Getting history stats starting from last friday

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1

Hi,
in the past HA was tracking the computer time of our son on a daily basis (and after reaching the limit, the computer is shutting down).

Now, we want to change that to a weekly budget. But it shouldn’t start on monday, but on every friday.

I saw the example for shifting to sunday in the documenation:

start: "{{ today_at('00:00') - timedelta(days=now().weekday()) }}"
end: "{{ now() }}"

That is easy, because weekday sunday is zero and the only thing needed is to subtract the weekday. But with friday the weekday will have a jump back to zero on sunday.

What is the easiest way to get the computer time since last friday?

Many thanks in advance!

2

I’ve tried to get last Friday date, but as it is Saturday, not sure if it will make the job.

{{ (today_at('00:00') - timedelta(days=(now().weekday() + 2) % 7 + 1) ) }}

That gives 2024-10-18 00:00:00+02:00 in my timezone

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3

Monday is 0. Even says so in the example you are referring to.

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The whole time I was doing this I thought there must be an easier way with modular arithmetic.

start: >
  {% set weekday = now().weekday() %}
  {% set offset = 3 if weekday < 5 else -4 %}
  {{ today_at('00:00') - timedelta(days = weekday - offset) }}

Looks like Olivier worked that out. I tested his formula in the template editor with all the integer days. It works as long as you want a whole week instead of 0 days on Friday.

4

Sorry, maybe my brain made monday to 1, because that was how I understood the start.

Let’s say it’s monday, that would mean weekday() is 0. But how does the example work?
Isn’t today_at(‘00:00’) - 0 the same as today and not yesterday?

Many thanks Olivier and Tom for your solution. I went through Oliviers solution with pen and paper. Here it is -1 for Saturday, which feels to be right. But for friday it will be -7 and shouldn’t that be -0?
I guess that is the same misunderstanding as already with the example :slight_smile:

Edit: friday will of course be -7. I forgot to add the +1

5

It will actually be -7, that’s why I said:

Mine will give you 0.

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6

Hi Tom,

is your idea really giving the correct offsets?

start: >
  {% set weekday = now().weekday() %}
  {% set offset = 3 if weekday < 5 else -4 %}
  {{ today_at('00:00') - timedelta(days = weekday - offset) }}

Today we have saturday which is weekday 5. Therefore, it isn’t smaller than 5 and the offset will be set to -4.
5 - (-4) is 9. and should be 1. Changing the timedelta to week + offset will correct it for saturday, but will be incorrect for the other days.

I now ended up with a pragmatic lookup table.

    start: >
      {% set offset = [3, 4, 5, 6, 0, 1, 2] %}
      {{ today_at('00:00') - timedelta(days = offset[now().weekday()]) }}

Thanks a lot! This discussion was extremely helpful for finding a solution.

7

Maybe not, it’s late here. I like your solution. Very neat.

8

This would be the modular arithmetic to get you the same answer:

{{ today_at() - timedelta(days=(now().weekday() + 3) % 7) }}
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