That's a good solution, but if you're really, really sure that the timestamps
are in the format you gave, it's quite a bit faster to use substr and paste,
because you don't have to do any searching in the string.
HTH
Rex
> x = rep("09:30:00.000.633",100)
> system.time(y<-paste(substr(x,1,12)
Feng,
Your matrix is *not* (practically) singular; its inverse is.
The message said that the *system* was singular, not the matrix.
Remember Cramer's Rule: xi = |Ai| / |A|
The really, really large determinant of your matrix is going to appear in the
denominator of your solutions, so, essentially,
Cheer up! R is a step closer to that concept than the old FORTRAN compilers
that couldn't even guarantee that 37 was a constant if used repeatedly in a
subroutine call.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Uwe Ligges
Se
I think you need to read an introduction to R.
For starters, read.table returns its results as a value, which you are not
saving.
The probable answer to your question:
Read the whole file with read.table, and select columns you need, e.g.:
tab <- read.table(myfile, skip=2)[,1:5]
-Original Mes
Hi Jon, I read your question differently. Is the answer? - Rex
> ch=scan(stdin(),what=character(0),n=1)
1: f
Read 1 item
> ch
[1] "f"
>
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Tuesday, March 15, 2011 10
Tsjerk,
It seems to me that memory and not time is your big efficiency problem, and
I've showed you how to avoid storing your entire input.
If you want to avoid doing each multiplication twice, you can replace the
"outer" with a function that computes each product only once and accumulate
sums o
Tjerk,
This is just a pseudo code outline of what you need to do:
M = matrix(0, number of variables, number of variables)
V = rep(0, number of variables)
N = 0
While (more observations to read) {
X <- next observation
V <- V + X
M <- M + outer(X,X)
N <- N+1
}
Compute covariance matrix
I think I need to retract the part about 3 iterations... not true if, e.g.,
the segments intersect and the angle is small.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of rex.dw...@syngenta.com
Sent: Friday, March 11, 2011 2:37 PM
I like Thomas's idea as a quick practical solution. Here is one more little
variation just in case you really do have millions of these distances. Pick
point P1 on line segment L1 (e.g., an endpoint). Pick 101 evenly spaced points
on line segment L2. Find the nearest to P1 and call it P2. N
But no one answered Kushan's question about performance implications of
for-loop vs lapply.
With apologies to George Orwell:
"for-loops BAAD, no loops GOOD."
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Uwe Ligges
Sent:
It sounds like you want a bunch of random permutations of 1:7.
Try order(runif(7))
If you need, say, 10 of them:
as.vector(sapply(1:10,function(i) order(runif(7
Is it more complicated than that?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
Or, if X1 Y1 X2 Y2... are really your column names
m[, grep("X",colnames(m)) ]
or
m[, grepl("X",colnames(m)) ]
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Dimitris Rizopoulos
Sent: Wednesday, March 09, 2011 9:36 AM
To: Nixon, Ma
Cesar, I think your basic misconception is that you believe 'sample' returns a
list of indices into the original vector. It does not; it returns actual
elements of the vector:
> sample(runif(100),3)
[1] 0.4492988 0.0336069 0.6948440
I'm not sure why you keep resetting the seed, but if it's imp
Hi Thomas,
Several of us explained this in different ways just last week, so you might
search the archive. Floating point numbers are an approximate representation
of real numbers. Things that can be expressed exactly in powers of 10 can't be
expressed exactly in powers of 2. So the sum 0.6+0
Harsh, not to worry, but you were wrong to assert that I engaged in any name
calling, let alone constant name calling.
I also didn't and don't claim to be an authority on survey design.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf
You still don't say what organization you are associated with. Your domain
name and e-mail address give no hint. How do we know that "Harsh Singhal" is
even a real person? An e-mail address at a university (for example) would go a
long way to establish that. Gmail doesn't cut it for me.
The p
Carrie,
If your while-loop condition depends only on dt, and you don't change dt in
your loop, your loop won't terminate.
The only thing inside your loop is "next".
Perhaps you mean to write:
temp=rep(NA, 10)
for(i in 1:10)
{
dt=sum(rbinom(10, 5, 0.5))
while (dt<25) {
dt=sum(rbinom(10, 5, 0.5))
Hi Michael,
In floating point calculation, 1.0-.9 is not exactly 0.1. This is easily seen
by subtracting.
> (1.0-.9)-0.1
[1] -2.775558e-17
> (1.0-.9)==0.1
[1] FALSE
David is right, you can't "correct" this. You can only compensate by taking
care that you never, ever test whether 2 FP numbers
Just out of curiosity : It is possible to get name, employer name, location,
usage information and academic background details when searching for R users on
LinkedIn and the many R related groups there.
