Dear Nicolette,
You can always use the bruit force solution which works for every discrete
distribution with finite number of states: let p0,p1,...,pK be the
probabilities of 0,1,...,K (such that they sum up to 1).
Let P - c(p0,p1,...,pK) and P1 - c(cumsum(P),1)
Now let x = runif() (uniform in
Hi Daniele,
One possibility would be to make two runs. In the first run you are not
building the matrix but just calculating the number of rows you need (in a
loop). Then you allocate such matrix (only once) and fill it in the second run.
Regards,
Moshe.
--- On Wed, 24/2/10, Daniele Amberti
Check read.table (?read.table).
--- On Wed, 24/2/10, RagingJim nowa0...@flinders.edu.au wrote:
From: RagingJim nowa0...@flinders.edu.au
Subject: [R] reading surfer files
To: r-help@r-project.org
Received: Wednesday, 24 February, 2010, 3:23 PM
To the R experts,
I am currently playing
You can compute the conditional probability that your variable equals k given
that it is non-zero. For example, if X has poisson distribution with parameter
lambda then
P(X=k/X!=0) = P(X=k)/(1-P(X=0)) = (exp(-lambda)/(1-exp(-lambda))*lambda^k/k!
Now you can find lambda for which the sum of
Hi Sergio,
Having singular Dmat is certainly a problem.
I can see two possibilities:
1) try to eliminate X1,...,X9, so that you are left with P1,...,P6 only.
2) if you can not do this, add eps*X1^+...+eps*X9^2 to your matrix Dmat so that
it is positive definite (eps is a small positive number).
Yes, this can be easily computed analytically (even though my result is a bit
different).
--- On Fri, 12/2/10, dav...@rhotrading.com dav...@rhotrading.com wrote:
From: dav...@rhotrading.com dav...@rhotrading.com
Subject: Re: [R] Integral of function of dnorm
To: Greg Snow
Hi Jonathan,
If minDate = min(Condition1) - max(Condition2) and maxDate = max(Condition1) -
min(Condition2) then all your differences would be between minDay and maxDay,
and hopefully this is not a very big range (unless you are going many thousands
years into the past or the future). So
Hi,
I believe that the reason is that even though the first 4 elements of your
fmodel look equal (when rounded to 4 decimal places) they are actually not.
To check this try
fmodel[1:4]-fmodel[1]
--- On Thu, 11/2/10, Something Something mailinglist...@gmail.com wrote:
From: Something Something
One possibility I can see is to replace - by NA and use mean with
na.rm=TRUE.
--- On Wed, 10/2/10, Steve Murray smurray...@hotmail.com wrote:
From: Steve Murray smurray...@hotmail.com
Subject: [R] Resampling a grid to coarsen its resolution
To: r-help@r-project.org
Received: Wednesday,
Hi Chris,
You can use lm with poly (look ?lm, ?poly).
If x and y are your arrays of points and you wish to fit a polynom of degree 4,
say, enter: model - lm(y~poly(x,4,raw=TRUE) and then summary(model)
The raw=TRUE causes poly to use 1,x,x^2,x^3,... instead of orthogonal
polynomials (which are
unable to
find out the
equation of the trendline from the summary table. Besides,
how do I fit the
trendline on the graph?
I intend to put the first column of data onto x axis and
the second column
onto y axis. Are they the x and y in your example?
Many thanks,
Chris
Moshe Olshansky-2
Dear list,
I have r towns, T1,...,Tr where town i has population Ni. For each town I
randomly sampled Mi individuals and found that Ki of them have a certain
property. So Pi = Ki/Mi is an unbiased estimate of the proportion of people in
town i having that property and the weighted average of
Hi Roslina,
I believe that you can ignore the warning.
Alternatively, you may add a very small random noise to pairs with ties, i.e.
something like
xobs[which(duplicated(xobs))] - xobs[which(duplicated(xobs))] +
1.0e-6*sd(xobs)*rnorm(length(which(duplicated(xobs
Regards,
Moshe.
