Re: [R] lmom package - Resending the email
Dear Dalgaard sir, Thanks a lot for detailed clarification. It indeed is very enlightening and will be very useful for me in future. And your suggestion is well taken. Thanks again. Regards Katherine On Thu, 4/12/14, peter dalgaard wrote: Subject: Re: [R] lmom package - Resending the email To: "Simon Zehnder" Date: Thursday, 4 December, 2014, 2:04 PM lmom is based on L-moments, which are different from ordinary moments, except for the 1st one. It would be truly miraculous if it gave the same result as the ordinary method of moments or maximum likelihood. Estimates of any distributional parameter requires that the model actually fits the data, and in your case a qqnorm(amounts) shows that they are certainly not normal. In such cases, the L-moment estimator of the std.dev. is not necessarily an estimate of the std.dev. of the actual distribution. A lognormal distribution seems to fit the data better. However, the L-moments suggest a value for zeta (the lower bound) of 3226 which is well inside the range of the actual data. In fact there are 16 observations that are less than 3226. Maximum likelihood would never do that, but the same sort of effect is well-known for the ordinary method of moments. In short, you need to study the theory before you appply its results. - Peter D. On 03 Dec 2014, at 10:57 , Simon Zehnder wrote: > Katherine, > > for a deeper understanding of differing values it makes sense to provide the list at least with an online description of the corresponding functions used in Minitab and SPSS… > > Best > Simon > On 03 Dec 2014, at 10:45, Katherine Gobin via R-help wrote: > >> Dear R forum >> I sincerely apologize as my earlier mail with the captioned subject, since all the values got mixed up and the email is not readable. I am trying to write it again. >> My problem is I have a set of data and I am trying to fit some distributions to it. As a part of this exercise, I need to find out the parameter values of various distributions e.g. Normal distribution, Log normal distribution etc. I am using lmom package to do the same, however the parameter values obtained using lmom pacakge differ to a large extent from the parameter values obtained using say MINITAB and SPSS as given below - >> _ >> >> amounts = c(38572.5599129508,11426.6705314315,21974.1571641187,118530.32782443,3735.43055996748,66309.5211176106,72039.2934132668,21934.8841708626,78564.9136114375,1703.65825161293,2116.89180930203,11003.495671332,19486.3296339113,1871.35861218795,6887.53851253407,148900.978055447,7078.56497101651,79348.1239806592,20157.6241066905,1259.99802108593,3934.45912233674,3297.69946631591,56221.1154121067,13322.0705174134,45110.2498756567,31910.3686613912,3196.71168501252,32843.0140437202,14615.1499458453,13013.9915051561,116104.176753387,7229.03056392023,9833.37962177814,2882.63239493673,165457.372543821,41114.066453219,47188.1677766245,25708.5883755617,82703.7378298092,8845.04197017415,844.28834047836,35410.8486123933,19446.3808445684,17662.2398792892,11882.8497070776,4277181.17817307,30239.0371267968,45165.7512343364,22102.8513746687,5988.69296597127,51345.0146170238,1275658.35495898,15260.4892854214,8861.76578480635,37647.1638704867,4979.53544046949,7012.48134772332 ,3385.20612391205,1911.03114395959,66886.5036605189,2223.47536156462,814.947809578378,234.028589468841,5397.4347625133,13346.3226579065,28809.3901352898,6387.69226236731,5639.42730553242,2011100.92675507,4150.63707173462,34098.7514446498,3437.10672573502,289710.315303182,8664.66947305203,13813.3867161134,208817.521491857,169317.624400274,9966.78447705792,37811.1721605562,2263.19211279927,80434.5581206454,19057.8093104899,24664.5067589624,25136.5042354789,3582.85741610706,6683.13898432794,65423.9991390846,134848.302304064,3018.55371579808,546249.641168158,172926.689143006,3074.15064180208,1521.70624812788,59012.4248281661,21226.928522236,17572.5682970983,226.646947337851,56232.2982652019,14641.0043361533,6997.94414914865) >> >> library(lmom) >> lmom = samlmu(amounts) >> # __ >> # Normal Distribution parameters >> parameters_of_NOR <- pelnor(lmom); parameters_of_NOR >> >> mu sigma 115148.4 175945.8 >> Location Scale Minitab 115148.4 485173SPSS 115148.4 485173 >> # __ >> # Log Normal (3 Parameter) Distribution parameters >> zeta mu sigma 3225.798890 9.114879 2.240841 >> Location Scale
[R] VGAM package : Frechet distribution - 2 parameter estimation
Dear R forum, I am trying to execute following code (Page no 259 - VGAM.pdf) # . library(VGAM) set.seed(123) fdata <- data.frame(y1 = rfrechet(nn <- 1000, shape = 2 + exp(1))) with(fdata, hist(y1)) fit2 <- vglm(y1 ~ 1, frechet, data = fdata, trace = TRUE) # . However, I receive following error Error in vglm(y1 ~ 1, frechet, data = fdata, trace = TRUE) : object 'frechet' not found Earlier there used to be a function called "frechet3" which I guess has been withdrawn by VGAM. Kindly guide Katherine [[alternative HTML version deleted]] __ 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.
Re: [R] VGAM package : Frechet distribution - 2 parameter estimation
Dear Mr Michael, Thanks a lot for your guidance. The pdf file describing VGAM package has mentioned 'frechet' in the example, so I got the error. Regards Katherine On Thursday, 6 November 2014 2:54 PM, Michael Dewey wrote: On 06/11/2014 06:04, Katherine Gobin wrote: > Dear R forum, > > I am trying to execute following code (Page no 259 - VGAM.pdf) > > # > . > > library(VGAM) > > set.seed(123) > fdata <- data.frame(y1 = rfrechet(nn <- 1000, shape = 2 + exp(1))) > with(fdata, hist(y1)) > fit2 <- vglm(y1 ~ 1, frechet, data = fdata, trace = TRUE) > > # > . > Is it not called frechet2? > > However, I receive following error > > Error in vglm(y1 ~ 1, frechet, data = fdata, trace = TRUE) : >object 'frechet' not found > > > Earlier there used to be a function called "frechet3" which I guess has been > withdrawn by VGAM. > > Kindly guide > > Katherine > [[alternative HTML version deleted]] > > __ > 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. > > > - > No virus found in this message. > Checked by AVG - www.avg.com > Version: 2015.0.5557 / Virus Database: 4189/8518 - Release Date: 11/05/14 > > -- Michael http://www.dewey.myzen.co.uk [[alternative HTML version deleted]] __ 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.
[R] Double return statement
Dear R forum, I have a function which generates say two outputs, say output_1 and output_2. Output_1 is a single row output whereas Output_2 is a dataframe having multiple records. Is it possible to use two return statements in function. Output_2 uses some records from output_1, hence I need to have these outputs generated from the same function. Thanking in advance With warm regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Display of data points in the Scatterplot
Respected R forum I am learning R and relatively quite new to R. I am generating a scatter-plot as given below. (My actual table is much larger). # Sample data frame y = c(20, 23, 17, 31, 68) x = c(200, 300, 400, 500, 600) plot(x, y, type = 'l') If I plot this scatter-plot in excel, the data values are displayed if I place the cursor at some desired place of the graph. E.g. if I place the cursor say at the point (400, 31), then the value (400, 31) is displayed. My question is (A) once I plot a graph in R, is it possible to display a particular (x, y) co-ordinate by placing the cursor there? (B) Suppose I have 100 pairs of (x, y ). then is it possible to display in the graph (irrespective of the curosr position) the values of (x, y) corresponding to say 10th, 20th, 30th, 40th etc. observations in the graph. Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] How to count the nos. in a range?
Dear R forum I have a following vector of random no.s x = runif(100, 0.01, 0.99) [1] 0.47212037 0.77867992 0.33947474 0.93369035 [5] 0.03720073 0.79307831 0.81801835 0.92710688 . I need to count the random no. falling in the range (0 - 0.10), (0.10 - 0.20), (0.20 - 0.30)..upto (0.90 - 1) Thus, I need to have a data frame as range frequency 0 - 0.10 ... 0.10 - 0.20 ... .. 0.90 - 1 . I understand I need to write my code and ask for some help if the need be. But I am simply clueless at the moment. Kindly guide. Katherine [[alternative HTML version deleted]] __ 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.
[R] Can data.frame be saved as image?
Dear R forum
I have one stupid question, but I have no other solution to it in sight?
Suppose some R process creates graphs etc alongwith main output as data.frame
e.g
output1 = data.frame(bands = c("A", "B", "C"), results = c(74, 108, 65))
I normally save this output as some csv file.
But I need to save this output as some image (I understand this is weird, but I
need to find out some way to do so) e.g. for graph, I use 'png' as
png("histogram.png", width=480,height=480)
.
..
dev.off()
Please advise.
Regards
Katherine
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Re: [R] Can data.frame be saved as image?
Dear Sir,
Thanks a lot for your suggestion. In the meantime I came across
http://stackoverflow.com/questions/10587621/how-to-print-to-paper-a-nicely-formatted-data-frame
and got to know about the package "gridExtra"
So I used following code
png(filename = "output1.png", width=480,height=480)
grid.table(output1)
dev.off()
And that solved it.
Thanks again Sir for suggesting other two pacakages 'lattice' or 'ggplot2' as
definitely I will like to decipher these two.
I understand before posting the mail to the forum, I should have tried old
mails etc, but I was bit desperate to know the solution and somehow I felt it's
a stupid thing to do so. I will remember it next time.
Regards
Katherine
--- On Fri, 21/12/12, jim holtman wrote:
From: jim holtman
Subject: Re: [R] Can data.frame be saved as image?
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Friday, 21 December, 2012, 2:39 PM
do you want to save the dataframe used in the plot and then the plot
itself? If so consider using 'lattice' or 'ggplot2' which create an
object for "print" and this would allow you to use 'save' to save both
objects in a file.
If you want to generate the 'png' file, the you would have to 'save'
the dataframe and then 'zip' the .RData and png file into a new file.
So what is it that you intend to do with the data that is saved in the
common file?
On Fri, Dec 21, 2012 at 8:59 AM, Katherine Gobin
wrote:
> Dear R forum
>
> I have one stupid question, but I have no other solution to it in sight?
>
> Suppose some R process creates graphs etc alongwith main output as data.frame
> e.g
>
> output1 = data.frame(bands = c("A", "B", "C"), results = c(74, 108, 65))
>
> I normally save this output as some csv file.
>
> But I need to save this output as some image (I understand this is weird, but
> I need to find out some way to do so) e.g. for graph, I use 'png' as
>
> png("histogram.png", width=480,height=480)
>
> .
>
> ..
>
> dev.off()
>
> Please advise.
>
> Regards
>
> Katherine
>
>
> [[alternative HTML version deleted]]
>
>
> __
> 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.
>
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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[R] lmomco package - Random number generation using Wakeby distribution
Dear R forum
>From the given data, I have estimated the parameters of Wakeby distribution
>using lmomco package as
library(lmomco)
(amounts <- read.csv("input_S.csv")$amount)
# ___
# Wakeby distribution - Parameter estimation
N =
length(amounts)
lmr = lmom.ub(amounts)
parameters_of_Wakeby = parwak(lmr)
> parameters_of_Wakeby
$type
[1]
"wak"
$para
xi alpha
1.18813927666405e+04 0.00e+00
beta gamma
0.00e+00 8.11391042554567e+04
delta
9.57554297149062e-01
This means the scale parameters are 0.
However, assuming, all the five parameters of Wakeby distribution (viz.
location parameter m (xi), the scale parameters a, b, and shape parameters g
and d are available.
Then, how do I generate say 100 random no.s using Wakeby distribution w.r.t.
these
5 available parameters.
I couldn't find any information about this in lmomco. Kindly guide if random
no.s can be generated or not and if yes, how it can be done in r.
Thanking in advance
Katherine
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Re: [R] lmomco package - Random number generation using Wakeby distribution
Dear Sir,
Thanks a lot for your eye-opener reply. I was just thinking of our usual
commands like rnorm, runif etc. So I was wondering if there exists something
like rwakeby etc.
And lastly, I have calculated the parameters using
> lmr = lmom.ub(amounts)
> parameters_of_Wakeby = parwak(lmr)
whereas you have mentioned lmom2par(), Will it create different set of
parameters? Actually I am travelling and don't have R installed on the laptop I
am carrying with me to verify ther results.
Regards
Katherine
--- On Mon, 21/1/13, David Winsemius wrote:
From: David Winsemius
Subject: Re: [R] lmomco package - Random number generation using Wakeby
distribution
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Monday, 21 January, 2013, 7:46 PM
On Jan 21, 2013, at 10:30 AM, Katherine Gobin wrote:
> Dear R forum
>
>> From the given data, I have estimated the parameters of Wakeby distribution
>> using lmomco package as
>
> library(lmomco)
>
> (amounts <- read.csv("input_S.csv")$amount)
>
> # ___
>
> # Wakeby distribution - Parameter estimation
>
> N =
> length(amounts)
> lmr = lmom.ub(amounts)
> parameters_of_Wakeby = parwak(lmr)
It appears you have a) not included the code that produced that output and b)
failed to read the Index page for that package
help(package="lmomco")
help(package="lmomco")
?rlmomco # Random Deviates of a Distribution
So on the assumption that you have an object in your workspace named
"parameters_of_Wakeby" and it is an lmomco produced object like that returned
by lmom2par() I would try:
rlmomco(100, parameters_of_Wakeby)
>
>> parameters_of_Wakeby
>
> $type
> [1]
> "wak"
>
> $para
> xi alpha
> 1.18813927666405e+04 0.00e+00
> beta gamma
> 0.00e+00 8.11391042554567e+04
> delta
> 9.57554297149062e-01
>
> This means the scale parameters are 0.
>
> However, assuming, all the five parameters of Wakeby distribution (viz.
> location parameter m (xi), the scale parameters a, b, and shape parameters g
> and d are available.
>
> Then, how do I generate say 100 random no.s using Wakeby distribution w.r.t.
> these
> 5 available parameters.
>
> I couldn't find any information about this in lmomco. Kindly guide if random
> no.s can be generated or not and if yes, how it can be done in r.
