[R] Bivariate ReLU Distribution

2020-07-10 Thread Arun Kumar Saha via R-help
Hi,
I would rather have a Statistics related question hope experts here can provide 
some suggestions. I have posted this request in some other forum but failed to 
generate meaningful response
I am looking for some technical document on deriving the Distribution function 
for sum of 2 ReLU(푋)=max{0,푋} distributions i.e max{0,푋1} + max{0,푋2} where X1 
and X2 jointly follow some bivariate Nomal distribution.
There are few technical notes available for univariate ReLU distribution, 
however I failed to find any spec for bivariate/multivariate setup.
Any pointer on above subject will be highly helpful.
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Re: [R] Help with Regular expression

2018-01-05 Thread Arun Kumar Saha
Hi Bogaso,

I see your ultimate goal is to extract the Date-time part from your
expression, then below should help :

> as.POSIXlt(gsub("[^0-9a-zA-Z]", "", "\":\"03-JAN-2018 16:00:00\""),
format = "%d%b%Y%H%M%OS")

[1] "2018-01-03 16:00:00 GMT"
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
 LinkedIn: http://in.linkedin.com/in/ArunFRM
 Personal : http://WWW.ARUNSAHA.IN  <http://WWW.ARUNSAHA.IN>
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[R] multi variable analysis

2017-10-08 Thread Arun Kumar
Hi All,

I have been given three raw data csv files related to an advertising
campaign. Impressions, clicks and conversion. They asked me to suggest
optimisation suggestions using uni variable and multi variable analysis.
The clicks or conversions are spread across various parameters like device,
country, city, ad type, browser, postal code etc. How do I approach the
analysis and modelling in this regard.

Thanks,
Arun

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Re: [R] Model selection in nonstationary VAR

2013-02-23 Thread Arun Kumar Saha
which method in statistics is completely free from model misspecification?

Thanks and regards,
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
_

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Re: [R] Spammer radhi

2012-12-24 Thread Arun Kumar Saha
Can the Statisticians here develop some good Statistical tool to stop this
Spamming keeping Type-I error almost zero?

Thanks and regards,
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
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Re: [R] how calculate seasonal component cyclic component of time series?

2012-03-21 Thread Arun Kumar Saha
I think you need to pass a time series object, as given in the help
page of decompose() function. From your text file it seems that you
are passing a numerical vector which is missing the key element of
Index as date for any legitimate time series data.
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM

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[R] VECM simulation

2012-03-07 Thread Arun Kumar Saha
Dear Mamush, you do not need to change the VAR simulation codes to VECM!!!
What you really need to change is your VECM model itself to correaponding
VAR representation. I guess there is some function vec2var() or something
like that (or perhaps you can do it manually.) Once you are done you may
then use VAR simulation code.

HTH,

Thanks and regards,

_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
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Re: [R] How to generate heterosced​astic random numbers?

2011-07-08 Thread Arun Kumar Saha
Answering 
http://r.789695.n4.nabble.com/How-to-generate-heteroscedastic-random-numbers-td3654534.html

I think best way to generated Heterosced​astic data would be, first
fix or assume some DGP of your choice, you may assume any arbitrary
model parameter(s) provided forms of those parameters are legitimate
under the model space. Then just simulate numbers starting from the
Residual (the form of that residual you already must have assumed).
Given some initial value(s), you would then get some Time series. You
may chose to generate a lengthy TS as most of the parameters
estimations happen within Asymptotic concept. Then apply your
statistical test on that realized series.

You may repeatedly generate TS and check the distribution of estimated
model parameters etc.

HTH
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM

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Re: [R] Running R from windows command prompt

2011-06-28 Thread Arun Kumar Saha
Hi Siddharth, many experts already answered your query, however I
would like to share how I run R in command prompt:

1. open command prompt
2. change working directory: cd C:\\R-2.13.0\bin\i386
(put the entire path here, however many people might find this step
weird, you can have better management setting window's path variable
appropriately)
3. type R.exe

You can use R within command prompt with same efficiency. However most
awkward thing I find in this process is you can never copy-paste any
code. So everything you need to type there manually!