--- On
test[which(test[,total] %in% needed),]
--- On Fri, 25/9/09, Dimitri Liakhovitski ld7...@gmail.com wrote:
From: Dimitri Liakhovitski ld7...@gmail.com
Subject: [R] keeping all rows with the same values, and not only unique ones
To: R-Help List r-h...@stat.math.ethz.ch
Received: Friday, 25
Assuming that at the end all of them are dead, you can do the following:
sum(deaths)-cumsum(deaths)
Regards,
Moshe.
--- On Wed, 2/9/09, Frostygoat frostyg...@gmail.com wrote:
From: Frostygoat frostyg...@gmail.com
Subject: [R] Basic population dynamics
To: r-help@r-project.org
Received:
You can do
for (i in 1:ncol(x)) {names -
rownames(x)[which(x[,i]==1)];eval(parse(text=paste(V,i,.ind-names,sep=)));}
--- On Thu, 27/8/09, Steven Kang stochastick...@gmail.com wrote:
From: Steven Kang stochastick...@gmail.com
Subject: [R] Help on efficiency/vectorization
To:
Hi Deb,
Based on your last note (and after briefly looking at Rserve) I believe that
you should install R with all the packages you need on the server and then use
it like you are using any workstation, i.e. log in to it and do whatever you
need.
Regards,
Moshe.
--- On Thu, 27/8/09,
One possible (but not very elegant) solution is:
aa - paste(1:12,:10:2009,sep=)
dd-as.Date(aa,format=%m:%d:%Y)
mon - format(dd,%b)
mon
[1] Jan Feb Mar Apr May Jun Jul Aug Sep Oct Nov Dec
--- On Thu, 20/8/09, Liviu Andronic landronim...@gmail.com wrote:
From: Liviu Andronic
Hi Misha,
Since PCA is a linear procedure and you have only 6000 observations, you do not
need 68000 variables. Using any 6000 of your variables so that the resulting
6000x6000 matrix is non-singular will do. You can choose these 6000 variables
(columns) randomly, hoping that the resulting
My guess is that 6. comes for 6.0 - something which comes from programming
languages where 6 represents 6 as integer while 6. (or 6.0) represents 6 as
floating point number.
--- On Fri, 21/8/09, kfcnhl zhengchenj...@hotmail.com wrote:
From: kfcnhl zhengchenj...@hotmail.com
Subject: [R]
Hi Tim,
As far as I know you can not weigh predictors (and I believe that you really
should not). You may weigh classes (and, in a sense, cases), but this is an
entirely different issue.
--- On Wed, 5/8/09, Häring, Tim (LWF) tim.haer...@lwf.bayern.de wrote:
From: Häring, Tim (LWF)
Hi,
What do you mean by outer product?
If you have two vectors, say x and y, of lenght n and you define matrix A by
A(i,j) = x(i)*y(j) then your matrix has rank one and it is VERY singular (in
exact arithmetics).
Is this is what you mean by outer product?
--- On Sun, 16/8/09, Stephan Lindner
You could do the following:
y - apply(dat,1,function(a) t.test(a[1:10],a[11:30])$p.value)
This will produce an array of 2 p-values.
--- On Fri, 14/8/09, Gina Liao yi...@hotmail.com wrote:
From: Gina Liao yi...@hotmail.com
Subject: [R] problem about t test
To: r-h...@stat.math.ethz.ch
Is your system of equations linear?
--- On Fri, 14/8/09, Moreno Mancosu nom...@tiscali.it wrote:
From: Moreno Mancosu nom...@tiscali.it
Subject: [R] Solutions of equation systems
To: r-help@r-project.org
Received: Friday, 14 August, 2009, 2:29 AM
Hello all!
Maybe it's a newbie
Try
tempFun - function(x) sum(!is.na(x))
nonZeros - aggregate(pollution[pol],format(pollution[date],%Y-%j), FUN
= tempFun)
--- On Wed, 12/8/09, Tim Chatterton tim.chatter...@uwe.ac.uk wrote:
From: Tim Chatterton tim.chatter...@uwe.ac.uk
Subject: [R] Counting the number of non-NA values per
Alternatively download the xlsReadWrite package from
http://treetron.googlepages.com/
install it an proceed as in older version of R.
