[R-SIG-Finance] use rows and cols of a matrix as dates

[Joshua Ulrich]

On Thu, Mar 14, 2013 at 10:05 AM, Sven D <sduve at hotmail.com> wrote:
>
> On Thu, Mar 14, 2013 at 3:21 AM, Sven D <sduve@> wrote:
>> Dear All,
>>
>> I am trying to evaluate a gas storage contract and I am trying to use R do
>> the work for me. What I am trying to do is to generate a matrix like:
>>
>> gy <- matrix(0, nrow=13, ncol=13)
>> gy <- xts(gy, order.by = timeBasedSeq('201304/201404'))
>>
>> # these numbers are an imaginariy gas forward curve
>> forwardcurve <-
>> c(26.5,26.45,26.4,26.25,26.25,26.25,26.21,27.54,27.91,28.14,28.17,27.91)
>>
>> # populate first row and first col with the forward curve
>>
>> gy[1,c(2:13)] <- forwardcurve
>> gy[c(2:13),1] <- forwardcurve
>>
>> # calculate the monthly spreads
>>
>> for(i in 2:13) gy[i,c(i:13)] <- (gy[1, c(i:13)] - gy[i,1])
>>
> This throws an error (that you didn't mention) because arithmetic
> operations on zoo/xts objects merge by index before performing the
> operation.  In other words, you can't perform arithmetic operations on
> different rows without using coredata() to drop the index attribute.
> You need something like this instead:
>
> for(i in 2:13) {
>   gy[i, i:13] <- gy[1, i:13] - drop(coredata(gy[i,1]))
> }
>
>> # the matrix should look like this
>>> gy
> <snip... the output is incorrect anyway>
>>
>
> apologies, yes, I was performing the inital loop before I turned the matrix
> into an xts. I kept on trying to find a solution while writing the question.
>
>> I now need to xts the columns as well, is that possibel?
>>
> No.  xts is a specific class of object, it's not something you do to any
> object.
>
>> I tried
>>
>> gy <- xts(colnames(gy), order.by = timeBasedSeq('201304/201404'))
>>
>> but this doesnt give me the required result at all.
>>
> What is the required result?  You only give a vague description; an
> example would help others help you.  You could set the column names of
> 'gy' to the index values, but I have no idea if that's what you
> actually want...
>
> colnames(gy) <- index(gy)
>
> thank you, as things brings me closer to what I would like to do:
>
> as.Date(as.yearmon(colnames(gy)[1], format="%B %Y"))
>
> will return a date, and from that date value I will be able to calculate the
> time to maturity for each single spreadoption. Like:
>
> as.numeric(as.Date(as.yearmon(colnames(gy)[2], format="%B %Y")) -
> as.Date(as.yearmon(colnames(gy)[1], format="%B %Y"))) / 365
>
Thanks for including context even though you're posting via Nabble.

No need to do all the conversions.  Your matrix is symmetric, which
means you can use the index values when you want to reference the
column "Dates".

indexDate <- as.Date(index(gy))
(indexDate[2]-indexDate[1])/365
c(indexDate[2]-indexDate[1])/365  # not a difftime object

>
> In other words thank you for the input, this was helpful, apologies for
> being a bit vague, wasnt intended.
>
>
> Thank you.
>
> Sven
>

Best,
--
Joshua Ulrich  |  about.me/joshuaulrich
FOSS Trading  |  www.fosstrading.com

R/Finance 2013: Applied Finance with R  | www.RinFinance.com