This is a nice trick, Jeff, thank you. I think this is what I was looking
for.
Thank you all.
On May 25, 2012 12:18 PM, "Jeff Newmiller" <jdnew...@dcn.davis.ca.us> wrote:
> This calls for a trick I have seen before on this list. Once you
> understand it, you will be able to apply it to many similar problems.
> The key is the "ave" function, which applies a function to various groups
> of values in a vector.
>
> a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10)
> f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1)
>
> at <- ifelse( is.na(a), f, a )
> lt <- cumsum( !is.na( a ) )
> cbind( lt, at ) # see the pattern of levels that will control ave
> ave( at, lt, FUN=cumprod )
>
> or in one statement
>
> ave( ifelse( is.na(a), f, a ), cumsum( !is.na( a ) ), FUN=cumprod )
>
> When learning, the trickiest step is defining the vector of levels.
> Usually a cumsum of booleans that mark transitions is involved. Sometimes
> rev(test(rev(data)))) can be useful.
>
> On Fri, 25 May 2012, David L Carlson wrote:
>
> This will loop only as many times as the largest number of consecutive
>> NA's
>> but uses vectorization within the loop. As currently defined, it will loop
>> forever if the first value is NA.
>>
>> a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10)
>> f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1)
>>
>> a1 <- a
>> alag <- c(NA, a1[1:length(a1)-1])
>> # change NA to the value to use if the first value in a is NA
>>
>> while (sum(is.na(a1)) > 0) {
>> a1 <- ifelse(is.na(a1), f*alag, a1)
>> alag <- c(NA, a1[1:length(a1)-1])
>> }
>>
>> ------------------------------**----------------
>> David L Carlson
>> Associate Professor of Anthropology
>> Texas A&M University
>> College Station, TX 77843-4352
>>
>>
>> -----Original Message-----
>>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>>> project.org] On Behalf Of Igor Reznikovsky
>>> Sent: Friday, May 25, 2012 9:08 AM
>>> To: Petr Savicky
>>> Cc: r-help@r-project.org
>>> Subject: Re: [R] Filling NA with cumprod?
>>>
>>> Hello Petr,
>>>
>>> Yes, I was hoping to avoid using loops. If nothing else works, I will
>>> take
>>> approach as the last resort.
>>>
>>> Thank you,
>>> Igor.
>>> On May 25, 2012 2:26 AM, "Petr Savicky" <savi...@cs.cas.cz> wrote:
>>>
>>> On Thu, May 24, 2012 at 08:24:38PM -0700, igorre25 wrote:
>>>>
>>>>> Hello,
>>>>>
>>>>> I need to build certain interpolation logic using R.
>>>>>
>>>> Unfortunately, I
>>>
>>>> just
>>>>
>>>>> started using R, and I'm not familiar with lots of advanced or just
>>>>> convenient features of the language to make this simpler. So I
>>>>>
>>>> struggled
>>>
>>>> for few days and pretty much reduced the whole exercise to the
>>>>>
>>>> following
>>>
>>>> problem, which I cannot resolve:
>>>>>
>>>>> Assume we have a vector of some values with NA:
>>>>> a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10)
>>>>>
>>>>> and some coefficients as a vector of the same length:
>>>>>
>>>>> f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1)
>>>>>
>>>>> I need to come up with function to get the following output
>>>>>
>>>>> o[1] = a[1]
>>>>> o[2] = a[2]
>>>>> o[3] = a[3]
>>>>> o[4] = o[3]*[f3] # Because a[3] is NA
>>>>> o[5] = o[4]*[f4] # Because a[4] is NA; This looks like recursive
>>>>> calculations; If the rest of the elements we NA, I would use a *
>>>>>
>>>> c(rep(1,
>>>>
>>>>> 3), cumprod(f[3:9])), but that's not the case
>>>>> o[6] = a[6] # Not NA anymore
>>>>> o[7] = a[7]
>>>>> o[8] = o[7]*f[7] # Again a[8] is NA
>>>>> o[9] = o[8]*f[8]
>>>>> o[10] = a[10] # Not NA
>>>>>
>>>>> Even though my explanation may seems complex, in reality the
>>>>>
>>>> requirement
>>>
>>>> is
>>>>
>>>>> pretty simple and in Excel is achieved with a very short formula.
>>>>>
>>>>> The need to use R is to demonstrate capabilities of the language
>>>>>
>>>> and
>>>
>>>> then to
>>>>
>>>>> expand to more complex problems.
>>>>>
>>>>
>>>> Hello:
>>>>
>>>> How is the output defined, if a[1] is NA?
>>>>
>>>> I think, you are not asking for a loop solution. However, in this
>>>>
>>> case,
>>>
>>>> it can be a reasonable option. For example
>>>>
>>>> a <- c(1, 2, 3, NA, NA, 6, 7, NA, NA, 10)
>>>> f <- c(0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1, 0.9, 1.1)
>>>> n <- length(a)
>>>> o <- rep(NA, times=n)
>>>>
>>>> prev <- 1
>>>> for (i in 1:n) {
>>>> if (is.na(a[i])) {
>>>> o[i] <- f[i]*prev
>>>> } else {
>>>> o[i] <- a[i]
>>>> }
>>>> prev <- o[i]
>>>> }
>>>>
>>>> A more straightforward translation of the Excel formulas is
>>>>
>>>> getCell <- function(i)
>>>> {
>>>> if (i == 0) return(1)
>>>> if (is.na(a[i])) {
>>>> return(f[i]*getCell(i-1))
>>>> } else {
>>>> return(a[i])
>>>> }
>>>> }
>>>>
>>>> x <- rep(NA, times=n)
>>>> for (i in 1:n) {
>>>> x[i] <- getCell(i)
>>>> }
>>>>
>>>> identical(o, x) # [1] TRUE
>>>>
>>>> Petr Savicky.
>>>>
>>>> ______________________________**________________
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>>>> PLEASE do read the posting guide
>>>> http://www.R-project.org/**posting-guide.html<http://www.R-project.org/posting-guide.html>
>>>> and provide commented, minimal, self-contained, reproducible code.
>>>>
>>>>
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>>
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>>
>>
> ------------------------------**------------------------------**
> ---------------
> Jeff Newmiller The ..... ..... Go Live...
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