I have a dataframe DF, with two columns A and B shown below:
A B
1 0
3 0
4 0
2 1
6 0
4 1
7 1
8 1
1 0
A sliding window approach is performed as shown below. The mean is calulated for column B in a sliding window of size 3 sliding by 1 using: rollapply(DF$B, width=3,by=1). The mean values for each window are shown on the left side.
A: 1 3 4 2 6 4 7 8 1
B: 0 0 0 1 0 1 1 1 0
[0 0 0] 0
[0 0 1] 0.33
[0 1 0] 0.33
[1 0 1] 0.66
[0 1 1] 0.66
[1 1 1] 1
[1 1 0] 0.66
output: 0 0.33 0.33 0.66 0.66 1 1 1 0.66
Now, for each row/coordinate in column A, all windows containing the coordinate are considered and should retain the highest mean value which gives the results as shown in column 'output'.
I need to obtain the output as shown above. The output should like:
A B Output
1 0 0
3 0 0.33
4 0 0.33
2 1 0.66
6 0 0.66
4 1 1
7 1 1
8 1 1
1 0 0.66
Any help in R?
Answer
Try this:
# form input data
library(zoo)
B <- c(0, 0, 0, 1, 0, 1, 1, 1, 0)
# calculate
k <- 3
rollapply(B, 2*k-1, function(x) max(rollmean(x, k)), partial = TRUE)
The last line returns:
[1] 0.0000000 0.3333333 0.3333333 0.6666667 0.6666667 1.0000000 1.0000000
[8] 1.0000000 0.6666667
If there are NA values you might want to try this:
k <- 3
B <- c(1, 0, 1, 0, NA, 1)
rollapply(B, 2*k-1, function(x) max(rollapply(x, k, mean, na.rm = TRUE)), partial = TRUE)
where the last line gives this:
[1] 0.6666667 0.6666667 0.6666667 0.5000000 0.5000000 0.5000000
Expanding it out these are formed as:
c(mean(B[1:3], na.rm = TRUE), ##
max(mean(B[1:3], na.rm = TRUE), mean(B[2:4], na.rm = TRUE)), ##
max(mean(B[1:3], na.rm = TRUE), mean(B[2:4], na.rm = TRUE), mean(B[3:5], na.rm = TRUE)),
max(mean(B[2:4], na.rm = TRUE), mean(B[3:5], na.rm = TRUE), mean(B[4:6], na.rm = TRUE)),
max(mean(B[3:5], na.rm = TRUE), mean(B[4:6], na.rm = TRUE)), ##
mean(B[4:6], na.rm = TRUE)) ##
If you don't want the k-1 components at each end (marked with ## above) drop partial = TRUE.