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 w
Check read.table (?read.table).
--- On Wed, 24/2/10, RagingJim wrote:
> From: RagingJim
> 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 with a program which was designed so
> th
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). Y
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 square
Hi,
As far as I understand, D is the value of (Kolmogorov-Smirnov) statistic and
p-value is the probability to get that (or greater) value for normally
distributed variables (so in your case you would most probably reject the
hypothesis that your data is normal).
--- On Tue, 23/2/10, Bosken w
Yes, this can be easily computed analytically (even though my result is a bit
different).
--- On Fri, 12/2/10, dav...@rhotrading.com wrote:
> From: dav...@rhotrading.com
> Subject: Re: [R] Integral of function of dnorm
> To: "Greg Snow" , "Trafim Vanishek"
> , "Peter Dalgaard"
> Cc: r-help@r
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 basica
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 wrote:
> From: Something Something
> Subject: [R] Questio
One possibility I can see is to replace - by NA and use mean with
na.rm=TRUE.
--- On Wed, 10/2/10, Steve Murray wrote:
> From: Steve Murray
> Subject: [R] Resampling a grid to coarsen its resolution
> To: r-help@r-project.org
> Received: Wednesday, 10 February, 2010, 3:20 AM
>
> Dear all,
>
> However, due to my limited stats background, I am 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
>
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
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 Pi
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 Tu
test[which(test[,"total"] %in% needed),]
--- On Fri, 25/9/09, Dimitri Liakhovitski wrote:
> From: Dimitri Liakhovitski
> Subject: [R] keeping all rows with the same values, and not only unique ones
> To: "R-Help List"
> Received: Friday, 25 September, 2009, 8:52 AM
> Dear R-ers,
>
> I have a
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 wrote:
> From: Frostygoat
> Subject: [R] Basic population dynamics
> To: r-help@r-project.org
> Received: Wednesday, 2 September, 2009, 4:48 AM
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 wrote:
> From: Steven Kang
> Subject: [R] Help on efficiency/vectorization
> To: r-help@r-project.org
> Received: Thursday, 27 August
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, Debabrat
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 wrote:
> From: kfcnhl
> Subject: [R] extra .
> To: r-help@r-project.org
> Received: Friday
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 matr
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 wrote:
> From: Liviu Andron
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)" wrote:
> From: "Häring, Tim (LWF)"
> Subject: [R] feature weighti
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
Is your system of equations linear?
--- On Fri, 14/8/09, Moreno Mancosu wrote:
> From: Moreno Mancosu
> 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 question(in fact I _am_, a newbie), bu
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 wrote:
> From: Gina Liao
> Subject: [R] problem about t test
> To: r-h...@stat.math.ethz.ch
> Received: Friday, 14 August, 20
Hi,
Is your matrix K symmetric? If yes, there is an "analytical" solution.
--- On Sat, 1/8/09, nhawrylyshyn wrote:
> From: nhawrylyshyn
> Subject: [R] Matrix Integral
> To: r-help@r-project.org
> Received: Saturday, 1 August, 2009, 12:15 AM
>
> Hi,
>
> Any help on this would be appreciated:
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 wrote:
> From: Inchallah Yarab
> Subject: [R] Re : How to Import Excel file into R 2.9.0 version
> To: r-help@r-project.o
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 wrote:
> From: Tim Chatterton
> Subject: [R] Counting the number of non-NA values per day
> To: r-help@r-project.org
> Recei
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 wrote:
> From: Chyden Finance
> 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 PhD program in Finance for stat
Try ?aggregate
--- On Wed, 15/7/09, Timo Schneider wrote:
> From: Timo Schneider
> Subject: [R] Grouping data in dataframe
> To: "r-help@r-project.org"
> Received: Wednesday, 15 July, 2009, 1:56 PM
> Hello,
>
> I have a dataframe (obtained from read.table()) which looks
> like
>
>
> Ex
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 the
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 wrote:
> From: Michael Knudsen
> Subject: [R] Nested for loops
> To: r-help@r-project.org
> Received: Tuesday, 14 July, 2009, 3:38 PM
> Hi,
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 wrote:
> From: Tish Robertson
> Subject: [R] averaging two matrices whilst ignoring missing values
> To: r-help@r-proje
Try A[1,,drop=FALSE] - see help("\[")
--- On Mon, 13/7/09, Weiwei Shi wrote:
> From: Weiwei Shi
> 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"
> Received: Monday, 13 July, 2009, 1:55 PM
> Hi, there:
>
> Assume I hav
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))
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 wrote:
> From: Olivella
> Subject: [R] Substituting numerical values using `apply'
> To: r-help@r-project.org
> Received: Thursday, 9 July, 2009, 6:25 AM
One possibility is to use sink (see ?sink).
