flip | ⏳ The online version of the classic flip clock | Animation library
kandi X-RAY | flip Summary
kandi X-RAY | flip Summary
Flip is an advanced and beautiful flip counter plugin. Easy to use and highly flexible, you can set up a custom counter on your website in minutes. Display visitor counts, countdown to a special date or celebrate progress. Whatever you’re planning, the options are endless. Made with By Rik Schennink.
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Top functions reviewed by kandi - BETA
- Transform an async generator
- Creates a FlipCard instance .
- Resolves a given type .
- Initialize a Tick .
- Resumes a yieldable
- Send a request
- call function
- Defines properties on an object .
- Determines if a tick event is valid .
- Check if a method is setter
flip Key Features
flip Examples and Code Snippets
Lines of Code : 8
License : No License
Copyconst flip = fn => (first, ...rest) => fn(...rest, first);
let a = { name: 'John Smith' };
let b = {};
const mergeFrom = flip(Object.assign);
let mergePerson = mergeFrom.bind(null, a);
mergePerson(b); // == b
b = {};
Object.assign(b, a); // = def _random_flip(image, flip_index, random_func, scope_name):
"""Randomly (50% chance) flip an image along axis `flip_index`.
Args:
image: 4-D Tensor of shape `[batch, height, width, channels]` or 3-D Tensor
of shape `[height, width, c def random_flip_left_right(image, seed=None):
"""Randomly flip an image horizontally (left to right).
With a 1 in 2 chance, outputs the contents of `image` flipped along the
second dimension, which is `width`. Otherwise output the image as-is def random_flip_up_down(image, seed=None):
"""Randomly flips an image vertically (upside down).
With a 1 in 2 chance, outputs the contents of `image` flipped along the first
dimension, which is `height`. Otherwise, output the image as-is.
W Community Discussions
Trending Discussions on flip
QUESTION
I'm using MSVC 2019 v16.11.12.
When I tried compiling my code with /fp:fast instead of /fp:precise, my tests started failing.
The simplest case is:
...ANSWER
Answered 2022-Apr-16 at 08:51One of the optimizations which is enabled by gcc's -ffast-math option (and probably msvc's /fp:fast option) is converting a "divide by constant" into a "multiply by reciprocal", as floating point divides are quite slow -- on some machines more than 10x as expensive as a multiply, as multipliers are commonly pipelined while dividers are less commonly pipelined.
With this, the / 1000.f would get turned into a * .001 of some precision, and .001 cannot be exactly represented in floating point, so some imprecision will occur.
More precisely, the closest 32-bit FP value to .001 is 0x1.0624dep-10, while the closest 64-bit FP is 0x1.0624dd2f1a9fcp-10. If that 32-bit value is multiplied by 3000 you'll get 0x1.80000132p+1 or about 3.0000001425. If you round that to 32 bits, you'll get 0x1.800002p+1 or about 3.0000002384. Interestingly, if you use the 64-bit value and multiply by 3000, you'll get 0x1.800000000000024p+1, which when rounded to 64 bits is the exact 0x1.8p+1 value.
QUESTION
For practice I've implemented the qoi specification in rust. In it there is a small hash function to store recently used pixels:
index_position = (r * 3 + g * 5 + b * 7 + a * 11) % 64
where r, g, b, and a are the red, green, blue and alpha channels respectively.
I assume this works as a hash because it creates a unique prime factorization for the numbers with the mod to limit the number of bytes. Anyways I implemented it naively in my code.
While looking at other implementations I came across this bit hack to optimize the hash calculation:
...ANSWER
Answered 2022-Apr-08 at 02:28If you think about the way the math works, you want this flipped order, because it means all the results from each of the "logical" multiplications cluster in the same byte. The highest byte in the first value multiplied by the lowest byte in the second produces a result in the highest byte. The lowest byte in the first value's product with the highest byte in the second value produces a result in the same highest byte, and the same goes for the intermediate bytes.
Yes, the 0x78... and 0x03... are also multiplied by each other, but they overflow way past the top of the value and are lost. Having the order "backwards" means the result of the multiplications we care about all ends up summed in the uppermost byte (the total shift of the results we want is always 56 bits, because the 56th bit offset value is multiplied by the 0th, the 40th by the 16th, the 16th by the 40th, and the 0th by the 56th), with the rest of the multiplications we don't want having their results either overflow (and being lost) or appearing in lower bytes (which we ignore). If you flipped the bytes in the second value, the 0x78 * 0x0B (alpha value & multiplier) component would be lost to overflow, while the 0x12 * 0x03 (red value & multiplier) component wouldn't reach the target byte (every component we cared about would end up somewhat that wasn't the uppermost byte).
