racecar | Racecar : a simple framework for Kafka consumers in Ruby | Pub Sub library
kandi X-RAY | racecar Summary
kandi X-RAY | racecar Summary
Racecar is a friendly and easy-to-approach Kafka consumer framework. It allows you to write small applications that process messages stored in Kafka topics while optionally integrating with your Rails models. The framework is based on rdkafka-ruby, which, when used directly, can be a challenge: it's a flexible library with lots of knobs and options. Most users don't need that level of flexibility, though. Racecar provides a simple and intuitive way to build and configure Kafka consumers. NOTE: Racecar requires Kafka 0.10 or higher.
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Top functions reviewed by kandi - BETA
- Run the consumer .
- Build the option parser
- Creates a new consumer
- Deliver the messages to be delivered .
- Make long - polling request
- Initialize a consumer .
- Wait for the worker to finish .
- Performs a consumer .
- Create a new message .
- Gets the configuration for the subscription
racecar Key Features
racecar Examples and Code Snippets
Lines of Code : 20License : Permissive (MIT License)
Copydef is_palindrome(s: str) -> bool:
"""
Determine whether the string is palindrome
:param s:
:return: Boolean
>>> is_palindrome("a man a plan a canal panama".replace(" ", ""))
True
>>> is_palindrome("He public int racecar(int target) {
Queue queue = new LinkedList<>();
queue.add(new int[] {0, 1, 0});
while (!queue.isEmpty()) {
int[] arr = queue.poll();
if (arr[0] == target) {
retu Community Discussions
Trending Discussions on racecar
QUESTION
I am trying to determine that whether any string is palindrome or not. For example, carrace should return true, since it can be rearranged to form racecar, which is a palindrome. daily should return false since there's no rearrangement that can form a palindrome. How can I solve this ?
Code:
...ANSWER
Answered 2021-Aug-09 at 10:35from itertools import permutations
def palindrome(s: str) -> bool:
return ''.join(reversed(s)) == s
def permutated_palindrome(s: str) ->:
return any(map(palindrome,
map(''.join, permutations(s))
))
QUESTION
I created the following palindrome function using a Deque class. The program should return True and False. However, nothing is returned when I run the program. Can anyone ascertain why?
...ANSWER
Answered 2022-Jan-28 at 10:48The return value of the palindrome function is a boolean, but in your code the function is just invoked without the result being used anywhere, thus the result is discarded and the program will just exit before outputting anything. If you want to display the result on a terminal/console after running the code, wraps the palindrome function call in print.
QUESTION
I am trying to solve the Leet Code challenge 14. Longest Common Prefix:
Write a function to find the longest common prefix string amongst an array of strings.
If there is no common prefix, return an empty string
Example 1: ..."".
ANSWER
Answered 2021-Aug-09 at 00:42this:
QUESTION
How can I implement a boolean tag search in SQL?
This question is about as close as I can find, but there's a few.
The only real solution I know is to generate a query like this through backend code and put it into SQL, But I imagine it is slow and I'm also wondering if there is any other way to do it (such as having one main query instead of multiple).
There's also solutions that probably use IN or something like:
How to query data based on multiple 'tags' in SQL?
I cannot use the typical GROUP BY HAVING COUNT Solution as it cannot operate on the context of having a list of tags, as this user points out: Implementing a tag search with operands
I should specify most of the existing solutions do not work as I'm looking for things that are capable of more complex queries such as parenthesis grouping and nested operands.
Schema is "Toxi" http://howto.philippkeller.com/2005/04/24/Tags-Database-schemas/
...ANSWER
Answered 2021-Nov-10 at 08:47Schema prep
QUESTION
I'm writing a program to check if a given user input is a palindrome or not. if it is the program should print "Yes", if not "no". I realize that this program is entirely too complex since I actually only needed to check the whole word using the reversed() function, but I ended up making it quite complex by splitting the word into two lists and then checking the lists against each other.
Despite that, I'm not clear why the last conditional isn't returning the expected "Yes" when I pass it "racecar" as an input. When I print the lists in line 23 and 24, I get two lists that are identical, but then when I compare them in the conditional, I always get "No" meaning they are not equal to each other. can anyone explain why this is? I've tried to convert the lists to strings but no luck.