HTH
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM

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Re: [R] strength of seasonal component

2011-05-12 Thread Arun Kumar Saha
Answering question as in 
http://r.789695.n4.nabble.com/strength-of-seasonal-component-td3517033.html
.

Possibly your code MASS::rlm(mean price ~ month) will result in Spurious
regression; you would make wrong impression of your estimated beta coef. as
in presence of the spurious regression they no longer have t-distribution.
May be you should try with MASS::rlm(diff(log(mean price)) ~ month)?

And secondly decomposition using standard approach is some sort of
**deterministic** act therefore, you would not get any measure of strength
in Statistical inference sense. To get that, I think above regression
approach would be handy.

HTH

Thanks and regards,
_

Arun Kumar Saha, FRM
QUANTITATIVE RISK AND HEDGE CONSULTING SPECIALIST
Visit me at: http://in.linkedin.com/in/ArunFRM
_

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[R] using lapply

2011-03-09 Thread Arun Kumar Saha
On reply to the post
http://r.789695.n4.nabble.com/using-lapply-td3345268.html

Dear Kushan, this may be a good start:

## assuming 'instr.list' is  your list object and you are applying
my.strat() function on each element of that list, you can use lapply
function as
lapply(instr.list, function(x) return(my.strat(x)))

Here resulting element will again be another list with length is same as the
length of your original list 'instr.list.'

Instead if the returned object for my.strat() function is a single number
then you might want to create a vector instead list, in that case just use
'sapply'

HTH

Arun,

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Re: [R] Help -normal distribution

2010-08-01 Thread Arun Kumar Saha
I gave you hint only, however whatever you done is correct.



On Mon, Aug 2, 2010 at 4:30 AM, n.a.s nn.r...@gmail.com wrote:
 Hi,

 But i didn't understand why you did it like this qnorm(40/200)??

 I already found the answer  qnorm(.20,181,7.3) and qnorm(.80,181,7.3)


 Thanks.


 On Mon, Aug 2, 2010 at 12:33 AM, Arun.stat arun.kumar.s...@gmail.com
 wrote:

 try
 qnorm(40/200)
 --
 View this message in context:
 http://r.789695.n4.nabble.com/Help-normal-distribution-tp2309730p2309735.html
 Sent from the R help mailing list archive at Nabble.com.

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[R] Best way to replace :SS with :00

2009-07-16 Thread Arun Kumar Saha
probably :
a - rep(HH:MM:SS, 5)
substring(a, 7) = 00
a


Not sure if there is an R way to do this or a regular express way, but here
is what I am trying to do.

I've got lots of data where the format is HH:MM:SS, but I need to format it
like HH:MM:00, i.e. round the second down to zero.

What is the best way to do this?

Thanks again.

Jason

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[R] Detecting Break

2009-07-05 Thread Arun Kumar Saha
Hi all,

say, I have following vector :

x - c(rep(5, 5), rep(3,4), rep(5,10))

Now I want to get the index numbers where elements of that vector changes
i.e. in above example I want to get a vector with elements : 6, 10. Because
at that indices, element of original vector changes value.

Is there any R function to do that?

Regards,

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[R] Calling R from .net environment

2009-05-13 Thread Arun Kumar Saha
Hi,  Currently I am a .net programmer and would like to use R for my
statistical computations engine. I already have installed RServer250.exe so
that I could call R from my .net programming environment, however
unfortunately, i could not be able to find RServer250.exe in the R-(D) COM
Interface region. If someone guide me how to add these COM components and
call the R-code through my application, it would be very good to me.
Regards,


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[R] Skewness

2008-09-12 Thread Arun Kumar Saha
I would like to know about the skewness function bundled in fBasics package.
Is it population measure of skewness or sample estimate?

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[R] Package for financial options calculation

2008-09-11 Thread Arun Kumar Saha
Hi all,

Is there any R package on European/American oprions pricing? And on
calculation of it's sensitivities?

Regards,

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Re: [R] Weighted variance function?

2008-07-27 Thread Arun Kumar Saha
ur prog gives following result:

weighted.var(c(1,-1), c(0.5,0.5))
[1] 2

is it ok?

