unWrap | Unwrap allsky images into rectilinear in Python | Computer Vision library
kandi X-RAY | unWrap Summary
kandi X-RAY | unWrap Summary
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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Top functions reviewed by kandi - BETA
- Unwraps an image
unWrap Key Features
unWrap Examples and Code Snippets
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
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
Trending Discussions on unWrap
QUESTION
The following code runs correctly.
...ANSWER
Answered 2022-Apr-02 at 18:41This 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)
, andSome { 0: &mut 3 }
are all extending expressions. The borrows in&0 + &1
andSome(&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):
QUESTION
The subject says it all. Specifically, say I created Channels
this way.
ANSWER
Answered 2022-Mar-30 at 14:41I 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.
QUESTION
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 07:47Edit:
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
:
QUESTION
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 19:08The 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
:
QUESTION
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 14:38Rust 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:
QUESTION
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 14:25I 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).
QUESTION
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 01:57The problem is here:
QUESTION
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 06:17The 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:
QUESTION
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 11:40The 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 lastArc
. 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.
QUESTION
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 19:35The 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.
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install unWrap
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.
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