You should have been able to find this with:
help.search("random", package="lmomco")
--
David Winsemius
Alameda, CA, USA
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[R] How to extract values of results in gamlss.tr
Dear R helpers,
I have following loss data and I need to fit LEFT truncated Log Normal
distribution to this data which is Truncated at 100.
dat =
c(1333834,5710254,9987567,7809469,6940935,3473671,1270209,1102523,1124002,
5830159,4302300,3925242,2638409,2324421,7238436,9088709,7439250,4976551,4864319,
8741334,1863770,7098310,4942288,4971829,4986372)
library(gamlss.tr)
gen.trun(5, LOGNO)
result <- gamlss(dat~1, family=LOGNOtr)
# THIS GIVES
> result
Family: c("LOGNOtr", "left truncated Log Normal")
Fitting method: RS()
Call: gamlss(formula = dat ~ 1, family = LOGNOtr)
Mu Coefficients:
(Intercept)
15.23
Sigma Coefficients:
(Intercept)
-0.3977
Degrees of Freedom for the fit: 2 Residual Deg. of Freedom 23
Global Deviance: 812.568
AIC: 816.568
SBC: 819.006
My problem is how do I extract these values of Mu Coefficients and Sigma
Coefficients, if I want to use these values for further analyses?
Kindly guide
Katherine Gobin
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[R] Counting various elemnts in a vactor
Dear R forum
I have a vector say as given below
df = c("F", "C", "F", "B", "D", "A", "D", "D", "A", "F", "D", "F", "B", "C")
I need to find
(1) how many times each element occurs? e.g. in above vector F occurs 4 times,
C occurs 2 times etc.
(2) Depending on the number of occurrences, I need to repeat the element 100
times of the occurrences e.g. I need to repeat F 6 * 100 = 600 times, C 2*100 =
200 times.
I can manage the second part i.e. repeating but I am not able to count the
number of times the element is appearing in a given vector.
Kindly guide
Katherine
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Counting various elemnts in a vactor
Dear Sir,
Thanks a lot for your great help. I couldn't have figured it out.
Thanks again.
Regards
Katherine
--- On Tue, 26/3/13, D. Rizopoulos wrote:
From: D. Rizopoulos
Subject: Re: [R] Counting various elemnts in a vactor
To: "Katherine Gobin"
Cc: "r-help@r-project.org"
Date: Tuesday, 26 March, 2013, 8:23 AM
try this:
df <- c("F", "C", "F", "B", "D", "A", "D", "D", "A", "F", "D", "F", "B",
"C")
tab <- table(df)
tab
rep(names(tab), 100 * tab)
I hope it helps.
Best,
Dimitris
On 3/26/2013 9:12 AM, Katherine Gobin wrote:
> Dear R forum
>
> I have a vector say as given below
>
> df = c("F", "C", "F", "B", "D", "A", "D", "D", "A", "F", "D", "F", "B",
> "C")
>
> I need to find
>
> (1) how many times each element occurs? e.g. in above vector F occurs 4
> times, C occurs 2 times etc.
>
> (2) Depending on the number of occurrences, I need to repeat the element 100
> times of the occurrences e.g. I need to repeat F 6 * 100 = 600 times, C 2*100
> = 200 times.
>
> I can manage the second part i.e. repeating but I am not able to count the
> number of times the element is appearing in a given vector.
>
> Kindly guide
>
> Katherine
>
>
>
>
>
>
>
>
>
>
>
> [[alternative HTML version deleted]]
>
>
>
> __
> 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.
>
--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
Web: http://www.erasmusmc.nl/biostatistiek/
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[R] Archieve of mails from R forum
Dear R helpers, Everyday I do receive many many mails from R forum and after some period of times, INBOX is filled with numerous mails. At times if for some period of time, I haven't accessed mails, it becomes difficult to keep track of mails and many times simply due to the volume (and owing to the lack of time due to office constraints), I have to simply delete the mails without opening them and I understand this is a huge loss. If in case I wish to refer to all the old emails that have been appeared in the R forum, where do I get these? Is there any list where I will get subject-wise of thread-wise archive of old emails? I understand that will be an ocean of quality information and one can learn a lot from these old mails and I don't need to keep track of my emails all the time. Kindly guide. Regards Katherine [[alternative HTML version deleted]] __ 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.
Re: [R] Archieve of mails from R forum
Dear Sir, Thanks a lot for the input. I am sure it will go a long way for me to understand R. Thanks again. Regards Katherine --- On Wed, 27/3/13, Mason wrote: From: Mason Subject: Re: [R] Archieve of mails from R forum To: "Marc Schwartz" Cc: "Katherine Gobin" , "r-help@r-project.org help" Date: Wednesday, 27 March, 2013, 7:30 PM http://r-help.markmail.org/ has a nice interface for searching the archives, too. On Wed, Mar 27, 2013 at 12:18 PM, Marc Schwartz wrote: On Mar 27, 2013, at 1:58 PM, Katherine Gobin wrote: > Dear R helpers, > > Everyday I do receive many many mails from R forum and after some period of > times, INBOX is filled with numerous mails. At times if for some period of > time, I haven't accessed mails, it becomes difficult to keep track of mails > and many times simply due to the volume (and owing to the lack of time due to > office constraints), I have to simply delete the mails without opening them > and I understand this is a huge loss. > > If in case I wish to refer to all the old emails that have been appeared in > the R forum, where do I get these? Is there any list where I will get > subject-wise of thread-wise archive of old emails? I understand that will be > an ocean of quality information and one can learn a lot from these old mails > and I don't need to keep track of my emails all the time. > > > Kindly guide. > > Regards > > Katherine > The official archives for R-Help are here: https://stat.ethz.ch/pipermail/r-help/ and these are mirrored in various locations, such as: http://www.mail-archive.com/r-help@stat.math.ethz.ch/ http://dir.gmane.org/gmane.comp.lang.r.general You can also search the archives for all R lists at: http://rseek.org/ http://finzi.psych.upenn.edu/search.html http://tolstoy.newcastle.edu.au/R/ My recommendation would be to set up a mail filter or rule (using r-help@r-project.org in the the sender and cc: address fields) so that the list e-mails are automatically moved from your main inbox to a folder just for these e-mails and you can then browse them as your schedule permits, rather than having them interspersed with other e-mails in the same location. I do this with a number of the R related lists and have a folder for each one to keep them separated. Most e-mail clients and/or online services have some type of filtering or rule configuration available to do this. Regards, Marc Schwartz __ 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. [[alternative HTML version deleted]] __ 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.
[R] How to delete Identical columns
Dear R forum
Suppose I have a data.frame
df = data.frame(id = c(1:6), x = c(15, 21, 14, 21, 14, 38), y = c(36, 38, 55,
11, 5, 18), x.1 = c(15, 21, 14, 21, 14, 38), z = c("D", "B", "A", "F", "H",
"P"))
> df
id x y x.1 z
1 1 15 36 15 D
2 2 21 38 21 B
3 3 14 55 14 A
4 4 21 11 21 F
5 5 14 5 14 H
6 6 38 18 38 P
Clearly columns x and x.1 are identical. In reality, I have a large data.frame
and can't make out which columns are identical, but I am sure that column with
name say x is repeated as x.1, x.2 etc.
How to automatically identify and retain only one column (in this example
column x) among the identical columns besides other non-identical columns (viz.
id, y and z).
Regards
Katherine
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Re: [R] How to delete Identical columns
Dear Sir,
Thanks a lot for your wonderful solution. When I applied it my data.frame,
however, it was deleting many other columns also having repeated type of column
names i.e. suppose I wanted only to delete say ABC.1, ABC.2 etc. and retain
XYZ, XYZ.1, XYZ2 etc. But this was not happening and alongwith ABC series, it
was deleting XYZ series too. So I changed the command you had given as -
df[ -grep( "\\.", names( df))] to
df[ -grep( "XYZ\\.", names( df))]
And it lead me to the desired result.
Thanks again sir.
Regards
Katherine
--- On Thu, 28/3/13, Gerrit Eichner wrote:
From: Gerrit Eichner
Subject: Re: [R] How to delete Identical columns
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Thursday, 28 March, 2013, 8:58 AM
Hi, Katherine,
IF the naming scheme of the columns of your data frame is consistently
and if duplicated columns
appear THEN (something like)
df[ -grep( "\\.", names( df))]
could help. (But it's maybe more efficient to avoid - a priori - producing
duplicated columns, if the data frame is large, as you say.)
Regards -- Gerrit
On Thu, 28 Mar 2013, Katherine Gobin wrote:
> Dear R forum
>
> Suppose I have a data.frame
>
> df = data.frame(id = c(1:6), x = c(15, 21, 14, 21, 14, 38), y = c(36, 38, 55,
> 11, 5, 18), x.1 = c(15, 21, 14, 21, 14, 38), z = c("D", "B", "A", "F", "H",
> "P"))
>
>
>> df
> id x y x.1 z
> 1 1 15 36 15 D
> 2 2 21 38 21 B
> 3 3 14 55 14 A
> 4 4 21 11 21 F
> 5 5 14 5 14 H
> 6 6 38 18 38 P
>
>
> Clearly columns x and x.1 are identical. In reality, I have a large
> data.frame and can't make out which columns are identical, but I am sure that
> column with name say x is repeated as x.1, x.2 etc.
>
> How to automatically identify and retain only one column (in this example
> column x) among the identical columns besides other non-identical columns
> (viz. id, y and z).
>
>
> Regards
>
> Katherine
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[R] Better way of writing R code
Dear R forum,
(Pl note this is not a finance problem)
I have two data.frames as
currency_df = data.frame(current_date = c("3/4/2013", "3/4/2013", "3/4/2013",
"3/4/2013"), issue_date = c("27/11/2012", "9/12/2012", "14/01/2013",
"28/02/2013"), maturity_date = c("27/04/2013", "3/5/2013", "14/6/2013",
"28/06/2013"), currency = c("USD", "USD", "GBP", "SEK"), other_currency =
c("EURO", "CAD", "CHF", "USD"), transaction = c("Buy", "Buy", "Sell", "Buy"),
units_currency = c(10, 25000, 15, 4), units_other_currency =
c(78000, 25350, 99200, 6150))
rate_df =
data.frame(date =
c("28/3/2013","27/3/2013","26/3/2013","25/3/2013","28/3/2013","27/3/2013","26/3/2013",
"25/3/2013","28/3/2013","27/3/2013","26/3/2013","25/3/2013","28/3/2013","27/3/2013","26/3/2013",
"25/3/2013","28/3/2013","27/3/2013","26/3/2013","25/3/2013","28/3/2013","27/3/2013","26/3/2013",
"25/3/2013","28/3/2013","27/3/2013","26/3/2013","25/3/2013","28/3/2013","27/3/2013","26/3/2013",
"25/3/2013","28/3/2013","27/3/2013","26/3/2013","25/3/2013"),
currency = c("USD","USD","USD","USD", "USD", "USD", "USD","USD","USD","USD",
"USD","USD", "GBP","GBP","GBP","GBP","GBP","GBP","GBP","GBP", "GBP","GBP",
"GBP","GBP", "EURO","EURO","EURO","EURO","EURO","EURO","EURO", "EURO",
"EURO","EURO", "EURO","EURO"),
tenor = c("1 day","1 day","1 day","1 day","1 week","1 week","1 week","1
week","2 weeks","2 weeks","2 weeks","2 weeks","1 day","1 day","1 day","1
day","1 week","1 week","1 week","1 week","2 weeks","2 weeks","2 weeks","2
weeks","1 day","1 day","1 day","1 day","1 week","1 week","1 week","1 week","2
weeks","2 weeks","2 weeks","2 weeks"),
rate =
c(0.156,0.157,0.157,0.155,0.1752,0.1752,0.1752,0.1752,0.1752,0.1752,0.1752,
0.1752,0.48625,
0.485,0.48625,0.4825,0.49,0.49125,0.4925,0.49,0.49375,0.49125,0.4925,
0.49125,0.02643,0.02214,
0.02214,0.01929,0.034,0.034,0.034125,0.034,0.044,0.044, 0.041,0.045))
# ___
# 1st data.frame
> currency_df
current_date issue_date maturity_date currency
1 3/4/2013 27/11/2012 27/04/2013 USD
2 3/4/2013 9/12/2012 3/5/2013 USD
3 3/4/2013 14/01/2013 14/6/2013 GBP
4 3/4/2013 28/02/2013 28/06/2013 SEK
other_currency transaction units_currency
1
EURO Buy 10
2 CAD Buy 25000
3 CHF Sell 15
4 USD Buy 4
units_other_currency
1 78000
2
25350
3 99200
4 6150
#
...
