You probably need to get your users permissions (in win 7) set properly...
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
Dear all,
Being a newbie to R, I've trawled through many old posts on this list
looking for a solution to my problem, but unfortunately couldn't quite
figure it out myself. I'd be very grateful if someone here on this
list could perhaps help me out.
I have a lattice plot with several panels and
Dear R people
Could you please help.
Basically, there are two variables in my data set. Each patient ('id')
may have one or more diseases ('diagnosis'). It looks like
id diagnosis
1 ah
2 ah
2 ihd
2 im
3 ah
3 stroke
4 ah
4 ihd
4 angina
5
This is causing me great consternation, and I've spent too much time
floundering around on it.
My data is in the form of columns in Excel, with the first column being in
m/dd/ hh:mm format. The spreadsheet is complicated (headers, merged
cells, lines w/o data); so I've tried various ways of
Dear Dr.Goslee and anyone may intrested in matrix manipulate,
I am using your ecodist to do mantel and partial mantel test, I have
locality data and shape variation data, and the two distance matrixs are
given as belowings. When I run the analysis, it is always report that the
matrix is not
Hi!
I think you should read the intro to R, as well as ?[ and ?subset. It
should help you to understand.
Let's say your data is in a data.frame called df:
# 1. ah and ihd
df_ah_ihd - df[df$diagnosis==ah | df$diagnosis==ihd, ] ## the |
is the boolean OR (you want one OR the other). Note the
Dear R users experts,
I'd like to create a model using lm (or glm) under some constraints of
how coefficients for each component could look like (sort of a range of
coefficients that should be allowed).
So let's go for an example :
model=lm(age ~ eyecolor + height, data=inputdata)
So let's
On Thu, 20 Jan 2011, Tal Galili wrote:
You probably need to get your users permissions (in win 7) set properly...
I see no evidence that is the issue
That site seems to be offering only source packages. So in Windows
you need
install.packages(ei,repos=http://r.iq.harvard.edu;,
Hello list.
Another 'puzzle' for which I don't have a clean solution.
Say I have a multidimensional object, e.g.:
Mm-matrix(1:6, nrow=2, dimnames=list(c(a,b), c(g,h,i)))
And on the other hand I have a list
Ind-list(b,g)
This holds, for each dimension, an indexer for that dimension.
Now I would
On Thu, 20 Jan 2011, Nick Sabbe wrote:
Hello list.
Another 'puzzle' for which I don't have a clean solution.
Say I have a multidimensional object, e.g.:
Mm-matrix(1:6, nrow=2, dimnames=list(c(a,b), c(g,h,i)))
And on the other hand I have a list
Ind-list(b,g)
This holds, for each dimension, an
Fitting a linear model with constraints is a completely different task
from fitting one without -- and it is the fit you want to constrain,
not the formula.
See CRAN package nnls to fit a linear model with sign constraints. If
you have range constraints you can use nls(algorithm=port) since
Hm. I got somewhat further:
Ind2-list(Mm,b,g)
do.call([,Ind2)
Seems to work.
However, now I need it one step beyond: in fact, my actual multidimensional
object holds one dimension more than my list holds indexes.
i.e.: I want the equivalent of Mm[a,].
I tried some variants of
On 01/20/2011 09:19 AM, claudia tebaldi wrote:
Hi all
I'm plotting colored contour maps using filled.contour. My levels
are very unevenly spaced, with, say, high resolution in the small
numbers but ranges that can be an order of magnitude or two larger in
absolute value compared to where the
I don't think Ivan's solution meets the OP's needs.
I think you could do it using %in% and the approriate logical operations
e.g.
aDF - data.frame(id=c(1,2,2,2,3,3,4,4,4,5),
diagnosis=c(ah, ah, ihd, im, ah, stroke, ah, ihd,
angina, ihd))
aDF[with(aDF,(id %in% id[diagnosis==ah]) (id %in%
Try this:
lapply(list(c('ah', 'ihd'), 'ah', 'ihd'), function(x)subset(aDF, diagnosis
== x))
On Thu, Jan 20, 2011 at 6:53 AM, Den d.kazakiew...@gmail.com wrote:
Dear R people
Could you please help.