On Thu, Jul 24, 2008 at 7:57 PM, Gavin Simpson [EMAIL PROTECTED]wrote:

 On Thu, 2008-07-24 at 02:25 +0530, Arun Kumar Saha wrote:
  There is a R function to calculate weighted mean : weighted.mean() under
  stats package. Is there any direct R function for calculating weighted
  variance as well?

 Here are two ways; weighted.var() is via the usual formula and
 weighted.var2() uses a running sums approach. The formulae for which are
 both on the weighted mean entry page on wikipedia for example.

 The removal of NA is as per weighted.mean, but I have not included any
 of the sanity checks that that functions contains.

 weighted.var - function(x, w, na.rm = FALSE) {
if (na.rm) {
w - w[i - !is.na(x)]
x - x[i]
}
sum.w - sum(w)
sum.w2 - sum(w^2)
mean.w - sum(x * w) / sum(w)
(sum.w / (sum.w^2 - sum.w2)) * sum(w * (x - mean.w)^2, na.rm =
 na.rm)
 }

 weighted.var2 - function(x, w, na.rm = FALSE) {
if (na.rm) {
w - w[i - !is.na(x)]
x - x[i]
}
sum.w - sum(w)
(sum(w*x^2) * sum.w - sum(w*x)^2) / (sum.w^2 - sum(w^2))
 }

 ## from example section in ?weighted.mean
 ## GPA from Siegel 1994
 wt - c(5,  5,  4,  1)/15
 x - c(3.7,3.3,3.5,2.8)
 weighted.mean(x,wt)

 weighted.var(x, wt)

 weighted.var2(x, wt)

 And some timings:

  system.time(replicate(10, weighted.var(x, wt)))
   user  system elapsed
  2.679   0.014   2.820
  system.time(replicate(10, weighted.var2(x, wt)))
   user  system elapsed
  2.224   0.010   2.315

 HTH

 G
 --
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
  Dr. Gavin Simpson [t] +44 (0)20 7679 0522
  ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
  Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
  Gower Street, London  [w] 
 http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/
  UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%



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and provide commented, minimal, self-contained, reproducible code.


Re: [R] Weighted variance function?

2008-07-27 Thread Arun Kumar Saha
I feel there is something wrong in definition in weighted variance.

Say for a fair dice the variance of outcome should be : 2.92 (
http://en.wikipedia.org/wiki/Variance)

However if I use cov.wt() or weighted.var() by Gavin, I get :

cov.wt(as.data.frame(1:6), rep(0.6, 6))
$cov
1:6
1:6 3.5

$center
1:6
3.5

$n.obs
[1] 6

$wt
[1] 0.167 0.167 0.167 0.167 0.167 0.167

i.e. 3.5

Therefore if I want to calculate Variance for a r.v. with different prob for
different values then should not use those formulae. Is it the case?

On Sun, Jul 27, 2008 at 2:30 PM, Patrick Burns [EMAIL PROTECTED]wrote:

 Have you seen 'cov.wt'?


 Patrick Burns
 [EMAIL PROTECTED]
 +44 (0)20 8525 0696
 http://www.burns-stat.com
 (home of S Poetry and A Guide for the Unwilling S User)

 Arun Kumar Saha wrote:

 ur prog gives following result:

 weighted.var(c(1,-1), c(0.5,0.5))
 [1] 2

 is it ok?

 On Thu, Jul 24, 2008 at 7:57 PM, Gavin Simpson [EMAIL PROTECTED]
 wrote:



 On Thu, 2008-07-24 at 02:25 +0530, Arun Kumar Saha wrote:


 There is a R function to calculate weighted mean : weighted.mean() under
 stats package. Is there any direct R function for calculating weighted
 variance as well?


 Here are two ways; weighted.var() is via the usual formula and
 weighted.var2() uses a running sums approach. The formulae for which are
 both on the weighted mean entry page on wikipedia for example.

