unWrap | Unwrap allsky images into rectilinear in Python | Computer Vision library

 by   rankinstudio Python Version: Current License: GPL-3.0

kandi X-RAY | unWrap Summary

kandi X-RAY | unWrap Summary

unWrap is a Python library typically used in Artificial Intelligence, Computer Vision, OpenCV applications. unWrap has no bugs, it has no vulnerabilities, it has a Strong Copyleft License and it has low support. However unWrap build file is not available. You can download it from GitHub.

Unwrap allsky (fisheye) images into rectilinear in Python. Simple to use python script to unwrap an allsky or 360 deg RGB fisheye image into a panoramic rectilinear image. Assumes square image. May need to crop square before running. Dependencies: imageio numpy opencv. Tested: Raspberry Pi 3/4.
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            kandi-support Support

              unWrap has a low active ecosystem.
              It has 6 star(s) with 0 fork(s). There are 2 watchers for this library.
              OutlinedDot
              It had no major release in the last 6 months.
              There are 1 open issues and 1 have been closed. On average issues are closed in 3 days. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of unWrap is current.

            kandi-Quality Quality

              unWrap has 0 bugs and 0 code smells.

            kandi-Security Security

              unWrap has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.
              unWrap code analysis shows 0 unresolved vulnerabilities.
              There are 0 security hotspots that need review.

            kandi-License License

              unWrap is licensed under the GPL-3.0 License. This license is Strong Copyleft.
              Strong Copyleft licenses enforce sharing, and you can use them when creating open source projects.

            kandi-Reuse Reuse

              unWrap releases are not available. You will need to build from source code and install.
              unWrap has no build file. You will be need to create the build yourself to build the component from source.
              Installation instructions are not available. Examples and code snippets are available.
              It has 97 lines of code, 8 functions and 3 files.
              It has low code complexity. Code complexity directly impacts maintainability of the code.

            Top functions reviewed by kandi - BETA

            kandi has reviewed unWrap and discovered the below as its top functions. This is intended to give you an instant insight into unWrap implemented functionality, and help decide if they suit your requirements.
            • Unwraps an image
            Get all kandi verified functions for this library.

            unWrap Key Features

            No Key Features are available at this moment for unWrap.

            unWrap Examples and Code Snippets

            Unwrap input_tensors .
            pythondot img1Lines of Code : 9dot img1License : Non-SPDX (Apache License 2.0)
            copy iconCopy
            def unnest_if_single_tensor(input_tensors):
              # Preserve compatibility with older configs
              flat_input_tensors = nest.flatten(input_tensors)
              # If this is a single element but not a dict, unwrap. If this is a dict,
              # assume the first layer expects  
            Unwrap a wrapped tensor .
            pythondot img2Lines of Code : 7dot img2License : Non-SPDX (Apache License 2.0)
            copy iconCopy
            def _unwrap_or_tile(self, wrapped_tensor):
                """Given a wrapped tensor, unwrap if stacked. Otherwise, tiles it."""
                output, is_stacked = wrapped_tensor.t, wrapped_tensor.is_stacked
                if is_stacked:
                  return output
                else:
                  return _s  

            Community Discussions

            QUESTION

            Why do these ways of creating a reference behave differently?
            Asked 2022-Apr-02 at 18:41

            The following code runs correctly.

            ...

            ANSWER

            Answered 2022-Apr-02 at 18:41

            This is a concept in Rust called temporary lifetime extension. There are rules that govern when a temporary's lifetime is extended, though. It doesn't happen any time a borrow of a temporary occurs; the expression on the right side of a let statement needs to be a so-called "extending expression." This part of the document linked above neatly explains the first and third examples in your question:

            So the borrow expressions in &mut 0, (&1, &mut 2), and Some { 0: &mut 3 } are all extending expressions. The borrows in &0 + &1 and Some(&mut 0) are not: the latter is syntactically a function call expression.

            Since the expression assigned to a in the third example isn't an extending expression, the lifetime of the temporary is not extended beyond the let statement, and that causes the lifetime problem.

            Why does this work for Some(&3) then? Because of constant promotion:

            Promotion of a value expression to a 'static slot occurs when the expression could be written in a constant and borrowed, and that borrow could be dereferenced where the expression was originally written, without changing the runtime behavior.

            Since you borrow immutably, Rust allocates a hidden, static i32 and borrows that instead, giving you an Option<&'static i32>, which is obviously valid for the whole life of the program itself. This is not technically temporary lifetime extension because after the i32 is promoted to have static lifetime it's no longer a temporary.