Basically, there are two variables in my data set. Each patient ('id')
may have one or more
Thank you.
That seems to work - also on my much larger data set.
I'm not sure I understand why it has to be defined as a factor, but if it
works...
Sandy
Dennis Murphy wrote:
Hi Sandy:
I can reproduce your problem given the data provided. When I change
ecd_rhythm
Dear all,
I'm puzzled with matrix indices in plotmath.
I'm plotting matrix elements: Z [i, i], and I'd like to put that as label. I'll
describe what I want and what I get in LaTeX-notation.
The output should look like Z_{i, i}, and my first try was
plot (1, 1, ylab = expression (Z[i, i]))
sorry, I forgot my sessionInfo: please see below.
Original Message
Subject: puzzled with plotmath
Date: Thu, 20 Jan 2011 12:48:18 +0100
From: Claudia Beleites cbelei...@units.it
To: R Help r-help@r-project.org
Dear all,
I'm puzzled with matrix indices in plotmath.
I'm
Hi,
If you read the help for mantel, it says that the data must be either
vectors of dissimilarities or objects of class dist. Since you are
reading in your data as full symmetric matrices, you have two options:
mantel(as.dist(distancematrix) ~ as.dist(distancematrix1))
mantelr pval1
Try
plot (1, 1, ylab = expression (Z[list(i,i)]))
Hth -- Gerrit
Original Message
Subject: puzzled with plotmath
Date: Thu, 20 Jan 2011 12:48:18 +0100
From: Claudia Beleites cbelei...@units.it
To: R Help r-help@r-project.org
Dear all,
I'm puzzled with matrix indices in
On 18.01.2011 07:42, fayazvf wrote:
I ve installed wordnet 2.1 and R 2.11.0 on windows 7.
Whenver i try to load wordnet in R, i get an error
initDict()
FALSE
cannot find wordnet 'dict' directory, Please set the WNHOME variable to its
parent.
I have tried setting WNHOME to C:\Program Files
Dear,
I am trying to visualise a time-progressing line (it's supposed to
represent spread patterns) using brew package and Google Earth.
The idea is to have a function which takes start and end point
geographic coordinates, as well as number of intervals to chop the path
up, and returns the
Gerrit,
thanks viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do with an
index list)
for future searches: probably the easiest cheat is, of course,
plot (1, 1, ylab = expression (Z[i, i]))
On 20.01.2011 14:08, Claudia Beleites wrote:
Gerrit,
thanks viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Works for me with a recent R version.
for future
Is this what you are trying to do:
x - read.table(textConnection(3/23/2010 20:55, -0.10655,
-0.121454561, -0.12032, -0.111680001,-0.122429997
+ 3/23/2010 21:25, -0.099166997, -0.114189997, -0.11287,
-0.104647976,-0.114720002), sep = ',', as.is = TRUE)
closeAllConnections()
On 2011-01-20 04:09, Gerrit Eichner wrote:
Try
plot (1, 1, ylab = expression (Z[list(i,i)]))
Hth -- Gerrit
Original Message
Subject: puzzled with plotmath
Date: Thu, 20 Jan 2011 12:48:18 +0100
From: Claudia Beleitescbelei...@units.it
To: R Helpr-help@r-project.org
You have to tell R that a column has date/time content. Here is an example:
test-3/23/2010 20:55 #this is your format
as.POSIXct(test, format=%m/%d/%Y %H:%M) #with the format parameter you
tell R how to interpret your data
___
Moritz Grenke
http://www.360mix.de
Hi to all members of this group,
how can i download a webpage using R (html), which is generated using
javasript when trying to open that webpage in the browser?
When i try to save the page as text file i get only some java code inside,
not the actual data.
When opening website in the browser
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each patient can have
more then one diagnosis and I take that you want to isolate patients based on
particular conditions.