--- On Tue, 11/8/09, Inchallah Yarab inchallahya...@yahoo.fr wrote:
From: Inchallah Yarab inchallahya...@yahoo.fr
Subject: [R] Re : How to Import Excel file
Hi,
Is your matrix K symmetric? If yes, there is an analytical solution.
--- On Sat, 1/8/09, nhawrylyshyn nichlas.hawrylys...@gmail.com wrote:
From: nhawrylyshyn nichlas.hawrylys...@gmail.com
Subject: [R] Matrix Integral
To: r-help@r-project.org
Received: Saturday, 1 August, 2009, 12:15 AM
Another possibility, if the total length of your intervals is small in
comparison to the big interval is to choose the starting points of all your
intervals randomly and to dismiss the entire set if some of the intervals
overlap. Most probably you will not have too many such cases (assuming,
?outer
--- On Thu, 16/7/09, Chyden Finance fina...@chyden.net wrote:
From: Chyden Finance fina...@chyden.net
Subject: [R] searching for elements
To: r-help@r-project.org
Received: Thursday, 16 July, 2009, 3:00 AM
Hello!
For the past three years, I have been using R extensively
in my
Make it
for (i in 1:9)
This is not the general solution, but in your case when i=10 you do not want to
do anything.
--- On Tue, 14/7/09, Michael Knudsen micknud...@gmail.com wrote:
From: Michael Knudsen micknud...@gmail.com
Subject: [R] Nested for loops
To: r-help@r-project.org
Received:
Hi Stephen,
The error message clearly says what is wrong.
Big Endian and Little Endian are two ways of storing data (mostly often double
precision numbers) in memory. A double precision number occupies two blocks of
4 bytes each. On Big Endian machines (most machines which are not Intel) if
Try ?aggregate
--- On Wed, 15/7/09, Timo Schneider timo.schnei...@s2004.tu-chemnitz.de wrote:
From: Timo Schneider timo.schnei...@s2004.tu-chemnitz.de
Subject: [R] Grouping data in dataframe
To: r-help@r-project.org r-help@r-project.org
Received: Wednesday, 15 July, 2009, 1:56 PM
Hello,
One (awkward) way to do this is:
x - matrix(c(c(test),c(test2)),ncol=2)
y - rowMeans(x,na.rm=TRUE)
testave - matrix(y,nrow=nrow(test))
--- On Tue, 14/7/09, Tish Robertson tishrobert...@hotmail.com wrote:
From: Tish Robertson tishrobert...@hotmail.com
Subject: [R] averaging two matrices
Try A[1,,drop=FALSE] - see help(\[)
--- On Mon, 13/7/09, Weiwei Shi helprh...@gmail.com wrote:
From: Weiwei Shi helprh...@gmail.com
Subject: [R] how to keep row name if there is only one row selected from a
data frame
To: r-h...@stat.math.ethz.ch r-h...@stat.math.ethz.ch
Received: Monday,
If df is your dataframe then names(df) contains the column names and so
names(df)[i] is the name of i-th column.
--- On Thu, 9/7/09, mister_bluesman mister_blues...@hotmail.com wrote:
From: mister_bluesman mister_blues...@hotmail.com
Subject: [R] Extracting a column name in loop?
To:
One possibility is to use sink (see ?sink).
--- On Thu, 9/7/09, Steve Jaffe sja...@riskspan.com wrote:
From: Steve Jaffe sja...@riskspan.com
Subject: [R] print() to file?
To: r-help@r-project.org
Received: Thursday, 9 July, 2009, 5:03 AM
I'd like to write some objects (eg arrays) to a
Let M be your matrix.
Do the following:
B - t(matrix(colnames(a),nrow=ncol(M),ncol=nrow(M)))
B[M==0] - NA
--- On Thu, 9/7/09, Olivella olive...@wustl.edu wrote:
From: Olivella olive...@wustl.edu
Subject: [R] Substituting numerical values using `apply'
To: r-help@r-project.org
Received:
Hi Mary,
Your data.frame has just one column (not 2)! You can check this by
dim(tresult2).
What appears to you to be the first column (names) are indeed rownames.
If you really want to have two columns do something like
tresult2 - cbind(colnames(tresult),data.frame(t(tresult),row.names=NULL))
Hi Antje,
Are your measurements taken every minute (i.e. 30 minutes correspond to 30
consecutive values)?