 The removal of NA is as per weighted.mean, but I have not included any
 of the sanity checks that that functions contains.

 weighted.var - function(x, w, na.rm = FALSE) {
   if (na.rm) {
   w - w[i - !is.na(x)]
   x - x[i]
   }
   sum.w - sum(w)
   sum.w2 - sum(w^2)
   mean.w - sum(x * w) / sum(w)
   (sum.w / (sum.w^2 - sum.w2)) * sum(w * (x - mean.w)^2, na.rm =
 na.rm)
 }

 weighted.var2 - function(x, w, na.rm = FALSE) {
   if (na.rm) {
   w - w[i - !is.na(x)]
   x - x[i]
   }
   sum.w - sum(w)
   (sum(w*x^2) * sum.w - sum(w*x)^2) / (sum.w^2 - sum(w^2))
 }

 ## from example section in ?weighted.mean
 ## GPA from Siegel 1994
 wt - c(5,  5,  4,  1)/15
 x - c(3.7,3.3,3.5,2.8)
 weighted.mean(x,wt)

 weighted.var(x, wt)

 weighted.var2(x, wt)

 And some timings:



 system.time(replicate(10, weighted.var(x, wt)))


  user  system elapsed
  2.679   0.014   2.820


 system.time(replicate(10, weighted.var2(x, wt)))


  user  system elapsed
  2.224   0.010   2.315

 HTH

 G
 --
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
  Dr. Gavin Simpson [t] +44 (0)20 7679 0522
  ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
  Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
  Gower Street, London  [w] 
 http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/
 http://www.ucl.ac.uk/%7Eucfagls/
  UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%





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 __
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 https://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.







--

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and provide commented, minimal, self-contained, reproducible code.


Re: [R] Weighted variance function?

2008-07-27 Thread Arun Kumar Saha
oh now I got, R is taking unbiased estimator, anyways thanks

On Sun, Jul 27, 2008 at 2:54 PM, Arun Kumar Saha
[EMAIL PROTECTED]wrote:

 I feel there is something wrong in definition in weighted variance.

 Say for a fair dice the variance of outcome should be : 2.92 (
 http://en.wikipedia.org/wiki/Variance)

 However if I use cov.wt() or weighted.var() by Gavin, I get :

 cov.wt(as.data.frame(1:6), rep(0.6, 6))
 $cov
 1:6
 1:6 3.5

 $center
 1:6
 3.5

 $n.obs
 [1] 6

 $wt
 [1] 0.167 0.167 0.167 0.167 0.167 0.167

 i.e. 3.5

 Therefore if I want to calculate Variance for a r.v. with different prob
 for different values then should not use those formulae. Is it the case?


 On Sun, Jul 27, 2008 at 2:30 PM, Patrick Burns [EMAIL PROTECTED]wrote:

 Have you seen 'cov.wt'?


 Patrick Burns
 [EMAIL PROTECTED]
 +44 (0)20 8525 0696
 http://www.burns-stat.com
 (home of S Poetry and A Guide for the Unwilling S User)

 Arun Kumar Saha wrote:

 ur prog gives following result:

 weighted.var(c(1,-1), c(0.5,0.5))
 [1] 2

 is it ok?

 On Thu, Jul 24, 2008 at 7:57 PM, Gavin Simpson [EMAIL PROTECTED]
 wrote:



 On Thu, 2008-07-24 at 02:25 +0530, Arun Kumar Saha wrote:


 There is a R function to calculate weighted mean : weighted.mean()
 under
 stats package. Is there any direct R function for calculating weighted
 variance as well?


 Here are two ways; weighted.var() is via the usual formula and
 weighted.var2() uses a running sums approach. The formulae for which are
 both on the weighted mean entry page on wikipedia for example.

 The removal of NA is as per weighted.mean, but I have not included any
 of the sanity checks that that functions contains.

 weighted.var - function(x, w, na.rm = FALSE) {
   if (na.rm) {
   w - w[i - !is.na(x)]
   x - x[i]
   }
   sum.w - sum(w)
   sum.w2 - sum(w^2)
   mean.w - sum(x * w) / sum(w)
   (sum.w / (sum.w^2 - sum.w2)) * sum(w * (x - mean.w)^2, na.rm =
 na.rm)
 }

 weighted.var2 - function(x, w, na.rm = FALSE) {
   if (na.rm) {
   w - w[i - !is.na(x)]
   x - x[i]
   }
   sum.w - sum(w)
   (sum(w*x^2) * sum.w - sum(w*x)^2) / (sum.w^2 - sum(w^2))
 }