            It's basically equivalent to this (except that HIDDEN has no name):

            Source https://stackoverflow.com/questions/71719617

            QUESTION

            How to correctly dispose System.Threading.Channels
            Asked 2022-Mar-30 at 14:41

            The subject says it all. Specifically, say I created Channels this way.

            ...

            ANSWER

            Answered 2022-Mar-30 at 14:41

            I think your dispose method is correct because TryComplete returns true only once and it is thread safe.

            But there is a problem, you expose the channel to subscribers, so they are able to explicitly call TryComplete/Complete to finish the writer. It's better to conceal it and just expose a write method.

            Source https://stackoverflow.com/questions/71612218

            QUESTION

            How to hash a binary file in Rust
            Asked 2022-Mar-24 at 17:06

            In rust, using sha256 = "1.0.2" (or similar), how do I hash a binary file (i.e. a tar.gz archive)?

            I'm trying to get the sha256 of that binary file.

            This doesn't work:

            ...

            ANSWER

            Answered 2021-Nov-01 at 07:47

            Edit: Upgrading to sha256 = "1.0.3" should fix this

            The issue is that digest_file is internally reading the file to a String, which requires that it contains valid UTF-8, which is obviously not what you want in this case.

            Instead, you could read the file in as bytes and pass that into sha256::digest_bytes:

            Source https://stackoverflow.com/questions/69787906

            QUESTION

            Different behavior between match and unwrap
            Asked 2022-Mar-15 at 18:39

            I have done a small program, and it shows a weird behavior that I cannot explain. I am using rodio crate to try out some audio stuff.

            I have done two programs that, in my opinion, should give the same result.

            The first one I use matches to handle errors:

            ...

            ANSWER

            Answered 2021-Sep-30 at 19:08

            The issue is one of scoping and an implementation detail of rodio: the one critical item here is OutputStream::try_default(), it doesn't really matter how you handle Sink::try_new(&handle) it'll always behave the same, not so try_default, if you match or if let it it'll work fine, if you unwrap it it'll fail.

            But why would that be, the two should be equivalent. The answer is in the details of rodio, specifically of OutputStreamHandle:

            Source https://stackoverflow.com/questions/69393226

            QUESTION

            Change elements in vector using multithreading in Rust
            Asked 2022-Feb-19 at 16:43

            I'm new in Rust and i'm trying to allocate computational work to threads.

            I have vector of strings, i would want to create to each string one thread to do his job. There's simple code:

            ...

            ANSWER

            Answered 2022-Feb-19 at 14:38

            Rust doesn't know that your strings will last as long as your threads, so it won't pass a reference to them to the threads. Imagine if you passed a reference to a string to another thread, then the original thread thought it was done with that string and freed its memory. That would cause undefined behavior. Rust prevents this by requiring that the strings either be kept behind a reference-counted pointer (ensuring their memory doesn't get freed while they are still referenced somewhere) or that they have a 'static lifetime, meaning they are stored in the executable binary itself.

            Also, Rust won't allow you to share a mutable reference across threads because it is unsafe (multiple threads could try to alter the referenced data at once). You want to use an std::sync::Arc in conjunction with a std::sync::Mutex. Your strings vector would become a Vec>>. Then, you could copy the Arc (using .clone()) and send that across threads. Arc is a pointer that keeps a reference count that gets incremented atomically (read: in a thread-safe way). The mutex allows a thread to lock the string temporarily so other threads can't touch it then unlock the string later (the thread can safely change the string while it is locked).

            Your code would look something like this:

            Source https://stackoverflow.com/questions/71182117

            QUESTION

            Spring Boot WebClient stops sending requests
            Asked 2022-Feb-18 at 14:42

            I am running a Spring Boot app that uses WebClient for both non-blocking and blocking HTTP requests. After the app has run for some time, all outgoing HTTP requests seem to get stuck.

            WebClient is used to send requests to multiple hosts, but as an example, here is how it is initialized and used to send requests to Telegram:

            WebClientConfig:

            ...

            ANSWER

            Answered 2021-Dec-20 at 14:25

            I would propose to take a look in the RateLimiter direction. Maybe it does not work as expected, depending on the number of requests your application does over time. From the Javadoc for Ratelimiter: "It is important to note that the number of permits requested never affects the throttling of the request itself ... but it affects the throttling of the next request. I.e., if an expensive task arrives at an idle RateLimiter, it will be granted immediately, but it is the next request that will experience extra throttling, thus paying for the cost of the expensive task." Also helpful might be this discussion: github or github

            I could imaginge there is some throttling adding up or other effect in the RateLimiter, i would try to play around with it and make sure this thing really works the way you want. Alternatively, consider using Spring @Scheduled to read from your queue. You might want to spice it up using embedded JMS for further goodies (message persistence etc).

