On 21/09/2010 1:29 PM, baptiste auguie wrote:
I see, thank you.
I'm still worried by the very dramatic error I obtained just from
shifting so slightly the support of the integrand, it took me a while
to figure what happened even with this basic example (I knew the
integral couldn't be so small!
I see, thank you.
I'm still worried by the very dramatic error I obtained just from
shifting so slightly the support of the integrand, it took me a while
to figure what happened even with this basic example (I knew the
integral couldn't be so small!).
For a general integration in [0, infty), ther
On Tue, 21 Sep 2010, baptiste Auguié wrote:
Thanks, I'll do that too from now on.
It strikes me that in a case such as this one it may be safer to use a
truncated, finite interval around the region where the integrand is non-zero,
rather than following the advice of ?integrate to use Inf as in
> of 1.e-08 (roughly, sqrt(machine epsilon)) in my computations, and I also
> increase subdivisions to 500.
>
> Ravi.
>
> From: baptiste Auguié [mailto:baptiste.aug...@googlemail.com]
> Sent: Tuesday, September 21, 2010 9:58 AM
> To: Ravi Varadhan
> Cc: 'bap
rel.tol of 1.e-08 (roughly, sqrt(machine epsilon)) in my computations, and I
also increase subdivisions to 500.
Ravi.
From: baptiste Auguié [mailto:baptiste.aug...@googlemail.com]
Sent: Tuesday, September 21, 2010 9:58 AM
To: Ravi Varadhan
Cc: 'baptiste auguie'; 'r-help'
Sub
ilto:r-help-boun...@r-project.org] On
> Behalf Of baptiste auguie
> Sent: Tuesday, September 21, 2010 8:38 AM
> To: r-help
> Subject: [R] puzzle with integrate over infinite range
>
> Dear list,
>
> I'm calculating the integral of a Gaussian function from 0 to
>
ct.org] On
> Behalf Of baptiste auguie
> Sent: Tuesday, September 21, 2010 8:38 AM
> To: r-help
> Subject: [R] puzzle with integrate over infinite range
>
> Dear list,
>
> I'm calculating the integral of a Gaussian function from 0 to
> infinity. I understand from
-07)$value }
shift <- seq(500, 800, by=10)
plot(shift, sapply(shift, shiftedGauss))
Hope this helps,
Ravi.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of baptiste auguie
Sent: Tuesday, September 21, 2010 8:38 AM
To: r-help
Subje
Dear list,
I'm calculating the integral of a Gaussian function from 0 to
infinity. I understand from ?integrate that it's usually better to
specify Inf explicitly as a limit rather than an arbitrary large
number, as in this case integrate() performs a trick to do the
integration better.
However,
Look for a vectorized solution such as ?table or ?aggregate to obtain a list
of combination counts, followed by logical indexing or ?subset to get a set of
valid combinations, and then use ?sample to get your random selections of
locations/surveyors and then process only those combinations from
Ben Holt bio.ku.dk> writes:
> I have data similar to this:
>
> Location Surveyor Result
> A1 83
> A2 76
> A3 45
> B1 71
> B4 67
> C2 23
> C5 12
> D3 34
> E4
I have data similar to this:
Location Surveyor Result
A1 83
A2 76
A3 45
B1 71
B4 67
C2 23
C5 12
D3 34
E4 75
F4 46
G5 90
et
Many many thanks to all of you. The beer cleared the air of doubts!
Pls look at the following lines of code. This is taken from the example of
tradesys documentation. When I run the given example using the data.frame
spx it works just very fine but while I use some other data.frame (here
nifty) it
Hi
I do not use any of mentioned libraries so I can not directly answer it. I
would try to use debug(expr.frame) to see at what time the error is
thrown.
I have no idea why did you obtain error. Try to evaluate code in peaces
e.g. what is result of
list(MAf=quote(SMA(Last, 20)), MAs=quote(SMA
Hi
r-help-boun...@r-project.org napsal dne 12.07.2010 16:09:30:
> When I just run a for loop it works. But if I am going to run a for loop
> every time for large vectors I might as well use C or any other
language.
> The reason R is powerful is becasue it can handle large vectors without
each
>
I wanted to point out one thing that Ted said, about initializing the
vectors ('s' in your example). This can make a dramatic speed
difference if you are using a for loop (the difference is neglible
with vectorized computations).
Also, a lot of benchmarks have been flying around, each from a
diff
tate University
> Corvallis, OR 97331
> ph: 541-737-6232
> fx: 541-737-1393
>
>
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of Ted Harding
> Sent: Monday, July 12, 2010 9:36 AM
> To: r-help@r-p
iety
Oregon State University
Corvallis, OR 97331
ph: 541-737-6232
fx: 541-737-1393
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Ted Harding
Sent: Monday, July 12, 2010 9:36 AM
To: r-help@r-project.org
Cc: Raghu
Subject: Re: [R] in
On 12-Jul-10 14:09:30, Raghu wrote:
> When I just run a for loop it works. But if I am going to
> run a for loop every time for large vectors I might as well
> use C or any other language.
> The reason R is powerful is becasue it can handle large vectors
> without each element being manipulated? Pl
On Jul 12, 2010, at 10:09 AM, Raghu wrote:
When I just run a for loop it works. But if I am going to run a for
loop
every time for large vectors I might as well use C or any other
language.
The reason R is powerful is becasue it can handle large vectors
without each
element being manipula
> The reason R is powerful is becasue it can handle large vectors without
each
> element being manipulated? Please let me know where I am wrong.
