Trying to integrate over discrete points from a data frame - r

Question

I have several months of weather data; an example day is here:
Hour Avg.Temp
1 1 11
2 2 11
3 3 11
4 4 10
5 5 10
6 6 11
7 7 12
8 8 14
9 9 15
10 10 17
11 11 19
12 12 21
13 13 22
14 14 24
15 15 23
16 16 22
17 17 21
18 18 18
19 19 16
20 20 15
21 21 14
22 22 12
23 23 11
24 24 10
I need to figure out the total number of hours above 15 degrees by integrating in R. I'm analyzing for degree days, a concept in agriculture, that gives valuable information about relative growth rate. For example, hour 10 is 2 degree hours and hour 11 is 4 degree hours above 15 degrees. This can help predict when to harvest fruit. How can I write the code for this?
Another column could potentially work with a simple subtraction. Then I would have to make a cumulative sum after canceling out all negative numbers. That is the approach I'm setting out to do right now. Is there an integral I could write and have an answer in one step?

Answer 1

This solution subtracts your threshold (i.e., 15°), fits a function to the result, then integrates this function. Note that if the temperature is below the threshold this contribute zero to the total rather than a negative value.
df <- read.table(text = "Hour Avg.Temp
1 1 11
2 2 11
3 3 11
4 4 10
5 5 10
6 6 11
7 7 12
8 8 14
9 9 15
10 10 17
11 11 19
12 12 21
13 13 22
14 14 24
15 15 23
16 16 22
17 17 21
18 18 18
19 19 16
20 20 15
21 21 14
22 22 12
23 23 11
24 24 10", header = TRUE)
with(df, integrate(approxfun(Hour, pmax(Avg.Temp-15, 0)),
lower = min(Hour), upper = max(Hour)))
#> 53.00017 with absolute error < 0.0039
Created on 2019-02-08 by the reprex package (v0.2.1.9000)

Answer 2

The OP has requested to figure out the total number of hours above 15 degrees by integrating in R.
It is not fully clear to me what the espected result is. Does the OP want to count the number of hours above 15 degrees or does the OP want to sum up the degrees greater 15 ("integrate").
However, the code below creates both figures. Supposed the data is sampled at each hour without gaps (as suggested by OP's sample dataset), cumsum() and sum() can be used, resp.:
library(data.table)
setDT(DT)[, c("deg_hrs_sum", "deg_hrs_cnt") :=
.(cumsum(pmax(0, Avg.Temp - 15)), cumsum(Avg.Temp > 15))]
Hour Avg.Temp deg_hrs_sum deg_hrs_cnt
1: 1 11 0 0
2: 2 11 0 0
3: 3 11 0 0
4: 4 10 0 0
5: 5 10 0 0
6: 6 11 0 0
7: 7 12 0 0
8: 8 14 0 0
9: 9 15 0 0
10: 10 17 2 1
11: 11 19 6 2
12: 12 21 12 3
13: 13 22 19 4
14: 14 24 28 5
15: 15 23 36 6
16: 16 22 43 7
17: 17 21 49 8
18: 18 18 52 9
19: 19 16 53 10
20: 20 15 53 10
21: 21 14 53 10
22: 22 12 53 10
23: 23 11 53 10
24: 24 10 53 10
Hour Avg.Temp deg_hrs_sum deg_hrs_cnt
Alternatively,
setDT(DT)[, .(deg_hrs_sum = sum(pmax(0, Avg.Temp - 15)),
deg_hrs_cnt = sum(Avg.Temp > 15))]
returns only the final result (last row):
deg_hrs_sum deg_hrs_cnt
1: 53 10
Data
library(data.table)
DT <- fread("
rn Hour Avg.Temp
1 1 11
2 2 11
3 3 11
4 4 10
5 5 10
6 6 11
7 7 12
8 8 14
9 9 15
10 10 17
11 11 19
12 12 21
13 13 22
14 14 24
15 15 23
16 16 22
17 17 21
18 18 18
19 19 16
20 20 15
21 21 14
22 22 12
23 23 11
24 24 10", drop = 1L)