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Vincy Pyne

2010-Dec-09 10:40 UTC

[R] Sequence generation in a table

Dear R helpers

I have following input

f = c(257, 520, 110). I need to generate a decreasing sequence (decreasing by
100) which will give me an input (in a tabular form) like

257, 157, 57
520, 420, 320, 220, 120, 20
110, 10


I tried the following R code


f = c(257, 520, 110)
yy = matrix(data = NA, nrow = 3, ncol = 6)

for (i in 1:3)
     {
     value = NULL

     for (j in 1 : 6)
          {
          value = c(ans, seq(f[i], 1, by = -100))
          }
    yy[i,] = ans[i,j]
    }

I get following message

Error in ans[i, j] : incorrect number of dimensions. Also, I understand above
logic will generate a result in (3 by 6) matrix format, while I need to generate
only 3 numbers pertaining to first no. i.e. 257, 6 nos. beginning from 520, and
only 2 numbers beginning from 110. I also tried tapply etc.

Please guide

Vincy





	[[alternative HTML version deleted]]

Jan van der Laan

2010-Dec-09 10:57 UTC

head link

[R] Sequence generation in a table

Vincy,

I suppose the following does what you want. yy is now a list which 
allows for differing lengths of the vectors.

 > yy <- lapply(c(257, 520, 110), seq, to=0, by=-100)
 > yy[[1]]
[1] 257 157  57
 > yy[[2]]
[1] 520 420 320 220 120  20

Regards,
Jan

On 9-12-2010 11:40, Vincy Pyne wrote:> c(257, 520, 110)

Vincy Pyne

2010-Dec-09 12:24 UTC

head link

[R] Sequence generation in a table

Dear Sirs,

I understand these already are numeric values. Sir, Basically I am working on
Value at Risk for the Bond portfolio using the historical simulation and for
this I need to find out Marked to Market (MTM) value suing the Present Value of
the coupon payments for each Bonds (here as an example I have taken only 3).

What I have done so far is for a given bond I have found no of days left for
maturity. E.g. in 1st case there are 257 days left for maturity. The bond pays
coupon twice a year and thus on 257th day the bond will mature and I will be
getting the Principal and final coupon payment. Since teh bond is paying the
coupons every 6 months, going backward from 257 th day, my earlier coupon
payment falls on (257 - 180) = 77 days. (However, in above example, I have just
taken 100 just for example purpose)

Thus, assuming 100 days, my coupons will be paid on 257, 157,  57days. I need to
convert these days in terms of years and so when I try to divide yy defined as

yy <- lapply(c(257, 520,
 110), seq, to=0, by=-100)

yy/360, I get following error.

Error in yy/360 : non-numeric argument to binary operator

On the other hand, 

yy[[1]]/365          fetches me 

[1] 0.7138889 0.4361111 0.1583333

Thus, I am trying to obtain the result yy <- lapply(c(257, 520,
 110), seq, to=0, by=-100) in such a form, so taht I should be able to further
analysis. What I was trying to say is since here I am taking only three bonds,
so I can do it individually, however if there are number of bonds (say 1000) in
the portfolio, my method of converting the days individually is not practical.

I am extremely sorry for the inconvenience caused. I tried to keep my problem
short in oder not to consume your valuable time.

Regards

Vince Pyne

--- On Thu, 12/9/10, Petr PIKAL <petr.pikal@precheza.cz> wrote:

From: Petr PIKAL <petr.pikal@precheza.cz>
Subject: Re: [R] Sequence generation in a table
To: "Vincy Pyne" <vincy_pyne@yahoo.ca>
Cc: r-help@r-project.org
Received: Thursday, December 9, 2010, 12:03 PM

Hi

r-help-bounces@r-project.org napsal dne 09.12.2010 12:41:47:
> Dear Sir,
> 
> Sorry to bother you again. Sir, the R code provided by you gives me 
following output.> 
> > yy <- lapply(c(257, 520,
 110), seq, to=0, by=-100)> > yy
> [[1]]
> [1] 257 157  57
> 
> [[2]]
> [1] 520 420 320 220 120  20
> 
> [[3]]
> [1] 110  10
> 
> The biggest constraint for me is here as an example I have taken only 
three > cases i.e. c(257, 520, 110), however in reality I will be dealing with 
no of > cases and that number is unknown. But your code will certainly generate 
me the> required numbers. In above case for doing further calculations, I can 
define say > 
> yy1 = as.numeric(yy[[1]])
> yy2 = as.numeric(yy[2]])
> yy3 = as.numeric(yy[[3]])
Why? Those values are already numeric.

