ordo | Ordered Representation for Distinguished Objects | Cryptography library

by cryptosphere Ruby Version: Current License: MIT

X-Ray Key Features Code Snippets Community Discussions(8)Vulnerabilities Install Support

kandi X-RAY | ordo Summary

ordo is a Ruby library typically used in Security, Cryptography applications. ordo has no bugs, it has no vulnerabilities, it has a Permissive License and it has low support. You can download it from GitHub.

Ordered Representation for Distinguished Objects: A Certificate Format

Support

Quality

Security

License

Reuse

Support

ordo has a low active ecosystem.

It has 19 star(s) with 1 fork(s). There are 3 watchers for this library.

It had no major release in the last 6 months.

ordo has no issues reported. There are no pull requests.

It has a neutral sentiment in the developer community.

The latest version of ordo is current.

Quality

ordo has 0 bugs and 0 code smells.

Security

ordo has no vulnerabilities reported, and its dependent libraries have no vulnerabilities reported.

ordo code analysis shows 0 unresolved vulnerabilities.

There are 0 security hotspots that need review.

License

ordo is licensed under the MIT License. This license is Permissive.

Permissive licenses have the least restrictions, and you can use them in most projects.

Reuse

ordo releases are not available. You will need to build from source code and install.

Installation instructions are not available. Examples and code snippets are available.

ordo saves you 78 person hours of effort in developing the same functionality from scratch.

It has 201 lines of code, 12 functions and 7 files.

It has medium code complexity. Code complexity directly impacts maintainability of the code.

Top functions reviewed by kandi - BETA

kandi has reviewed ordo and discovered the below as its top functions. This is intended to give you an instant insight into ordo implemented functionality, and help decide if they suit your requirements.

Serializes the message .

Get all kandi verified functions for this library.

ordo Key Features

No Key Features are available at this moment for ordo.

ordo Examples and Code Snippets

No Code Snippets are available at this moment for ordo.

Community Discussions

Trending Discussions on ordo

Unique trend Curve Fitting

How to change value in the select option of an array in the Yii2 view?

Add item to Python dictionary while looping over

Matplotlib plot Pandas df, fill between index values if condition

Find an algorithm for sorting integers with time complexity O(n + k*log(k))

What causes "OpenCV(4.0.1) Error : Assertion failed(m.dims <= 2)"

XNA how to write a good collision algorithm?

What is the time complexity for this specific algorithm?

QUESTION

Unique trend Curve Fitting

Asked 2021-Mar-20 at 22:05

I have data like this:

...

ANSWER

Answered 2021-Mar-20 at 22:05

To do this, you should use your analytical function (with parameters) based on some assumption (not only polynomial functions). You can use curve_fit form scipy.optimize to find the unknown parameters of your analytic function that best fit your input data.

For example:

Source https://stackoverflow.com/questions/66714923

QUESTION

How to change value in the select option of an array in the Yii2 view?

Asked 2021-Feb-13 at 17:46

I use Yii2 Framework and I have this array filter for the 'Ordine' column:

...

ANSWER

Answered 2021-Feb-13 at 17:46

You have mapped the same column for id and description

assuming your Elenchi models contains at least two column ordine and description with values :

Source https://stackoverflow.com/questions/66187455

QUESTION

Add item to Python dictionary while looping over

Asked 2020-Dec-04 at 09:28

I have a SPARQL query output which I want to save as a Python dictionary. Where subject will be a key and the predicate will be key inside the subject containing respective object. The data look like as follows:

...

ANSWER

Answered 2020-Dec-03 at 23:10

In [1]: f = [["http://www.orpha.net/ORDO/Orphanet_2514", "http://www.geneontology.org/formats/oboInOwl#hasExactSynonym", "microcephaly (disease), autosomal dominant"],
   ...: ["http://www.orpha.net/ORDO/Orphanet_2514", "http://www.geneontology.org/formats/oboInOwl#hasRelatedSynonym", "microcephaly, autosomal dominant"],
   ...: ["http://www.orpha.net/ORDO/Orphanet_2514", "http://www.geneontology.org/formats/oboInOwl#hasRelatedSynonym", "autosomal dominant microcephaly"],
   ...: ["http://www.ebi.ac.uk/efo/EFO_0000722", "http://www.w3.org/2000/01/rdf-schema#comment", "syn of disease free survival"],
   ...: ["http://www.ebi.ac.uk/efo/EFO_0000722", "http://purl.obolibrary.org/obo/IAO_0100001",  "http://www.ebi.ac.uk/efo/EFO_0004920"],
   ...: ["http://www.ebi.ac.uk/efo/EFO_0000722", "http://purl.obolibrary.org/obo/IAO_0000117",  "James Malone"],
   ...: ["http://www.ebi.ac.uk/efo/EFO_0000722", "http://www.w3.org/2000/01/rdf-schema#label",  "obsolete_time before disease progression"],
   ...: ["http://www.ebi.ac.uk/efo/EFO_0000722", "http://www.ebi.ac.uk/efo/reason_for_obsolescence", "created as a synonym of progression free survival"]]

In [2]: d = {}
   ...: for row in f:
   ...:     if not row[0] in d:
   ...:         d[row[0]] = dict()
   ...:     d[row[0]][row[1]] = row[2]
   ...: 