 ## from example section in ?weighted.mean
 ## GPA from Siegel 1994
 wt - c(5,  5,  4,  1)/15
 x - c(3.7,3.3,3.5,2.8)
 weighted.mean(x,wt)

 weighted.var(x, wt)

 weighted.var2(x, wt)

 And some timings:



 system.time(replicate(10, weighted.var(x, wt)))


  user  system elapsed
  2.679   0.014   2.820


 system.time(replicate(10, weighted.var2(x, wt)))


  user  system elapsed
  2.224   0.010   2.315

 HTH

 G
 --
 %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
  Dr. Gavin Simpson [t] +44 (0)20 7679 0522
  ECRC, UCL Geography,  [f] +44 (0)20 7679 0565
  Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
  Gower Street, London  [w] 
 http://www.ucl.ac.uk/~ucfagls/http://www.ucl.ac.uk/%7Eucfagls/
 http://www.ucl.ac.uk/%7Eucfagls/
  UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
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[R] Weighted variance function?

2008-07-23 Thread Arun Kumar Saha
There is a R function to calculate weighted mean : weighted.mean() under
stats package. Is there any direct R function for calculating weighted
variance as well?

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[R] looking for alternative of 'if-else'

2008-07-06 Thread Arun Kumar Saha
There is if-else loop if I have to choose 1 item from a 2-item list.
However if I have a list of 4 items (let say) then how i can choose a single
item without employing 'if-else' loop? I mean in VBA I can use
select-case, is there any equivalent in R as well?

Regards,

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Re: [R] looking for alternative of 'if-else'

2008-07-06 Thread Arun Kumar Saha
I am not sure whether I could understand all things. Here is my code :

library(zoo)
# an reproducible example
SD - 1
date1 - seq(as.Date(2001-01-01), as.Date(2002-12-1), by = day)
len1 - length(date1)
set.seed(1) # to make it reproducible
data1 - zoo(rnorm(len1), date1)
plot(data1)

# calculation for monthly or quarterly mean
calc = function(frequ = c(M S, Q S))
  {
   switch(frequ
   M S = aggregate(data1, as.yearmon, mean)
   Q S = aggregate(data1, as.yearqtr, mean))
  }

However I am getting lot of errors while executing. Any further suggestion?

Regards,



On Sun, Jul 6, 2008 at 3:38 PM, [EMAIL PROTECTED] wrote:

 Why don't you try switch Let me assume that you want to calculate a=3 and
 b =5 by using one of MEAN, SUM,MIN, and MAX functions. Then you could
 write your code: (If you want to calculate MEAN)

  example-function(fun=c(MEAN, SUM, MIN, MAX)){

 +   fun-match.arg(fun);a=3;b=5
 +   switch(  fun,
 +MEAN=(a+b)/2,
 +SUM=a+b,
 +MIN=min(a,b),
 +MAX=max(a,b))
 +}

example(MEAN)

 [1] 4


  Hope this is helpful,
 Chunhao Tu






 Quoting Arun Kumar Saha [EMAIL PROTECTED]:

  There is if-else loop if I have to choose 1 item from a 2-item list.
 However if I have a list of 4 items (let say) then how i can choose a
 single
 item without employing 'if-else' loop? I mean in VBA I can use
 select-case, is there any equivalent in R as well?

 Regards,

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 http://www.R-project.org/posting-guide.html
 and provide commented, minimal, self-contained, reproducible code.






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[R] Converting variance covariance matrix to correlation matrix

2008-05-19 Thread Arun Kumar Saha
Suppose I have a Variance-covariance matrix A. Is there any fast way to
calculate correlation matrix from 'A' and vice-versa without emplying any
'for' loop?

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[R] Simulation study in R

2008-04-29 Thread Arun Kumar Saha
Here I am in a simulation study where I want to find different values
of x and y such that f(x,y)=c (some known constant) w.r.t. x, y 0,
y=x and x=c1 (another known constant). Can anyone please tell me how
to do it efficiently in R. One way I thought that I will draw
different random numbers from uniform dist according to that
constraints and pick those which satisfy f(x,y)=c. However it is not I
think computationally efficient. Can anyone here suggest me any other
efficient approach?

