abracadabra | Automated refactorings for VS Code ✨ It 's magic ✨
kandi X-RAY | abracadabra Summary
kandi X-RAY | abracadabra Summary
Refactoring (noun): a change made to the internal structure of software to make it easier to understand and cheaper to modify without changing its observable behavior. — "Refactoring: Improving the Design of Existing Code" by Martin Fowler. With Abracadabra, you can quickly and safely refactor existing code in VS Code.
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abracadabra Examples and Code Snippets
Lines of Code : 37License : Permissive (MIT License)
Copydef z_function(input_str: str) -> list[int]:
"""
For the given string this function computes value for each index,
which represents the maximal length substring starting from the index
and is the same as the prefix of the same size def find_pattern(pattern: str, input_str: str) -> int:
"""
Example of using z-function for pattern occurrence
Given function returns the number of times 'pattern'
appears in 'input_str' as a substring
>>> find_pattern Community Discussions
Trending Discussions on abracadabra
QUESTION
I've been trying to use RegExp in JS to test a string for a certain count of a substrings. I would like a purely RegExp approach so I can combine it with my other search criteria, but have not had any luck.
As a simple example I would like to test a word if it has exactly 2-3 as.
cat
false
2
shazam
true
3
abracadabra
false
Most of my guesses at regex fail case 3. Example: ^(?
...ANSWER
Answered 2022-Mar-09 at 00:31Could use this regex.
With any other character, including whitespace.
^[^a]*(?:a[^a]*){2,3}$
or if using multi-lines and don't want to span.
^[^a\r\n]*(?:a[^a\r\n]*){2,3}$
QUESTION
I am creating simple function to count all existing vowels in given string. I wrote function to do it but I cannot compile this with Typescript version higher than 3. I am curious how this function should be written to pass Typescript versions higher than 3.
...ANSWER
Answered 2022-Feb-01 at 17:52This is because .match() will return null if there is no match found.
If it is null then length cannot be used. You could save the outcome of str.match(/[aeiou]/gi) to a variable and then check it is not null. The resulting function could look like this:
QUESTION
Sort a character array lexicographically with an additional condition that all c's comes before all b's. This could be done manually but I want to code it via inbuilt sorting using comparators. The code I wrote -
...ANSWER
Answered 2021-Dec-17 at 11:06The problem is that if passing 'c' and 'b' makes the comparison return a negative value, then passing 'b' and 'c' should return a positive one (And your first version returns a negative number in that case instead). If the function doesn't return consistent results no matter the order of arguments, the sort algorithm is going to produce a garbage order.
Consider this version:
QUESTION
I have two lists:
...ANSWER
Answered 2021-Aug-09 at 19:14We can see that the set difference method is faster using timeit:
QUESTION
I have a library that for various reasons I want to build up in pieces (mainly so I can document them, work on them and test them independently in a notebook).
Suppose I have an owner class Parse - and I want to add a bunch of static methods to it.
This code works:
...ANSWER
Answered 2021-Jul-15 at 08:32So my first somewhat crude answer would be to define the helper function, set it as an attribute of Parse and then delete the original reference.
QUESTION
In firebase I have the following children:
...ANSWER
Answered 2021-Jun-21 at 04:19What does your database structure looks like? You need to enter the path of resource you are looking for in ref().
QUESTION
I need help writing a function that returns the length of an array of characters that match "a" || "e" || "i" || "o" || "u".
...ANSWER
Answered 2021-Jun-01 at 21:35Sorry but you're not doing the comparison right.
QUESTION
I want to achieve something like this:
if array [ "a", "b", "c"] includes any of the characters of const word = "abracadabra" give me that character and its position in const word
I have tried something like this:
...ANSWER
Answered 2021-Apr-25 at 18:32You have to use brackets when accessing a field of an array. Parentheses are for calling functions.
QUESTION
I am trying to achieve something like this:
...ANSWER
Answered 2021-Apr-26 at 01:52you can do that...
QUESTION
I am trying to achieve something like this:
const word = "abracadabra" let usedCharacter = ["a", "b", "c", "d", "r"]
if word consists of the available usedCharacter return true
let usedCharacter = ["a", "b", "c", "d", "r", "z"] should also return true
let usedCharacter = ["b", "c", "d", "r", "z"] should return false
should i iterate through word with a for loop and compare every word(i) with the used characters or is there an more easy way?
Thanks for your thoughts!
...ANSWER
Answered 2021-Apr-25 at 22:43You can easily achieve this using splitting the word with each character and check if every character is there in usedCharacter or not.
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