Thus, simply looking for ah or idh as Ivan suggests will yield
Dear all,
I think I have a rather strange question, but I'd like to give it a try:
I want to perform a simulation numerous times, thats why I can't do it by
hand. I sample a small dataset from a very large one, and use backward
selection to select significant predictors for some arbitrary
On 01/20/2011 02:11 PM, Uwe Ligges wrote:
On 20.01.2011 14:08, Claudia Beleites wrote:
Gerrit,
thanks viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Works for me
I did try it. It gave me
[[1]]
id diagnosis
1 1ah
5 3ah
7 4ah
8 4 ihd
10 5 ihd
[[2]]
id diagnosis
1 1ah
2 2ah
5 3ah
7 4ah
[[3]]
id diagnosis
3 2 ihd
8 4 ihd
10 5 ihd
Which isn't what
Hi Taras,
Indeed, I've overlooked the problem. Anyway, I'm not sure I would have
been able to give a complete answer like you did!
Ivan
Le 1/20/2011 11:05, Taras Zakharko a écrit :
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each
Prof. Brian Ripley and Tal Galili,
Thanks for your replies.
I cannot make the
*install.packages(ei,repos=http://r.iq.harvard.edu;, type='source')*
command work.
Alternatively, I can find the way (thanks to stackoverflow guys) I can
install that file from a local repo. For the sake of record,
On Thu, Jan 20, 2011 at 10:53:01AM +0200, Den wrote:
Dear R people
Could you please help.
Basically, there are two variables in my data set. Each patient ('id')
may have one or more diseases ('diagnosis'). It looks like
iddiagnosis
1 ah
2 ah
2 ihd
2 im
3 ah
Hi
I want to do correlation of genes but I need to know p-values to
construct netwrok based on correlation. But I,m getting too many zero
for some reasons. I read one paper and they did analysed their data
by ::
1- R values
2-P-values
3-Spearman p values - FDR corrected (Bejamini-Hochberg)
Do
Claudia,
Mittelhessen! ;-)
thanks viele Grüße nach Oberhessen :-)
plot (1, 1, ylab = expression (Z[list(i,i)]))
though that cannot be evaluated, either (due to [ not knowing what to do
with an index list)
Something is missing; this last sentence of yours appears to be not
complete,
On 2011-01-20 02:05, Taras Zakharko wrote:
Hello Den,
your problem is not as it may seem so Ivan's suggestion is only a partial
answer. I see that each patient can have
more then one diagnosis and I take that you want to isolate patients based on
particular conditions.
Thus, simply looking
Robinson, David G drobin at sandia.gov writes:
I have an mcmc object and I'm trying to plot the quantiles of the
variables - and not as a function of the
iterations as in cumuplot.
I cannot seem to find the right combination of indexing to
access the variables; after which I'm sure I
Hi,
On Tue, Jan 18, 2011 at 3:48 PM, ADias diasan...@gmail.com wrote:
I have done like this to get the result I need more directly
A-c(Tell me how many different letter this vector has?)
prop.table(table(strsplit(A,)))
? a c d e f h i l m n o r
On 2011-01-19 08:48, Paul Ossenbruggen wrote:
Hi,
I have missing values in my time series. na.action = na.pass works
for acf and pacf. Why do I get the following error for the ccf?
ts(matrix(c(dev$u[1:10],dev$q[1:10]),ncol=2),start=1,freq=1)
Time Series:
Start = 1
End = 10
Peter,
Look for 'comma-separated list' on the help page!!
Yes, seeing the solution I also understand why list is the solution.