Regards,

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Re: [R] Simulation study in R

2008-04-29 Thread Arun Kumar Saha
x, y are cont. variable, and f also have to be cont.. And your second
suggestion is correct of course, it actually should be |f(x,y) - c| 
epsilon

Thanks

On Tue, Apr 29, 2008 at 12:34 PM, Moshe Olshansky [EMAIL PROTECTED] wrote:
 Are the pairs (x,y) belong to some lattice or can
 change continuously?
 Does f assume some discrete values (or is constant on
 sets of positive measure)? If not then it will be hard
 to randomly select x and y which satisfy the exact
 equality (this still can happen since there are
 finitely many computer numbers, but their number is
 quite large!). So if f change continuously you may
 need the condition |f(x,y) - c|  epsilon for some
 epsilon  0.

 Regards,

 Moshe.


 --- Arun Kumar Saha [EMAIL PROTECTED] wrote:

  Here I am in a simulation study where I want to find
  different values
  of x and y such that f(x,y)=c (some known constant)
  w.r.t. x, y 0,
  y=x and x=c1 (another known constant). Can anyone
  please tell me how
  to do it efficiently in R. One way I thought that I
  will draw
  different random numbers from uniform dist according
  to that
  constraints and pick those which satisfy f(x,y)=c.
  However it is not I
  think computationally efficient. Can anyone here
  suggest me any other
  efficient approach?
 
  Regards,
 
  __
  R-help@r-project.org mailing list
  https://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.
 





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[R] Writing list object to a file

2008-04-23 Thread Arun Kumar Saha
Hi all,

I am wondering how to write a 'list' object to a file. I already gone
through some threads like
http://mail.python.org/pipermail/python-list/2001-April/080639.html, however
could not trace out any reliable solution. I tried following :

 write.table(calc, file=c:/data.csv)
Error in data.frame(200501 = c(-0.000387071806652095,
-0.000387221689252648,  :
  arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27, 19,
23, 20, 11
Anybody can help?

Regards,

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Re: [R] Writing list object to a file

2008-04-23 Thread Arun Kumar Saha
Previous question that I asked was originated from this problem :

suppose I have following time series :

library(zoo)
date1 = seq(as.Date(01/01/01, format = %m/%d/%y), as.Date(12/31/08,
format = %m/%d/%y), by = 1)
len1 = length(date1); data1 = zoo(matrix(rnorm(len1, mean=0, sd=0.5), nrow =
len1),  date1)
Now I group those time series observation month wise :
data2 = split(as.data.frame(data1), format(index(data1), %Y%m)

Now I want to perform  Levene test (
http://en.wikipedia.org/wiki/Levene's_test) to test whether the variance for
each group (here month) is significantly different or not.

Can anyone suggest me any easiest way how to construct the test statistic,
described above?

Thanks




On Wed, Apr 23, 2008 at 12:21 PM, Arun Kumar Saha [EMAIL PROTECTED]
wrote:

 Hi all,

 I am wondering how to write a 'list' object to a file. I already gone
 through some threads like
 http://mail.python.org/pipermail/python-list/2001-April/080639.html,
 however could not trace out any reliable solution. I tried following :

  write.table(calc, file=c:/data.csv)
 Error in data.frame(200501 = c(-0.000387071806652095,
 -0.000387221689252648,  :
   arguments imply differing number of rows: 25, 24, 16, 17, 18, 26, 27,
 19, 23, 20, 11
 Anybody can help?

 Regards,




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[R] action executing twice while loading tiles

2008-02-26 Thread arun kumar
*hi

I aam facing following problem

my action executing twice when i try to load the tileslayout page as
actionforward success
*

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[R] Loading user defined functions autometically each time I start R

2008-02-26 Thread Arun Kumar Saha
Hi all,

I wrote some user defined function for my own. Now I want to get a mechanism
so that every time I start R, those function will automatically be loaded in
R without manually copying pasting. Can gurus here pls tell me how to do
that? Or I have to build my own packages bundled with those functions.
However I am not familiar in writing R package yet.