How fast is your transition?
If you had 30 minures of upper temperature, then 1000 minutes of room
temperature and then 30 minutes of lower temperature - would you count this as
a cycle?
As mentioned by somebody before, there is no problem for the normal case - use
mvrnorm function from MASS package with any mu and make Sigma be any diagonal
matrix (with strictly positive diagonal). Note that even though all the
correlations are 0, the SAMPLE correlations won't be 0. If you
One way is to use convolution (?convolve):
If A(x) = a_p*x^p + ... + a_1*x + a_0
and B(x) = b_q*x^q + ... + b_1*x + b_0
and if C(x) = A(x)*B(x) = c_(p+q)*x^(p+q) + ... + c_0
then c = convolve(a,rev(b),type=open)
where c is the vector (c_(p+q),...,c_0), a is (a_p,...,a_0) and b is
(b_q,...,b_0).
First of all, we must define what is a run of length r: is it a tail, then
EXACTLY r heads and a tail again or is it AT LEAST r heads.
Let's assume that we are looking for a run of EXACTLY r heads (and we toss the
coin n times).
Let X[1],X[2],...,X[n-r+1] be random variables such that Xi = 1 if
Try
AA - apply(A,1,function(x) paste(x,collapse=))
and work with AA.
--- On Tue, 30/9/08, Jose Luis Aznarte M. [EMAIL PROTECTED] wrote:
From: Jose Luis Aznarte M. [EMAIL PROTECTED]
Subject: [R] ordering problem
To: [EMAIL PROTECTED]
Received: Tuesday, 30 September, 2008, 8:43 PM
Hi
I think that you can use read.csv with nrows and skip arguments (see
?read.table).
--- On Mon, 22/9/08, DS [EMAIL PROTECTED] wrote:
From: DS [EMAIL PROTECTED]
Subject: [R] design question on piping multiple data sets from 1 file into R
To: r-help@r-project.org
Received: Monday, 22
One possibility is:
x - data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2))
names - t(matrix(rep(names(x),times=nrow(x)),nrow=ncol(x)))
m - as.matrix(x)
ind - order(m)
df - data.frame(name=names[ind],value=m[ind])
df
name value
1 x1 1
2 x3 2
3 x2 4
4 x2 6
5 x1 7
6 x3
Hi Mark,
stock-/opt/limsrv/mark/research/equity/projects/testDL/stock_data/fhdb/US/BLC.NYSE
gsub(.*/([^/]+)$, \\1,stock)
[1] BLC.NYSE
--- On Tue, 23/9/08, [EMAIL PROTECTED] [EMAIL PROTECTED] wrote:
From: [EMAIL PROTECTED] [EMAIL PROTECTED]
Subject: [R] perl expression question
To:
Hi Sonia,
If I did not make a mistake, the conditional distribution of X given that X 0
is very close to exponential distribution with parameter lambda = 40, so you
can sample from this distribution.
--- On Mon, 15/9/08, Daniel Davis [EMAIL PROTECTED] wrote:
From: Daniel Davis [EMAIL
Well, I made a mistake - your lambda should be 400 and not 40!!!
--- On Thu, 18/9/08, Moshe Olshansky [EMAIL PROTECTED] wrote:
From: Moshe Olshansky [EMAIL PROTECTED]
Subject: Re: [R] help on sampling from the truncated normal/gamma
distribution on the far end (probability is very low
Hi Ramya,
Assuming that the problem is well defined (i.e. the values in col1 of the
data.frames are unique and every value in D.F.sub.2[,1] appears also in
D.F1[,1]) you can do the following:
ind - match(D.F.sub.2[,1],D.F1[,1])
D.F1[ind,] - D.F.sub.2
--- On Thu, 18/9/08, Rajasekaramya [EMAIL
One possibility is as follows:
If r$userid is your array of (2000) ID's then
s - paste(r$userid,sep=,)
s- paste(select t.userid, x, y, z from largetable t where t.serid in
(,s,),sep=)
and finally
d - sqlQuery(connection,s)
Regards,
Moshe.