lapply(yy, is.numeric)

[[1]]
[1] TRUE

[[2]]
[1] TRUE

[[3]]
[1] TRUE

and you can use the same construction to perform almost any operation on 
list.

lapply(yy, max)
lapply(yy,
 mean)
lapply(yy, sd)
lapply(yy, t.test)

Regards
Petr

> 
> But when the number of cases are unknown, perhaps this is not the 
practical > way of me defining individually. So is there any way that I can have all 
the > sequence numbers generated can be accommodated in a single dataframe. I 
> sincerely apologize for disturbing you Sir and hope I am able to put up 
my > problem in a proper manner.
> 
> Regards
> 
> Vincy Pyne
> 
> 
> --- On Thu, 12/9/10, Jan van der Laan <rhelp@eoos.dds.nl> wrote:
> 
> From: Jan van der Laan <rhelp@eoos.dds.nl>
> Subject: Re: [R] Sequence generation in a table
> To: r-help@r-project.org, vincy_pyne@yahoo.ca
> Received: Thursday, December 9, 2010, 10:57 AM
> 
> Vincy,
> 
> I suppose the following does what you want. yy is now a list which 
allows for > differing lengths of the vectors.
> 
> > yy <- lapply(c(257, 520, 110), seq, to=0, by=-100)
> > yy[[1]]
> [1] 257 157  57
> > yy[[2]]
> [1] 520 420 320 220 120  20
> 
> 
> Regards,
> Jan
> 
> 
> On 9-12-2010 11:40, Vincy Pyne wrote:
> > c(257, 520, 110)
> 
> 
> 
>    [[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.

	[[alternative HTML version deleted]]

Liviu Andronic

2010-Dec-09 12:30 UTC

head link

[R] Sequence generation in a table

On Thu, Dec 9, 2010 at 1:24 PM, Vincy Pyne <vincy_pyne at yahoo.ca>
wrote:> yy <- lapply(c(257, 520, 110), seq, to=0, by=-100)
>
> yy/360, I get following error.
>
> Error in yy/360 : non-numeric argument to binary operator
>
> On the other hand,
>
> yy[[1]]/365????????? fetches me
>
> [1] 0.7138889 0.4361111 0.1583333
>
What about this?> bond.fun <- function(x, days=360) {x/days}
> lapply(yy, bond.fun)[[1]]
[1] 0.7138889 0.4361111 0.1583333

[[2]]
[1] 1.44444444 1.16666667 0.88888889 0.61111111 0.33333333 0.05555556

[[3]]
[1] 0.30555556 0.02777778
> lapply(yy, bond.fun, days=366)[[1]]
[1] 0.7021858 0.4289617 0.1557377

[[2]]
[1] 1.42076503 1.14754098 0.87431694 0.60109290 0.32786885 0.05464481

[[3]]
[1] 0.30054645 0.02732240

Petr PIKAL

2010-Dec-09 13:07 UTC

head link

[R] Sequence generation in a table

Hi

r-help-bounces at r-project.org napsal dne 09.12.2010 13:24:17:
> Dear Sirs,
> 
> I understand these already are numeric values. Sir, Basically I am 
working on > Value at Risk for the Bond portfolio using the historical simulation and 
for > this I need to find out Marked to Market (MTM) value suing the Present 
Value > of the coupon payments for each Bonds (here as an example I have taken 
only 3). > 
> What I have done so far is for a given bond I have found no of days left 
for > maturity. E.g. in 1st case there are 257 days left for maturity. The 
bond pays> coupon twice a year and thus on 257th day the bond will mature and I 
will be > getting the Principal and final coupon payment. Since teh bond is paying 
the > coupons every 6 months, going backward from 257 th day, my earlier 
coupon > payment falls on (257 - 180) = 77 days. (However, in above example, I 
have > just taken 100 just for example purpose)
> 
> Thus, assuming 100 days, my coupons will be paid on 257, 157,  57days. I 
need > to convert these days in terms of years and so when I try to divide yy 
defined as > 
> yy <- lapply(c(257, 520,
>  110), seq, to=0, by=-100)
> 
> yy/360, I get following error.
> 
> Error in yy/360 : non-numeric argument to binary operator
Did you bother to read ?lapply or some intro
documentation?> 
> On the other hand, 
> 
> yy[[1]]/365          fetches me 
> 
> [1] 0.7138889 0.4361111 0.1583333
I do not believe it. I got
> yy[[1]]/365[1] 0.7041096 0.4301370 0.1561644

and as I already told you you can use lapply for almost any kind of 
computation. Just try
> lapply(yy, "/", 365)[[1]]
[1] 0.7041096 0.4301370 0.1561644

