Rolling lowest mean and stdev of n consecutive minutes of each time period - r

Question

New R user here looking for guidance. I am working with a 15-minute data set and looking to parse out the following by one variable (buildings in my case) for each day of the year:
(1) lowest mean of "value" for n consecutive rows (preferably 2 or 3 hours worth)
(2) standard deviation of the same period
Sample df:
variable year month day hr min date value
building_a 2018 6 2 0 0 6/2/2018 19
building_a 2018 6 2 0 15 6/2/2018 19
building_a 2018 6 2 0 30 6/2/2018 19
building_a 2018 6 2 0 45 6/2/2018 17
building_a 2018 6 2 1 0 6/2/2018 17
building_a 2018 6 2 1 15 6/2/2018 15
building_a 2018 6 2 1 30 6/2/2018 15
building_a 2018 6 2 1 45 6/2/2018 14
building_a 2018 6 2 2 0 6/2/2018 14
building_a 2018 6 2 2 15 6/2/2018 13
building_a 2018 6 2 2 30 6/2/2018 13
building_a 2018 6 2 2 45 6/2/2018 13
building_a 2018 6 2 3 0 6/2/2018 12
building_a 2018 6 2 3 15 6/2/2018 14
building_a 2018 6 2 3 30 6/2/2018 13
building_a 2018 6 2 3 45 6/2/2018 13
building_b 2018 6 2 0 0 6/2/2018 37
building_b 2018 6 2 0 15 6/2/2018 41
building_b 2018 6 2 0 30 6/2/2018 38
building_b 2018 6 2 0 45 6/2/2018 39
building_b 2018 6 2 1 0 6/2/2018 37
building_b 2018 6 2 1 15 6/2/2018 36
building_b 2018 6 2 1 30 6/2/2018 34
building_b 2018 6 2 1 45 6/2/2018 34
building_b 2018 6 2 2 0 6/2/2018 35
building_b 2018 6 2 2 15 6/2/2018 35
building_b 2018 6 2 2 30 6/2/2018 29
building_b 2018 6 2 2 45 6/2/2018 32
building_b 2018 6 2 3 0 6/2/2018 30
building_b 2018 6 2 3 15 6/2/2018 33
building_b 2018 6 2 3 30 6/2/2018 30
building_b 2018 6 2 3 45 6/2/2018 32
I've been able to perform this for one-hour segments using the following approach, but cannot figure out how to adapt this to a larger window (e.g., lowest 135 minute mean instead of 60 min).
tmp <- aggregate(value~variable+date+hour, df,
function(x)
c(mean = mean(x), sd = sd(x)))
tmp2 <- do.call("data.frame",tmp)
tmp2$value.mean <- as.numeric(tmp2$value.mean)
tmp2$value.sd <- as.numeric(tmp2$value.sd)
tmp2_flat <- tmp2 %>%
group_by(variable, date) %>%
filter(value.mean == min(value.mean)) %>%
arrange(variable, date, value.sd) %>%
slice(1)
Thank you in advance for any advice

Answer 1

I played a little around and this is what I came up with:
UPDATE: The last answer wasn't very practicable. There was no feedback but I'm changing it nevertheless.
library(zoo)
library(dplyr)
df %>%
group_by(variable, date) %>%
mutate(minimum = rollapply(value, width = 4, FUN = mean, fill = NA, align = "right"),
sd = rollapply(value, width = 4, FUN = sd, fill = NA, align = "right")) %>%
slice(which.min(minimum))
# A tibble: 2 x 10
# Groups: variable, date [2]
variable year month day hr min date value minimum sd
<fct> <int> <int> <int> <int> <int> <fct> <int> <dbl> <dbl>
1 building_a 2018 6 2 3 0 6/2/2018 12 12.8 0.5
2 building_b 2018 6 2 2 30 6/2/2018 29 33.2 2.87
The idea remains the same however. In the rollapply() function one can specifiy the n of consecutive rows via as width= argument. 4 means in this case 4 * 15 minutes = 1 hour, but can be any number of quarter hours.
And it calculates a "moving average" of valueat each row by looking back width rows.
That should do it I hope.