>
> for(i in 1:length(news1o)){
> + if(news1o[i]>s2o[i])
> + s[i]<-1
> + else
> + s[i]<--1
> + }
You might give ifelse() a shot here.
s <- ifelse(news
I don't know what is wrong with your code but I believe you should use
ifelse instead of a for loop:
s <- ifelse(news1o > s2o, 1 , -1 )
Alain
On 12-Jul-10 16:09, Raghu wrote:
When I just run a for loop it works. But if I am going to run a for loop
every time for large vectors I might as wel
When I just run a for loop it works. But if I am going to run a for loop
every time for large vectors I might as well use C or any other language.
The reason R is powerful is becasue it can handle large vectors without each
element being manipulated? Please let me know where I am wrong.
for(i in 1
On 10/14/2009 2:29 PM, Adrian Dragulescu wrote:
Thank you.
If I use
gsub(" \xad", "-", x)
[1] "NEW YORK-NEW ENGLAND"
I get what I want.
Right, that's simpler than what I suggested.
Duncan Murdoch
Adrian
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=En
Thank you.
If I use
gsub(" \xad", "-", x)
[1] "NEW YORK-NEW ENGLAND"
I get what I want.
Adrian
sessionInfo()
R version 2.9.2 (2009-08-24)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=
On 10/14/2009 2:16 PM, Adrian Dragulescu wrote:
I get the same results (not working) using R 2.9.2 and R.10.0 beta.
But it is working: the dash is an "ad" in x, not a "2d". You need to
ask to substitute for the "ad" character, e.g. by
spacelongdash <- rawToChar(as.raw(c(0x20, 0xad)))
gsub(
I get the same results (not working) using R 2.9.2 and R.10.0 beta.
Thank you for looking at this.
On Wed, 14 Oct 2009, Duncan Murdoch wrote:
On 10/14/2009 1:41 PM, Adrian Dragulescu wrote:
charToRaw(x)
[1] 4e 45 57 20 59 4f 52 4b 20 ad 4e 45 57 20 45 4e 47 4c 41 4e 44
charToRaw(y)
On Wed, 14 Oct 2009, Adrian Dragulescu wrote:
charToRaw(x)
[1] 4e 45 57 20 59 4f 52 4b 20 ad 4e 45 57 20 45 4e 47 4c 41 4e 44
charToRaw(y)
[1] 4e 45 57 20 59 4f 52 4b 20 2d 4e 45 57 20 45 4e 47 4c 41 4e 44
So they are different.
We really do need the 'at a minimum' information we asked
On 10/14/2009 1:41 PM, Adrian Dragulescu wrote:
charToRaw(x)
[1] 4e 45 57 20 59 4f 52 4b 20 ad 4e 45 57 20 45 4e 47 4c 41 4e 44
charToRaw(y)
[1] 4e 45 57 20 59 4f 52 4b 20 2d 4e 45 57 20 45 4e 47 4c 41 4e 44
So they are different.
Adrian
I use R 2.8.1 on WinXP
But that's ancient.
charToRaw(x)
[1] 4e 45 57 20 59 4f 52 4b 20 ad 4e 45 57 20 45 4e 47 4c 41 4e 44
charToRaw(y)
[1] 4e 45 57 20 59 4f 52 4b 20 2d 4e 45 57 20 45 4e 47 4c 41 4e 44
So they are different.
Adrian
I use R 2.8.1 on WinXP
On Wed, 14 Oct 2009, Duncan Murdoch wrote:
On 10/14/2009 1:30 PM, A
On 10/14/2009 1:30 PM, Adrian Dragulescu wrote:
Hello,
Below is some output that shows my issue.
I have a variable x that I read from a file (more on this below)
x
[1] "NEW YORK NEW ENGLAND"
gsub(" -", "-", x)# this does not work!
[1] "NEW YORK NEW ENGLAND"
It looks as though
Hello,
Below is some output that shows my issue.
I have a variable x that I read from a file (more on this below)
x
[1] "NEW YORK NEW ENGLAND"
gsub(" -", "-", x)# this does not work!
[1] "NEW YORK NEW ENGLAND"
Encoding(x) # is x in a special encoding? no
[1]
Peter Dalgaard skrev:
> Prof Brian Ripley skrev:
>
>> -0.5*(A+B) is not a contrast, which is the seat of your puzzlement.
>>
>> All you can get from y ~ x is an intercept (a column of ones) and a
>> single 'contrast' column for 'x'.
>>
>> If you use y ~ 0+x you can get two columns for 'x', but R
Prof Brian Ripley skrev:
> -0.5*(A+B) is not a contrast, which is the seat of your puzzlement.
>
> All you can get from y ~ x is an intercept (a column of ones) and a
> single 'contrast' column for 'x'.
>
> If you use y ~ 0+x you can get two columns for 'x', but R does not
> give you an option of w
-0.5*(A+B) is not a contrast, which is the seat of your puzzlement.
All you can get from y ~ x is an intercept (a column of ones) and a single
'contrast' column for 'x'.
If you use y ~ 0+x you can get two columns for 'x', but R does not give
you an option of what columns in the case: see the
Hi,
I'm trying to redefine the contrasts for a linear model.
With a 2 level factor, x, with levels A and B, a two level
factor outputs A and B - A from an lm fit, say
lm(y ~ x). I would like to set the contrasts so that
the coefficients output are -0.5 (A + B) and B - A,
but I can't get the sign
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