Zero-downtime Django migrations
52 points • 9 comments
@cryvate1284
From 10/13/2020, 4:35:33 PM till now, @cryvate1284 has achieved 60 Karma Points with the contribution count of 23.
Recent @cryvate1284 Activity
It's not that binary though. Imagine you belief Twitter is worth 45.01USD but it's trading 45USD, would you throw your life savings at it (disregarding fees), no, of course not, way too risky!
The way I see it, people (and other entities!) don't think Twitter is worth X, they think there is a distribution of worth.
This means there is risk in their bet, so if I thought Twitter was worth (on average) 100USD, that wouldn't mean I would sink my life savings into it but would put in some, based on the risk and also what else I need capital for.
Also, how much one thinks Twitter is worth probably also depends on your time horizon: like the old adagium that the market can stay stupid for longer than you can stay solvent, e.g. when shorting a scam.
It might be a work address (easily done with john.doe@acompany.com), but 40 is not very much though regardless...
Are you sure that's not just the sales website?
Quite common for the sales website to be wordpress and separate from the frontend of the product.
Does PHP say it follows SemVer? I don't know as I don't use it but Python doesn't.
You aren't very explicit what you mean, but I disagree I think.
The main thing is willing things into existence (like ZFC, the Axiom of Choice) is equally consistent with not having it (ZF + not AC) or not deciding it (ZF) as shown by forcing.
Russel's Paradox in particular was worked around by not allowing to "build sets" quantifying over all sets (see axiom schema of specification).
If you are talking about avoiding "unprovable statements" (i.e. Godel's incompleteness theorem), you have to strip things way back further: it applies to systems that only have e.g. Peano arithmetic. An actual practical statement that is unprovable in Peano arithmetic is Goodstein's theorem: - write out a number in its base 2, including its exponents (i.e. recursive) - replace all the 2s with 3s - subtract 1 - write out a number in base 3, including its exponents - replace all the 3s with 4s - subtract 1 - ... The question is whether for all starting numbers this sequence eventually ends at 1. The answer is yes, trivially if you use ordinals (and replace all your base numbers with omega: the +1 will basically do nothing and -1 means it must be finite because ordinals well-ordered), but this cannot be used in Peano arithmetic.
Fundamentally, like Paris-Harrington Theorem, I think the way to think about why this cannot be proved, is because these sequences get big before they end up at 1.
Not that I am taking a stance here, but you didn't really engage with the GP saying that having functions instead of lambdas allows one to name, test and document them: in fact, they might not be called f1 to f20, in fact they might have docstrings (I guess one could comment lambdas, but see it rarely) and you couldn't test lambdas unless you assign them, i.e. give them a name i.e. turn them into functions in all but name.
EasyJet
Carmichael's Function is the name.
"apparently I object to the very fact that mathematicians like to fix things and make inferences about them": that's not really what's going on: time can be present in mathematics (e.g. defining a sequence iteratively) and the like.
What you do have to fix is the definition of things: your reasoning is very shaky, and once you would start formalizing things you will find you cannot "defeat" the Halting problem or incompleteness theorem.
Not sure it's strong, but The Netherlands has DigiD with 2FA?
The comment you are replying to said "usually" and that was in response to the GP that said "I don't believe the London Underground has ever been profitable."
The government did not bail out the London Underground in particular last year but TfL, though I would be surprised if the underground was profitable last year.
Anyway, unsure what your comment was adding.
Similar here, though we did narrow it down to the cause/brownout at the time.
This won't work as a deploy key can only be used in one repo.
Is he wrong though? If you went to a shop and spent $2,000 but forgot to pay for 3 items (say worth $5) by accident, I do not think the shop manager should be upset, or that you should feel bad about it? If it was a smaller/non-chain store, it would probably be free! If anything, the main annoyance is probably their stock numbers being out of date.
Making mistakes is human, and I do not think they should feel bad for what they have done (and for not correcting it), and this is coming from someone who cannot steal the bags that now cost 20 pence at Tesco and even went to buy imaginary food at Waitrose after I forgot to pay another time...
It seems this is run by the "Office of the French Ambassador for Digital Affairs": cool initiative, see their readme on GitHub:
It is harder than x-y + down, however I don't think this video is impressive really, having slowed down the video it doesn't look special to me and I did work on robotics/machine vision around that time.
What's the relevance of it being older than `pip`?
This depends on what you're doing. For example, if you are changing an intermediary (abstraction) layer, then the (unit) tests (TDD or not) for those will have to change and the "guarantee that you're not inadvertently changing behaviour" is kinda moot.
If you do not have TDD, I guess the reasoning is that either you do not have tests for this (probably bad) and if you do, less of them (and so more easily changed).
Not sure it's an argument against TDD, but I guess if management/the programmer do not know about sunk cost fallacy, it might make them hold on to bad abstractions/layers.
Because the integral is then over ln(1) and the log of 1 is 0.
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