--- On Fri, 12/9/08, Avram Aelony [EMAIL PROTECTED]
Just a small correction:
start with
s - paste(r$userid,collapse=,)
and not
s - paste(r$userid,sep=,)
--- On Fri, 12/9/08, Moshe Olshansky [EMAIL PROTECTED] wrote:
From: Moshe Olshansky [EMAIL PROTECTED]
Subject: Re: [R] database table merging tips with R
To: [EMAIL PROTECTED], Avram
Let X be normally distributed with mean 0 and let f be it's density. Now the
density of X+a will be f shifted right by a. Since the density is symmetric
around mean it follows that the area of overlap of the two densities is exactly
P(Xa) + P(X-a).
So if X~N(0,1), we want P(Xa) + P(X-a) =
Just a correction:
if we take X+2a then everything is OK (the curves intersect at a), so a =
0.9345893 is correct but one must take X ~ N(0,1) and Y ~N(2*a,1).
--- On Tue, 9/9/08, Moshe Olshansky [EMAIL PROTECTED] wrote:
From: Moshe Olshansky [EMAIL PROTECTED]
Subject: Re: [R] densities
I do not see why you can not use regression even in this case.
To make things more simple suppose that the exact model is:
y = a + b*x, i.e.
y1 = a + b*x1
...
yn = a + b*xn
But you can not observe y and x. Instead you observe
ui = xi + ei (i=1,...,n) and
vi = yi + di (i=1,...,n)
Now you have
This can be done analytically: after changing a variable (2*t - t) and some
scaling we need to compute
f(x) = integral from 0 to 20 of (t^x*exp(-t))dt/factorial(x)
f(0) = int from 0 to 20 of exp(-t)dt = 1 - exp(-20)
and integration by parts yields (for x=1,2,3,...)
f(x) =
If you look at your sech(pi*x/2) you can write it as
sech(pi*x/2) = 2*exp(pi*x/2)/(1 + exp(pi*x))
For x -15, exp(pi*x) 10^-20, so for this interval you can replace
sech(pi*x/2) by 2*exp(pi*x/2) and so the integral from -Inf to -15 (or even -10
- depends on your accuracy requirements) can be
You commands are correct and the interpretation is that the probability that a
normal random variable with mean 1454.190 and
standard deviation 162.6301 achieves a value of 417 or less is 8.99413e-11
--- On Wed, 27/8/08, rr400 [EMAIL PROTECTED] wrote:
From: rr400 [EMAIL PROTECTED]
Subject:
I was too optimistic - the complexity is O(E*log(V)) where V is the number of
nodes, but since log(25000) 20 this is still reasonable.
--- On Mon, 25/8/08, Moshe Olshansky [EMAIL PROTECTED] wrote:
From: Moshe Olshansky [EMAIL PROTECTED]
Subject: Re: [R] Igraph library: How to calculate APSP
One possibility is:
y - rep( ,6)
y[6] -
y[c(2,4)] - \n
res - paste(paste(x,y,sep=),collapse=)
--- On Tue, 26/8/08, remko duursma [EMAIL PROTECTED] wrote:
From: remko duursma [EMAIL PROTECTED]
Subject: [R] paste: multiple collapse arguments?
To: r-help@r-project.org
Received: Tuesday, 26
Hi Wolf,
Without noise you could use FFT, i.e. FFT of a convolution is the product of
the individual FFTs and so you get the FFT of your input signal and using
inverse FFT you get the signal itself.
When there is noise you must experiment. You may want to filter the response
before doing FFT.
As far as I know/remember, if your graph is connected and contains E edges then
you can find the shortest distance from any particular vertex to all other
vertices in O(E) operations. You can repeat this procedure starting from every
node (out of the 500). If you have 100,000 edges this will
The phenomenon is most likely caused by numerical errors. I do not know how
'integrate' works but numerical integration over a very long interval does not
look a good idea to me.
I would do the following:
f1-function(x){
return(dchisq(x,9,77)*((13.5/x)^5)*exp(-13.5/x))
}
f2-function(y){
Hi Nitin,
I believe that you can not have null hypothesis to be that A and B come from
different distributions.
Asymptotically (as both sample sizes go to infinity) KS test has power 1, i.e.
it will reject H0:A=B for any case where A and B have different distributions.