In [3]: d
Out[3]: 
{'http://www.orpha.net/ORDO/Orphanet_2514': {'http://www.geneontology.org/formats/oboInOwl#hasExactSynonym': 'microcephaly (disease), autosomal dominant',
  'http://www.geneontology.org/formats/oboInOwl#hasRelatedSynonym': 'autosomal dominant microcephaly'},
 'http://www.ebi.ac.uk/efo/EFO_0000722': {'http://www.w3.org/2000/01/rdf-schema#comment': 'syn of disease free survival',
  'http://purl.obolibrary.org/obo/IAO_0100001': 'http://www.ebi.ac.uk/efo/EFO_0004920',
  'http://purl.obolibrary.org/obo/IAO_0000117': 'James Malone',
  'http://www.w3.org/2000/01/rdf-schema#label': 'obsolete_time before disease progression',
  'http://www.ebi.ac.uk/efo/reason_for_obsolescence': 'created as a synonym of progression free survival'}}

Source https://stackoverflow.com/questions/65135043

QUESTION

Matplotlib plot Pandas df, fill between index values if condition

Asked 2020-Sep-24 at 06:36

I have a Pandas df:

...

ANSWER

Answered 2020-Sep-24 at 06:15

Python in keyword checks for an element in a iterable, but you are using it to check if an array (plotMonths.index) is in a iterable ['January', "February", 'March'].
The proper way to vectorize the conditional is to use the built in method .isin().
So, you should use:

Source https://stackoverflow.com/questions/64040364

QUESTION

Find an algorithm for sorting integers with time complexity O(n + k*log(k))

Asked 2020-Apr-20 at 16:04

Design an algorithm that sorts n integers where there are duplicates. The total number of different numbers is k. Your algorithm should have time complexity O(n + k*log(k)). The expected time is enough. For which values of k does the algorithm become linear?

I am not able to come up with a sorting algorithm for integers which satisfies the condition that it must be O(n + k*log(k)). I am not a very advanced programmer but I was in the problem before this one supposed to come up with an algorithm for all numbers xi in a list, 0 ≤ xi ≤ m such that the algorithm was O(n+m), where n was the number of elements in the list and m was the value of the biggest integer in the list. I solved that problem easily by using counting sort but I struggle with this problem. The condition that makes it the most difficult for me is the term k*log(k) under the ordo notation if that was n*log(n) instead I would be able to use merge sort, right? But that's not possible now so any ideas would be very helpful.

Thanks in advance!

...

ANSWER

Answered 2020-Apr-20 at 16:04

Here is a possible solution:

Using a hash table, count the number of unique values and the number of duplicates of each value. This should have a complexity of O(n).
Enumerate the hashtable, storing the unique values into a temporary array. Complexity is O(k).
Sort this array with a standard algorithm such as mergesort: complexity is O(k.log(k)).
Create the resulting array by replicating the elements of the sorted array of unique values each the number of times stored in the hash table. complexity is O(n) + O(k).
Combined complexity is O(n + k.log(k)).

For example, if k is a small constant, sorting an array of n values converges toward linear time as n becomes larger and larger.

If during the first phase, where k is computed incrementally, it appears that k is not significantly smaller than n, drop the hash table and just sort the original array with a standard algorithm.

Source https://stackoverflow.com/questions/60981554

QUESTION

What causes "OpenCV(4.0.1) Error : Assertion failed(m.dims <= 2)"

Asked 2019-Jan-08 at 15:03

What causes this error?

OpenCV: terminate handler is called!The last OpenCV error is : OpenCV(4.0.1) Error : Assertion failed(m.dims <= 2) in cv::FormattedImpl::FormattedImpl, file c : \build\master_winpack - build - win64 - vc15\opencv\modules\core\src\out.cpp, line 87

...

ANSWER

Answered 2019-Jan-08 at 15:03

This error is on this line:

Source https://stackoverflow.com/questions/54094026

QUESTION

XNA how to write a good collision algorithm?

Asked 2017-Jun-18 at 17:00

I'm currently working on a new game in XNA and I'm just setting basic stuff up like sprites/animations, input, game objects etc.

Meanwhile I'm trying to think of a good way to detect collisions for all the game objects, but I can't really think of a fast algorithm, which limits the game to very few objects. This is what I did in my last project that was a school assignment

...

ANSWER

Answered 2017-Jun-18 at 17:00

There are quite a few things you can do:

The nested for loops produce a quadratic running time, which grows rather quickly as the number of objects increase. Instead, you could use some acceleration data structures (e.g. grids, kd-trees, BVHs). These allow you to reduce the intersection checks to only the entities that can potentially intersect.

If you can order your entities, you can reduce the collision checks by half by just checking entity pairs, where the second is "greater" (with respect to the order) than the first. I.e. you do not need to check b-a if you already checked a-b.

This part can be potentially slow:

Source https://stackoverflow.com/questions/44616027

QUESTION

What is the time complexity for this specific algorithm?

Asked 2017-Jan-25 at 20:18

Given this algorithm:

...

ANSWER

Answered 2017-Jan-25 at 18:49

that algorithm is O(n) linear time complexity

Source https://stackoverflow.com/questions/41859292

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

Vulnerabilities

No vulnerabilities reported

Install ordo

You can download it from GitHub.
On a UNIX-like operating system, using your system’s package manager is easiest. However, the packaged Ruby version may not be the newest one. There is also an installer for Windows. Managers help you to switch between multiple Ruby versions on your system. Installers can be used to install a specific or multiple Ruby versions. Please refer ruby-lang.org for more information.

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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