Irrigation scheduling
Question:
I’m trying to code an irrigation controller. I have a preliminary tkinter UI that uses checkbuttons to get the days of the week when 4 valves will be turned on:
After I’ve checked the boxes, the data looks like this:
raw_data= [
[1, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 1, 0, 0],
[0, 0, 1, 0, 0, 1, 0],
[0, 0, 0, 1, 0, 0, 1]
]
Where ‘1’s represent days that the valve will be turned on. What I need to know (and don’t know how to write the code for) is a way to calculate the number of days since the last irrigation. It’s easy for the second day of the week, I just subtract the weekday index for the second day from the first. For Valve 1 in the present example that would be Thursday and Monday (3 days). But…on the first day of the week, I need to know how many days since the valve was turned on the previous week. Again, for Valve 1, on Monday I need to know how many days since the previous Thursday (4 days). It’s a trivial calculation for me, but I can’t seem to get how to code it so a computer would understand.
What approach can I take (what logic can I use) to solve this problem?
Answers:
I think you can do it in the following way.
Let’s take as an example the valve 1 whose array would be: [1, 0, 0, 1, 0, 0, 0].
As you say, the easiest thing to do is to subtract the indexes, so we simply add the same array at the end to repeat the week with the same data.
a = [1, 0, 0, 1, 0, 0, 0]
b = a+a
aux = []
for i,v in enumerate(b):
if v == 1:
aux.append(i)
print(aux)
This code adds in aux index each time it finds a 1 in the array. Now we only need to subtract. In this case the last value does not interest us, so if you want you can remove it with aux.pop().
Output:
[0, 3, 7, 10]
Where 3-0=3 and 7-3=4
It appears that I now have two solutions to the problem I raised. The one above and one that I came up with myself based on a answer to my previous question. Unless I’m mistaken (which is possible, since I’m a beginner Python coder), the answer to the previous question didn’t include today’s date as a benchmark. Once I put that into the code:
if today > day1:
days_since_last_irrigation = day2 - day1
else:
days_since_last_irrigation = ((day1 - day2) + 7)
I started to get results that made sense. This exercise is an illustration that there’s more than one way to solve most problems.
Easier way from @FG94.
a = [1, 0, 0, 1, 0, 0, 0]
b = a + a
aux = [i for i,v in enumerate(b) if v]
print(aux)
I’m trying to code an irrigation controller. I have a preliminary tkinter UI that uses checkbuttons to get the days of the week when 4 valves will be turned on:
After I’ve checked the boxes, the data looks like this:
raw_data= [
[1, 0, 0, 1, 0, 0, 0],
[0, 1, 0, 0, 1, 0, 0],
[0, 0, 1, 0, 0, 1, 0],
[0, 0, 0, 1, 0, 0, 1]
]
Where ‘1’s represent days that the valve will be turned on. What I need to know (and don’t know how to write the code for) is a way to calculate the number of days since the last irrigation. It’s easy for the second day of the week, I just subtract the weekday index for the second day from the first. For Valve 1 in the present example that would be Thursday and Monday (3 days). But…on the first day of the week, I need to know how many days since the valve was turned on the previous week. Again, for Valve 1, on Monday I need to know how many days since the previous Thursday (4 days). It’s a trivial calculation for me, but I can’t seem to get how to code it so a computer would understand.
What approach can I take (what logic can I use) to solve this problem?
I think you can do it in the following way.
Let’s take as an example the valve 1 whose array would be: [1, 0, 0, 1, 0, 0, 0].
As you say, the easiest thing to do is to subtract the indexes, so we simply add the same array at the end to repeat the week with the same data.
a = [1, 0, 0, 1, 0, 0, 0]
b = a+a
aux = []
for i,v in enumerate(b):
if v == 1:
aux.append(i)
print(aux)
This code adds in aux index each time it finds a 1 in the array. Now we only need to subtract. In this case the last value does not interest us, so if you want you can remove it with aux.pop().
Output:
[0, 3, 7, 10]
Where 3-0=3 and 7-3=4
It appears that I now have two solutions to the problem I raised. The one above and one that I came up with myself based on a answer to my previous question. Unless I’m mistaken (which is possible, since I’m a beginner Python coder), the answer to the previous question didn’t include today’s date as a benchmark. Once I put that into the code:
if today > day1:
days_since_last_irrigation = day2 - day1
else:
days_since_last_irrigation = ((day1 - day2) + 7)
I started to get results that made sense. This exercise is an illustration that there’s more than one way to solve most problems.
Easier way from @FG94.
a = [1, 0, 0, 1, 0, 0, 0]
b = a + a
aux = [i for i,v in enumerate(b) if v]
print(aux)