            Source https://stackoverflow.com/questions/70357582

            QUESTION

            Returning a higher-kinded closure that captures a reference
            Asked 2022-Feb-12 at 01:57

            I'm trying to write a method on a struct that returns a closure. This closure should take a &[u8] with an arbitrary lifetime 'inner as an argument and return the same type, &'inner [u8]. To perform its function, the closure also needs a (shared) reference to a member of the struct &self. Here is my code:

            ...

            ANSWER

            Answered 2022-Feb-12 at 01:57

            QUESTION

            Preferring shift over reduce in parser for language without statement terminators
            Asked 2022-Jan-04 at 06:17

            I'm parsing a language that doesn't have statement terminators like ;. Expressions are defined as the longest sequence of tokens, so 5-5 has to be parsed as a subtraction, not as two statements (literal 5 followed by a unary negated -5).

            I'm using LALRPOP as the parser generator (despite the name, it is LR(1) instead of LALR, afaik). LALRPOP doesn't have precedence attributes and doesn't prefer shift over reduce by default like yacc would do. I think I understand how regular operator precedence is encoded in an LR grammar by building a "chain" of rules, but I don't know how to apply that to this issue.

            The expected parses would be (individual statements in brackets):

            ...

            ANSWER

            Answered 2022-Jan-04 at 06:17

            The issue you're going to have to confront is how to deal with function calls. I can't really give you any concrete advice based on your question, because the grammar you provide lacks any indication of the intended syntax of functions calls, but the hint that print(5) is a valid statement makes it clear that there are two distinct situations, which need to be handled separately.

            Consider:

            Source https://stackoverflow.com/questions/70571344

            QUESTION

            Move `Var` out from `Arc>`
            Asked 2021-Dec-14 at 01:59
            use std::ops::Deref;
            use std::sync::{Arc, Mutex, MutexGuard};
            
            struct Var {}
            
            fn multithreading() -> Var {
                let shared_var = Arc::new(Mutex::new(Var {}));
                /*
                multithreading job
                 */
            
                return *(shared_var.lock().unwrap().deref());
            }
            
            ...

            ANSWER

            Answered 2021-Dec-13 at 11:40

            The problem here is that if you remove your Var from the shared variable, what would be left there? What happens if any other copy of your Arc is left somewhere and it tries to access the now removed object?

            There are several possible answers to that question:

            1. I'm positively sure there is no other strong reference, this is the last Arc. If not, let it panic.

            If that is the case, you can use Arc::try_unwrap() to get to the inner mutex. Then another into_inner() to get the real value.

            Source https://stackoverflow.com/questions/70333509

            QUESTION

            Why are JS / React much quicker than WebAssembly / Rust?
            Asked 2021-Dec-06 at 19:35

            I'm doing some tests with Wasm generated by Rust (wasm_bindgen). What dazzles me is that the JavaScript implementation seems to be much quicker than the Rust implementation.

            The program is creating n items of a dict / object / map and pushing it to an array.

            The JavaScript implementation is very easy:

            ...

            ANSWER

            Answered 2021-Dec-06 at 19:35

            The answer is that you are likely not doing anything wrong. Essentially, WASM has the potential to be faster, but will not always be faster.

            I really liked this quick read from Winston Chen in which they run performance testing comparing web assembly to vanilla JS.

            If you're interested in more in-depth conversation, this talk by Surma at Google I/O '19 is very informative. From the video:

            Both JavaScript and Web Assembly have the same peak performance. They are equally fast. But it is much easier to stay on the fast path with Web Assembly than it is with JavaScript.

            Source https://stackoverflow.com/questions/70250002

            Community Discussions, Code Snippets contain sources that include Stack Exchange Network

            Vulnerabilities

            No vulnerabilities reported

            Install unWrap

            You can download it from GitHub.
            You can use unWrap like any standard Python library. You will need to make sure that you have a development environment consisting of a Python distribution including header files, a compiler, pip, and git installed. Make sure that your pip, setuptools, and wheel are up to date. When using pip it is generally recommended to install packages in a virtual environment to avoid changes to the system.

            Support

            For any new features, suggestions and bugs create an issue on GitHub. If you have any questions check and ask questions on community page Stack Overflow .
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