date - R: replace NA by mean of non-NA values based on a multiple conditions (Hour, Day of Week and Temperature range) -


i have data frame 3 columns: "date/time", "power", "temperature", represents values of power measured, 10 10 minutes, , temperature of outdoor air. entries correspond whole year of data acquiring, result on large number of strings. sampling example:

df1 <- data.frame("date/hour" = c("2016-08-09 12:00", "2016-08-09 12:10", "2016-08-09 12:15", "2016-08-09 12:20"), "power" = c(500.21, 500.23, 500.24, 500.30), "day_week" = c("tuesday", "tuesday", "tuesday", "tuesday"), "temp" = c(21.1, 21.5, 21.3, 21.0))

the problem @ periods, due equipment problem, power value got na values. fill gap, want use mean of non-na values satisfy next 3 conditions:

  1. occur @ same hour.
  2. occur @ same day of week.
  3. (the intriging one) occur @ range of temperature of +/- 0.5, , increases 0.1 if there's no non-na values replace.

to solve problem, first of all, created fourth column in df1, transforms values of "date/hour" column "hour_minute", ignoring date. 

df1$hour_minute <- format(df1$`date/hour`, "%h:%m")

temperature range calculated based on fixed range +/- 0.5 stored in fifth , sixth column of dataframe:

df$upper_temp <- dado_total$temp+0.5 df$lower_temp <- dado_total$temp-0.5

due impossibility of data frame mean calculation, created numeric vector (index) df1 power column: 

index <- subset(df1, select = c(power)) %>% unlist() %>% as.numeric(as.character())

so created seventh column (df1$eq_power) gets mean of index vector based on 3 mentioned conditions using loop:

for (i in 1:length(df1$power)) {   df1$eq_power[i] <- mean(index[which(     df1$hour_minute==df1$hour_minute[i] &     df1$day_week==df1$day_week[i] &     df1$temp<=df1$upper_temp[i] &     df1$temp>=df1$lower_temp[i])]) }

for command returns me profitable result, though large time expenditure , missing values didn't manage use incremental temperature ranges. if me, appreciate lot.


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