Averages per time-steps-CodePudding

I am working with 24-hour time-use data. A time-step has a 1 if the measurement was taken and 0 otherwise. I want to calculate the average measurement counts over time. Based on the below example his means first I need to sum per column and multiply by time-step such as: t1 x (1 0 1) t2 X (0 1 0) t3 x (1 1 1) and second to divide with the sum of counts ((1 0 1) (0 1 0) (1 1 1)). With regards to the time-steps t1=1, t2=2 and t3=3.Therefore [(1 x 2) (2 x 1) (3 x 3)] / (2 1 3) = 13 / 6 = 2.16. How can I implement this for n timesteps?

Example:

id   t1 t2 t3
 10   1  0  1
 12   0  1  1
 14   1  0  1

Output: AvgC=2.16

Sample data:

df<-structure(list(id = c("10", "12", "14"), t1 = c(1, 0, 1), t2 = c(0, 
1, 0), t3 = c(1, 1, 1)), class = "data.frame", row.names = c(NA, 
-3L))

CodePudding user response：

Another solution with Base R using a self-written function:

calcAverage <- function(df, dt){
  vector4df <- 1:(ncol(df)-1)*dt
  nom <- sum(as.matrix(df[,(2:ncol(df))]) %*% vector4df)
  denom <- sum(as.matrix(df[,(2:ncol(df))]))
  return(nom/denom)
}

calcAverage(df, 1)
2.166667

I've added the option to define dt which means "equidistant time steps between measurement". So when dt=1, a measurement is taken every hour, when dt =2, measurements are taken every second hour (2,4,6,...).

CodePudding user response：

With Base R,

time_step = c(1,2,3) # or 1:3
sum(time_step * colSums(df[,-1])) / sum(df[,-1])

# 2.166667