CheckiO | Solutions for tasks from http : //www.checkio.org/
kandi X-RAY | CheckiO Summary
kandi X-RAY | CheckiO Summary
Solutions for tasks from http://www.checkio.org/
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Top functions reviewed by kandi - BETA
- 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
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Community Discussions
Trending Discussions on CheckiO
QUESTION
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 19:50match.toString()
doesn't change the value of match
variable. You need to return it from the function.
QUESTION
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 01:59I 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.
QUESTION
Why the if statement stay after the expression in this code:
...ANSWER
Answered 2019-Dec-02 at 20:23This is a ternary operator (or conditional expression), here a very underperformant way of writing:
QUESTION
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 08:25You can accomplish that using a single liner
QUESTION
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 09:21You'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.
QUESTION
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 09:46The 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.
QUESTION
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 10:54If 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"
QUESTION
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 20:17you'd be better off splitting your sentence, then count the words, not the substrings:
QUESTION
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 15:22You may use collections.Counter to do it directly.
Consider the following example.
QUESTION
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 13:52Here's one solution using zip
with itertools.islice
:
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