The special meaning of list () in plotmath was only in my passive vocabulary -
and after this discussion I think it is upgraded to active ;-)
I have to admit that
Hello
I would like to obtain the coefficients for a quadratic function (ax^2 + bx +
c) given three sets of points on the quadratic curve. For instance:
Y X
0.1595290
0.5 0.773019
1 1
Is there a function in R to
Dear R helpers
I am having recovery rates as given below and I am trying to estimate the Loss
Given Default (LGD) and for this I am using Kernel Density estimation method.
recovery_rates =
c(0.61,0.12,0.10,0.68,0.87,0.19,0.84,0.81,0.87,0.54,0.08,0.65,0.91,
Hi Barth,
Here is an option fitting a linear model toa second order polynomial
and extracting the coefficients. The Intercept corresponds to c in
your email, then poly(...)1 to b and poly(...)2 to a.
dat - read.table(textConnection(
Y X
0.159529 0
0.5 0.773019
1 1), header =
I'm attempting to generalise a function that reads individual list components,
in this case they are matrices, and converts them into 3 dimensional array. I
can input each matrix individually, but want to do it for about 1,000 of them
...
This works
array2 -
Hi everybody.
I want to identify duplicate numbers and to increase a value of 0.01 for each
time that it is duplicated.
Example:
x=c(1,2,3,5,6,2,8,9,2,2)
I want to do this:
1
2 + 0.01
3
5
6
2 + 0.02
8
9
2 + 0.03
2 + 0.04
I am trying to get something like this:
1
2.01
3
5
6
2.02
8
9
2.03
Hello,
I used to use RMySQL but as there is no more package for windows, I decided
to move to RODBC.
I installed ODBC driver for MySQL (downloaded on the MySQL website) and then
the RODBC package.
I finally discovered that it was not needed to register your database with
ODBC before using it.
Hi Ben Bolker,
Just to say thank you VERY much for the reply below and for taking the time to
go through my code. I think you're absolutely right and I have been using the
wrong formula completely. I have been trying out several more examples (am
still struggling with this) and will submit any
Hi, my name is Guilherme and I'm trying to solve an optimization problem in
R, regarding reliability and survival time of equipments. I Have to write
the function in the image attached to this email, please take a look, where
the f(t) is weibull's distribution density function, Ca=1000 and Cb=100
I think you need poly(X, 2, raw = TRUE) to interpret the coefficients
in the manner described below.
poly uses orthogonal polynomials by default:
poly package:stats R Documentation
Compute Orthogonal Polynomials
Description:
Returns or evaluates
Many thanks for the correction David.
Josh
On Thu, Jan 20, 2011 at 7:17 AM, David Scott d.sc...@auckland.ac.nz wrote:
I think you need poly(X, 2, raw = TRUE) to interpret the coefficients in
the manner described below.
poly uses orthogonal polynomials by default:
poly
Hi,
On Thu, Jan 20, 2011 at 6:59 AM, Maas James Dr (MED) j.m...@uea.ac.uk wrote:
I'm attempting to generalise a function that reads individual list
components, in this case they are matrices, and converts them into 3
dimensional array. I can input each matrix individually, but want to do it
Thanks Josh and David
Barth
PRIVILEGED AND CONFIDENTIAL INFORMATION
This transmittal and any attachments may contain PRIVILEGED AND
CONFIDENTIAL information and is intended only for the use of the
addressee. If you are not the designated recipient, or an
If you haven't got so much data a loop should do:
while(sum(duplicated(x))0) #if this condition is TRUE then there are still
duplicates in x
{
x[duplicated(x)] - x[duplicated(x)]+0.01 #using duplicated(x) to
index the x vector
}
Hope this helps,
Regards
Moritz
Dear list,
I need to convert this data.frame
names(codesM)
[1] keyAMR.pa1.M AMR.pa2.M AMR.pa3.M AMR.pa4.M
[6] AMR.pa5.M AMR.pa6.M AMR.pa7.M AMR.pa8.M AMR.pa9.M