Does this also provide potential opportunities for misuse and "outrageous"
analyses, since
Eval it. This works at my house:
plot(0)
title(eval(parse(text=paste("expression(paste(delta^13,'C Station ',",i,"))"
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Filoche
Sent: Thursday, March 03, 2011 9:39 AM
To: r-help@r-
Harsh, "Suitably analyzed" for whose purposes? One man's "suitable" is
another's "outrageous". That's why people want to see the gowns at the Oscars.
Under what auspices are you conducting this survey? What do you intend to do
with it? You don't give any assurance that the results you post w
mytitle = parse(text=paste("expression(paste(delta^13,'C Station ',",i,"))"))
title(mytitle)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Filoche
Sent: Thursday, March 03, 2011 8:16 AM
To: r-help@r-project.org
Subject: [R] Greek
Perl seems like a 10x better choice for the task, but try looking at the
examples in ?strsplit to get started.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of antujsrv
Sent: Thursday, March 03, 2011 4:23 AM
To: r-help@r-project.org
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of dms
Sent: Wednesday, March 02, 2011 3:16 PM
To: r-help@r-project.org
Subject: [R] merge( , by='row.names') slowness
I noticed that joining two data.frames in R using the "merge"
fun
Don't you expect it to be a lot faster if you cluster 20 items instead of 25000?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Maxim
Sent: Wednesday, March 02, 2011 4:08 PM
To: r-help@r-project.org
Subject: [R] clustering problem
Hi Ivo,
It might be useful for you to study the examples below.
The key from a programming language point of view is that functions like ifelse
are functions of whole vectors, not elements of vectors. You either evaluate
an argument or you don't; you don't evaluate only part of argument. (Someb
No, that's not what I meant, but maybe I didn't understand the question.
What I suggested would involve sorting y, not x: "sort the *distances*".
If you want to minimize the sd of a subset of numbers, you sort the numbers and
find a subset that is clumped together.
If the numbers are a function of
?cor answers that question. If Housing is a dataframe, cor(Housing) should do
it. Surprisingly, ??correlation doesn't point you to ?cor.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of JoonGi
Sent: Tuesday, March 01, 2011 5:41 A
You determine the variance explained by *any* unit vector by taking its inner
product with the data points, then finding the variance of the results. In the
case of FastICA, the variance explained by the ICs collectively is exactly the
same as the variance explained by the principal components
James,
It seems the 2*mean(x) term is irrelevant if you are seeking to minimize sd.
Then you want to sort the distances from smallest to largest. Then it seems
clear that your five values will be adjacent in the list, since if you have a
set of five adjacent values, exchanging any of them for
I have to agree that it's pretty hard to take something that works and figure
out why it doesn't work :)
The only other suggestion is that sometimes I find that this sort of error goes
away if I add "drop=FALSE" to the subsetting, and, if so, that usually lets me
figure out why.
-Original
Generally, you can save your excel spreadsheet as comma-separated values, and
then read with read.csv function: ?read.csv
Or, tab-separated values and use read.delim.
Then look at ?barplot
Possibly you would like to read the Intro to R on the CRAN website. Go to
www.r-project.org , find Docume
Does it work for FUN=mean? If yes, you need to print out the results of f
before you return them to find the anomalous value.
BTW "Error" is not a very good subject line. I don't see many posts from
people reporting how well things are going :)
-Original Message-
From: r-help-boun...@
Are you clear about the question you are asking? Do you want to know whether
there are 6 balls or at least 6 balls? (It sounds like "at least".) Do you
want to know whether there are at least 6 balls in the first box, or at least 6
balls in exactly one box or at least 6 balls in at least one
One way to do Dirichlet triangulations is to map point (x,y) to point
(x,y,x^2+y^2) (I think, it's been a while) and then find the convex hull of
these points in 3 dimensions. You can do the Voronoi diagram of circles by
mapping (x,y,r) to (x,y,x^2+y^2-r^2) I would try assigning an r to each
You don't need to write functions to "source" files:
source("code1.R")
source("code2.R")
source("code3.R")
When you source a file with a bunch of function definitions, the definitions
are just assignment statements:
f <- function (x)...
g <- function (x,y,z) ...
Did you think you would break y
If you're in a hurry, it's way easier than that:
t <- c(1,2,3,7,8,9,11,12,13)
x <- rnorm(length(t))
new.t <- min(t):max(t)
new.x <- NULL
new.x[t-min(t)+1] <- x
plot(new.t, new.x, type='l')
This is wastes max(t)-min(t)-length(t)+1 vector entries, but presumably you
won't be wasting a lot of rea
How is the curve is represented? That's more important that its
organ-of-origin. If you have values of y=f(x) at discrete time points, then
y-(x+2) will change sign sometimes... the intersection point is at some time x'
in between. Am I missing something subtle here?