[[2]]
[1] 1.42465753 1.15068493 0.87671233 0.60273973 0.32876712 0.05479452

[[3]]
[1] 0.30136986 0.02739726

Regards
Petr


> 
> 
> Thus, I am trying to obtain the result yy <- lapply(c(257, 520,
>  110), seq, to=0, by=-100) in such a form, so taht I should be able to 
further> analysis. What I was trying to say is since here I am taking only three 
bonds,> so I can do it individually, however if there are number of bonds (say 
1000) > in the portfolio, my method of converting the days individually is not 
practical.> 
> I am extremely sorry for the inconvenience caused. I tried to keep my 
problem > short in oder not to consume your valuable time.
> 
> Regards
> 
> Vince Pyne
> 
> 
> 
> --- On Thu, 12/9/10, Petr PIKAL <petr.pikal at precheza.cz> wrote:
> 
> From: Petr PIKAL <petr.pikal at precheza.cz>
> Subject: Re: [R] Sequence generation in a table
> To: "Vincy Pyne" <vincy_pyne at yahoo.ca>
> Cc: r-help at r-project.org
> Received: Thursday, December 9, 2010, 12:03 PM
> 
> Hi
> 
> r-help-bounces at r-project.org napsal dne 09.12.2010 12:41:47:
> 
> > Dear Sir,
> > 
> > Sorry to bother you again. Sir, the R code provided by you gives me 
> following output.
> > 
> > > yy <- lapply(c(257, 520,
>  110), seq, to=0, by=-100)
> > > yy
> > [[1]]
> > [1] 257 157  57
> > 
> > [[2]]
> > [1] 520 420 320 220 120  20
> > 
> > [[3]]
> > [1] 110  10
> > 
> > The biggest constraint for me is here as an example I have taken only 
> three 
> > cases i.e. c(257, 520, 110), however in reality I will be dealing with
> no of 
> > cases and that number is unknown. But your code will certainly 
generate > me the
> > required numbers. In above case for doing further calculations, I can 
> define say 
> > 
> > yy1 = as.numeric(yy[[1]])
> > yy2 = as.numeric(yy[2]])
> > yy3 = as.numeric(yy[[3]])
> 
> Why? Those values are already numeric.
> 
> lapply(yy, is.numeric)
> 
> [[1]]
> [1] TRUE
> 
> [[2]]
> [1] TRUE
> 
> [[3]]
> [1] TRUE
> 
> and you can use the same construction to perform almost any operation on 
> list.
> 
> lapply(yy, max)
> lapply(yy,
>  mean)
> lapply(yy, sd)
> lapply(yy, t.test)
> 
> Regards
> Petr
> 
> 
> > 
> > But when the number of cases are unknown, perhaps this is not the 
> practical 
> > way of me defining individually. So is there any way that I can have 
all > the 
> > sequence numbers generated can be accommodated in a single dataframe. 
I > > sincerely apologize for disturbing you Sir and hope I am able to put 
up > my 
> > problem in a proper manner.
> > 
> > Regards
> > 
> > Vincy Pyne
> > 
> > 
> > --- On Thu, 12/9/10, Jan van der Laan <rhelp at eoos.dds.nl>
wrote:
> > 
> > From: Jan van der Laan <rhelp at eoos.dds.nl>
> > Subject: Re: [R] Sequence generation in a table
> > To: r-help at r-project.org, vincy_pyne at yahoo.ca
> > Received: Thursday, December 9, 2010, 10:57 AM
> > 
> > Vincy,
> > 
> > I suppose the following does what you want. yy is now a list which 
> allows for 
> > differing lengths of the vectors.
> > 
> > > yy <- lapply(c(257, 520, 110), seq, to=0, by=-100)
> > > yy[[1]]
> > [1] 257 157  57
> > > yy[[2]]
> > [1] 520 420 320 220 120  20
> > 
> > 
> > Regards,
> > Jan
> > 
> > 
> > On 9-12-2010 11:40, Vincy Pyne wrote:
> > > c(257, 520, 110)
> > 
> > 
> > 
> >    [[alternative HTML version deleted]]
> > 
> > ______________________________________________
> > R-help at 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 at 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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