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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