CheckiO | Solutions for tasks from http : //www.checkio.org/

 by   therin Python Version: Current License: No License

kandi X-RAY | CheckiO Summary

kandi X-RAY | CheckiO Summary

CheckiO is a Python library. CheckiO has no bugs, it has no vulnerabilities and it has low support. However CheckiO build file is not available. You can download it from GitHub.

Solutions for tasks from http://www.checkio.org/
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            kandi-support Support

              CheckiO has a low active ecosystem.
              It has 4 star(s) with 3 fork(s). There are 1 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. There are no pull requests.
              It has a neutral sentiment in the developer community.
              The latest version of CheckiO is current.

            kandi-Quality Quality

              CheckiO has no bugs reported.

            kandi-Security Security

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

            kandi-License License

              CheckiO does not have a standard license declared.
              Check the repository for any license declaration and review the terms closely.
              OutlinedDot
              Without a license, all rights are reserved, and you cannot use the library in your applications.

            kandi-Reuse Reuse

              CheckiO releases are not available. You will need to build from source code and install.
              CheckiO has no build file. You will be need to create the build yourself to build the component from source.

            Top functions reviewed by kandi - BETA

            kandi has reviewed CheckiO and discovered the below as its top functions. This is intended to give you an instant insight into CheckiO implemented functionality, and help decide if they suit your requirements.
            • Return the rank of a given hand
            • Return the first rank of a given rank
            • Return the rank of a card
            • Returns True if there is only one match
            • Returns a set of all connected connections
            • Remove a connection
            • Check if a matrix is in the correct order
            • Test if the best wild
            • Find the weak point of a matrix
            • Return the clock angle from a time
            • Adds a connection to the pool
            • Returns the corners of the bounding box
            • Count all the neighbours of a grid
            • Check if a line is in the current line
            • Calculate the simple matrices
            • Process a list of commands
            • Decode a message from a message
            • Turn a list of commands into a sum
            • Takes a pyramid
            • Generate a life counter
            • Find the index of an array n
            Get all kandi verified functions for this library.

            CheckiO Key Features

            No Key Features are available at this moment for CheckiO.

            CheckiO Examples and Code Snippets

            No Code Snippets are available at this moment for CheckiO.

            Community Discussions

            QUESTION

            Checkio Common Words Challenge: How do i continue?
            Asked 2020-Mar-24 at 19:50

            I need your help since i'm stuck at the challenge from checkio. What am i missing? I get back:

            Your result:"one,two,three" Right result:"one,three,two"

            The Challenge: You are given two string with words separated by commas. Try to find what is common between these strings. The words are not repeated in the same string. Your function should find all of the words that appear in both strings. The result must be represented as a string of words separated by commas in alphabetic order.

            UPDATE

            this is my code:

            ...

            ANSWER

            Answered 2020-Mar-24 at 19:50

            match.toString() doesn't change the value of match variable. You need to return it from the function.

            Source https://stackoverflow.com/questions/60838125

            QUESTION

            How to iterate through the Cartesian product of ten lists (ten elements each) faster? (Probability and Dice)
            Asked 2020-Jan-18 at 07:17

            I'm trying to solve this task. I wrote function for this purpose which uses itertools.product() for Cartesian product of input iterables:

            ...

            ANSWER

            Answered 2020-Jan-16 at 01:59

            I guess the problem is to find the distribution of the sum of dice. An efficient way to do that is via discrete convolution. The distribution of the sum of variables is the convolution of their probability mass functions (or densities, in the continuous case). Convolution is an n-ary operator, so you can compute it conveniently just two pmf's at a time (the current distribution of the total so far, and the next one in the list). Then from the final result, you can read off the probabilities for each possible total. The first element in the result is the probability of the smallest possible total, and the last element is the probability of the largest possible total. In between you can figure out which one corresponds to the particular sum you're looking for.

            The hard part of this is the convolution, so work on that first. It's just a simple summation, but it's just a little tricky to get the limits of the summation correct. My advice is to work with integers or rationals so you can do exact arithmetic.

            After that you just need to construct an appropriate pmf for each input die. The input is just [1, 1, 1, ... 1] if you're using integers (you'll have to normalize eventually) or [1/n, 1/n, 1/n, ..., 1/n] if rationals, where n = number of faces. Also you'll need to label the indices of the output correctly -- again this is just a little tricky to get it right.

            Convolution is a very general approach for summations of variables. It can be made even more efficient by implementing convolution via the fast Fourier transform, since FFT(conv(A, B)) = FFT(A) FFT(B). But at this point I don't think you need to worry about that.

            Source https://stackoverflow.com/questions/59751892

            QUESTION

            Why if statement stay after the expression
            Asked 2019-Dec-02 at 20:23

            Why the if statement stay after the expression in this code:

            ...

            ANSWER

            Answered 2019-Dec-02 at 20:23

            This is a ternary operator (or conditional expression), here a very underperformant way of writing:

            Source https://stackoverflow.com/questions/59146065

            QUESTION

            Non-unique Elements
            Asked 2019-Nov-17 at 13:53

            I am doing an exercise on checkio about writing a function to exclude the unique elements in a list, and only keep the non-unique ones. I wrote the following lines.