To work with a finite
Hi Miao,
I can write a function which takes an integer and produces a float number whose
binary representation equals to that of the integer, but this would be an
awkward solution.
So if nobody suggests anything better I will write such a function for you, but
let's wait for a better solution.
How about
d[sample(length(d),10)]
--- On Wed, 20/8/08, Lauri Nikkinen [EMAIL PROTECTED] wrote:
From: Lauri Nikkinen [EMAIL PROTECTED]
Subject: [R] Random sequence of days?
To: [EMAIL PROTECTED]
Received: Wednesday, 20 August, 2008, 4:04 PM
Dear list,
I tried to find a solution for this
Use toupper or tolower (see ?toupper, ?tolower)
--- On Wed, 20/8/08, suman Duvvuru [EMAIL PROTECTED] wrote:
From: suman Duvvuru [EMAIL PROTECTED]
Subject: [R] Conversion - lowercase to Uppercase letters
To: r-help@r-project.org
Received: Wednesday, 20 August, 2008, 2:19 PM
I would like to
Hi Jeff,
If I understand correctly, the overhead of a loop is that at each iteration the
command must be interpreted, and this time is independent of the number of rows
N. So if N is small this overhead may be very significant but when N is large
this should be very small compared to the time
Hi Alberto,
Please disregard my previous note - I probably had a black-out!!!
--- On Tue, 19/8/08, Alberto Monteiro [EMAIL PROTECTED] wrote:
From: Alberto Monteiro [EMAIL PROTECTED]
Subject: [R] A doubt about lm and the meaning of its summary
To: r-help@r-project.org
Received: Tuesday, 19
Hi Jose,
If you are only interested in the expected duration, the problem can be solved
analytically - no simulation is needed.
Let P be the probability to get total.capital (and then 1-P is the probability
to loose all the money) when starting with initial.capital. This probability P
is well
Since 0 can be represented exactly as a floating point number, there is no
problem with something like x[x==0].
What you can not rely on is something like 0.1+0.2 == 0.3 to be TRUE.
--- On Thu, 14/8/08, Roland Rau [EMAIL PROTECTED] wrote:
From: Roland Rau [EMAIL PROTECTED]
Subject: Re: [R]
The problem is that if x is either NA or NaN then x != 0 is NA (and not FALSE
or TRUE) and the function is.nan tests for a NaN but not for NA, i.e.
is.nan(NA) returns FALSE.
You can do something like:
mat_zeroless[!is.na(mat) mat != 0] - mat[!is.na(mat) mat != 0]
--- On Thu, 14/8/08,
Just interchange rows 2 and 3 and then columns 2 and 3 of the original
covariance matrix.
--- On Fri, 8/8/08, Zhang Yanwei - Princeton-MRAm [EMAIL PROTECTED] wrote:
From: Zhang Yanwei - Princeton-MRAm [EMAIL PROTECTED]
Subject: [R] Covariance matrix
To: r-help@r-project.org
Hi Benjamin,
Creating 0 correlations is easier and always possible, but creating arbitrary
correlations can be done as well (when possible - see below).
Suppose that x1,x2,x3,x4 have mean 0 and suppose that the desired correlations
are r = (r1,r2,r3,r4). Let A be an orthogonal 4x4 matrix such
Hello Jason,
You are not specific enough. What do you mean by significant difference?
Let's assume that indeed the incidence in A is 6% and in B is 10% and we are
looking for Na and Nb such that with probability of at least 80% the mean of Nb
sample from B will be at least, say, 0.03 (=3%)
You can use uniroot (see ?uniroot).
As an example, suppose you have a $100 bond which pays 3% every half year (6%
coupon) and lasts for 4 years. Suppose that it now sells for $95. In such a
case your time intervals are 0,0.5,1,...,4 and the payoffs are:
-95,3,3,...,3,103.
To find internal rate
This is something that is easier done in C than in R (to the best of my very
limited knowledge).
To do this in R you could do something like:
x - 082-232-232-1
y -unlist(strsplit(x,))
i - which(y != 0)[1]-1
paste(y[-(1:i)],collapse=)
[1] 82-232-232-1
--- On Fri, 1/8/08, calundergrad
y - 2 - (x[,1] x[,2])
you can also do
cbind(x,y)
if you wish.