# 2nd data.frame
> rate_df
date currency tenor rate
1 28/3/2013 USD 1 day 0.156000
2 27/3/2013 USD 1 day 0.157000
3 26/3/2013 USD 1 day 0.157000
4 25/3/2013 USD 1 day 0.155000
5 28/3/2013 USD 1 week 0.175200
6 27/3/2013 USD 1 week
0.175200
7 26/3/2013 USD 1 week 0.175200
8 25/3/2013 USD 1 week 0.175200
9 28/3/2013 USD 2 weeks 0.175200
10 27/3/2013 USD 2 weeks 0.175200
11 26/3/2013 USD 2 weeks 0.175200
12 25/3/2013 USD 2 weeks 0.175200
13 28/3/2013 GBP 1 day 0.486250
14 27/3/2013 GBP 1 day 0.485000
15 26/3/2013 GBP 1 day 0.486250
16 25/3/2013 GBP 1 day 0.482500
17 28/3/2013 GBP 1 week 0.49
18 27/3/2013 GBP 1 week 0.491250
19 26/3/2013 GBP 1 week 0.492500
20
25/3/2013 GBP 1 week 0.49
21 28/3/2013 GBP 2 weeks 0.493750
22 27/3/2013 GBP 2 weeks 0.491250
23 26/3/2013 GBP 2 weeks 0.492500
24 25/3/2013 GBP 2 weeks 0.491250
25 28/3/2013 EURO 1 day 0.026430
26 27/3/2013 EURO 1 day 0.022140
27 26/3/2013 EURO 1 day 0.022140
28 25/3/2013 EURO 1 day 0.019290
29 28/3/2013 EURO 1 week 0.034000
30 27/3/2013 EURO 1 week 0.034000
31 26/3/2013 EURO 1 week 0.034125
32 25/3/2013 EURO 1 week 0.034000
33 28/3/2013 EURO 2 weeks 0.044000
34
27/3/2013 EURO 2 weeks 0.044000
35 26/3/2013 EURO 2 weeks 0.041000
36 25/3/2013 EURO 2 weeks 0.045000
# ___
Using plyr and reshape libraries, I have converted the rate_df into tabular
form as
date USD_1 day USD_1 week USD_2 weeks GBP_1 day
1 25/3/2013 0.155 0.1752 0.1752 0.48250
2 26/3/2013 0.157 0.1752 0.1752 0.48625
3 27/3/2013 0.157 0.1752 0.1752 0.48500
4 28/3/2013 0.156 0.1752 0.1752 0.48625
GBP_1 week GBP_2 weeks EURO_1 day EURO_1 week
1 0.49000 0.49125 0.01929 0.034000
2 0.49250 0.49250 0.02214 0.034125
3 0.49125 0.49125 0.02214 0.034000
4 0.49000 0.49375 0.02643 0.034000
EURO_2 weeks
1 0.045
2 0.041
3 0.044
4 0.044
# __
Depending on the maturity period, I hav
Re: [R] Better way of writing R code
Dear Sirs, I sincerely apologize for the blunder at my end. Problem is I was told that one cannot or should not send any ATTACHMENTS. In the past, when I had tried to attach some files and the message was displayed less the attachment. Also, at times it becomes very difficult to attach the csv file. As my input files contain the csv files and since I was under the impression that we cannot attach the files to this forum. I once again apologize to all of you for the inconvenience caused. Regards Katherine --- On Thu, 4/4/13, Gabor Grothendieck wrote: From: Gabor Grothendieck Subject: Re: [R] Better way of writing R code To: "Adams, Jean" Cc: "Katherine Gobin" , "R help" Date: Thursday, 4 April, 2013, 2:48 PM On Thu, Apr 4, 2013 at 9:32 AM, Adams, Jean wrote: > Katherine, > > You should cc the R-help on all correspondence. > The more eyes that see your query, the quicker and probably the better the > response will be. > Send your message as plain text with no attachments ... so, include your > code, and use dput() to share some example data. > Although many types of attachments are not allowed it seems that .txt, .R, .png, .pdf and possibly certain other types are accepted. [[alternative HTML version deleted]] __ 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.
Re: [R] Better way of writing R code
Dear Sir,
Thanks a lot for your great help. Do appreciate it a lot. In my earlier mail,
where I had attached some files,
I have realized yesterday that instead of sending the R code customized by me
based on your guidance, I had by mistake attached the contents of email. I do
apologize to you for the same. Thanks once again and sorry for the
inconvenience caused by me.
Regards
Katherine
--- On Fri, 5/4/13, Adams, Jean wrote:
From: Adams, Jean
Subject: Re: [R] Better way of writing R code
To: "Katherine Gobin"
Cc: "R help"
Date: Friday, 5 April, 2013, 2:40 PM
Katherine,
To preserve the original order, you could create a new variable for the
currency data frame (BEFORE the merges), then use this variable to reorder at
the end.
currency_df$orig.order <- 1:dim(currency_df)[1]
You can do another merge for the other currency, you just need to specify the
columns that you want to merge by. The rate information will be called rate.x
for the first currency (from the first merge) and rate.y for the other currency
(from the second merge).
both2 <- merge(both, rate_df, by.x=c("other_currency", "tenor"),
by.y=c("currency", "tenor"), all.x=TRUE)
Then reorder.
both2 <- both2[order(both2$orig.order), ]
Jean
On Thu, Apr 4, 2013 at 3:19 AM, Katherine Gobin
wrote:
Dear Mr Adams,
I sincerely apologize for taking the liberty of writing to you. I
wholeheartedly thank you for the wonderful solution you had provided me
yesterday. I have customized the R code you had provided and it's yielding the
results. I can't imagine me repeating the 1 lines code after receving such
a powerful solution from you. In future it will save lots of efforts from my
side as I always deal with such situation.
There is one small problem though -
I am dealing with pair of currencies
e.g. currency other_currency transaction
USD EURO Buy
USD CAD
Buy
GBP CHF Sell
SEK USD Buy
The R code gives me the currency rates (w.r.t. appropriate "tenor"), however, I
need the corresponding rates pertaining to the other currency too i.e. in the
first case, the maturity period applicable is one month so the R - code gives
me one month LIBOR wr.t. USD, but I need the corresponding one month LIBOR
w.r.t. the other currency i.e. EURO in this case.
I tried to improve upon the merge statement and used "?merge", but couldn't.
Another problem is the order of the original portfolio is not mainteained , but
I think I can manage the order.
With warm regards
Katherine
--- On Wed, 3/4/13, Adams,
Jean wrote:
From: Adams, Jean
Subject: Re: [R] Better way of writing R code
To: "Katherine Gobin"
Cc: "R help"
Date: Wednesday, 3 April, 2013, 2:08 PM
Katherine,
You don't need to convert rate_df into tabular form. You just need to
categorize each row in currency_df into a "tenor". Then you can merge the two
data frames (by currency and tenor). For example ...
# convert dates to R dates, to calculate the number of days to maturity# I am
assuming this is the number of days from the current date to the maturity date
currency_df$maturity <- as.Date(currency_df$maturity_date,
"%d/%m/%Y")currency_df$current <- as.Date(currency_df$current_date,
"%d/%m/%Y")currency_df$days2mature <- as.numeric(currency_df$maturity -
currency_df$current)
# categorize the number of days to maturity as you wish# you may need to change
the breaks= option to suit your needs# read about the cut function to make sure
you get the cut points included in the proper category, ?cut
currency_df$tenor <- cut(currency_df$days2mature, breaks=c(0, 1, 7, 14,
seq(from=30.5, length=12, by=30.5)),labels=c("1 day", "1 week", "2 weeks",
"1 month", paste(2:12, "months")))
# merge the currency_df and rate_df# this will work better with real data,
since the example data you provided didn't have matching tenorsboth <-
merge(currency_df, rate_df, all.x=TRUE)
Jean
On Wed, Apr 3, 2013 at 5:21 AM, Katherine Gobin
wrote:
Dear R forum,
(Pl note this is not a finance problem)
I have two data.frames as
currency_df = data.frame(current_date = c("3/4/2013", "3/4/2013", "3/4/2013",
"3/4/2013"), issue_date = c("27/11/2012", "9/12/2012", "14/01/2013",
"28/02/2013"), maturity_date = c("27/04/2013", "3/5/2013", "14/6/2013",
"28/06/2013"), currency = c("USD", "USD", "GBP", "SEK"), other_currency =
c("EURO", "CAD", "CHF", "USD"), transactio
[R] lmomco - Three-Parameter Pearson 5 Distribution
Dear R forum, I am bit confused and please guide me - (1) Is "Pearson Type III Distribution" as given in lmomco package same as Three Parameter Pearson 5 Distribution? If not, how do I estimate the parameters of Three Parameter Pearson 5 Distribution? (2) Is there any other R forum dealing with only Statistical queries? Kindly guide Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Package ‘FAdist’ - Log-Pearson Type III Distribution
Dear Sir, I am referring to your package "FAdist". I wish to know how to estimate the parameters of the distribution - "Log-Pearson Type III Distribution"? Will it be possible for you to guide me or inform the package in R, I can use to estimate the parameters. Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Sorting data.frame and again sorting within data.frame
Dear R forum,
I have a data.frame as defied below -
df = data.frame(names = c("C", "A", "A", "B", "C", "B", "A", "B", "C"), dates =
c("4/15/2013", "4/13/2013", "4/15/2013", "4/13/2013", "4/13/2013", "4/15/2013",
"4/14/2013", "4/14/2013","4/14/2013" ),values = c(10, 31, 31, 17, 11, 34, 102,
47, 29))
> df
names dates values
1 C 4/15/2013 10
2 A 4/13/2013 31
3 A 4/15/2013 31
4 B 4/13/2013 17
5 C 4/13/2013 11
6 B
4/15/2013 34
7 A 4/14/2013 102
8 B 4/14/2013 47
9 C 4/14/2013 29
I need to sort df first on "names" in increasing order and then further on
"dates" in a decreasing order i.e. I need
names dates values
A 4/15/2013 31
A 4/14/2013 102
A 4/13/2013 31
B 4/15/2013
34
B 4/14/2013 47
B 4/13/2013 17
C 4/15/2013 10
C 4/14/2013 29
C 4/13/2013 11
I tried
df_sorted = df[order(df$names, (as.Date(df$dates, "%m/%d/%Y")), decreasing =
TRUE),]
> df_sorted
names dates values
1 C 4/15/2013 10
9 C 4/14/2013 29
5 C 4/13/2013
11
6 B 4/15/2013 34
8 B 4/14/2013 47
4 B 4/13/2013 17
3 A 4/15/2013 31
7 A 4/14/2013 102
2 A 4/13/2013 31
I need A to appear first with all three corresponding dates in decreasing
order, then B and so on.
Please guide.
With regards
Katherine
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Re: [R] Sorting data.frame and again sorting within data.frame
Dear Sir,
Thanks a lot for your valuable input and guidance.
Regards
Katherine
--- On Mon, 15/4/13, Jeff Newmiller wrote:
From: Jeff Newmiller
Subject: Re: [R] Sorting data.frame and again sorting within data.frame
To: "David Winsemius" , "Katherine Gobin"
Cc: r-help@r-project.org
Date: Monday, 15 April, 2013, 5:33 PM
Yes, that would be because she converted to Date on the fly in her example, and
so apparently did not need this reminder.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
David Winsemius wrote:
>
>On Apr 14, 2013, at 11:01 PM, Katherine Gobin wrote:
>
>> Dear R forum,
>>
>> I have a data.frame as defied below -
>>
>> df = data.frame(names = c("C", "A", "A", "B", "C", "B", "A", "B",
>"C"), dates = c("4/15/2013", "4/13/2013", "4/15/2013", "4/13/2013",
>"4/13/2013", "4/15/2013", "4/14/2013", "4/14/2013","4/14/2013" ),values
>= c(10, 31, 31, 17, 11, 34, 102, 47, 29))
>>
>>> df
>> names dates values
>> 1 C 4/15/2013 10
>> 2 A 4/13/2013 31
>> 3 A 4/15/2013 31
>> 4 B 4/13/2013 17
>> 5 C 4/13/2013 11
>> 6 B
>> 4/15/2013 34
>> 7 A 4/14/2013 102
>> 8 B 4/14/2013 47
>> 9 C 4/14/2013 29
>>
>> I need to sort df first on "names" in increasing order and then
>further on "dates" in a decreasing order i.e. I need
>>
>
>So far no one has pointed out that these are not really "Dates" in the
>R sense and will not sort correctly if any of the proposed methods are
>applied to sequences that extend beyond6 months, i.e, until October
>forward. You would be advised to convert to real Date-classed
>variables.
>
>?strptime
>?as.Date
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[R] Splitting the Elements of character vector
Dear R forum
I have a data.frame
df = data.frame(currency_type = c("EURO_o_n", "EURO_o_n", "EURO_1w", "EURO_1w",
"USD_o_n", "USD_o_n", "USD_1w", "USD_1w"), rates = c(0.47, 0.475, 0.461, 0.464,
1.21, 1.19, 1.41, 1.43))
currency_type rates
1 EURO_o_n 0.470
2 EURO_o_n 0.475
3 EURO_1w 0.461
4 EURO_1w 0.464
5 USD_o_n 1.210
6 USD_o_n 1.190
7 USD_1w 1.410
8 USD_1w 1.430
I need to split the values appearing under currency_type to obtain following
data.frame in the "original order"
currency tenor rates
EURO o_n 0.470
EURO o_n 0.475
EURO 1w 0.461
EURO 1w 0.464
USD o_n 1.210
USD o_n 1.190
USD 1w 1.410
USD 1w 1.430
Basically I need to split the currency name and tenors.
I tried
strsplit(df$currency_type, "_")
Error in strsplit(df$currency_type, "_") : non-character argument
Kindly guide
Katherine
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[R] Creating a vector with repeating dates
Dear R forum
I have a data.frame
df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"),
values = c(47, 38, 56, 92))
I need to to create a vector by repeating the dates as
"Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, "Current_date",
4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date, 4/15/2013, 4/14/2013,
4/13/2013, 4/12/2013
i.e. I need to create a new vector as given below which I need to use for some
other purpose.
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Is it possible to construct such a
column?
Regards
Katherine
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Re: [R] Creating a vector with repeating dates
Dear Andrija Djurovic,
Thanks for the suggestion. Ia m aware of "rep". However, here I need to repeat
not only dates, but a string "Current_date". Thus, I need to create a vector (
to be included in some other data.frame) with the name say "dt" which will
contain
dt
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
So this is combination of dates and a string. Hence, I am just wondering if it
is possible to create such a vector or not?
Regards
Katherine
--- On Wed, 17/4/13, andrija djurovic wrote:
From: andrija djurovic
Subject: Re: [R] Creating a vector with repeating dates
To: "Katherine Gobin"
Cc: "r-help@r-project.org"
Date: Wednesday, 17 April, 2013, 10:14 AM
?rep
On Wed, Apr 17, 2013 at 11:11 AM, Katherine Gobin
wrote:
Dear R forum
I have a data.frame
df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013", "4/12/2013"),
values = c(47, 38, 56, 92))
I need to to create a vector by repeating the dates as
"Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, "Current_date",
4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date, 4/15/2013, 4/14/2013,
4/13/2013, 4/12/2013
i.e. I need to create a new vector as given below which I need to use for some
other purpose.
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Current_date
4/15/2013
4/14/2013
4/13/2013
4/12/2013
Is it possible to construct such a
column?
Regards
Katherine
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Re: [R] Creating a vector with repeating dates
Dear Sir,
Thanks a lot for your valuable suggestions and help.
Regards
Katherine
--- On Wed, 17/4/13, Jim Lemon wrote:
From: Jim Lemon
Subject: Re: [R] Creating a vector with repeating dates
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Wednesday, 17 April, 2013, 10:35 AM
On 04/17/2013 07:11 PM, Katherine Gobin wrote:
> Dear R forum
>
> I have a data.frame
>
> df = data.frame(dates = c("4/15/2013", "4/14/2013", "4/13/2013",
> "4/12/2013"), values = c(47, 38, 56, 92))
>
> I need to to create a vector by repeating the dates as
>
> "Current_date", 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, "Current_date",
> 4/15/2013, 4/14/2013, 4/13/2013, 4/12/2013, Current_date, 4/15/2013,
> 4/14/2013, 4/13/2013, 4/12/2013
>
> i.e. I need to create a new vector as given below which I need to use for
> some other purpose.