...ANSWER
Answered 2021-Nov-10 at 00:22Despite your comments first_half and second_half are substrings of your input, not lists. When you print them out, you're converting them to lists, but in the comparison, you do not convert first_half or reversed(second_half). Thus you are comparing a string to an iterator (returned by reversed), which will always be false.
So a basic fix is to do the conversion for the if, just like you did when printing the lists out:
QUESTION
Forgive the lack of documentation. I'm a novice, so my code isn't the best, but I always appreciate positive feedback. So my problem is trying to get the shape to return to the original starting position after appearing off canvas and to continuously repeat the process. Any suggestions? Also, is there a way I could bind the up and down arrow keys to increase/decrease the shape's speed?
...ANSWER
Answered 2021-Sep-07 at 03:52You can attach a common tag to the four items of the car, then
- save the starting position of the car using the common tag
- move the car using the common tag instead of moving each of the items
- check whether the car is out of canvas by checking the first value of
canvas.coords(...), if yes move the car back to its starting position
QUESTION
I'm trying to write a query to convert a single character to lower or upper case in MySQL if it occurs in a string Like: "racecar" and I want to convert 'a' to upper should output "rAcecAr";
Here is my complicated query
...ANSWER
Answered 2021-Apr-27 at 16:28Unless I've missed a complexity that isn't obvious, this is as simple as using REPLACE to replace all a characters with uppercase:
QUESTION
spoiler alert: I'm a noob at Java.
With that disclosure out of the way, I have a homework assignment (yup I understand answers will be limited) where I'm supposed to "race" a hard-coded array of RaceCar objects, and I need to find out how I can update the Odometer/RaceCar class in order to increment miles of each of the raceCar objects by their respective MPH.
I think I've got a handle on most of it, but there's one, maybe two sections tripping me up in what syntax I need to use in order to update the Odometer's miles. My hunches:
- I probably didn't create the constructors or something on the Odometer/RaceCar class correctly in order for it to update the miles for each car (and I'm not sure how to do it correctly)
- I'm not calling it the right way and/or have not implemented the math properly.
Please help me understand how/where I'm going wrong and suggestions/direction to proceed forward because I'm pretty stuck and I'm not sure what to search for in order correct my mistakes.
Note: Odometer class is a "has-a" relationship with RaceCar class.
I'm having issue making this particular section work:
...ANSWER
Answered 2021-Apr-18 at 16:15Replace these two lines
QUESTION
Can you write a method that checks whether a word or a phrase is a palindrome?
NOTE: Palindrome is a word which reads the same backwards. Eg = madam, racecar or phrases like "nurses run".
...ANSWER
Answered 2021-Apr-12 at 22:01You can do it like this:
QUESTION
Such as in the 'racecar' example, could I set a lower and upper limit for the 'mass' design_parameter and then optimise the vehicle mass while solving the optimal control problem?
I see that there is an "opt" argument for phase.add_design_parameter() but when I run the problem with opt=True the value stays static. Do I need another layer to the solver that optimises this value?
This feature would be useful for allocating budgets to design decisions (e.g. purchasing a lighter chassis), and tuning parameters such as gear ratio.
...ANSWER
Answered 2021-Apr-06 at 13:54It's absolutely possible, and in fact that is the intent of the opt flag on design parameters.
Just to make sure things are working as expected, when you have a design parameter with opt=True, make sure it shows up as one of the optimizer's design variables by invoking list_problem_vars on the problem instance after run_model. The documentation for list_problem_vars is here.
If it shows up as a design variable but the optimizer is refusing to change it, it could be that it sees no sensitivity wrt that variable. This could be due to
- incorrectly defined derivatives in the model (wrong partials)
- poor scaling (the sensitivity of the objective/constraints wrt the design parameter may be miniscule in the optimizer's units
- sometimes by nature of the problem, a certain input has little to no impact on the result (this is probably the least likely here).
Things you can try:
- run problem.check_totals (make sure to call problem.run_model first) and see if any of the total derivatives appear to be incorrect.
- run problem.driver.scaling_report and verify that the values are not negligible in the units in which the optimizer sees them. If they're really small at the starting point, then it may be appropriate to scale the design parameter smaller (set ref to a smaller number like 0.01) so that a small change from the optimizer's perspective results in a larger change within the model.
If things don't appear to be working after trying this and I'll work with you to figure this out.
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