[11] AMR.pa10.M AMR.ta1.M AMR.ta2.M AMR.ta3.M AMR.ta4.M
[16] AMR.ta5.M AMR.ta6.M AMR.ta7.M AMR.ta8.M AMR.ta9.M
Hi John,
If you only have one duplicated number (e.g., just 2), then this will work:
x - c(1,2,3,5,6,2,8,9,2,2)
xd - duplicated(x)
x[xd] - x[xd] + seq(sum(xd))/100
x
otherwise, I think a different framework than duplicated() will be
necessary, because it will matter not just if the number is
I would think that the following code should work:
newcodesM = reshape(codesM, id=1)
If other variables in the data.frame are factors, reshape thinks all of them
are ID variables and tries to use all of them as keys. Specifying the id
variable you want to keep (I used id=1 since key is in the
As for your second question, you could certainly do
newcodesM = transform(newcodesM, variable1 =
unlist(strsplit(variable,'\\.'))[1], variable2 = unlist(strsplit(variable,
'\\.'))[2], variable3 = unlist(strsplit(variable,'\\.'))[3])
though I'm sure there is a more efficient use of strsplit in
On 2011-01-19 20:15, Max Kuhn wrote:
Hello everyone,
I'm stumped. I'd like to create a scatterplot matrix with circular
reference lines. Here is an example in 2d:
library(ellipse)
set.seed(1)
dat- matrix(rnorm(300), ncol = 3)
colnames(dat)- c(X1, X2, X3)
dat- as.data.frame(dat)
grps-
On Jan 20, 2011, at 10:51 AM, Fredrik Karlsson wrote:
Dear list,
I need to convert this data.frame
names(codesM)
[1] keyAMR.pa1.M AMR.pa2.M AMR.pa3.M AMR.pa4.M
[6] AMR.pa5.M AMR.pa6.M AMR.pa7.M AMR.pa8.M AMR.pa9.M
[11] AMR.pa10.M AMR.ta1.M AMR.ta2.M AMR.ta3.M AMR.ta4.M
-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of Ortiz, John
Sent: Thursday, January 20, 2011 7:13 AM
To: r-help@r-project.org
Subject: [R] Identify duplicate numbers and to increase a value
Hi everybody.
I want to
Try this:
replace(x + ave(x, x, FUN = seq) * .01, !(duplicated(x) | duplicated(x,
fromLast = TRUE)), x)
On Thu, Jan 20, 2011 at 1:12 PM, Ortiz, John ort...@si.edu wrote:
Hi everybody.
I want to identify duplicate numbers and to increase a value of 0.01 for
each time that it is duplicated.
The following worked for me recently:
library(RMySQL)
MySQL. - MySQL()
MySQLcon - dbConnect(MySQL., user='thisuser', password='thispassword',
dbname='desiredDB')
I have the following suggestions and questions for you:
1. Have you tried
Dear list,
My query follows on from a question I posted a few days ago. I have the
following 2 sets of data:
wetMeans[1] 9.904762[2] 6.344828[3] 6.346154[4] 6.855769[5] 9.074324[6]
9.953988[7] 13.482966[8] 14.546053[9] 10.841584[10] 9.752033[11] 6.739336[12]
8.955056burnMeans[1] 0.06214286[2]
Dear all,
This may be a simple question, but I cannot find the solution. I have a
point shapefile containing several hundreds of points, and a grid file
with a certain pixels size. Certain grid cells have several points
falling into them. I am looking for a way to randomly select one point
Hello John,
If many numbers are duplicated, then one way is to coerce to a factor
and use the levels() function. For instance:
x - c(1,1,2,2,2,3,3,4,1,1,2,4)
X - factor(x)
for (i in levels(X))
{
loc - (X==i); len = length(loc)
x[loc] - x[loc] + 0.01 * (1:len)
}
x
[1] 1.01 1.02
Stoker, Ben wrote:
A reproducible example would be very useful. In general
I find that I can just index an mcmc object as though it
were a matrix, e.g.:
z - cbind(a=runif(20),b=runif(20),c=runif(20))
library(coda)
m - mcmc(data=z)
t(apply(m,2,quantile,c(0.025,0.975)))
If that sort
Hi everyone,
I'm trying to perform a linear regression y = b1x1 + b2x2 + b3x3 + b4x4 +
b5x5 while constraining the coefficients such that -3 = bi = 3, and the
sum of bi =1. I've searched R-help and have found solutions for
constrained regression using quadratic programming (solve.QP) where the
I've gotten good results using the sine function to map colors.