            ...

            ANSWER

            Answered 2019-Nov-17 at 08:25

            You can accomplish that using a single liner

            Source https://stackoverflow.com/questions/58898669

            QUESTION

            SDL2 - SDL_PollEvent slower on Ubuntu/Mac OS X than Windows
            Asked 2019-Oct-14 at 07:25

            I wrote a Space Invaders emulator in C++ using SDL2 only for creating the game window and for playing sounds (sdl2_mixer). On Windows the emulator works at 60 FPS (I can change this value to whatever I want and it works without any problem) but if I build it on Ubuntu or Mac OS X the game is unplayable (maybe 10% of the wanted FPS).

            Is there an explanation for this?

            • Renderer name on Ubuntu: opengl

            • Renderer name on Windows: direct3d

            • Renderer flags is 0x0a both on Ubuntu than Windows.

            Compiled with:

            ...

            ANSWER

            Answered 2019-Oct-01 at 09:21

            You're creating a new surface and a new texture every frame. Yeah it's going to be slow, creating textures is a ridiculously slow operation. Don't do it so often.

            If you need to write pixels directly, create a texture of the right size once (with SDL_TEXTUREACCESS_STREAMING) and then use SDL_LockTexture and SDL_UnlockTexture to update it (note: don't use SDL_UpdateTexture for this, it's not really much - if at all - better than recreating the texture performance wise). I'm guessing the windows version is fast "by accident" because your driver implementation notices you're doing something strange and quietly ignores what you're telling it to do and does something faster.

            Of course without looking at the code it's hard to say this is the only problem with it, but this is pretty likely to be a (or maybe even the) bottleneck.

            Source https://stackoverflow.com/questions/58180023

            QUESTION

            Code problem - Flatten dictionary using generator
            Asked 2019-Oct-07 at 09:46

            First, this post is a solution not using generators: Flatten dictionary of dictionaries

            I have a dictionary of dictionaries.

            For example:

            ...

            ANSWER

            Answered 2019-Oct-07 at 09:46

            The first problem is that with if isinstance(a[i],type(dict)) you are testing whether the item is an instance of type(dict), which is type, not dict, but that does not fix all the problems.

            Source https://stackoverflow.com/questions/58266689

            QUESTION

            How to find repeated letters in the same word in Python?
            Asked 2019-May-19 at 11:09

            I was trying to solve a challenge for finding some pattern in a given string. First idea came to my mind is to loop over the characters and find the pattern.

            (A "stressful" subject line means that all letters are in uppercase, and/or ends by at least 3 exclamation marks, and/or contains at least one of the following “red” words: "help", "asap", "urgent". Any of those "red" words can be spelled in different ways - "HELP", "help", "HeLp", "H!E!L!P!", "H-E-L-P", even in a very loooong way "HHHEEEEEEEEELLP")

            Someone submitted below code for that and I don't understand what's going on. How does this work?

            ...

            ANSWER

            Answered 2019-Mar-14 at 10:54

            If we take the example of the word "help":

            subj.isupper() returns true if subj="HELP"

            subj.endswith('!!!') returns true if subj="help!!!"

            subj.lower() forces the string into lowercase subj="HelP"-> subj="help"

            re.search('+[.!-]*'.join(c for c in word) joins/removes occurences of the same characters next to each other, for instance, it will transform "heeellp" into "help"

            Source https://stackoverflow.com/questions/55160536

            QUESTION

            Check for (whole only) words in string
            Asked 2018-Sep-12 at 20:58

            Training on Checkio. The task is called Popular words. The task is to search for words from a list (of strings) in a given string.

            For example:

            ...

            ANSWER

            Answered 2018-Sep-12 at 20:17

            you'd be better off splitting your sentence, then count the words, not the substrings:

            Source https://stackoverflow.com/questions/52302509

            QUESTION

            How to create a dictionary from existing list? Key Value must be the amount of repeats for that key
            Asked 2018-Sep-09 at 16:03

            I'm trying to create a "most wanted letter" program which takes a string as input and output should be the most repeated letter in that string. Currently, I can't figure out how to create a dictionary (e.g. dict.items():[("a", 2), ("b", 5)...]).

            ...

            ANSWER

            Answered 2018-Sep-09 at 15:22

            You may use collections.Counter to do it directly.
            Consider the following example.

            Source https://stackoverflow.com/questions/52245821

            QUESTION

            Check if the string contains three words in succession
            Asked 2018-Jul-02 at 13:52

            I am given a string with words and numbers separated by whitespaces (one space). The words contains only letters. I should check if the string contains three words in succession. For example, the string "start 5 one two three 7 end" contains three words in succession, so it should return True.

            Example:

            ...

            ANSWER

            Answered 2018-Jul-02 at 13:52

            Here's one solution using zip with itertools.islice:

            Source https://stackoverflow.com/questions/51137021

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

            Vulnerabilities

            No vulnerabilities reported

            Install CheckiO

            You can download it from GitHub.
            You can use CheckiO 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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