Regards,

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Re: [R] Loading user defined functions autometically each time I start R

2008-02-26 Thread Arun Kumar Saha
I am using R 2.6.2. under windows

On Wed, Feb 27, 2008 at 12:43 PM, [EMAIL PROTECTED] wrote:

 From: Arun Kumar Saha [EMAIL PROTECTED]
 Date: 2008/02/27 Wed AM 01:03:26 CST
 To: [EMAIL PROTECTED] [EMAIL PROTECTED]
 Subject: [R] Loading user defined functions autometically each time I
 start R

 i'm on linux so if you are not it's
 probably slightly different on windows
 but not much. I just put my .Rprofile
 file ( you need the . in front but I had to take it
 off when I sent you ) in my /opt/mark ( hiome
 dorectory on linux ) directory.

 generally speaking, when you ask questions
 like that, be sure to specify your OS or
 people ( not me ) will get annoyed.

 i you start up R, the .Rprofile should
 get called and the functions in your
 arun.R file should be available to you.
 good luck.


 I';m no expert so hopefully someone else
 will reply also ( that's why i emailed
 privately ).



 Hi all,
 
 I wrote some user defined function for my own. Now I want to get a
 mechanism
 so that every time I start R, those function will automatically be loaded
 in
 R without manually copying pasting. Can gurus here pls tell me how to do
 that? Or I have to build my own packages bundled with those functions.
 However I am not familiar in writing R package yet.
 
 Regards,
 
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-- 
--
Arun Kumar Saha, M.Sc.
Statistical Arbitrage Quant
RISK  MANAGEMENT  DIVISION
Transgraph Consulting [www.transgraph.com]
Hyderabad, INDIA
Contact #  Home: (91-033) 25558038 [CALCUTTA]
   Home: (91-040) 40033010 [HYDERABAD]
   Office: (91-040) 30685012 Ext. 17 [email preferred]
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[R] Fetching datapoints from a Time Series

2007-10-24 Thread Arun Kumar Saha
Hi all,

I have a time series like that :

DateValue

01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166
01/10/05 -0.004174282
01/11/05 0.01027384
01/12/05 -0.006099558
01/13/05 -0.009811054
01/14/05 0.000657073


I import this dataset as a zoo object. Now I want to fetch the dataset
between two particular dates. For example if I want to get data point
between 01/02/05 and 01/09/05 I should get :

01/03/05 -0.008471364
01/04/05 -0.008153802
01/05/05 -0.000780031
01/06/05 -0.000130064
01/07/05 -0.000650576
01/08/05 -0.000130166


Can anyone please tell me how to do that? Is there any function to do that?

Regards,

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Re: [R] Need help on date

2007-09-20 Thread Arun Kumar Saha
This mail is continuation of my previous one. i have some raw data from
Excel which was carried to R:

data = read.delim(file=clipboard, header=T)
 data
   Date Price
1  09/01/05   365
2  09/02/05   360
3  09/03/05   360
4  09/05/05   370
5  09/06/05   370
6  09/08/05   365
7  09/09/05   365
8  09/10/05   365
9  09/12/05   365
10 09/13/05   360
11 09/14/05   360
12 09/15/05   360

and, using the input from R-help I exctracted day, month, and year, from
Date

year = as.numeric(format(as.Date(data[,1],%m/%d/%y),%Y))
month = as.numeric(format(as.Date(data[,1],%m/%d/%y),%m))
day = as.numeric(format(as.Date(data[,1],%m/%d/%y),%d))

Now I want to create a date-variable and put it in actual data:

library(date)
data1 = cbind(mdy.date(month, day, year), data[,-1])
 data1
   [,1] [,2]
 [1,] 16680  365
 [2,] 16681  360
 [3,] 16682  360
 [4,] 16684  370
 [5,] 16685  370
 [6,] 16687  365
 [7,] 16688  365
 [8,] 16689  365
 [9,] 16691  365
[10,] 16692  360
[11,] 16693  360
[12,] 16694  360

However this is not that thing what I wanted, first column has been jumbled,
it is not in actual date format.

Can anyone tell me what should i do here?

Regards,






On 9/18/07, Arun Kumar Saha [EMAIL PROTECTED] wrote:

 Dear all,

 I have a variable 'x' like that:

  x
 [1] 2005-09-01

 Here, 2005 represents year, 09 month and 01 day.

 Now I want to create three variables naming: y, m, and d such that:

 y = 2005
 m = 09
 d = 01

 can anyone tell me how to do that?

 Regards,



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