For example, when plotting x, map the range(x) to (-pi/2,pi/2)
which the sine will transform to (-1,1), then add 1 and multiply by
half the desired number of colors. Now the integer values will
pick the colors and give a pleasing
Hi Andy,
this is my way of doing it:
#first data:
wetMeans-c(9.904762,6.344828,6.346154,6.855769,9.074324,9.953988,13.482966,
14.546053,10.841584,9.752033,6.739336,8.955056)
burnMeans-c(0.06214286,0.05396552,0.04096154,0.05302885,0.05831081,0.073926
On Thu, Jan 20, 2011 at 10:12 AM, Ortiz, John ort...@si.edu wrote:
Hi everybody.
I want to identify duplicate numbers and to increase a value of 0.01 for each
time that it is duplicated.
Example:
x=c(1,2,3,5,6,2,8,9,2,2)
I want to do this:
1
2 + 0.01
3
5
6
2 + 0.02
8
9
2 + 0.03
Dear list,
is there a function in R that returns the product of a vector?
E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
Cheers
Jannis
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do
Jeremy,
I don't seem to remember signing up to have R-help be my personal e-mail :-)
It is common to use phi to represent the standard normal distribution (with
mean 0, and variance 1), the script/small/lower case version is for the
distribution (height of the bell curve) and the
Is this what you want:
prod(c(1,2,3,4))
[1] 24
On Thu, Jan 20, 2011 at 12:18 PM, Jannis bt_jan...@yahoo.de wrote:
Dear list,
is there a function in R that returns the product of a vector?
E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
Cheers
Jannis
Take a look in prod function:
prod(1:4)
On Thu, Jan 20, 2011 at 3:18 PM, Jannis bt_jan...@yahoo.de wrote:
Dear list,
is there a function in R that returns the product of a vector?
E.g. if the vector is c(1,2,3,4) it should return 1*2*3*4=24
Cheers
Jannis
I getting Error in matrix(0, n, n) : too many elements specified
while building randomForest model, which looks like memory allocation
error.
Software versions are: randomForest 4.5-25, R version 2.7.1
Dataset is big (~90K rows, ~200 columns), but this is on a big machine (
~120G RAM)
and I call
://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__ Information from ESET NOD32 Antivirus, version of virus signature
database 5803 (20110120
This did the trick:
panel.circ3 - function(...)
{
args - list(...)
circ1 - ellipse(diag(rep(1, 2)), t = 1)
panel.xyplot(circ1[,1], circ1[,2],
type = l,
lty = trellis.par.get(reference.line)$lty,
col =
Hello, I have fit a simple spline model to the following data.
Data
x y
0 1.298
2 0.605
3 0.507
4 0.399
5 0.281
6 0.203
7 0.150
8 0.101
Model
Sp.1=lm(y~bs(x,df=4))
Now I wish to inverse predict the x for y=.75, say. Optimize works fine
for
How does one then interpret kernel density distributions with values greater
than one?
My output from the density function.
---
density(delt.m[[1]][,6], na.rm=TRUE)
Call:
density.default(x = delt.m[[1]][, 6], na.rm = TRUE)
Data: delt.m[[1]][, 6] (171 obs.); Bandwidth 'bw' =
I'm new to R and some what new to the world of stats. I got frustrated
with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
ds - read.csv(book2.csv)
df - data.frame(ds)
I then
Dear List,
I recently started using R and I have a simple question. I am running R (v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI.
combine them and then use acf:
x - ts(rnorm(10))
y - ts(x+ rnorm(10))
u - ts.union(x, y)
(acf(u, na.action=na.pass))
I don't know if it's correct, but it gives an answer... I'm too afraid to
check if it's correct.
--
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The same way you interpret densities less than one?