--- On Thu, 9/7/09, Steve Jaffe wrote:
> From: Steve Jaffe
> 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 log file.
> cat() flattens them
>
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 wrote:
> From: mister_bluesman
> Subject: [R] Extracting a column name in loop?
> To: r-help@r-project.org
> Received: Thursday, 9 July, 2009,
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 wan
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?
C
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 t
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 P
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 S
Hi Mark,
stock<-"/opt/limsrv/mark/research/equity/projects/testDL/stock_data/fhdb/US/BLC.NYSE"
> gsub(".*/([^/]+)$", "\\1",stock)
[1] "BLC.NYSE"
--- On Tue, 23/9/08, [EMAIL PROTECTED] <[EMAIL PROTECTED]> wrote:
> From: [EMAIL PROTECTED] <[EMAIL PROTECTED]>
> Subject: [R] perl expression questi
One possibility is:
> x <- data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2))
> names <- t(matrix(rep(names(x),times=nrow(x)),nrow=ncol(x)))
> m <- as.matrix(x)
> ind <- order(m)
> df <- data.frame(name=names[ind],value=m[ind])
> df
name value
1 x1 1
2 x3 2
3 x2 4
4 x2 6
5 x1
Hi Ana,
There are two problems:
First of all, if you want your matrix to have 4 columns it's number of elements
should not be 17!
Secondly, and this is what causes your error message, you should not call your
second function matrix. Call it matrix1, my_matrix, whatever. Otherwise R
thinks tha
Well, I made a mistake - your lambda should be 400 and not 40!!!
--- On Thu, 18/9/08, Moshe Olshansky <[EMAIL PROTECTED]> wrote:
> From: Moshe Olshansky <[EMAIL PROTECTED]>
> Subject: Re: [R] help on sampling from the truncated normal/gamma
> distribution on the far end
Hi Sonia,
If I did not make a mistake, the conditional distribution of X given that X > 0
is very close to exponential distribution with parameter lambda = 40, so you
can sample from this distribution.
--- On Mon, 15/9/08, Daniel Davis <[EMAIL PROTECTED]> wrote:
> From: Daniel Davis <[EMAIL P
Hi Ramya,
Assuming that the problem is well defined (i.e. the values in col1 of the
data.frames are unique and every value in D.F.sub.2[,1] appears also in
D.F1[,1]) you can do the following:
ind <- match(D.F.sub.2[,1],D.F1[,1])
D.F1[ind,] <- D.F.sub.2
--- On Thu, 18/9/08, Rajasekaramya <[EMA
Just a small correction:
start with
s <- paste(r$userid,collapse=",")
and not
s <- paste(r$userid,sep=",")
--- On Fri, 12/9/08, Moshe Olshansky <[EMAIL PROTECTED]> wrote:
> From: Moshe Olshansky <[EMAIL PROTECTED]>
> Subject: Re: [R] dat
One possibility is as follows:
If r$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 P
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]>
&g
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(X>a) + P(X<-a).
So if X~N(0,1), we want P(X>a) + P(X<-a) = 2P
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) = -exp(-20)*20^x/facto
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]>
> Subje
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
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
> Rec
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] I
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 req
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.
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.
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 samp
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){
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 solutio
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
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: Tue
Hi Alberto,
In your second case the linear model
y = a*x + b + error
does not hold.
--- 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
> Recei
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 n
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 k
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, rco
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:
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"
> Recei
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 th
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,
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 Aug
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, c
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
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%) abo
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 tha
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 eige
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]> wr
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
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 su
Or, as suggested by Duncan Murdoch,
qnorm(runif(500,pnorm(-1.5),pnorm(1.5)))
--- On Fri, 25/7/08, jim holtman <[EMAIL PROTECTED]> wrote:
> From: jim holtman <[EMAIL PROTECTED]>
> Subject: Re: [R] simple random number generation
> To: "dxc13" <[EMAIL PROTECTED]>
> Cc: r-help@r-project.org
> Rec
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
D]> wrote:
> From: Prasenjit 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 yahoo.com>
> writes:
>
> > How large is your matr
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 exists
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: G
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]"
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 (NIH
method is also correct except
> it is based on
> the conditioning.
> On 2008-7-8, at 下午1:58, Shubha Vishwanath 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.
> >
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
binomial distribution with size = N and p = 1/n, i.e.
P(Xi=k/Sn=N) = (N over k)*
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
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:
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 PROT
t: Re: [R] Lots of huge 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 t
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 mean(
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