--- On Fri, 1/8/08, Gundala Viswanath [EMAIL PROTECTED] wrote:
From: Gundala Viswanath [EMAIL PROTECTED]
Subject: [R] Grouping Index of Matrix Based on Certain Condition
To: [EMAIL PROTECTED]
Received: Friday, 1 August,
Yes, this is how it should be done!
--- On Fri, 1/8/08, Christos Hatzis [EMAIL PROTECTED] wrote:
From: Christos Hatzis [EMAIL PROTECTED]
Subject: Re: [R] cutting out numbers from vectors
To: 'calundergrad' [EMAIL PROTECTED], r-help@r-project.org
Received: Friday, 1 August, 2008, 2:11 PM
Hi Yunlei,
Is your problem constrained or not?
If it is unconstrained and your matrix is not positive definite, the minimum is
unbounded (unless you are extremely lucky and the matrix is positive
semi-definite and the vector which multiplies the unknowns is exactly
perpendicular to all the
I am not sure that this is well defined.
For a multivariate normal distribution (which is well defined), the covariance
matrix (and the means vector) fully determine the distribution. In the
exponential case, what is multivariate (bivariate) exponential distribution? I
believe that knowing
Assuming that the number of rows is even and that your matrix is A,
element-wise product of pairs of rows can be calculated as
A[seq(1,nrow(A),by=2),]*A{seq(2,nrow(A),by=2),]
--- On Mon, 28/7/08, rcoder [EMAIL PROTECTED] wrote:
From: rcoder [EMAIL PROTECTED]
Subject: [R] product of
Try
abs(outer(xk,x,-))
(see ?outer)
--- On Wed, 30/7/08, dxc13 [EMAIL PROTECTED] wrote:
From: dxc13 [EMAIL PROTECTED]
Subject: [R] finding a faster way to do an iterative computation
To: r-help@r-project.org
Received: Wednesday, 30 July, 2008, 4:12 AM
useR's,
I am trying trying to
If v is your vector of sample variances (and assuming that their distribution
is chi-square) you can define
f(df) - sum(dchisq(v,df,log=TRUE))
and now you need to maximize f, which can be done using any optimization
function (like optim).
--- On Sat, 26/7/08, Julio Rojas [EMAIL PROTECTED]
This problem can be easily solved analytically:
we want to minimize sum(res(i) -a*st(i) -b*mod(i))^2 subject to a+b=1,a,b=0,
so we want to minimize
f(a) = sum((res(i)-mod(i)) - a*(st(i)-mod(i)))^2 for 0=a=1
Define Xi = res(i) - mod(i), Yi = st(i) - mod(i), then
f(a) = sum(Xi - a*Yi)^2
f(0)
How large is your matrix?
Are the very small eigenvalues well separated?
If your matrix is not very small and the lower eigenvalues are clustered, this
may be a really hard problem! You may need a special purpose algorithm and/or
higher precision arithmetic.
If your matrix is A and there
Kapat [EMAIL PROTECTED]
Subject: Re: [R] spectral decomposition for near-singular pd matrices
To: [EMAIL PROTECTED]
Received: Thursday, 17 July, 2008, 10:56 AM
Moshe Olshansky m_olshansky at yahoo.com
writes:
How large is your matrix?
Right now I am looking at sizes between 30x30
The problem is that neither 0.55 nor 2.55 are exact machine numbers (the
computer uses binary representation), so it may happen that the machine
representation of 0.55 is slightly less than 0.55 while the machine
representation of 2.55 is slightly above 2.55.
--- On Fri, 11/7/08, Korn, Ed
is below 255, so
that x is less than 2.55 and should have been rounded to 2.5.
--- On Fri, 11/7/08, Moshe Olshansky [EMAIL PROTECTED] wrote:
From: Moshe Olshansky [EMAIL PROTECTED]
Subject: Re: [R] rounding
To: [EMAIL PROTECTED], Korn, Ed (NIH/NCI) [E] [EMAIL PROTECTED]
Received: Friday, 11 July
It looks like SR, SU and ST are strongly correlated to each other, as well as
DR, DU and DT.