For a possibly more intuitive example, imagine doing the same work, but where all the bytes of one input except a single component are zero. If you multiply:
QUESTION
I want to write a function that checks if the first list is longer than the second list and one of them can be infinite. However I can't find a working solution.
...ANSWER
Answered 2022-Mar-22 at 20:54Plain old natural numbers will not do the trick, because you can't calculate the natural number length of an infinite list in finite time. However, lazy natural numbers can do it.
QUESTION
I'm trying to create a triangular grid with HTML and CSS which involves offsetting each successive triangle in the grid to the left by larger and larger amounts so that each triangle fits neatly next to the previous one. Since the amount that each triangle needs to move is based on it's index in the parent container, I'm currently using JS to set this offset. I'm looking for a way to do this with pure CSS. Using JS like this feels like a hack and I'm wondering if I'm missing something in CSS that would let me access each triangle div's index or perhaps there's another way altogether in CSS to achieve what I'm doing.
...ANSWER
Answered 2022-Mar-16 at 08:16I created the same result with a negative margin. So the triangles don't have to move an increasing space to the left.
QUESTION
I am amazed at how many topics on StackOverflow deal with finding out the endianess of the system and converting endianess. I am even more amazed that there are hundreds of different answers to these two questions. All proposed solutions that I have seen so far are based on undefined behaviour, non-standard compiler extensions or OS-specific header files. In my opinion, this question is only a duplicate if an existing answer gives a standard-compliant, efficient (e.g., use x86-bswap), compile time-enabled solution.
Surely there must be a standard-compliant solution available that I am unable to find in the huge mess of old "hacky" ones. It is also somewhat strange that the standard library does not include such a function. Perhaps the attitude towards such issues is changing, since C++20 introduced a way to detect endianess into the standard (via std::endian), and C++23 will probably include std::byteswap, which flips endianess.
In any case, my questions are these:
Starting at what C++ standard is there a portable standard-compliant way of performing host to network byte order conversion?
I argue below that it's possible in C++20. Is my code correct and can it be improved?
Should such a pure-c++ solution be preferred to OS specific functions such as, e.g., POSIX-
htonl? (I think yes)
I think I can give a C++23 solution that is OS-independent, efficient (no system call, uses x86-bswap) and portable to little-endian and big-endian systems (but not portable to mixed-endian systems):
ANSWER
Answered 2022-Feb-06 at 05:48compile time-enabled solution.
Consider whether this is useful requirement in the first place. The program isn't going to be communicating with another system at compile time. What is the case where you would need to use the serialised integer in a compile time constant context?
- Starting at what C++ standard is there a portable standard-compliant way of performing host to network byte order conversion?
It's possible to write such function in standard C++ since C++98. That said, later standards bring tasty template goodies that make this nicer.
There isn't such function in the standard library as of the latest standard.
- Should such a pure-c++ solution be preferred to OS specific functions such as, e.g., POSIX-htonl? (I think yes)
Advantage of POSIX is that it's less important to write tests to make sure that it works correctly.
Advantage of pure C++ function is that you don't need platform specific alternatives to those that don't conform to POSIX.
Also, the POSIX htonX are only for 16 bit and 32 bit integers. You could instead use htobeXX functions instead that are in some *BSD and in Linux (glibc).
Here is what I have been using since C+17. Some notes beforehand:
Since endianness conversion is always1 for purposes of serialisation, I write the result directly into a buffer. When converting to host endianness, I read from a buffer.
I don't use
CHAR_BITbecause network doesn't know my byte size anyway. Network byte is an octet, and if your CPU is different, then these functions won't work. Correct handling of non-octet byte is possible but unnecessary work unless you need to support network communication on such system. Adding an assert might be a good idea.I prefer to call it big endian rather than "network" endian. There's a chance that a reader isn't aware of the convention that de-facto endianness of network is big.
Instead of checking "if native endianness is X, do Y else do Z", I prefer to write a function that works with all native endianness. This can be done with bit shifts.
Yeah, it's constexpr. Not because it needs to be, but just because it can be. I haven't been able to produce an example where dropping constexpr would produce worse code.
QUESTION
Basically, I need to define infix operator for function composition, flipped g . f manner.