>
> Current_date
> 4/15/2013
> 4/14/2013
> 4/13/2013
> 4/12/2013
> Current_date
> 4/15/2013
> 4/14/2013
> 4/13/2013
> 4/12/2013
> Current_date
> 4/15/2013
> 4/14/2013
> 4/13/2013
> 4/12/2013
>
> Is it possible to construct such a
> column?
>
Hi Katherine,
How about:
rep(c("Current date",paste(4,15:12,2013,sep="/")),3)
Jim
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[R] Error with function
Dear R forum,
I have a data.frame as given below:
df = data.frame(tran = c("tran1", "tran2", "tran3", "tran4"), tenor = c("2w",
"1m", "7m", "3m"))
Also, I define
libor_tenor_labels = as.character(c("o_n", "1w", "2w",
"1m", "2m", "3m", "4m", "5m", "6m", "7m", "8m", "9m", "10m", "11m",
"12m"))
#
> df
tran tenor
1 tran1 2w
2 tran2 1m
3 tran3 7m
4 tran4 3m
# __
# libor_tenor_labels can be anything and need not be 15. Also, df need not be
consisting of only 4 record. Basically, I can't HARD CODE anything.
In df, first tenor is 2w. So I need to define a previous tenor as "1w" and nest
tenor as "1m" i.e. I need the output
> df_new
tran tenor prev_tenor nxt_tenor
1 tran1 2w 1w 1m
2 tran2 1m 2w 2m
3 tran3 7m 6m 8m
4 tran4 3m 2m 4m
# ___
# I have two special cases also. If the tenor is "o_n" or "12m" i.e. extremes,
I needed to adjust the rates as given in code.
# My code
# ==
tenor_function = function(tran, tenor)
{
if (tenor == libor_tenor_labels[1])
{
prev_tenor = libor_tenor_labels[1]
nxt_tenor = libor_tenor_labels[2]
}
for (i in 2:(length(libor_tenor_labels)-1))
{
if (tenor == libor_tenor_labels[i])
{
prev_tenor = libor_tenor_labels[i-1]
nxt_tenor = libor_tenor_labels[i+1]
}
}
if (tenor == libor_tenor_labels[length(libor_tenor_labels)])
{
prev_tenor = libor_tenor_labels[(length(libor_tenor_labels)-1)]
nxt_tenor = libor_tenor_labels[length(libor_tenor_labels)]
}
return(data.frame(tran = tran, prev_tenor = prev_tenor, tenor = tenor,
nxt_tenor = nxt_tenor)
}
(tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
tenor_function(tran = x$tran, tenor = x$tenor)))
# __
# ERROR - I get following error
Error: unexpected '}' in:
"
}"
>
> (tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
> tenor_function(tran = x$tran, tenor = x$tenor)))
Error in .fun(piece, ...) : could not find function "tenor_function"
# __
Kindly guide
With warn regards
Katherine
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[R] Fw: Error with function - USING library(plyr)
Dear R forum,
Please refer to my query regarding "Error with function". I forgot to mention
that I am using "plyr" library.
Sorry for inconvenience.
Regards
Katherine
--- On Tue, 23/4/13, Katherine Gobin wrote:
From: Katherine Gobin
Subject: [R] Error with function
To: r-help@r-project.org
Date: Tuesday, 23 April, 2013, 7:06 AM
Dear R forum,
I have a data.frame as given below:
df = data.frame(tran = c("tran1", "tran2", "tran3", "tran4"), tenor = c("2w",
"1m", "7m", "3m"))
Also, I define
libor_tenor_labels = as.character(c("o_n", "1w", "2w",
"1m", "2m", "3m", "4m", "5m", "6m", "7m", "8m", "9m", "10m", "11m",
"12m"))
#
> df
tran tenor
1 tran1 2w
2 tran2 1m
3 tran3 7m
4 tran4 3m
# __
# libor_tenor_labels can be anything and need not be 15. Also, df need not be
consisting of only 4 record. Basically, I can't HARD CODE anything.
In df, first tenor is 2w. So I need to define a previous tenor as "1w" and nest
tenor as "1m" i.e. I need the output
> df_new
tran tenor prev_tenor nxt_tenor
1 tran1 2w 1w 1m
2 tran2 1m 2w 2m
3 tran3 7m 6m 8m
4 tran4 3m 2m 4m
# ___
# I have two special cases also. If the tenor is "o_n" or "12m" i.e. extremes,
I needed to adjust the rates as given in code.
# My code
# ==
tenor_function = function(tran, tenor)
{
if (tenor == libor_tenor_labels[1])
{
prev_tenor = libor_tenor_labels[1]
nxt_tenor = libor_tenor_labels[2]
}
for (i in 2:(length(libor_tenor_labels)-1))
{
if (tenor == libor_tenor_labels[i])
{
prev_tenor = libor_tenor_labels[i-1]
nxt_tenor = libor_tenor_labels[i+1]
}
}
if (tenor == libor_tenor_labels[length(libor_tenor_labels)])
{
prev_tenor = libor_tenor_labels[(length(libor_tenor_labels)-1)]
nxt_tenor = libor_tenor_labels[length(libor_tenor_labels)]
}
return(data.frame(tran = tran, prev_tenor = prev_tenor, tenor = tenor,
nxt_tenor = nxt_tenor)
}
(tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
tenor_function(tran = x$tran, tenor = x$tenor)))
# __
# ERROR - I get following error
Error: unexpected '}' in:
"
}"
>
> (tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
> tenor_function(tran = x$tran, tenor = x$tenor)))
Error in .fun(piece, ...) : could not find function "tenor_function"
# __
Kindly guide
With warn regards
Katherine
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[R] Fw: " PROBLEM SOLVED" - Error with function
Dear R forum
Please refer to my query captioned Error with function.
I had missed in bracket ")" in the return statement and hence I was getting the
error. I has struggled for more than 2 hours to find out the problem and only
then has posted to the forum. I sincerely apologize to all for consuming your
valuable time.
Thanks for the efforts at your end.
Regards
Katherine
--- On Tue, 23/4/13, Katherine Gobin wrote:
From: Katherine Gobin
Subject: [R] Error with function
To: r-help@r-project.org
Date: Tuesday, 23 April, 2013, 7:06 AM
Dear R forum,
I have a data.frame as given below:
df = data.frame(tran = c("tran1", "tran2", "tran3", "tran4"), tenor = c("2w",
"1m", "7m", "3m"))
Also, I define
libor_tenor_labels = as.character(c("o_n", "1w", "2w",
"1m", "2m", "3m", "4m", "5m", "6m", "7m", "8m", "9m", "10m", "11m",
"12m"))
#
> df
tran tenor
1 tran1 2w
2 tran2 1m
3 tran3 7m
4 tran4 3m
# __
# libor_tenor_labels can be anything and need not be 15. Also, df need not be
consisting of only 4 record. Basically, I can't HARD CODE anything.
In df, first tenor is 2w. So I need to define a previous tenor as "1w" and nest
tenor as "1m" i.e. I need the output
> df_new
tran tenor prev_tenor nxt_tenor
1 tran1 2w 1w 1m
2 tran2 1m 2w 2m
3 tran3 7m 6m 8m
4 tran4 3m 2m 4m
# ___
# I have two special cases also. If the tenor is "o_n" or "12m" i.e. extremes,
I needed to adjust the rates as given in code.
# My code
# ==
tenor_function = function(tran, tenor)
{
if (tenor == libor_tenor_labels[1])
{
prev_tenor = libor_tenor_labels[1]
nxt_tenor = libor_tenor_labels[2]
}
for (i in 2:(length(libor_tenor_labels)-1))
{
if (tenor == libor_tenor_labels[i])
{
prev_tenor = libor_tenor_labels[i-1]
nxt_tenor = libor_tenor_labels[i+1]
}
}
if (tenor == libor_tenor_labels[length(libor_tenor_labels)])
{
prev_tenor = libor_tenor_labels[(length(libor_tenor_labels)-1)]
nxt_tenor = libor_tenor_labels[length(libor_tenor_labels)]
}
return(data.frame(tran = tran, prev_tenor = prev_tenor, tenor = tenor,
nxt_tenor = nxt_tenor)
}
(tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
tenor_function(tran = x$tran, tenor = x$tenor)))
# __
# ERROR - I get following error
Error: unexpected '}' in:
"
}"
>
> (tenor_libors = ddply(.data = df, .variables = "tran", .fun = function(x)
> tenor_function(tran = x$tran, tenor = x$tenor)))
Error in .fun(piece, ...) : could not find function "tenor_function"
# __
Kindly guide
With warn regards
Katherine
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[R] Linear Interpolation : Missing rates
Dear R forum
I have data.frame as
df = data.frame(rate_name = c("USD_1w", "USD_1w", "USD_1w", "USD_1w", "USD_1m",
"USD_1m", "USD_1m", "USD_1m", "USD_2m", "USD_2m", "USD_2m", "USD_2m",
"GBP_1w", "GBP_1w", "GBP_1w", "GBP_1w", "GBP_1m", "GBP_1m", "GBP_1m", "GBP_1m",
"GBP_2m", "GBP_2m", "GBP_2m", "GBP_2m", "EURO_1w", "EURO_1w", "EURO_1w",
"EURO_1w", "EURO_2w", "EURO_2w", "EURO_2w", "EURO_2w", "EURO_2m", "EURO_2m",
"EURO_2m", "EURO_2m"), rates = c(2.05, 2.07, 2.06, 2.06, 2.22, 2.24, 2.23,
2.23, 2.31, 2.33, 2.33, 2.31, 1.06, 1.08, 1.08, 1.08, 1.21, 1.21, 1.23, 1.21,
1.41, 1.39, 1.39, 1.37, 1.82, 1.82, 1.81, 1.80, 1.98, 1.98, 1.97, 1.97, 2.1,
2.09, 2.09, 2.11))
currency = c("EURO", "GBP", "USD")
tenor = c("1w", "2w", "1m", "2m", "3m")
# _
> df
rate_name rates
rate_name rates
1 USD_1w 2.05
2 USD_1w 2.07
3 USD_1w 2.06
4 USD_1w 2.06
5 USD_1m 2.22
6 USD_1m 2.24
7 USD_1m 2.23
8 USD_1m 2.23
9 USD_2m 2.31
10 USD_2m 2.33
11 USD_2m 2.33
12 USD_2m 2.31
13 GBP_1w 1.06
14 GBP_1w 1.08
15 GBP_1w 1.08
16 GBP_1w 1.08
17 GBP_1m 1.21
18 GBP_1m 1.21
19 GBP_1m 1.23
20 GBP_1m 1.21
21 GBP_2m 1.41
22 GBP_2m 1.39
23 GBP_2m 1.39
24 GBP_2m 1.37
25 EURO_1w 1.82
26 EURO_1w 1.82
27 EURO_1w 1.81
28 EURO_1w 1.80
29 EURO_2w 1.98
30 EURO_2w 1.98
31 EURO_2w 1.97
32 EURO_2w 1.97
33 EURO_2m 2.10
34 EURO_2m 2.09
35 EURO_2m 2.09
36 EURO_2m 2.11
As can be seen that USD_2w, GBP_2w and EURO_1m are missing and I need to
INTERPOLATE these rates, which can be done using approx or approxfun. In
reality I can have many currencies with many tenors. Problem is when the
data.frame "df" is read or accessed in R, I am not aware which tenor is
missing. For a given currency, it is possible that mare than 1 consecutive
tenors may be missing e.g. in case of EURO, I may have EURO_1w, EURO_2w and
then EURO_4m. So EURO_1m, EURO_2m and EURO_3m are missing.
I understand it's sort of vague question from me and do apologize for the same.
Any suggestion please.
Regards
Katherine
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Re: [R] Linear Interpolation : Missing rates
Dear Mr Adams,
Thanks a lot for your solution. I understand it was very tricky and needed lot
of application. Thanks again and do appreciate your efforts.
Regards
Katherine
--- On Thu, 25/4/13, Adams, Jean wrote:
From: Adams, Jean
Subject: Re: [R] Linear Interpolation : Missing rates
To: "Katherine Gobin"
Cc: "R help"
Date: Thursday, 25 April, 2013, 2:23 PM
Katherine,
Split the rate names into their currency and tenor parts and assign a numeric
value to each tenor. Choose a model to do your approximations (I used linear
regression in the example below). Use this model to generate estimates for all
combinations of currency and tenor.
