I have managed to reduce it to two statements using the trick that the
elements of x-x are NA or 0 according to whether the corresponding
element of x is NA or not. You could probably extend this to your
situation too.
> library(zoo)
> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, NA), seq(7)^2), as.Date(1:7))
>
> r <- replace(m[,1], TRUE,
+ na.locf(m[,1]) * m[,2] / na.locf(m[,2] + (m[,1]-m[,1])))
> cbind(V1 = r, V2 = m[,2])
V1 V2
1970-01-02 1.0 1
1970-01-03 2.0 4
1970-01-04 4.5 9
1970-01-05 8.0 16
1970-01-06 5.0 25
1970-01-07 7.2 36
1970-01-08 9.8 49
On Wed, May 12, 2010 at 4:58 PM, Abiel X Reinhart
<abiel.x.reinh...@jpmchase.com> wrote:
> Thanks Gabor, this looks like it serves my needs. I've extended the code to
> work with an example where we have two multicolumn zoo objects, one with the
> original data and another that has the growth rates.
>
> # mat1 = zoo object to extend
> # mat2 = zoo object whose growth rate is used to extend mat1
> mergeGrowth <- function(mat1, mat2)
> {
> ix <- is.na(mat1)
> mat1.locf <- na.locf(mat1, na.rm=F)
> mat2.locf <- mat2
> mat2.locf[ix] <- NA
> mat2.locf <- na.locf(mat2.locf, na.rm=F)
> coredata(mat1)[ix] <- coredata(mat1.locf * mat2 / mat2.locf)[ix]
> mat1
> }
>
>
> Abiel Reinhart
>
> -----Original Message-----
> From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Sent: Wednesday, May 12, 2010 12:00 PM
> To: Abiel X Reinhart
> Cc: r-help@r-project.org
> Subject: Re: [R] Vectorized expression to extrapolate matrix columns with
> columns of another matrix
>
> Yes, that is what it does. Note that na.approx interpolates and does
> not work precisely as you discussed but its easy, does use m[,2] and
> may be good enough. If you really do want something precisely as you
> discussed try this. It NAs out the rows of m for which column 1 is NA
> and then uses na.locf to move the prior non-NA into it. Then we apply
> the formula:
>
>> library(zoo)
>> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, NA), seq(7)^2), as.Date(1:7))
>>
>> # mm will hold result; m.locf
>> m.locf <- mm <- m
>> ix <- is.na(mm[,1])
>> m.locf[ix,] <- NA
>> m.locf <- na.locf(m.locf)
>> mm[ix, 1] <- m.locf[ix, 1] * mm[ix,2] / m.locf[ix,2]
>> mm
>
> 1970-01-02 1.0 1
> 1970-01-03 2.0 4
> 1970-01-04 4.5 9
> 1970-01-05 8.0 16
> 1970-01-06 5.0 25
> 1970-01-07 7.2 36
> 1970-01-08 9.8 49
>
> 1970-01-02 1.0 1
> 1970-01-03 2.0 4
> 1970-01-04 4.5 9
> 1970-01-05 8.0 16
> 1970-01-06 5.0 25
> 1970-01-07 7.2 36
> 1970-01-08 9.8 49
>
>
>
>
>
> Yes, that is what it does. Please read the help file for na.approx and
> approx. If you want something different you will have to special case
> the end values.
>
> On Wed, May 12, 2010 at 11:11 AM, Abiel X Reinhart
> <abiel.x.reinh...@jpmchase.com> wrote:
>> Gabor,
>>
>> Maybe I am doing this wrong, but rule=2 does not look like it is growing the
>> series out, but rather just carrying the last value forward. It looks like
>> na.approx() followed by na.locf(). For instance:
>>
>> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, NA), seq(7)^2), as.Date(1:7))
>> na.approx(m[, 1], x = m[, 2], rule=2)
>>
>> 1970-01-02 1970-01-03 1970-01-04 1970-01-05 1970-01-06 1970-01-07 1970-01-08
>> 1.0 2.0 2.7 3.7 5.0 5.0 5.0
>>
>> Abiel Reinhart
>>
>> -----Original Message-----
>> From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
>> Sent: Wednesday, May 12, 2010 10:40 AM
>> To: Abiel X Reinhart
>> Cc: r-help@r-project.org
>> Subject: Re: [R] Vectorized expression to extrapolate matrix columns with
>> columns of another matrix
>>
>> Use rule = 2 as in the extrapolation examples in the na.approx help file.
>>
>> On Wed, May 12, 2010 at 10:10 AM, Abiel X Reinhart
>> <abiel.x.reinh...@jpmchase.com> wrote:
>>> Gabor,
>>>
>>> This comes close to solving my problem, but I am still left with the
>>> problem of how I can extrapolate, not just interpolate. In our example, if
>>> I define m as,
>>>
>>> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, NA), seq(7)^2), as.Date(1:7))
>>>
>>> instead of
>>>
>>> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, 7), seq(7)^2), as.Date(1:7))
>>>
>>> then I will only get five values back, when I really want m[,1] to be fully
>>> extrapolated so that there are seven values. Is there a workaround?
>>>
>>> Abiel Reinhart
>>>
>>> -----Original Message-----
>>> From: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
>>> Sent: Wednesday, May 12, 2010 8:37 AM
>>> To: Abiel X Reinhart
>>> Cc: r-help@r-project.org
>>> Subject: Re: [R] Vectorized expression to extrapolate matrix columns with
>>> columns of another matrix
>>>
>>> Try this using the zoo package. See ?na.approx for more and note that
>>> this functionality requires zoo 1.6-3 or later.
>>>
>>> .> m <- zoo(cbind(c(1, 2, NA, NA, 5, NA, 7), seq(7)^2), as.Date(1:7))
>>>> na.approx(m[, 1], x = m[, 2])
>>> 1970-01-02 1970-01-03 1970-01-04 1970-01-05 1970-01-06 1970-01-07 1970-01-08
>>> 1.000000 2.000000 2.714286 3.714286 5.000000 5.916667 7.000000
>>>> na.approx(m[, 1])
>>> 1970-01-02 1970-01-03 1970-01-04 1970-01-05 1970-01-06 1970-01-07 1970-01-08
>>> 1 2 3 4 5 6 7
>>>
>>>
>>> On Tue, May 11, 2010 at 4:48 PM, Abiel X Reinhart
>>> <abiel.x.reinh...@jpmchase.com> wrote:
>>>> I have two identically sized matrices of data that represent time series
>>>> (I am storing the data in zoo objects, but the idea should apply to any
>>>> matrix of data). The time series in the second matrix extend further than
>>>> in the first matrix, and I would like to use the data in matrix 2 to
>>>> extrapolate the data in matrix 1. In other words, if mat1[i,j] == NA, then
>>>> mat1[i,j] <- mat1[i-1, j]*mat2[i,j]/mat2[i-1,j]. Of course, before we can
>>>> calculate mat1[i,j] we may need to calculate mat1[i-1,j], and that in turn
>>>> may require the computation of mat1[i-2,j], etc. This could all clearly be
>>>> done with loops, but I am wondering if anyone can think of a vectorized
>>>> expression or other efficient method that would work.
>>>>
>>>> Thanks very much.
>>>>
>>>> Abiel Reinhart
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