R help - Jan 2012 - Conditionally adding a constant

I am trying to add a constant to the previous value of a variable based on
certain conditions. Maybe there is a simple way to do this that I am missing
completely. I have given an example below:

df <- data.frame(x = c(1,2,3,4,5), y = c(10,20,30,NA,NA))
> df
  x  y
1 1 10
2 2 20
3 3 30
4 4 NA
5 5 NA

I want to add 2 to the previous value of y, if x exceeds 3 (also will have
to handle NAs in the process). The resulting output would look like:

  x  y
1 1 10
2 2 20
3 3 30
4 4 32
5 5 34

Can someone please explain how to do it? Thank you.

Ravi










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Hello,

I believe this works.

f1 <- function(x){
	for(i in 2:length(x)) x[i] <- ifelse(x[i-1] > 3, x[i-1] + 2, x[i])
	x
}

f2 <- function(x){
	for(i in 2:length(x)) x[i] <- ifelse(is.na(x[i]) & (x[i-1] > 3),
x[i-1] +
2, x[i])
	x
}

df <- data.frame(x = c(1,2,3,4,5), y = c(10,20,30,NA,NA))

apply(df, 2, f1)      # df$x[4] > 3, df$x[5] also changes
apply(df, 2, f2)      # only df$y has NA's

Maybe there's a better way, avoiding the loop.

Rui Barradas


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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?

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





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