For example:
# split the rate names into currency and tenorsplitnames <- do.call(rbind,
strsplit(df$rate_name, "_"))df$currency <- as.factor(splitnames[, 1])
df$tenor <- splitnames[, 2]
# assign numeric value to each tenoruniquetenors <- c("1w", "2w", "1m",
"2m")uniquedays <- c(7, 14, 30.5, 61)
df$tenordays <- uniquedays[match(df$tenor, uniquetenors)]
# fit a linear model of rate on tenordays for each currencyfit <- lm(rates ~
currency*tenordays, data=df)
# estimate rates for all combinations of currency and tenorfulldf <-
expand.grid(tenordays=unique(df$tenordays),
currency=unique(df$currency))fulldf$est.rates = predict(fit, newdata=fulldf)
# merge observed rates with estimated ratesdfwithest <- merge(df, fulldf,
all=TRUE)
Jean
On Thu, Apr 25, 2013 at 12:33 AM, Katherine Gobin
wrote:
Dear R forum
I have data.frame as
df = data.frame(rate_name = c("USD_1w", "USD_1w", "USD_1w", "USD_1w", "USD_1m",
"USD_1m", "USD_1m", "USD_1m", "USD_2m", "USD_2m", "USD_2m", "USD_2m",
"GBP_1w", "GBP_1w", "GBP_1w", "GBP_1w", "GBP_1m", "GBP_1m", "GBP_1m", "GBP_1m",
"GBP_2m", "GBP_2m", "GBP_2m", "GBP_2m", "EURO_1w", "EURO_1w", "EURO_1w",
"EURO_1w", "EURO_2w", "EURO_2w", "EURO_2w", "EURO_2w", "EURO_2m", "EURO_2m",
"EURO_2m", "EURO_2m"), rates = c(2.05, 2.07, 2.06, 2.06, 2.22, 2.24, 2.23,
2.23, 2.31, 2.33, 2.33, 2.31, 1.06, 1.08, 1.08, 1.08, 1.21, 1.21, 1.23, 1.21,
1.41, 1.39, 1.39, 1.37, 1.82, 1.82, 1.81, 1.80, 1.98, 1.98, 1.97, 1.97, 2.1,
2.09, 2.09, 2.11))
currency = c("EURO", "GBP", "USD")
tenor = c("1w", "2w", "1m", "2m", "3m")
# _
> df
rate_name rates
rate_name rates
1 USD_1w 2.05
2 USD_1w 2.07
3 USD_1w 2.06
4 USD_1w 2.06
5 USD_1m 2.22
6 USD_1m 2.24
7 USD_1m 2.23
8 USD_1m 2.23
9 USD_2m 2.31
10 USD_2m 2.33
11 USD_2m 2.33
12 USD_2m 2.31
13 GBP_1w 1.06
14 GBP_1w 1.08
15 GBP_1w 1.08
16 GBP_1w 1.08
17 GBP_1m 1.21
18 GBP_1m 1.21
19 GBP_1m 1.23
20 GBP_1m 1.21
21 GBP_2m 1.41
22 GBP_2m 1.39
23 GBP_2m 1.39
24 GBP_2m 1.37
25 EURO_1w 1.82
26 EURO_1w 1.82
27 EURO_1w 1.81
28 EURO_1w 1.80
29 EURO_2w 1.98
30 EURO_2w 1.98
31 EURO_2w 1.97
32 EURO_2w 1.97
33 EURO_2m 2.10
34 EURO_2m 2.09
35 EURO_2m 2.09
36 EURO_2m 2.11
As can be seen that USD_2w, GBP_2w and EURO_1m are missing and I need to
INTERPOLATE these rates, which can be done using approx or approxfun. In
reality I can have many currencies with many tenors. Problem is when the
data.frame "df" is read or accessed in R, I am not aware which tenor is
missing. For a given currency, it is possible that mare than 1 consecutive
tenors may be missing e.g. in case of EURO, I may have EURO_1w, EURO_2w and
then EURO_4m. So EURO_1m, EURO_2m and EURO_3m are missing.
I understand it's sort of vague question from me and do apologize for the same.
Any suggestion please.
Regards
Katherine
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[R] Splitting data.frame and saving to csv files
Dear R Forum,
I have a data.frame as
df = data.frame(date = c("2013-04-15", "2013-04-14", "2013-04-13",
"2013-04-12", "2013-04-11"),
ABC_f = c(62.80739769,81.04525895,84.65712455,12.78237251,57.61345256),
LMN_d = c(21.16794336,54.6580401,63.8923307,87.59880367,87.07693716),
XYZ_p = c(55.8885464,94.1358684,84.0089114,98.99746696,64.71083712),
LMN_a = c(56.6768395,25.81530198,40.12268441,35.74175237,47.95892209),
ABC_e = c(11.36783959,62.29651784,47.63481552,32.27820673,52.12561419),
LMN_c = c(45.4484695,17.72362438,36.7690054,68.58912931,35.80767235),
XYZ_zz = c(85.74755089,63.48582415,81.61107212,58.1572924,27.44132817),
PQR = c(71.22867519,95.09994812,83.62437819,30.18524735,25.81804865),
ABC_d =
c(38.71089816,93.48216193,93.14432203,78.2738731,31.87170019),
ABC_m = c(40.28473769,43.97076327,47.38761559,97.33573412,22.06884976))
> df
date ABC_f LMN_d XYZ_p LMN_a ABC_e
1 2013-04-15 62.80740 21.16794 55.88855 56.67684 11.36784
2 2013-04-14 81.04526 54.65804 94.13587 25.81530 62.29652
3 2013-04-13 84.65712 63.89233 84.00891 40.12268 47.63482
4 2013-04-12 12.78237 87.59880 98.99747 35.74175 32.27821
5 2013-04-11 57.61345 87.07694 64.71084 47.95892 52.12561
LMN_c XYZ_zz PQR ABC_d ABC_m
1 45.44847 85.74755 71.22868 38.71090 40.28474
2 17.72362 63.48582 95.09995 93.48216 43.97076
3 36.76901 81.61107 83.62438 93.14432 47.38762
4 68.58913 58.15729 30.18525 78.27387
97.33573
5 35.80767 27.44133 25.81805 31.87170 22.06885
I need to identify columns with same labels and along-with the dates in the
first column, save the columns in different csv files.
E.g. in the above data frame, I have 4 columns beginning with ABC so I need to
save these four columns with the date in the first column as ABC.csv, then
LMN_d, LMN_a, LMN_c in the LMN.csv file as date, LMN_a, LMN_c, LMN_d and so on.
In my actual data.frame, I won't be aware how many such rates combinations are
available. If there is no matching column as "PQR", the PQR.csv file should
have only date and PQR column.
Kindly guide how do I split the data.frame and save the respective csv files.
Regards
Katherine
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[R] Adding elements in data.frame subsets and also subtracting an element from the rest elements in data.frame
Dear R forum
I have a data.frame as
cashflow_df = data.frame(instrument =
c("ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC","ABC",
"ABC", "PQR", "PQR", "PQR","PQR","PQR","PQR","PQR","PQR","PQR","PQR", "PQR",
"PQR", "PQR","PQR", "PQR","PQR","PQR","PQR", "PQR","PQR","UVWXYZ","UVWXYZ",
"UVWXYZ", "UVWXYZ", "UVWXYZ","UVWXYZ","UVWXYZ","UVWXYZ", "UVWXYZ", "UVWXYZ"),
id = c(1,1,1,2,2,2,3,3,3,4,4,4,5,5,5,1,1,1,1,2,2,2,2,3,3,3,3,4,4,4,4,5,5,5,5,
1,1,2,2,3,3,4,4, 5,5),
cashflow = c(5000,5000,505000,5000,5000,505000,5000,5000,505000, 5000,5000,
505000,
5000,5000,505000,500,500,500,102000,500,500,500,102000,500,500,500,102000,500,500,500,102000,500,500,500,102000,8000,808000,8000,808000,8000,808000,8000,808000,8000,808000),
cashflows_pv = c(4931.054, 4479.1116, 431160.8529,4931.9604, 4485.6393,
432064.0228,
4932.5438,4489.8451,432646.2398,4932.1548,4487.0404,432257.9551,4932.6087,4490.3129,432711.0084,493.6326,474.0524,455.2489,82252.0304,493.8083,474.7543,456.4356,82744.9157,493.6003,473.9235,455.031,82161.7368,493.8175,474.7913,456.4982,82770.9849,493.8592,474.9581,456.7804,82888.4556,7451.3118,681810.5522,7462.0148,684153.4992,7441.1294,679585.9186,7426.6407,676427.7274,7427.1225,676532.6262))
# __
> cashflow_df
instrument id cashflow cashflows_pv
1 ABC 1 5000 4931.0540
2 ABC 1 5000 4479.1116
3 ABC 1 505000 431160.8529
4 ABC 2 5000 4931.9604
5 ABC 2 5000 4485.6393
6 ABC 2 505000 432064.0228
7 ABC 3 5000 4932.5438
8 ABC 3 5000 4489.8451
9 ABC 3 505000 432646.2398
10 ABC 4 5000 4932.1548
11 ABC 4 5000 4487.0404
12 ABC 4 505000 432257.9551
13 ABC 5 5000 4932.6087
14 ABC 5 5000 4490.3129
15 ABC 5 505000 432711.0084
16 PQR 1 500 493.6326
17 PQR 1 500 474.0524
18 PQR 1 500 455.2489
19 PQR 1 102000 82252.0304
20 PQR 2 500 493.8083
21 PQR 2 500 474.7543
22 PQR 2 500 456.4356
23 PQR 2 102000 82744.9157
24 PQR 3 500 493.6003
25 PQR 3 500 473.9235
26 PQR 3 500 455.0310
27 PQR 3 102000 82161.7368
28 PQR 4 500 493.8175
29 PQR 4 500 474.7913
30 PQR 4 500 456.4982
31 PQR 4 102000 82770.9849
32 PQR 5 500 493.8592
33 PQR 5 500 474.9581
34 PQR 5 500 456.7804
35 PQR 5 102000 82888.4556
36 UVWXYZ 1 8000 7451.3118
37 UVWXYZ 1 808000 681810.5522
38 UVWXYZ 2 8000 7462.0148
39 UVWXYZ 2 808000 684153.4992
40 UVWXYZ 3 8000 7441.1294
41 UVWXYZ 3 808000 679585.9186
42 UVWXYZ 4 8000 7426.6407
43 UVWXYZ 4 808000 676427.7274
44 UVWXYZ 5 8000 7427.1225
45 UVWXYZ 5 808000 676532.6262
# ===
# My PROBLEM
For a given instrument and id, I need the totals of cashflow and cashflows_pv
and also the difference of (total_cashflow_pv pertaining to the first ID for
the given instrument from total_cashflow_pv for the same instrument) as shown
in the fourth column of following output.
output
instrument id total_cashflow total_cashflow_pv
1 ABC 1 515000 440571.02
2 ABC 2 515000 441481.62
3 ABC 3 515000 442068.63
4 ABC 4 515000 441677.15
5 ABC 5 515000 442133.93
6 PQR 1 103500 83674.96
7 PQR 2 103500 84169.91
8 PQR 3 103500 83584.29
9 PQR 4 103500 84196.09
10 PQR 5 103500 84314.05
11 UVWXYZ 1 816000 689261.86
12 UVWXYZ 2 816000 691615.51
13 UVWXYZ 3 816000 687027.05
14 UVWXYZ 4 816000 683854.37
15 UVWXYZ 5 816000 683959.75
cashflow_change
1 0. # This is (440571.02 - 440571.02) 1st ID value - 1st
ID value for ABC
2 910.6040 # This is (441481.62 - 440571.02) 2nd ID value - 1st ID
value for ABC
3 1497.6102 # This is (442068.63 - 440571.02) 3rd ID value - 1st ID
value for ABC
4 1106.1318
5 1562.9115
6 0. # This is (83674.96 - 83674.96) 1st ID value - 1st ID
value for PQR
7 494.9496
8 -90.6727
9 521.1276
10 639.0890
11 0.
12 2353.6500
13 -2234.8160
14 -5407.4959
15 -5302.1153 # This is (683959.75 -689261.86 ) 5th ID value - 1st ID
value for UVWXYZ
Kindly guide
Regards
Ka
[R] Clean Price of Bond : Can't install "RQuantLib" in R version 3.0.0
Dear Forum,
I have R version 3.0.0 installed and need to install RQuantLib pacakge. I tried
to install it from CRAN Mirror and I couldn't load it. I had saved the package
i zip format and tried to install it locally but I am getting following error.
> utils:::menuInstallLocal()
Error in read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) :
cannot open the connection
In addition: Warning message:
In read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) :
cannot open compressed file 'RQuantLib_0.3.10(1)/DESCRIPTION', probable
reason 'No such file or directory'
I need to install this package as I need to find out how the clean price of
bond is arrived at?
Kindly
guide
Regards
Katherine
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Re: [R] Clean Price of Bond : Can't install "RQuantLib" in R version 3.0.0
Dear Sir,
Thanks a lot. The "1 in 'RQuantLib_0.3.10(1)", I understand was appearing
because I had saved RQuantLib number of times in my local directory and each
time it renamed the installation file with no added to it.
Thanks again. Now I am able to install the package along-with Rcpp.
Regards
Katherine
--- On Thu, 2/5/13, Prof Brian Ripley wrote:
From: Prof Brian Ripley
Subject: Re: [R] Clean Price of Bond : Can't install "RQuantLib" in R version
3.0.0
To: "Katherine Gobin"
Cc: r-help@r-project.org
Date: Thursday, 2 May, 2013, 1:23 PM
On 02/05/2013 13:09, Katherine Gobin wrote:
> Dear Forum,
>
> I have R version 3.0.0 installed and need to install RQuantLib pacakge. I
> tried to install it from CRAN Mirror and I couldn't load it. I had saved the
> package i zip format and tried to install it locally but I am getting
> following error.
What is the name of the file you downloaded? I am guessing it did not
arrive with the name on the archive.
You seem to be using Windows without saying so. The file for Windows is
http://cran.r-project.org/bin/windows/contrib/r-release/RQuantLib_0.3.10.zip
without (1) in the name.
>
>> utils:::menuInstallLocal()
> Error in read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) :
> cannot open the connection
> In addition: Warning message:
> In read.dcf(file.path(pkgname, "DESCRIPTION"), c("Package", "Type")) :
> cannot open compressed file 'RQuantLib_0.3.10(1)/DESCRIPTION', probable
>reason 'No such file or directory'
>
> I need to install this package as I need to find out how the clean price of
> bond is arrived at?
>
> Kindly
> guide
>
> Regards
>
> Katherine
>
> [[alternative HTML version deleted]]
>
>
>
> __
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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.
>
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UK Fax: +44 1865 272595
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[R] Finding Beta
Dear R forum
I have a dataframe (of prices) as given below -
dat
= data.frame(company = rep(c("A", "B", "C", "D", "index"), each = 5),
prices = c(runif(5, 10, 12), runif(5, 108, 112), runif(5, 500, 510),
runif(5, 40, 50), runif(5, 1000, 1020)))
company prices
1 A 10.61727
2 A 10.51892
3 A 11.80495
4 A 11.15243
5 A 10.77543
6 B 111.23817
7 B 109.19825
8 B
108.80053
9 B 110.79876
10 B 108.84385
11 C 504.71801
12 C 504.11778
13 C 502.89416
14 C 500.65996
15 C 502.26748
16 D 42.35901
17 D 43.71947
18 D 46.46092
19 D 43.62220
20 D 48.47480
21 index 1017.24476
22 index 1002.88139
23 index 1005.16148
24 index 1014.54480
25 index 1014.12103
I need to find the beta
of A, B, C and D w.r.t index.
Beta between two variables X and Y (where Y is dependent) is given by,
beta = coef(lm(Y ~ X))[2]
Any guidance is appreciated.