ANSWER
Answered 2022-Feb-11 at 01:54You don't see a list of Haskell built-in operators for the same reason you don't see a list of all built-in functions. They're everywhere. Some are in Prelude, some are in Control.Monad, etc., etc. Operators aren't special in Haskell; they're ordinary functions with neat syntax. And Haskellers are generally pretty operator-happy in general. Spend any time inside your favorite lens library and you'll find plenty of amusing-looking operators.
In terms of (#), it might be best to avoid. I don't know of any built-in operators called that, but # can be a bit special when it comes to parsing. Specifically, a compiler extension enables # at the end of ordinary identifiers, and GHC defines a lot of built-in (primitive) types following this practice. For instance, Int is the usual (boxed) integer type, whereas Int# is a primitive integer. Most people don't need to interface with this directly, but it is a use that # has in Haskell that would be confused slightly by the addition of your operator.
Your (#) operator is called (>>>) in Haskell, and it works on all Category instances, including functions. Its companion is (<<<), the generalization of (.) to all Category instances. If three characters is too long for you, I've seen it called (|>) in some other languages, but Haskell already uses that operator for something else. If you're not using Data.Sequence, you could use that operator. But personally, I'd just go with (>>>). Any Haskeller will recognize it pretty quickly.
QUESTION
[I ran into the issues that prompted this question and my previous question at the same time, but decided the two questions deserve to be separate.]
The docs describe using destructuring assignment with my and our variables, but don't mention whether it can be used with has variables. But Raku is consistent enough that I decided to try, and it appears to work:
ANSWER
Answered 2022-Feb-10 at 18:47This is currently a known bug in Rakudo. The intended behavior is for has to support list assignment, which would make syntax very much like that shown in the question work.
I am not sure if the supported syntax will be:
QUESTION
Last night I saw the episode 7 of the Squid game tv series. The episode has a game with binomial distribution in the bridge.
Specifically there are 16 players and a bridge with 18 pair of glasses (one pure glass and one safe glass).If one player happened to choose the pure glass then the glass couldn't stand the weight of the player and the glass broke. The next player had the advantage that he/she was starting from the position that the last player had and continues the binomial search.At the end 3 players happened to cross the bridge.
So i was wondering: It is like, I have 16 euros in my pocket and I play head or tails with p = 1/2. Every time I bet on heads. If the coin flip is head then I earn 0 and if is tails I lose 1 euro. What is the probability of hitting 18 times (consecutive or not) heads and to be left 3 euros in my pocket.
I tried to simulate this problem in R:
...ANSWER
Answered 2021-Oct-16 at 13:02Here is how I think you can model the game in R. The first version is similar to what you have: there's a 50% chance of guessing correctly and if the guess is correct, the players advance a tile. Otherwise they do not, and the number of players decrements by 1. If the number of players reaches 0, or they advance to the end, the game ends. This is shown in squid_bridge1().
QUESTION
The definition of (>>) function is following:
(>>) :: Monad m => m a -> m b -> m b
But I would like to achieve this function flipped like following:
I have a function tabulate :: Int -> [Int] -> IO Int which prints the list as a table with the given number of columns and returns a sum of all the list items in the IO monad.
After that I want to have an explicit putStr "\n".
If I would use following:
tabulate >> (putStr "\n")
it would discard the result of the tabulate, the other way around it would not print newline after the table.
In case of doing this in do:
ANSWER
Answered 2022-Jan-20 at 23:31You can work with (<*) :: Applicative f => f a -> f b -> f a here:
QUESTION
I want to turn this halting, discontinuous trails in the particle on this simulation
to something worth staring at as in this beautiful field flow in here (not my work, but I don't remember where I got it from).
I have tried different permutations of the code in the accomplished field flow without getting anything remotely close to the smoothness in the transitions that I was aiming for. I suspect I am mishandling the updates or the placement of the black rectangle that seems to circumvent the need for a black background, which would erase the wake of the particles.
...ANSWER
Answered 2022-Jan-11 at 17:55You can get trials in multiple ways. The sketch you mentioned creates the trails by adding opacity to the background with the "fill( 0, 10 )" function.
If you want to know more about p5 functions you can always look them up here: https://p5js.org/reference/. The fill() page shows that the first argument is the color ( 0 for black ) and the second argument is the opacity ( 10 out of 255 ).
In the sketch you mentioned, in draw(), they wrote:
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