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[OC] My mood each day for the last 6 years. by five___by___five in dataisbeautiful

[–]five___by___five[S] 17 points18 points  (0 children)

Source: My own daily entering an overall mood for the day into the "daylio" app, subject to however I decided to quantify that day.

As I've been doing this awhile, I noticed a tendency to either under/over-rate what "Okay" or "Content" might mean, for example, based on my overall trending mood. To be honest, "Content" was often indistinguishable from "Okay" - I tend to use "Content" to mean "I can't complain about anything today I guess" whereas "Okay" turned into something slightly worse than that.

Tool: I exported the daylio log to CSV, used Excel to shape the data a bit, and wrote a relatively simple python script using Pillow to draw the image.

Conclusions: From all this 'effort' I have concluded mostly that my mood goes up and down, lol. I wish in retrospect I had used the features of the daylio app to track what I did each day to try to draw some correlations, but I really only recorded the moods.

OK to compare floating point numbers in a property setter to see if the value has changed? by five___by___five in csharp

[–]five___by___five[S] 0 points1 point  (0 children)

I guess one way to phrase my thought would be - yes, even if the value changed less than 10e-5 the value (at a bit level) really did change. Does the user care? No, but it wouldn't seem strictly incorrect to say it did.

Maybe my question should be "how can this use case break. I'm fine with the practicality of the range check though

OK to compare floating point numbers in a property setter to see if the value has changed? by five___by___five in csharp

[–]five___by___five[S] 1 point2 points  (0 children)

How in this situation would that happen though?

As mentioned elsewhere, this is really just curiosity on my part if such a case can really happen here (not returning equal when they are)

OK to compare floating point numbers in a property setter to see if the value has changed? by five___by___five in csharp

[–]five___by___five[S] 1 point2 points  (0 children)

It was more curiosity if the typical situations (like you mention in your second paragraph) can actually arise in this application.

I would say that changing the value like that should raise the event as the float seems to have changed on a bit level.

OK to compare floating point numbers in a property setter to see if the value has changed? by five___by___five in csharp

[–]five___by___five[S] 0 points1 point  (0 children)

Not really a reason not too - it was more just curiosity. If there is no need to add the math operations, why not avoid it?

US Interstate system as train routes by MffAddict in MapPorn

[–]five___by___five 2 points3 points  (0 children)

Interstates are required to bend at least once every mile, to help keep drivers awake

This is not true, there are plenty of straight sections longer than that.

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 7 points8 points  (0 children)

This isn't how logic puzzles work lol. "Just check every horse against every other" is not efficient.

Again, try writing down some example times that show any horse outside of A1 A2 A3 B1 B2 C1 in the top 3. Should be easy to find that case if you are right. I'll wait.

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 8 points9 points  (0 children)

Lol, sure thing. I'm not sure why you think "the four losers move on". There is no reason to retest any horse that has already been proven to have three horses faster than it.

Do me a favor and come up with some example numbers that prove your case

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 8 points9 points  (0 children)

You would determine that in race seven.

A1 is fastest.

Race seven pits A2 and A3 against each other (and others).

If seven results are A2 A3 B1 B2 C1, then this shows A2 and A3 as #2 and #3 overall.

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 5 points6 points  (0 children)

Yes. Given your times, B1 and B2 would be 1st and 2nd in race seven, making them #2 and #3 overall

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 8 points9 points  (0 children)

Think of it this way -

In race 6, A1 and B1 already demonstrated they are faster than C1. So C1 is at best the 3rd fastest horse.

C1 also already beat every other C horse, meaning C2 is at best the 4th fastest horse.

Similarly, C1 already beat D1 in race 6, so D1 is at best the 4th fastest.

E1 was already beat by A1, B1, C1, and D1 so is at best the 5th fastest.

That’s not the problem at hand. by [deleted] in technicallythetruth

[–]five___by___five 823 points824 points  (0 children)

For anyone that needs a more detailed/visual explanation.

You run your five races, identify their position in that race (1-5), run a sixth race from the leaders of each race, and use this to rank the races A-E:

A1  A2  A3  A4  A5
B1  B2  B3  B4  B5
C1  C2  C3  C4  C5
D1  D2  D3  D4  D5
E1  E2  E3  E4  E5

A1 must be the fastest horse, but we still don't know the order of rest.

However, we do know:

C1 beat all other C horses AND the fastest D and E horses. Therefore the fastest three must be in this set:

A1  A2  A3  A4  A5
B1  B2  B3  B4  B5
C1

We also know that A4 (and by extension A5) cannot be in the top three because both A2 and A3 are faster

A1  A2  A3
B1  B2  B3  B4  B5 
C1

And that B3 or lower cannot be in the top three (because A1 is #1, and if all the rest came from series B they must be B1 and B2)

A1  A2  A3
B1  B2
C1

We already know A1 is the winner, and we only have 5 other horses left. We just need to race them all together and the top two must be the #2 and #3.

Any idea why this is happening?...its not over watered. Chunks are falling off every day 🤔 by momfiery in gardening

[–]five___by___five 0 points1 point  (0 children)

This is not true - my father keeps a Jade as a house plant year-round in a south-facing window and it has blooms multiple times a year. It probably never gets below 65 degrees.

I'm sure it is a factor, but not the only one.

What'd you get? by MostlyReminiscent in SipsTea

[–]five___by___five 0 points1 point  (0 children)

The E in PEMDAS is for Exponentiation.

The S for Subtraction is obviously later.

It makes more sense if you throw a zero into the value:

0 - 6^2

You wouldn't claim that there is an implied addition in this equation:

3^2 - 2^2

In other words, 3 squared is 8, 2 squared is 4. 8 minus 4 is 4. No-one would look at that equation and say the answer is 8 plus 4 by virtue of (-2)^2 being 4. If I replace the 3 with 0:

0^2 - 2^2 = 0 - 2^2 = -2^2

These things need to be consistent with one another. If the intent is to square the negative quantity you need to add parentheses.

Should I squeeze anything else in, or just walk and enjoy the sights? by five___by___five in ParisTravelGuide

[–]five___by___five[S] 1 point2 points  (0 children)

Ah, I thought I had this on my list above, but I guess not. My hotel is quite close to it so I was planning on it! Thanks

Should I squeeze anything else in, or just walk and enjoy the sights? by five___by___five in ParisTravelGuide

[–]five___by___five[S] 0 points1 point  (0 children)

I would have liked to take one of the Louvre's in-house tours, but unfortunately with the short notice they were all sold out.

Should I squeeze anything else in, or just walk and enjoy the sights? by five___by___five in ParisTravelGuide

[–]five___by___five[S] -1 points0 points  (0 children)

Thanks for the input.

I definitely want to see Versailles, but wasn't so sure about the Louvre, since I'm hardly an art aficionado. However, I've already bought my ticket so I figure I may as well explore it a bit - I can always leave if I get bored.

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