With regards
Katherine
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[R] Removing "NA" from matrix
Dear R forum, I have a data frame dat = data.frame( ABC = c(25.28000732,48.33857234,19.8013245,10.68361461), DEF = c(14.02722251,10.57985168,11.81890316,21.40171514), GHI = c(1,1,1,1), JKL = c(45.96423231,44.52986236,16.56514176,32.14545122), MNO = c(45.38438063,15.54338206,18.78444777,24.29486984)) > dat ABC DEF GHI JKL MNO 1 25.28001 14.02722 1 45.96423 45.38438 2 48.33857 10.57985 1 44.52986 15.54338 3 19.80132 11.81890 1 16.56514 18.78445 4 10.68361 21.40172 1 32.14545 24.29487 When I try to find the correlation I get (which is obvious as my one column shows no variation) dat_cor = cor(dat) Warning message: In cor(dat) : the standard deviation is zero > dat_cor ABC DEF GHI JKL MNO ABC 1.000 -0.75600764 NA 0.55245223 -0.2735585 DEF -0.7560076 1. NA -0.06479082 0.2020781 GHI NA NA 1 NA NA JKL 0.5524522 -0.06479082 NA 1. 0.4564568 MNO -0.2735585 0.20207810 NA 0.45645683 1.000 In reality I am dealing with about 300 variables and don't know which variables don't vary. My query is how do I remove the columns and rows with NA's. So for example, I need the correlation matrix for ABC, DEF, JKL and MNO only. Kindly guide. Thanking in advance. Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Choosing subset of data.frame
Dear R Forum
I have a data frame as
beta_results = data.frame(instrument = c("ABC", "DEF", "JKL", "LMN", "PQR",
"STU", "UVW", "XYZ"),
beta_values = c(1.27, -0.22, 0.529, 0.011, 2.31, -1.08, -2.7, 0.42))
> beta_results
instrument beta_values
1 ABC 1.270
2 DEF -0.220
3 JKL 0.529
4 LMN 0.011
5 PQR 2.310
6 STU -1.080
7 UVW -2.700
8 XYZ 0.420
Through some other process, I am getting instrument names as say (which may
change each time I run this process
and hence I can't hard code it).
instru = c("JKL", "STU", "XYZ")
Now I want the subset of beta_results, (say beta_results_A) pertaining to only
instru i.e
beta_results_A =
instrument beta_values
3 JKL 0.529
6 STU -1.080
8 XYZ 0.420
I did try
beta_results_A = beta_results[instru]
or
beta_results_A = subset(beta_results, beta_results$instrument = instru]
but I guess it's failing.
Kindly guide
Regards
Katherine
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[R] How to store interim print results
Dear R forum,
Following is an customized extract of a code I am working on.
settlement = as.Date("2013-11-25")
maturity = as.Date("2015-10-01")
coupon = 0.066
yield = 0.1040
basis = 1
frequency = 2
redemption = 100
# __
add.months = function(date, n)
{
nC <- seq(date, by=paste (n, "months"), length = 2)[2]
fD <- as.Date(strftime(as.Date(date), format='%Y-%m-01'))
C <- (seq(fD, by=paste (n+1, "months"), length = 2)[2])-1
if(nC>C) return(C)
return(nC)
}
date.diff = function(end, start, basis=1) {
if (basis != 0 && basis != 4)
return(as.numeric(end - start))
e <- as.POSIXlt(end)
s <- as.POSIXlt(start)
d <- (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
return (d)
}
cashflows <- 0
last.coupon <- maturity
while (last.coupon > settlement) {
print(last.coupon) # I need to store these dates
last.coupon <- add.months(last.coupon, -12/frequency)
cashflows <- cashflows + 1
print(cashflows) # I need to store these cashflow numbers
}
The print command causes the following output
[1] "2015-10-01"
[1] 1
[1] "2015-04-01"
[1] 2
[1] "2014-10-01"
[1] 3
[1] "2014-04-01"
[1] 4
My problem is how do I store these print outputs or while the loop is getting
executed, how do I save these to some data.frame say
output_dat
cashflow_tenure cashflow_nos
1 2015-10-01 1
2 2015-04-01 2
3 2014-10-01 3
4 2014-04-01 4
Kindly advise
With regards
Katherine
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Re: [R] How to store interim print results
Dear Jean
Thanks a lot for this solution. Its very useful. I did only one small change
and defined
cashflow.tenure <- numeric(0) instead of character(0). This helps me in further
numerical calculations using these dates like finding the difference between
two dates etc.
Thanks again,
Regards
Katherine
On Thursday, 3 April 2014 6:58 PM, "Adams, Jean" wrote:
Katherine,
One easy way to do this for small data is by using the append() function (see
code below). But, if you have a lot of data, it may be too slow for you. In
that case, you can gain some efficiency if you determine in advance how long
the vectors will be, then use indexing to fill in the vectors without using the
append() function. Or, rewrite the code to be vectorized instead of using a
while() loop.
cashflows <- 0
last.coupon <- maturity
# create "empty" vectors
cashflow.tenure <- character(0)
cashflow.nos <- numeric(0)
while (last.coupon > settlement) {
print(last.coupon)
# store the dates
cashflow.tenure <- append(cashflow.tenure, last.coupon)
last.coupon <- add.months(last.coupon, -12/frequency)
cashflows <- cashflows + 1
print(cashflows)
# store the cashflow numbers
cashflow.nos <- append(cashflow.nos, cashflows)
}
output.dat <- data.frame(cashflow.tenure, cashflow.nos)
output.dat
Jean
On Thu, Apr 3, 2014 at 5:22 AM, Katherine Gobin
wrote:
Dear R forum,
>
>Following is an customized extract of a code I am working on.
>
>settlement = as.Date("2013-11-25")
>maturity = as.Date("2015-10-01")
>coupon = 0.066
>yield = 0.1040
>basis = 1
>frequency = 2
>redemption = 100
>
># __
>
>add.months = function(date, n)
>{
> nC <- seq(date, by=paste (n, "months"), length = 2)[2]
> fD <- as.Date(strftime(as.Date(date), format='%Y-%m-01'))
> C <- (seq(fD, by=paste (n+1, "months"), length = 2)[2])-1
> if(nC>C) return(C)
> return(nC)
>}
>
>date.diff = function(end, start, basis=1) {
> if (basis != 0 && basis != 4)
> return(as.numeric(end - start))
> e <- as.POSIXlt(end)
> s <- as.POSIXlt(start)
> d <- (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
>e$mday) - min(30, s$mday))
>
> return (d)
>}
>
> cashflows <- 0
> last.coupon <- maturity
> while (last.coupon > settlement) {
> print(last.coupon) # I need to store these dates
> last.coupon <- add.months(last.coupon, -12/frequency)
> cashflows <- cashflows + 1
>print(cashflows) # I need to store these cashflow numbers
> }
>
>The print command causes the following output
>
>[1] "2015-10-01"
>[1] 1
>[1] "2015-04-01"
>[1] 2
>[1] "2014-10-01"
>[1] 3
>[1] "2014-04-01"
>[1] 4
>
>My problem is how do I store these print outputs or while the loop is getting
>executed, how do I save these to some data.frame say
>
>output_dat
>
>cashflow_tenure cashflow_nos
>
>1 2015-10-01 1
>2 2015-04-01 2
>3 2014-10-01 3
>4 2014-04-01 4
>
>Kindly advise
>
>With regards
>
>Katherine
> [[alternative HTML version deleted]]
>
>
>__
>R-help@r-project.org mailing list
>https://stat.ethz.ch/mailman/listinfo/r-help
>PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>and provide commented, minimal, self-contained, reproducible code.
>
>
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Re: [R] How to store interim print results
Dear Sir,
Thanks a lot for your guidance and efforts. Appreciate it.
Thanks again.
Katherine
On Thursday, 3 April 2014 6:55 PM, jim holtman wrote:
This will get you close:
> settlement = as.Date("2013-11-25")
> maturity = as.Date("2015-10-01")
> coupon = 0.066
> yield = 0.1040
> basis = 1
> frequency = 2
> redemption = 100
>
> # __
>
> add.months = function(date, n)
+ {
+ nC <- seq(date, by=paste (n, "months"), length = 2)[2]
+ fD <- as.Date(strftime(as.Date(date), format='%Y-%m-01'))
+ C <- (seq(fD, by=paste (n+1, "months"), length = 2)[2])-1
+ if(nC>C) return(C)
+ return(nC)
+ }
>
> date.diff = function(end, start, basis=1) {
+ if (basis != 0 && basis != 4)
+ return(as.numeric(end - start))
+ e <- as.POSIXlt(end)
+ s <- as.POSIXlt(start)
+ d <- (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
e$mday) - min(30, s$mday))
+
+ return (d)
+ }
>
> output <- capture.output({ # collect the print output
+ cashflows <- 0
+ last.coupon <- maturity
+ while (last.coupon > settlement) {
+ print(last.coupon) # I need to store these dates
+ last.coupon <- add.months(last.coupon, -12/frequency)
+ cashflows <- cashflows + 1
+ print(cashflows) # I need to store these cashflow numbers
+ }
+
+ })
>
> # remove line numbers
> output <- sub("^", "", output)
>
> # remove extra quotes
> output <- gsub('"', '', output)
>
>
> # now read in the data
> report <- matrix(output, ncol = 2, byrow = TRUE)
>
> report
[,1] [,2]
[1,] "2015-10-01" "1"
[2,] "2015-04-01" "2"
[3,] "2014-10-01" "3"
[4,] "2014-04-01" "4"
>
>
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Thu, Apr 3, 2014 at 6:22 AM, Katherine Gobin
wrote:
Dear R forum,
>
>Following is an customized extract of a code I am working on.
>
>settlement = as.Date("2013-11-25")
>maturity = as.Date("2015-10-01")
>coupon = 0.066
>yield = 0.1040
>basis = 1
>frequency = 2
>redemption = 100
>
># __
>
>add.months = function(date, n)
>{
> nC <- seq(date, by=paste (n, "months"), length = 2)[2]
> fD <- as.Date(strftime(as.Date(date), format='%Y-%m-01'))
> C <- (seq(fD, by=paste (n+1, "months"), length = 2)[2])-1
> if(nC>C) return(C)
> return(nC)
>}
>
>date.diff = function(end, start, basis=1) {
> if (basis != 0 && basis != 4)
> return(as.numeric(end - start))
> e <- as.POSIXlt(end)
> s <- as.POSIXlt(start)
> d <- (360 * (e$year - s$year)) + (30 * (e$mon - s$mon )) + (min(30,
>e$mday) - min(30, s$mday))
>
> return (d)
>}
>
> cashflows <- 0
> last.coupon <- maturity
> while (last.coupon > settlement) {
> print(last.coupon) # I need to store these dates
> last.coupon <- add.months(last.coupon, -12/frequency)
> cashflows <- cashflows + 1
>print(cashflows) # I need to store these cashflow numbers
> }
>
>The print command causes the following output
>
>[1] "2015-10-01"
>[1] 1
>[1] "2015-04-01"
>[1] 2
>[1] "2014-10-01"
>[1] 3
>[1] "2014-04-01"
>[1] 4
>
>My problem is how do I store these print outputs or while the loop is getting
>executed, how do I save these to some data.frame say
>
>output_dat
>
>cashflow_tenure cashflow_nos
>
>1 2015-10-01 1
>2 2015-04-01 2
>3 2014-10-01 3
>4 2014-04-01 4
>
>Kindly advise
>
>With regards
>
>Katherine
> [[alternative HTML version deleted]]
>
>
>__
>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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and provide commented, minimal, self-contained, reproducible code.
[R] Conditional subtraction
Dear R forum
I have following data.frame
dat = data.frame(key = c("A", "B", "C", "D", "E", "E"), id = c("instru_A",
"instru_B", "instru_B", "instru_B", "instru_C", "instru_C"), price = c(101.38,
3.9306, 3.7488, 92.9624, 5.15, 96.1908), adj_factor = c(2.08, 2.5217, 2.5217,
2.5217, 3.08, 3.08))
> dat
key id price adj_factor
1 A instru_A 101.3800 2.0800
2 B instru_B 3.9306 2.5217
3 C instru_B 3.7488 2.5217
4 D instru_B 92.9624 2.5217
5 E instru_C 5.1500 3.0800
6 E instru_C 96.1908 3.0800
This is just a part of big database and ids can appear any no of times.
# MY PROBLEM
I need to subtract adj_factor from the price, however only from the first id
only.
In case of instru_A, there is only 1 id, so 2.08 should be subtracted from
101.38.
The id "instru_B" is appearing 3 times. So in this case, adj_factor = 2.5217
should be subtracted from 3.9306 and rest should remain same.
Similarly, id "instru_C" is appearing 2 times, hence the adj_factor = 3.08
should be subtracted from 5.15.
Effectively I am looking for
> dat_new
key id price adj_factor adjusted_price
1 A instru_A 101.3800 2.0800 99.3000 # price adjusted
2 B instru_B 3.9306 2.5217 1.4089 # price adjusted
3 C instru_B 3.7488 2.5217 3.7488
4 D instru_B 92.9624 2.5217 92.9624
5 E instru_C 5.1500 3.0800 2.0700 # price adjusted
6 E instru_C 96.1908 3.0800 96.1908
I tried something like
adj_price = function(id, price, adj_factor)
{
id_length = length(id)
if(id_length == 1)
{
(adjusted_price = price-adj_factor)
}
if(id_length == 2)
{
(adjusted_price = c(price[1]-adj_factor[1], price[2]))
}
if(id_length > 2)
{
(adjusted_price = c(price[1]-adj_factor[1],price[2:id_length]))
}
return(adjusted_price)
}
(final_price = adj_price(dat$id, dat$price, dat$adj_factor))
> (final_price = adj_price(dat$id, dat$price, dat$adj_factor))
[1] 99.3000 3.9306 3.7488 92.9624 5.1500 96.1908
Kindly advise
Regards
Katherine
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and provide commented, minimal, self-contained, reproducible code.
