Good afternoon,
I've encountered a little bit of a problem, would appreciate any help
here.
I made a small vector consisting of ones and zeros.
Something like this x - c(0,1,0,1,0,0,1,0), and all I need is to count
how many times 0 becomes 1.
Tried various, of what I
On Sat, Feb 18, 2012 at 11:51:39AM -0800, Pete Brecknock wrote:
maris478 wrote
Good afternoon,
I've encountered a little bit of a problem, would appreciate any help
here.
I made a small vector consisting of ones and zeros.
Something like this x - c(0,1,0,1,0,0,1,0), and all I
I made a small vector consisting of ones and zeros.
Something like this x - c(0,1,0,1,0,0,1,0), and all I need is to count
how many times 0 becomes 1.
Since x consists solely of 0's and 1's this is same as
sum(diff(x)==1) # 3 in your example
Bill Dunlap
Spotfire, TIBCO Software
wdunlap
try this:
x - c(0,1,0,1,0,0,0,0)
sum(diff(x) == 1)
[1] 2
On Sat, Feb 18, 2012 at 2:51 PM, Pete Brecknock peter.breckn...@bp.com wrote:
maris478 wrote
Good afternoon,
I've encountered a little bit of a problem, would appreciate any help
here.
I made a small vector consisting of ones
Simon,
Thanks for yout fast response. Thing is - i managed to get Version 0.9-1
installed and fully functional. And
$ locate jdk
returns too many entries to post here, so i'm pretty sure its on the
machine.
So i'd like to know how i can ensure it's registered in R. This i have
no idea
On 18 February 2012 13:13, Karl Brand k.br...@erasmusmc.nl wrote:
Thanks for yout fast response. Thing is - i managed to get Version 0.9-1
installed and fully functional. And
$ locate jdk
returns too many entries to post here, so i'm pretty sure its on the
machine.
What you want to look
Elai,
The - trick is of no use, and sourcing the script within the loop gives
me nothing, if i put it in .option all variables are loaded to my
workspace; though, they're empty and generates some errors. I'll put a
reproducible example:
Lets call this script MC:
library(foreach)
library(doMC)
Hi Everyone,
I have csv file which is in following format
xmin,xmax,ymin,ymax,zmin,zmax
I want to plot 3d graph with all the all octants being displayed as well.
Any idea? I have used scatterplot3d package but it does not seem to have
anything by which i can draw octants inside cube as well.
Dear all,
I need to generate numbers from multivariate normal with large dimensions
(5,000,000).
Below is my code and the error I got from R. Sigma in the code is the
covariance
matrix. Can anyone give some idea on how to take care of this error. Thank
you.
Hannah
m - 500
How can i file this isssue as an bugreport?
On Fri, Feb 03, 2012 at 02:01:02PM +0100, peter dalgaard wrote:
On Feb 2, 2012, at 21:24 , Frank Schwidom wrote:
Hi,
the following Code demonstrates an possibly Error in R
(or you can explain me, why this happens, thanks in advance)
Dear Etienne, I'm a colleauge of Gabriele and I'm more into R (but he is
more into GRASS).
I'll try to explain you what we didi so far
1) Our ASTER images, (B1, B2 and B3) have 8363134 pixels; we made a subset
in order to have training data sets: that is, for each band (B1,B2 and B3)
916 pixels
Good afternoon,
I've encountered a little bit of a problem, would appreciate any help here.
I made a small vector consisting of ones and zeros.
Something like this x - c(0,1,0,1,0,0,1,0), and all I need is to count how
many times 0 becomes 1.
Tried various, of what I thought, methods with
This error occurs because the == comparison operator doesn't allow
comparison of ordered and normal factors:
/df[df5$close_quarter == as.Date(2011-02-01),]/
Warning message:
In /`[.data.frame`(df, df$close_quarter == as.Date(2011-02-01)/, :
Incompatible methods (Ops.ordered, Ops.Date) for ==
It's not a matter of unordered ordered factors, but ordered factors
and Dates (as the warning says)
I can see at least one ambiguity -- should comparison be made from the
level or the internal code -- so the warning makes sense to me (though
an error might make even more sense). Generally, for
Elai,
Sooo maaanyy thanks... It worked, the problem was with out- foreach I
thought foreach wouldn't need to put the results in a variable, this did my
day.
On Sat, Feb 18, 2012 at 5:57 PM, ila...@gmail.com wrote:
Marcos,
This worked for me. Can't say why it didn't for you. Did you make sure
Just for clarity, I changed your x a bit - in your version, the 0-1 and 1-0
change occurred the same number of times.
If all your values are 0 and 1, this will work:
x - c(0,1,0,1,0,0,0,1,1,1)
table(diff(x))
-1 0 1
2 4 3
sum(diff(x) == 1)
[1] 3
If other values can occur, it would need
Marcos,
This worked for me. Can't say why it didn't for you. Did you make sure to
source AFTER you define the variables ? Not doing so is the only reason I
can think of why they would be empty.
Here I'm running your example (I called the script that needs to be
sourced src.R and put it in
On 19/02/12 08:40, Frank Schwidom wrote:
How can i file this isssue as an bugreport?
Please don't. Peter has just explained to you that this is not a bug,
it's a feature. You need to amend your understanding of this,
feature; R does not need to be amended.
cheers,
Rolf Turner
The train function in the caret package will do this. The trainControl function
would use method =repeatedcv and repeats = 100.
On Feb 18, 2012, at 2:15 PM, Greg Snow 538...@gmail.com wrote:
The validate function in the rms package can do cross validation of
ols objects (ols is similar to
Hi,
Is there a way to parse a postscript (*.ps) file with R (or
perhaps with some other command-line utility)?
E.g., I have a map in postscript format with lots of
features, but I just want to extract the coastline and it's
coordinates.
Any help very much appreciated! Cheers!
Nick
--
For completeness, if you want to count all possible four transitions:
x - c(0,1,0,1,0,0,0,1,1,1,0,0,0,1)
# lets keep count of the 4 different transitions that can happen
indx - cbind(head(x, -1), tail(x, -1)) %*% c(2, 1)
table(indx) # 0=0-0, 1=0-1, 2=1-0, 3=1-1
indx
0 1 2 3
4 4 3 2
On Sat,
Hi,
the grImport package provides some tools for this.
HTH,
b.
On 19 February 2012 16:06, Nick Matzke mat...@berkeley.edu wrote:
Hi,
Is there a way to parse a postscript (*.ps) file with R (or perhaps with
some other command-line utility)?
E.g., I have a map in postscript format with
Hi, Ista,
I found the graph were same as before recording
http://izahn.homedns.org/tmp/barplotOriginal.png
Anyway, I will try again,
Many thanks.
I am now trying to stack it horizontally and found that the axis text
overlap each other!!
New battle start again. X_X!!
VD.
--
View this message
Hi VD,
The plot should not be the same after reordering. Please show us what you did,
and what the result was.
Best,
Ista
On Saturday, February 18, 2012 06:22:00 PM vd3000 wrote:
Hi, Ista,
I found the graph were same as before recording
http://izahn.homedns.org/tmp/barplotOriginal.png
Dear R People:
I'm trying to replicate something that I saw on an R blog.
The first step is to load in the .rda file, which is fine.
However, some of the names of the columns in the data frame have
special characters, accents, and such.
How do I get around this on a basic keyboard, please?
On 19/02/2012 07:30, Erin Hodgess wrote:
Dear R People:
I'm trying to replicate something that I saw on an R blog.
The first step is to load in the .rda file, which is fine.
However, some of the names of the columns in the data frame have
special characters, accents, and such.
Most of the
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