You can try to do PCA on your 6 variables, pick the first 2 principal
components as your new variables and use them for regression.
--- On Fri, 11/7/08, Georg Ehret [EMAIL PROTECTED] wrote:
From:
Karanth
wrote:
On 2008-7-8, at 下午2:39, Moshe Olshansky wrote:
If they are really random you can not expect their sum
to be 100.
However, it is not difficult to get that given that
the sum of n
independent Poisson random variables equals N, any
individual one
has the conditional
The answer to your first question is
sum(x)8sum(y) - sum(x*y)
and for the second one
x %*% R %*% y - sum(x*y*diag(R))
--- On Thu, 3/7/08, Murali Menon [EMAIL PROTECTED] wrote:
From: Murali Menon [EMAIL PROTECTED]
Subject: [R] multiplication question
To: [EMAIL PROTECTED]
Received:
dnorm() computes the density, so it may be 1; pnorm() computes the
distribution function.
--- On Tue, 8/7/08, Mike Lawrence [EMAIL PROTECTED] wrote:
From: Mike Lawrence [EMAIL PROTECTED]
Subject: Re: [R] odd dnorm behaviour (?)
To: Rhelp [EMAIL PROTECTED]
Received: Tuesday, 8 July, 2008,
Another possibility is to use explicit formula, i.e. if you are doing linear
regression like y = a*x + b then the explicit formulae are:
a = (meanXY - meanX*meanY)/(meanX2 - meanX^2)
b = (meanY*meanX2 - meanX*meanXY)/(meanX2 - meanX^2)
where meanX is mean(x), meanXY is mean(x*y), meanX2 is
matrices, for-loops, speed
To: [EMAIL PROTECTED]
Cc: r-help@r-project.org, Zarza [EMAIL PROTECTED]
Received: Monday, 7 July, 2008, 9:40 AM
On 7/07/2008, at 11:05 AM, Moshe Olshansky wrote:
Another possibility is to use explicit formula, i.e.
if you are
doing linear regression like y = a*x + b
I know very little about graphics, so my primitive and brute force solution
would be
plot(density(x[1:30]),col=blue);lines(density(x[31:60]),col=red);lines(density(x[61:90]),col=green)
--- On Mon, 7/7/08, Gundala Viswanath [EMAIL PROTECTED] wrote:
From: Gundala Viswanath [EMAIL PROTECTED]
Do you have a reason to treat all 3 levels together and not have a separate
regression for each level?
--- On Tue, 1/7/08, rlearner309 [EMAIL PROTECTED] wrote:
From: rlearner309 [EMAIL PROTECTED]
Subject: [R] A regression problem using dummy variables
To: r-help@r-project.org
Received:
Let F be the distribution function of Y, PSI the standard normal distribution
anf IPSI it's inverse. Let f(x) = IPSI(F(x)). It is not difficult to see that
f(Y) has standard normal distribution.
So you can replace F with the empirical distribution and IPSI is the qnorm
function of R.
--- On
If the main diagonal element of matrix A is 1 and the off diagonal element is a
then for any vector x we get that t(x)*A*x = (1-a)*sum(x^2) +a*(sum(x))^2 . If
we want A to be positive (semi)definite we need this expression to be positive
(non-negative) for any x!= 0. Since sum(x)^2/sum(x*2) = n
What do you mean by A similar to X?
Do you mean norm of the difference, similar eigenvalues/vectors, anything else?
--- On Wed, 25/6/08, Gundala Viswanath [EMAIL PROTECTED] wrote:
From: Gundala Viswanath [EMAIL PROTECTED]
Subject: [R] Measuring Goodness of a Matrix
To: [EMAIL PROTECTED]
One possibility is:
x - c(30.9, 60.1 , 70.0 , 73.0 , 75.0 , 83.9 , 93.1 , 97.6 , 98.8 ,
113.9)
for (i in 1:9) assign(paste(PAIR,i,sep=),list(part1 = x[1:i],part2 =
x[-(1:i)]))
--- On Mon, 23/6/08, Gundala Viswanath [EMAIL PROTECTED] wrote:
From: Gundala Viswanath [EMAIL PROTECTED]
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