[R] R equivalent functions of some EXCEL functions
Dear R forum, EXCEL has some standard functions e.g. (1) PRICE function : Returns the price per $100 face value of a security that pays periodic interest. (2) COUPDAYBS : Returns the number of days from the beginning of the coupon period to the settlement date. (3) COUPDAYS : Returns the number of days in the coupon period that contains settlement date. 4) COUPDAYSNC : Returns the number of days from the settlement date to the next coupon date. Kindly guide if R has some inbuilt functions giving the results same as obtained from Excel functions mentioned above. With regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Splitting vector elements
Dear R forum
I have a vector as
dat = c("ABC 1", "ABC 2", "ABC 3", "DEF 10", "DEF 20")
> dat
[1] "ABC 1" "ABC 2" "ABC 3" "DEF 10" "DEF 20"
I need to split the names into two parts say
p1 p2
ABC 1
ABC 2
ABC 3
DEF 10
DEF 20
Kindly guide
Katherine
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and provide commented, minimal, self-contained, reproducible code.
[R] Paper on Analytics using R
Dear R Forum, I am looking for some write-up or paper on Use of R for Analytics or why R should be preferred over others for Analytics purpose. Tried google but got some info about some commercial vendors using R for analytics. I am looking for some paper where no commercial flavor is given, I mean it deals with R strictly and doesn't talk about some product using R for analytics. Kindly share if you are aware of some writeups or paper. Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Reversing the Equation to find value of variable
Dear R forum I have following variables - EAD = 1 LGD = 0.45 PD = 0.47 M = 3 # Equation 1 R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) + 0.24*(1-(1-exp(-50*PD))/(1-exp(-50))) b = (0.11852 - 0.05478 * log(PD))^2 K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 * qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b) RWA = K * 12.5 * EAD > RWA [1] 22845.07 # _ # MY Problem In the above part, knowing values of LGD, EAD, M and PD, the value of RWA was calculated. However, I need to go reverse way in the sense knowing the values of LGD, EAD, M and RWA, I need to find value of PD. So I have tried to use uniroot as (RWA - K * 12.5 * EAD and used the above equations i place of K and R) RWA = 22845.07 LGD = 0.45 EAD = 1 M = 3 f = function(x) RWA - (LGD*pnorm((1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50)^(-0.5)*qnorm(x)+((0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50/(1-(0.12*(1-exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50))^0.5*qnorm(0.999))-x*LGD) * (1-1.5*((0.11852-0.05478 * log(x))^2))^(-1)*(1+(M-2.5)*((0.11852-0.05478 * log(x))^2))*12.5*EAD uniroot(f, c(0,1), tol = 0.01) I get following error - > uniroot(f, c(0,1), tol = 0.01) Error in uniroot(f, c(0, 1), tol = 1e-10) : f.lower = f(lower) is NA Kindly guide as I am not sure if uniroot is the correct way of doing it or not. Ideally, I should be getting the PD value of 0.47. With regards Katherine [[alternative HTML version deleted]] __ 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.
Re: [R] Reversing the Equation to find value of variable
Dear Sir, Thanks a lot for your wonderful guidance. It gave me a new vision to look at teh equations. Really appreciate. Thanks a lot once again. Katherine On Monday, 6 January 2014 5:31 PM, Frede Aakmann Tøgersen wrote: Hi Reading the error message carefully you can see that f() is not defined at 0: > uniroot(f, c(0, 1)) Error in uniroot(f, c(0, 1)) : f.lower = f(lower) is NA > f(0) [1] NaN If you plot f() in the interval (0,1) then you'll see there is two solutions: > uniroot(f, c(0.0001, 1)) Error in uniroot(f, c(1e-04, 1)) : f() values at end points not of opposite sign > uniroot(f, c(0.0001, 0.2)) $root [1] 0.1533901 $f.root [1] 0.3414232 $iter [1] 6 $estim.prec [1] 6.103516e-05 > uniroot(f, c(0.3, 1)) $root [1] 0.4699984 $f.root [1] -0.04112121 $iter [1] 8 $estim.prec [1] 6.103516e-05 > Yours sincerely / Med venlig hilsen Frede Aakmann Tøgersen Specialist, M.Sc., Ph.D. Plant Performance & Modeling Technology & Service Solutions T +45 9730 5135 M +45 2547 6050 fr...@vestas.com http://www.vestas.com Company reg. name: Vestas Wind Systems A/S This e-mail is subject to our e-mail disclaimer statement. Please refer to www.vestas.com/legal/notice If you have received this e-mail in error please contact the sender. > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] > On Behalf Of Katherine Gobin > Sent: 6. januar 2014 12:42 > To: r-help@r-project.org > Subject: [R] Reversing the Equation to find value of variable > > Dear R forum > > I have following variables - > > EAD = 1 > LGD = 0.45 > PD = 0.47 > M = 3 > > # Equation 1 > > R = 0.12*(1-exp(-50*PD))/(1-exp(-50)) + 0.24*(1-(1-exp(-50*PD))/(1-exp(- > 50))) > > b = (0.11852 - 0.05478 * log(PD))^2 > > K = (LGD * pnorm((1 - R)^(-0.5) * qnorm(PD) + (R / (1 - R))^0.5 * > qnorm(0.999)) - PD * LGD) * (1 - 1.5 * b)^(-1) * (1 + (M - 2.5) * b) > > RWA = K * 12.5 * EAD > > > > RWA > [1] 22845.07 > > # > __ > ___ > > # MY Problem > > In the above part, knowing values of LGD, EAD, M and PD, the value of RWA > was calculated. However, I need to go reverse way in the sense knowing the > values of LGD, EAD, M and RWA, I need to find value of PD. > > So I have tried to use uniroot as (RWA - K * 12.5 * EAD and used the above > equations i place of K and R) > > RWA = 22845.07 > LGD = 0.45 > EAD = 1 > M = 3 > > f = function(x) RWA - (LGD*pnorm((1-(0.12*(1-exp(-50*x))/(1-exp(- > 50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50)^(-0.5)*qnorm(x)+((0.12*(1- > exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(-50/(1-(0.12*(1- > exp(-50*x))/(1-exp(-50))+0.24*(1-(1-exp(-50*x))/(1-exp(- > 50))^0.5*qnorm(0.999))-x*LGD) * (1-1.5*((0.11852-0.05478 * > log(x))^2))^(-1)*(1+(M-2.5)*((0.11852-0.05478 * log(x))^2))*12.5*EAD > > uniroot(f, c(0,1), tol = 0.01) > > I get following error - > > > uniroot(f, c(0,1), tol = 0.01) > Error in uniroot(f, c(0, 1), tol = 1e-10) : f.lower = f(lower) is NA > > Kindly guide as I am not sure if uniroot is the correct way of doing it or > not. > Ideally, I should be getting the PD value of 0.47. > > With regards > > Katherine > [[alternative HTML version deleted]] [[alternative HTML version deleted]] __ 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.
[R] Running the Loop
Dear R forum,
I have following data.frames
dat = data.frame(id = c(1:3), root = c(0.10, 0.20, 0.74), maturity_period =
c(20, 155, 428), mtm = c(1000, 1, 10), curve = c("USD", "USD", "USD"))
> dat
id root maturity_period mtm curve
1 1 0.10 20 1e+03 USD
2 2 0.20 155 1e+04 USD
3 3 0.74 428 1e+05 USD
standard_tenors = data.frame(T = c("1m", "3m", "6m", "12m", "5yr"), D = c(30,
91, 182, 365, 1825))
> standard_tenors
T D
1 1m 30
2 3m 91
3 6m 182
4 12m 365
5 5yr 1825
#
.
library(plyr)
T = standard_tenors$T
D = standard_tenors$D
n = length(standard_tenors$T)
mtm_split_function = function(maturity_period, curve, root, mtm)
{
for(i in 1:(n-1))
{
if (maturity_period < D[i])
{
N1 = paste(curve, T[i], sep ="_")
N2 = paste(curve, T[i], sep ="_")
PV1 = mtm
PV2 = 0
}else
if (maturity_period > D[i] & maturity_period < D[i+1])
{
N1 = paste(curve, T[i], sep ="_")
N2 = paste(curve, T[1+1], sep ="_")
PV1 = (mtm)*root
PV2 = (mtm)*(1-root)
}else
if (maturity_period > D[i+1])
{
N1 = paste(curve, T[i], sep ="_")
N2 = paste(curve, T[i], sep ="_")
PV1 = 0
PV2 = mtm
}
}
return(data.frame(Risk_factor1 = N1, Risk_factor2 = N2, Risk_factor1_mtm = PV1,
Risk_factor2_mtm = PV2))
}
#
.
splitted_mtm <- ddply(.data = dat, .variables = "id",
.fun=function(x) mtm_split_function(maturity_period =
x$maturity_period, curve = x$curve, root = x$root, mtm = x$mtm))
# OUTPUT I am getting
id Risk_factor1 Risk_factor2 Risk_factor1_mtm Risk_factor2_mtm
1 1 USD_12m USD_12m 1000 0
2 2 USD_12m USD_12m 1 0
3 3 USD_12m USD_3m 74000 26000
# My PROBLEM
However, My OUTPUT should be
id Risk_factor1 Risk_factor2 Risk_factor1_mtm Risk_factor2_mtm
1 1 USD_1m USD_1m 1000 0
2 2 USD_3m USD_6m 2000 8000
3 3 USD_12m USD_5yr 74000 26000
Kindly guide
With warm regards
Katherine
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and provide commented, minimal, self-contained, reproducible code.
[R] How to write an error to output
Dear R forum,
The example below is just an indicative one and I have constructed it. My real
life data and conditions are different.
I have a data.frame as given below
mydat = data.frame(A = c(19, 20, 19, 19, 19, 18, 16, 18, 19, 20), B = c(19, 20,
20, 19, 20, 18, 19, 18, 17, 16))
if (length(mydat$A) > 10)
{
stop("A has length more than 10")
}else
if (max(mydat$B) > 18)
{
stop("max B exceeds limit")
}else
{result = mydat$A + mydat$B
if (length(result) > 0)
{
write.csv(data.frame(result = result), 'result.csv', row.names = FALSE)
}
}
# -
When i execute above code, I get message
Error: max B exceeds limit
If all conditions are met, obviously I am getting an output as result.csv
If result.csv is generated, I am able to capture and show the output in front
end. However, if the process couldn't be run owing to the violation of
conditions, the error is produced. How do I capture this error (and express it
as csv file) so that I can show it as a comment in front end.
Kindly guide.
Katherine
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to write an error to output
Dear sir,
Thanks a lot for your wonderful suggestion.
Regards
Katherine
On Wednesday, 16 October 2013 5:28 PM, jim holtman wrote:
Will this work for you:
mydat = data.frame(A = c(19, 20, 19, 19, 19, 18, 16, 18, 19, 20), B =
c(19, 20, 20, 19, 20, 18, 19, 18, 17, 16))
if (length(mydat$A) > 10)
{
write.csv(data.frame(error = "A has length more than 10"),
'result.csv', row.names = FALSE)
stop("A has length more than 10")
}else
if (max(mydat$B) > 18)
{
write.csv(data.frame(error = "max B exceeds limit"), 'result.csv',
row.names = FALSE)
stop("max B exceeds limit")
}else
{result = mydat$A + mydat$B
if (length(result) > 0)
{
write.csv(data.frame(result = result), 'result.csv', row.names = FALSE)
}
}
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
On Wed, Oct 16, 2013 at 7:01 AM, Katherine Gobin
wrote:
> Dear R forum,
>
> The example below is just an indicative one and I have constructed it. My
> real life data and conditions are different.
>
> I have a data.frame as given below
>
> mydat = data.frame(A = c(19, 20, 19, 19, 19, 18, 16, 18, 19, 20), B = c(19,
> 20, 20, 19, 20, 18, 19, 18, 17, 16))
>
> if (length(mydat$A) > 10)
>
> {
> stop("A has length more than 10")
> }else
>
> if (max(mydat$B) > 18)
> {
> stop("max B exceeds limit")
> }else
>
> {result = mydat$A + mydat$B
>
> if (length(result) > 0)
>
> {
> write.csv(data.frame(result = result), 'result.csv', row.names =
>FALSE)
> }
> }
>
> # -
>
> When i execute above code, I get message
>
> Error: max B exceeds limit
>
> If all conditions are met, obviously I am getting an output as result.csv
>
> If result.csv is generated, I am able to capture and show the output in front
> end. However, if the process couldn't be run owing to the violation of
> conditions, the error is produced. How do I capture this error (and express
> it as csv file) so that I can show it as a comment in front end.
>
> Kindly guide.
>
>
> Katherine
> [[alternative HTML version deleted]]
>
>
> __
> 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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] Subseting a data.frame
Dear Forum, I have a data frame as mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) > mydat basel_asset_class defa_frequency 1 2 0.150 2 8 0.070 3 8 0.030 4 8 0.001 I need to get the subset of this data.frame where no of records for the given basel_asset_class is > 2, i.e. I need to obtain subset of above data.frame as (since there is only 1 record, against basel_asset_class = 2, I want to filter it) > mydat_a basel_asset_class defa_frequency 1 8 0.070 2 8 0.030 3 8 0.001 Kindly guide Katherine [[alternative HTML version deleted]] __ 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.
Re: [R] Subseting a data.frame
I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is > 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. > mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class > 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is > 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter wrote: "Kindly guide" ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin wrote: Dear Forum, > >I have a data frame as > >mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, >0.07, 0.03, 0.001)) > >> mydat > basel_asset_class defa_frequency >1 2 0.150 >2 8 0.070 >3 8 0.030 >4 8 0.001 > > >I need to get the subset of this data.frame where no of records for the given >basel_asset_class is > 2, i.e. I need to obtain subset of above data.frame as >(since there is only 1 record, against basel_asset_class = 2, I want to filter >it) > >> mydat_a > basel_asset_class defa_frequency >1 8 0.070 >2 8 0.030 >3 8 0.001 > >Kindly guide > >Katherine > [[alternative HTML version deleted]] > > >__ >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. > > -- Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 [[alternative HTML version deleted]] __ 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.
