R - =if max statement from excel to R - r

Question

Within a data frame. I want to compare today's value against a look back 'n' day period.
I know how to do it in excel for comparing today's value to see if it was higher than the previous 10 days.
=IF(A11>MAX(A1:A10),1,0)
How can I do the same logic within a function in R?
The output would look like this below:
Column Output
1 12 NA
2 13 NA
3 14 NA
4 15 NA
5 9 NA
6 9 NA
7 7 NA
8 8 NA
9 16 NA
10 17 NA
11 20 1
12 14 0
13 9 0
14 8 0
15 6 0
16 5 0
17 28 1
In row 11. Because value 20 is higher than the previous 10 days it denotes a 1 value.
In row 12, because value 14 is not the highest number in previous 10 days it receives a 0 value.
And it of course rolls on moving 10 day window.

Answer 1

P Lapointe's answer is great but anytime I'm doing a 'rolling' calculation my first instinct is to think of rollapply from the zoo package.
is_last_greatest <- function(x){
#' Given an input vector this returns
#' 1 if the last element is greater than
#' all of the other elements and 0 otherwise
ifelse(all(tail(x,1) > head(x,-1)), 1, 0)
}
# We want to compare to the previous 10 values but the function
# I wrote requires us to include the value we're using as
# comparison so I set the width to 11
output <- rollapply(dat,
width = 11,
FUN = is_last_greatest,
fill = NA,
align = "right")
cbind(dat, output)
which gives
dat vals
[1,] 12 NA
[2,] 13 NA
[3,] 14 NA
[4,] 15 NA
[5,] 9 NA
[6,] 9 NA
[7,] 7 NA
[8,] 8 NA
[9,] 16 NA
[10,] 17 NA
[11,] 20 1
[12,] 14 0
[13,] 9 0
[14,] 8 0
[15,] 6 0
[16,] 5 0
[17,] 28 1

Answer 2

Here's how to do that with roll_maxr from RcppRoll.
library(RcppRoll)
df$Output2 <- ifelse(df$Column>roll_maxr(lag(df$Column),11, na.rm = TRUE),1,0)
Column Output Output2
1 12 NA NA
2 13 NA NA
3 14 NA NA
4 15 NA NA
5 9 NA NA
6 9 NA NA
7 7 NA NA
8 8 NA NA
9 16 NA NA
10 17 NA NA
11 20 1 1
12 14 0 0
13 9 0 0
14 8 0 0
15 6 0 0
16 5 0 0
17 28 1 1
data
df <- read.table(text=" Column Output
1 12 NA
2 13 NA
3 14 NA
4 15 NA
5 9 NA
6 9 NA
7 7 NA
8 8 NA
9 16 NA
10 17 NA
11 20 1
12 14 0
13 9 0
14 8 0
15 6 0
16 5 0
17 28 1",header=TRUE,stringsAsFactors=FALSE)

Answer 3

Here is a base R method using embed to construct the comparison sets and performing the comparisons with apply.
# get a matrix for comparisons
myMat <- embed(df$Column, 11)
Here, this returns
myMat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
[1,] 20 17 16 8 7 9 9 15 14 13 12
[2,] 14 20 17 16 8 7 9 9 15 14 13
[3,] 9 14 20 17 16 8 7 9 9 15 14
[4,] 8 9 14 20 17 16 8 7 9 9 15
[5,] 6 8 9 14 20 17 16 8 7 9 9
[6,] 5 6 8 9 14 20 17 16 8 7 9
[7,] 28 5 6 8 9 14 20 17 16 8 7
So the goal is to compare the value in the first column with those in the remaining column for each row.
as.integer(max.col(myMat) == 1L)
[1] 1 0 0 0 0 0 1
Now, tack on the appropriate number of NA values, which is the number of columns in myMat minus one.
df$output2 <- c(rep(NA, ncol(myMat) - 1), as.integer(max.col(myMat) == 1L))
this returns
df
Column Output output2
1 12 NA NA
2 13 NA NA
3 14 NA NA
4 15 NA NA
5 9 NA NA
6 9 NA NA
7 7 NA NA
8 8 NA NA
9 16 NA NA
10 17 NA NA
11 20 1 1
12 14 0 0
13 9 0 0
14 8 0 0
15 6 0 0
16 5 0 0
17 28 1 1
An advantage of max.col is that it is quite fast. One of its biggest drawbacks is that it does not have an na.rm argument to remove missing values. In the case that there are missing values, here is a method using apply on myMat instead of max.col.
apply(myMat, 1, function(x) as.integer(all(head(x, 1) > tail(x, -1))))
The operating comparison function here is
all(head(x, 1) > tail(x, -1))
Functions that produce the same result include the following
head(x, 1) == max(x) # or
x[1] == max(x)
and
1L == which.max(x)