You could interpolate the
My surmise would be that you have not analyzed the situation correctly, and you
are making a false assumption about your code. Since you can't show the code,
it's pretty hard to figure out what that is. I think you're going to have to
produce a simple example that you can share that has the sa
How is the curve defined? If the curve is y=f(x) and the line is y=mx+b, you
look for the roots of f(x)-mx-b.
?polyroot
?uniroot
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of FMH
Sent: Tuesday, February 22, 2011 6:28 AM
To: r-he
?InternalMethods
?S3groupGeneric
?S4groupGeneric
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Erin Hodgess
Sent: Friday, February 18, 2011 10:45 PM
To: R help
Subject: [R] question about generics
Dear R People:
Is there a way
Well, you can lose B by just adding to X in the first for-loop, can't you?
For (...) X <- X + A[...]
But if you want elegance, you could try:
X = Reduce("+",lapply(1:(p+1), function(i) A[i:(n-p-1+i),i:(n-p-1+i)]))
I imagine someone can be even more eleganter than this.
rad
-Original Messag
I was going to say:
The problem with for-loops (as best I understand it) is that the R code gets
interpreted over and over; what you normally want to do is design the
computation so that you jump into the internals of R and stay there. But the
inner loop is in the R internals of the sort in th
Although I suggested to someone else that for-loops be avoided, they are not in
the inner loop in this code, and it's probably easier to understand than some
sort of apply:
a = array(round(100*runif(60)),dim=c(3,4,5))
a
for (i in 1:dim(a)[1])
for (j in 1:dim(a)[2])
a[i,j,] = sort(a[i,j,])
a
Yes, remove the call to intersect, and rely on the results of match to tell you
whether there is an overlap. If there are any matches, all(is.na(index)) will
be false. Read help for match.
?match
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.
I hate to sound like David "Have You Read The Posting Guide?" Winsemius, but
there's no way for anyone to know what you are trying to accomplish here
without a lot more information. You don't show us the output you expect and
the output you got. I would expected "relatedness" to be on a scale
Try putting d,e,f in a list:
Xxx = list(d,e,f)
For (I in 1:length(xxx))
For (j in 1:length(xxx))
If (i!=j) bigfunction(xxx[[i]],xxx[[j]])
(bad indentation, caps thanks to outlook)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of s
If y'all want to discuss this more, do it somewhere else, please.
This has little to do with R except that both depend on Peano's Axioms.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of jlu...@ria.buffalo.edu
Sent: Tuesday, February
It's only "awfully" inefficient if it's a bottleneck. You're not doing this
more than once per item fetched from the network, and the time is insignificant
relative to the fetch. If it were somehow in your inner loop, it would be
worth worrying about, but your purpose is to eliminate Ms and Bs
A quick way to do this is to replace \d and \D with character classes [0-9.]
and [^0-9.] . This assumes that there is no scientific notation and that there
is nothing like 123.45.678 in the string. You did not account for a leading
minus sign.
The book Mastering Regular Expressions is probably
If the graph has n nodes and is represented by an adjacency matrix, you can
square the matrix (log_2 n)+1 times. Then you can multiply the matrix
element-wise by its transpose. The positive entries in the 7th row will tell
you all nodes sharing a cycle with node 7. This assumes all edge weigh
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Marine Andersson
Sent: Thursday, February 10, 2011 3:53 PM
To: r-help@r-project.org
Subject: [R] Calculating rowMeans from different columns in each row?
Hello!
I have a dataset lik
If you want a random correlation matrix, why not just generate random data and
accept the correlation matrix that you get? The standard normal distribution
in k dimensions is (hyper)spherically symmetric. If you generate k standard
normal N(0,1) variates, you have a point in k-space with direc
ab = paste(a,b,sep=";~;~;~")
flag = length(ab)==length(unique(ab))
This should work unless you use 3 consecutive winking elephants in other places
in your program.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Nipesh Bajaj
Sen
Dear Marius,
Try this:
plot.list = lapply(1:10,
function(i) xyplot(i~i,type="p",xlim=c(0,11),panel=function(...) {
panel.xyplot(...); panel.abline(v=i)})
)
plot.list[[3]]
I imagine it will work for Mr Luftjammer, too.
Rex
-Original Message-
From: r-help-boun...@r-project.org [ma
Hi Marianne,
The quick-and-dirty solution is to add one character and make ns global:
ns <<- nrow(x)
Poor practice, but OK temporarily if you're just debugging.
This is an issue of "scope". You are assuming dynamic scope, whereas R uses
static scope. 'ns' was not defined when you said paste("n=
56 matches
Mail list logo
