[R] Conditionally adding a constant

[jim holtman]

If you are worried about an NA in the first, then use the following:

> y <- c(NA, 1, 2, NA, 4, NA)
> y <- na.locf(y, na.rm = FALSE)
> y
[1] NA  1  2  2  4  4
> y <- na.locf(y, fromLast = TRUE)
> y
[1] 1 1 2 2 4 4
>


On Mon, Jan 2, 2012 at 5:07 PM, Joshua Wiley <jwiley.psych at gmail.com> wrote:
> Good points, Rui.
>
> On Mon, Jan 2, 2012 at 12:48 PM, Rui Barradas <ruipbarradas at sapo.pt> wrote:
>> Hello again,
>>
>> I believe we are all missing something. Isn't it possible to have NAs as the
>> first values of 'y'?
>> And isn't it also possible to have x[1] > 3?
>
> Theoretically, yes, in the OPs data, maybe?  If the data is a time
> series (or time series like), the zoo package is not a bad environment
> to be working in anyways.  There are all sorts of handy functions (I
> had almost recommended na.approx() which replaces NAs with a linear
> interpolation) based on the OPs little example dataset.  Not sure if
> the +2 thing is just an attempt at interpolation though or something
> more general.
>
>>
>> Here is my point (I have changed function 'f2' to predict for such cases,
>> 'f1' is rubbish)
>>
>> # Rui
>> f3 <- function(x, y){
>>        inx <- which(x > 3)
>>        ynx <- which(is.na(y))
>>        for(i in which(inx %in% ynx)) y[ynx[i]] <- y[ynx[i]-1] + 2L
>>        y
>> }
>>
>> # Jim's, as a function, 'na.rm' option added or else 'df3' would produce an
>> error
>> require(zoo)
>> f4 <- function(x, y){
>>        y <- na.locf(y, na.rm=FALSE)
>>        inc <- cumsum(x > 3) * 2
>>        y + inc
>> }
>>
>> df <- data.frame(x = c(1,2,3,4,5), y = c(10,20,30,NA,NA))
>> df
>> df2 <- data.frame(x = c(1,2,3,4,5), y = c(10,20,NA,40,NA))
>> df2
>> df3 <- data.frame(x = c(1,2,3,4,5), y = rev(c(10,20,30,NA,NA)))
>> df3
>>
>> # Joshua
>> f(df$x, df$y)      # works
>> f(df2$x, df2$y)    # infinite loop
>> f(df3$x, df3$y)    # infinite loop
>>
>> # Rui
>> f3(df$x, df$y)     # works
>> f3(df2$x, df2$y)   # works as expected?
>> f3(df3$x, df3$y)   # works as expected?
>>
>> # Jim
>> f4(df$x, df$y)     # works
>> f4(df2$x, df2$y)   # works as expected?
>> f4(df3$x, df3$y)   # works as expected?
>>
>> If this makes sense, the performance tests are very much in favour of Jim's
>> solution.
>>
>>
>> # If this is what is asked for, test the performance
>> # with large enough N
>> N <- 1.e5
>> dftest <- data.frame(x=1:N, y=c(sample(c(rep(NA, 5), 10*1:5), N,
>> replace=TRUE)))
>>
>> sum(is.na(dftest))/N    # proportion of NAs in 'dftest'
>>
>> t2 <- system.time(invisible(apply(dftest, 2, f2)))[c(1, 3)]
>> t3 <- system.time(invisible(f3(dftest$x, dftest$y)))[c(1, 3)]
>> t4 <- system.time(invisible(f4(dftest$x, dftest$y)))[c(1, 3)]
>> rbind(t2=t2, t3=t3, t4=t4, t2.t3=t2/t3, t2.t4=t2/t4, t3.t4=t3/t4)
>>
>> Sample output
>>
>>      user.self   elapsed
>> t2      2.93000   2.95000
>> t3      0.22000   0.22000
>> t4      0.01000   0.01000
>> t2.t3  13.31818  13.40909
>> t2.t4 293.00000 295.00000
>> t3.t4  22.00000  22.00000
>>
>> A factor of 300 over the initial solution or 20+ over the other loop based
>> one.
>>
>> Downside, it needs an extra package loaded, but 'zoo' is rather common
>> place.
>>
>> Rui Barradas
>>
>>
>>
>>
>>
>> --
>> View this message in context: http://r.789695.n4.nabble.com/Conditionally-adding-a-constant-tp4253049p4254470.html
>> Sent from the R help mailing list archive at Nabble.com.
>>
>> ______________________________________________
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>> and provide commented, minimal, self-contained, reproducible code.
>
>
>
> --
> Joshua Wiley
> Ph.D. Student, Health Psychology
> Programmer Analyst II, Statistical Consulting Group
> University of California, Los Angeles
> https://joshuawiley.com/
>
> ______________________________________________
> 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.



-- 
Jim Holtman
Data Munger Guru

What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.