On Jan 20, 2011, at 2:28 PM, Paul Ramer wrote:
How does one then interpret kernel density distributions with values
greater
than one?
The same way you interpret densities less than one?
density != probability
My output from the
Hi again
thank you to those who replied. I actually figured out a simple way
(which does entail transforming the data, oh well...) and I thought I
throw it out there just in case someone needs to solve a similar
problem. Apologies if it is in fact obvious.
Once the right levels for the
On Jan 20, 2011, at 2:08 PM, Mojo wrote:
I'm new to R and some what new to the world of stats. I got
frustrated with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
ds -
On 1/20/2011 3:37 PM, David Winsemius wrote:
On Jan 20, 2011, at 2:08 PM, Mojo wrote:
I'm new to R and some what new to the world of stats. I got
frustrated with excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file
On Jan 20, 2011, at 2:37 PM, jeffrey.mor...@sanofipasteur.com jeffrey.mor...@sanofipasteur.com
wrote:
Hello, I have fit a simple spline model to the following data.
Data
x y
0 1.298
2 0.605
3 0.507
4 0.399
5 0.281
6 0.203
7 0.150
8
I think I should be able to do this using the reshape function, but
I cannot get it to work. I think I need some help to understand
this...
(If I could split the variable into three separate columns splitting
by ., that would be even better.)
Use strsplit and [
Or colsplit, from reshape,
Dear,
I am trying to visualise a time-progressing line (it's supposed to
represent spread patterns) using brew package and Google Earth.
The idea is to have a function which takes start and end point
geographic coordinates, as well as number of intervals to chop the path
up, and returns the
So many matrices are square symmetrical (i.e. variance-covariance matrices),
is there any way to get R to split the matrix on its diagonal and just
return one diagonal?
So if I have
mat-matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
is there anyway to get the lower right
On Thu, 20-Jan-2011 at 10:34AM +0200, E Hofstadler wrote:
| Dear all,
|
| Being a newbie to R, I've trawled through many old posts on this list
| looking for a solution to my problem, but unfortunately couldn't quite
| figure it out myself. I'd be very grateful if someone here on this
| list
Ok, that's a known bug:
https://github.com/hadley/ggplot2/issues/labels/facet#issue/96
Thanks for the reproducible example though!
Hadley
On Thu, Jan 20, 2011 at 3:46 AM, Sandy Small sandy.sm...@nhs.net wrote:
Thank you.
That seems to work - also on my much larger data set.
I'm not
Joe P King wrote:
So many matrices are square symmetrical (i.e. variance-covariance matrices),
is there any way to get R to split the matrix on its diagonal and just
return one diagonal?
mat-matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
is there anyway to get the lower
I'm not sure what a real ANOVA diagram is supposed to look like, nor
do I know what your data look like.
But this might get you started:
fakedata - runif(100)
fakegroups - sample(rep(letters[1:5], each=20))
boxplot(fakedata ~ fakegroups)
If that isn't what you're after, a clearer explanation
On Thu, 20 Jan 2011, Mojo wrote:
I'm new to R and some what new to the world of stats. I got frustrated with
excel and found R. Enough of that already.
I'm trying to test and correct for Heteroskedasticity
I have data in a csv file that I load and store in a dataframe.
ds -
Haseeb - thank you for republishing your solution.
Prof Brian Ripley - thanks for pointing to my error.
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) |
It looks like you might be asking for this. This gets the lower right
diagonal.
mat - matrix(c(1,4,3,4,1,2,3,2,1), nrow = 3, ncol=3, byrow=TRUE)
mat
[,1] [,2] [,3]
[1,]143
[2,]412
[3,]321
diag(mat[rev(2:nrow(mat)),-1])
[1] 2 2
This gets the upper
I recently started using R and I have a simple question. I am running R
(v.
2.12.1) and Rcmdr (v.1-6.3) on Mac (Snow Leopard).
I am using a data set I used before for practicing ANOVA with R, so I
know
what the results should look like. I can get ANOVA table using both Rcmdr
and GUI. However,
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