Re: [R] Subseting a data.frame
Correction. (2nd para first three lines) Pl read following line What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, as What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are THREE default frequencies w.r.t. basel aseet class 8, I alpologize for the incovenience. Regards KAtherine On , Katherine Gobin wrote: I am sorry perhaps was not able to put the question properly. I am not looking for the subset of the data.frame where the basel_asset_class is > 2. I do agree that would have been a basic requirement. Let me try to put the question again. I have a data frame as mydat = data.frame(basel_asset_class = c(4, 8, 8 ,8), defa_frequency = c(0.15, 0.07, 0.03, 0.001)) # Please note I have changed the basel_asset_class to 4 from 2, to avoid confusion. > mydat basel_asset_class defa_frequency 1 4 0.150 2 8 0.070 3 8 0.030 4 8 0.001 This is just an representative example. In reality, I may have no of basel asset classes. 4, 8 etc are the IDs can be anything thus I cant hard code it as subset(mydat, mydat$basel_asset_class > 2). What I need is to select only those records for which there are more than two default frequencies (defa_frequency), Thus, there is only one default frequency = 0.150 w.r.t basel_asset_class = 4 whereas there are default frequencies w.r.t. basel aseet class 4, similarly there could be another basel asset class having say 5 default frequncies. Thus, I need to take subset of the data.frame s.t. the no of corresponding defa_frequencies is greater than 2. The idea is we try to fit exponential curve Y = A exp( BX ) for each of the basel asset classes and to estimate values of A and B, mathematically one needs to have at least two values of X. I hope I may be able to express my requirement. Its not that I need the subset of mydat s.t. basel asset class is > 2 (now 4 in revised example), but sbuset s.t. no of default frequencies is greater than or equal to 2. This 2 is not same as basel asset class 2. Kindly guide With warm regards Katherine Gobin On Thursday, 17 October 2013 9:33 PM, Bert Gunter wrote: "Kindly guide" ... This is a very basic question, so the kindest guide I can give is to read an Introduction to R (ships with R) or a R web tutorial of your choice so that you can learn how R works instead of posting to this list. Cheers, Bert On Wed, Oct 16, 2013 at 11:55 PM, Katherine Gobin wrote: Dear Forum, > >I have a data frame as > >mydat = data.frame(basel_asset_class = c(2, 8, 8 ,8), defa_frequency = c(0.15, >0.07, 0.03, 0.001)) > >> mydat > basel_asset_class defa_frequency >1 2 0.150 >2 8 0.070 >3 8 0.030 >4 8 0.001 > > >I need to get the subset of this data.frame where no of records for the given >basel_asset_class is > 2, i.e. I need to obtain subset of above data.frame as >(since there is only 1 record, against basel_asset_class = 2, I want to filter >it) > >> mydat_a > basel_asset_class defa_frequency >1 8 0.070 >2 8 0.030 >3 8 0.001 > >Kindly guide > >Katherine > [[alternative HTML version deleted]] > > >__ >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. > > -- Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 [[alternative HTML version deleted]] __ 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.
Re: [R] Subseting a data.frame
Dear sir,
Thanks a lot for your guidance. I have been benefited immensely by this
discussion. Thanks again.
Regards
Katherine
On Friday, 18 October 2013 2:50 AM, Bert Gunter wrote:
Thanks, Bill.
But ?ave specifically says:
ave(x, ..., FUN = mean)
Arguments:
x
A numeric.
So that it should not be expected to work properly if the argument is
not (coercible to) numeric. Nevertheless, defensive programming is
always wise.
Cheers,
Bert
On Thu, Oct 17, 2013 at 1:34 PM, William Dunlap wrote:
> May I ask why:
> count_by_class <- with(dat, ave(numeric(length(basel_asset_class)),
> basel_asset_class, FUN=length))
>
> should not be more simply done as:
> count_by_class <- with(dat, ave(basel_asset_class, basel_asset_class,
> FUN=length))
>
> The way I did it would work if basel_asset_class were non-numeric.
>
> In ave(x, group, FUN=FUN), FUN's return value should be the same type as x
> (or
>
> you can get some odd type conversions). E.g.,
>
>
>
> > num <- c(2,3,2,2) ; char <- c("Two","Three","Two","Two")
>
> > ave(num, num, FUN=length) # good
>
> [1] 3 1 3 3
>
> > ave(char, char, FUN=length) # bad
>
> [1] "3" "1" "3" "3"
>
> > fac <- factor(char, levels=c("One","Two","Three"))
>
> > ave(fac, fac, FUN=length)
>
> [1]
>
> Levels: One Two Three
>
> Warning messages:
>
> 1: In `[<-.factor`(`*tmp*`, i, value = 0L) :
>
> invalid factor level, NA generated
>
> 2: In `[<-.factor`(`*tmp*`, i, value = 3L) :
>
> invalid factor level, NA generated
>
> 3: In `[<-.factor`(`*tmp*`, i, value = 1L) :
>
> invalid factor level, NA generated
>
> but x=integer(length(group)) works in all cases:
>
> > ave(integer(length(fac)), fac, FUN=length)
>
> [1] 3 1 3 3
>
> > ave(integer(length(char)), char, FUN=length)
>
> [1] 3 1 3 3
>
>
>
> Bill Dunlap
>
> Spotfire, TIBCO Software
>
> wdunlap tibco.com
>
>
>
> From: Bert Gunter [mailto:gunter.ber...@gene.com]
> Sent: Thursday, October 17, 2013 1:06 PM
> To: William Dunlap
> Cc: Katherine Gobin; r-help@r-project.org
> Subject: Re: [R] Subseting a data.frame
>
>
>
> May I ask why:
>
> count_by_class <- with(dat, ave(numeric(length(basel_
>
> asset_class)), basel_asset_class, FUN=length))
>
> should not be more simply done as:
>
> count_by_class <- with(dat, ave(basel_asset_class, basel_asset_class,
> FUN=length))
>
> ?
>
> -- Bert
>
>
>
> On Thu, Oct 17, 2013 at 12:36 PM, William Dunlap wrote:
>
>> What I need is to select only those records for which there are more than
>> two default
>> frequencies (defa_frequency),
>
> Here is one way. There are many others:
> > dat <- data.frame( # slightly less trivial example
> basel_asset_class=c(4,8,8,8,74,3,74),
> defa_frequency=(1:7)/8)
> > count_by_class <- with(dat, ave(numeric(length(basel_asset_class)),
> basel_asset_class, FUN=length))
> > cbind(dat, count_by_class) # see what we just computed
> basel_asset_class defa_frequency count_by_class
> 1 4 0.125 1
> 2 8 0.250 3
> 3 8 0.375 3
> 4 8 0.500 3
> 5 74 0.625 2
> 6 3 0.750 1
> 7 74 0.875 2
> > mydat[count_by_class>1, ] # I think this is what you are asking for
> basel_asset_class defa_frequency
> 2 8 0.250
> 3 8 0.375
> 4 8 0.500
> 5 74 0.625
> 7 74 0.875
>
> Bill Dunlap
> Spotfire, TIBCO Software
> wdunlap tibco.com
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
>> On Behalf
>> Of Katherine Gobin
>> Sent: Thursday, October 17, 2013 11:05 AM
>> To: Bert Gunter
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Subseting a data.frame
>>
>> Correction. (2nd para first three lines)
>>
>> Pl read following line
>>
>> What I need is to select only those records for which there are more than
>> two default
>> frequencies (defa_frequency), Thus, there is only one default frequency =
>> 0.150 w.r.t
>> basel_asse
[R] Yield to maturity in R
Dear R forum, Just want to know if there is any function / package in R which will calculate Yield to Maturity in R for a given bond? Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] Readjusting frequencies
Dear Forum, I have following data.frame as fraud_data = data.frame(no_of_frauds = c(1, 2, 4, 6, 7, 9, 10), frequency = c(3, 1, 7, 11, 13, 1, 4)) > fraud_data no_of_frauds frequency 1 1 3 2 2 1 3 4 7 4 6 11 5 7 13 6 9 1 7 10 4 I need to regroup the data in such a way that if the frequency is less than 5, the corresponding class data gets merged to next class i.e. the frequencies get added added till the added frequencies exceed 5. Thus, in above data.frame since frequencies pertaining to no_of_frauds 1 and 2 are 3 and 1 respectively, these get added to class 4 and the frequency of this class now becomes 3+1+7 = 11. Likewise, frequency of classes 9 and 10 are 1 and 4 and when these are added still it is 5 i.e. doesn't exceed 5. Thus, these should get added to the previous class i.e. 7. Thus I need to have no_of_frauds frequency 4 11 # ( 3 + 1 + 7) 6 11 7 18 # (13 + 1 + 4) Kindly guide Regards Katherine [[alternative HTML version deleted]] __ 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.
[R] lmom package
Dear R Forum I have a set of data say as given below and as an exercise of trying to fit statistical distribution to this data, I am estimating parameters. amounts = c(38572.5599129508,11426.6705314315,21974.1571641187,118530.32782443,3735.43055996748,66309.5211176106,72039.2934132668,21934.8841708626,78564.9136114375,1703.65825161293,2116.89180930203,11003.495671332,19486.3296339113,1871.35861218795,6887.53851253407,148900.978055447,7078.56497101651,79348.1239806592,20157.6241066905,1259.99802108593,3934.45912233674,3297.69946631591,56221.1154121067,13322.0705174134,45110.2498756567,31910.3686613912,3196.71168501252,32843.0140437202,14615.1499458453,13013.9915051561,116104.176753387,7229.03056392023,9833.37962177814,2882.63239493673,165457.372543821,41114.066453219,47188.1677766245,25708.5883755617,82703.7378298092,8845.04197017415,844.28834047836,35410.8486123933,19446.3808445684,17662.2398792892,11882.8497070776,4277181.17817307,30239.0371267968,45165.7512343364,22102.8513746687,5988.69296597127,51345.0146170238,1275658.35495898,15260.4892854214,8861.76578480635,37647.1638704867,4979.53544046949,7012.48134772332,3385.20612391205,1911.03114395959,66886.5036605189,2223.47536156462,814.947809578378,234.028589468841,5397.4347625133,13346.3226579065,28809.3901352898,6387.69226236731,5639.42730553242,2011100.92675507,4150.63707173462,34098.7514446498,3437.10672573502,289710.315303182,8664.66947305203,13813.3867161134,208817.521491857,169317.624400274,9966.78447705792,37811.1721605562,2263.19211279927,80434.5581206454,19057.8093104899,24664.5067589624,25136.5042354789,3582.85741610706,6683.13898432794,65423.9991390846,134848.302304064,3018.55371579808,546249.641168158,172926.689143006,3074.15064180208,1521.70624812788,59012.4248281661,21226.928522236,17572.5682970983,226.646947337851,56232.2982652019,14641.0043361533,6997.94414914865) library(lmom)lmom <- samlmu(amounts) # # Normal distribution parameters_of_NOR <- pelnor(lmom); parameters_of_NOR > parameters_of_NOR <- pelnor(lmom); parameters_of_NOR mu sigma > 115148.4 175945.8 # Minitab and SPSS parameter values Location Scale Minitab 115148.4 485173SPSS 115148.4 485173 # __ # Log normal 3 parameter distribution parameters_of_LN3 <- pelln3(lmom); parameters_of_LN3 > parameters_of_LN3 <- pelln3(lmom); parameters_of_LN3 zeta mu sigma 3225.798890 9.114879 2.240841 Location Scale ShapeMinitab 9.73361 1.76298 75.51864SPSS 9.7336 1.763 75.519 Similarly besides Generalized extreme Value distribution, all the parameter values vary significantly than parameter values obtained using Minitab and SPSS. In case of Normal distribution, the dispersion parameter is simply sample standard deviation and excel also gives the parameter value 485172.8 and varies significantly than what we get from R. And parameter values do differ even for many other distributions too viz. Gamma distribution etc. Is there any different algorithm or logic used in R? Can someone please guide.? Regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
[R] lmom package - Resending the email
Dear R forum I sincerely apologize as my earlier mail with the captioned subject, since all the values got mixed up and the email is not readable. I am trying to write it again. My problem is I have a set of data and I am trying to fit some distributions to it. As a part of this exercise, I need to find out the parameter values of various distributions e.g. Normal distribution, Log normal distribution etc. I am using lmom package to do the same, however the parameter values obtained using lmom pacakge differ to a large extent from the parameter values obtained using say MINITAB and SPSS as given below - _ amounts = c(38572.5599129508,11426.6705314315,21974.1571641187,118530.32782443,3735.43055996748,66309.5211176106,72039.2934132668,21934.8841708626,78564.9136114375,1703.65825161293,2116.89180930203,11003.495671332,19486.3296339113,1871.35861218795,6887.53851253407,148900.978055447,7078.56497101651,79348.1239806592,20157.6241066905,1259.99802108593,3934.45912233674,3297.69946631591,56221.1154121067,13322.0705174134,45110.2498756567,31910.3686613912,3196.71168501252,32843.0140437202,14615.1499458453,13013.9915051561,116104.176753387,7229.03056392023,9833.37962177814,2882.63239493673,165457.372543821,41114.066453219,47188.1677766245,25708.5883755617,82703.7378298092,8845.04197017415,844.28834047836,35410.8486123933,19446.3808445684,17662.2398792892,11882.8497070776,4277181.17817307,30239.0371267968,45165.7512343364,22102.8513746687,5988.69296597127,51345.0146170238,1275658.35495898,15260.4892854214,8861.76578480635,37647.1638704867,4979.53544046949,7012.48134772332,3385.20612391205,1911.03114395959,66886.5036605189,2223.47536156462,814.947809578378,234.028589468841,5397.4347625133,13346.3226579065,28809.3901352898,6387.69226236731,5639.42730553242,2011100.92675507,4150.63707173462,34098.7514446498,3437.10672573502,289710.315303182,8664.66947305203,13813.3867161134,208817.521491857,169317.624400274,9966.78447705792,37811.1721605562,2263.19211279927,80434.5581206454,19057.8093104899,24664.5067589624,25136.5042354789,3582.85741610706,6683.13898432794,65423.9991390846,134848.302304064,3018.55371579808,546249.641168158,172926.689143006,3074.15064180208,1521.70624812788,59012.4248281661,21226.928522236,17572.5682970983,226.646947337851,56232.2982652019,14641.0043361533,6997.94414914865) library(lmom) lmom = samlmu(amounts) # __ # Normal Distribution parameters parameters_of_NOR <- pelnor(lmom); parameters_of_NOR mu sigma 115148.4 175945.8 Location Scale Minitab 115148.4 485173SPSS 115148.4 485173 # __ # Log Normal (3 Parameter) Distribution parameters zeta mu sigma 3225.798890 9.114879 2.240841 Location Scale Shape MINITAB 9.73361 1.76298 75.51864SPSS 9.7336 1.763 75.519 # __ Besides Genaralized extreme Value distributions, all the other distributions e.g. Gamma, Exponential (2 parameter) distributions etc give different results than MINITAB and SPSS. Can some one guide me? Regards Katherine [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see 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.
