Is there an easier approach to convert an Intellij IDEA environment variable into a list of Tuples?
My environment variable for Intellij is
GROCERY_LIST=[("egg", "dairy"),("chicken", "meat"),("apple", "fruit")]
The environment variable gets accessed into Kotlin file as String.
val g_list = System.getenv("GROCERY_LIST")
Ideally I'd like to iterate over g_list, first element being ("egg", "dairy") and so on.
And then ("egg", "dairy") is a tuple/pair
I have tried to split g_list by comma that's NOT inside quotes i.e
val splitted_list = g_list.split(",(?=(?:[^\\\"]*\\\"[^\\\"]*\\\")*[^\\\"]*\$)".toRegex()).toTypedArray()
this gives me first element as [("egg", second element as "dairy")] and so on.
Also created a data class and tried to map the string into data class using jacksonObjectMapper following this link:
val mapper = jacksonObjectMapper()
val g_list = System.getenv("GROCERY_LIST")
val myList: List<Shopping> = mapper.readValue(g_list)
data class Shopping(val a: String, val b: String)
You can create a regular expression to match all strings in your environmental variable.
Regex::findAll()
Then loop through the strings while creating a list of Shopping objects.
// Raw data set.
val groceryList: String = "[(\"egg\", \"dairy\"),(\"chicken\", \"meat\"),(\"apple\", \"fruit\")]"
// Build regular expression.
val regex = Regex("\"([\\s\\S]+?)\"")
val matchResult = regex.findAll(groceryList)
val iterator = matchResult.iterator()
// Create a List of `Shopping` objects.
var first: String = "";
var second: String = "";
val shoppingList = mutableListOf<Shopping>()
var i = 0;
while (iterator.hasNext()) {
val value = iterator.next().value;
if (i % 2 == 0) {
first = value;
} else {
second = value;
shoppingList.add(Shopping(first, second))
first = ""
second = ""
}
i++
}
// Print Shopping List.
for (s in shoppingList) {
println(s)
}
// Output.
/*
Shopping(a="egg", b="dairy")
Shopping(a="chicken", b="meat")
Shopping(a="apple", b="fruit")
*/
data class Shopping(val a: String, val b: String)
Never a good idea to use regex to match parenthesis.
I would suggest a step-by-step approach:
You could first match the name and the value by
(\w+)=(.*)
There you get the name in group 1 and the value in group 2 without caring about any subsequent = characters that might appear in the value.
If you then want to split the value, I would get rid of start and end parenthesis first by matching by
(?<=\[\().*(?=\)\])
(or simply cut off the first and last two characters of the string, if it is always given it starts with [( and ends in )])
Then get the single list entries from splitting by
\),\(
(take care that the split operation also takes a regex, so you have to escape it)
And for each list entry you could split that simply by
,\s*
or, if you want the quote character to be removed, use a match with
\"(.*)\",\s*\"(.*)\"
where group 1 contains the key (left of equals sign) and group 2 the value (right of equals sign)
Related
I have a reference string on which the allowed characters are listed. Then I also have input strings, from which not allowed characters should be replaced with a fixed character, in this example "0".
I can use filter but it removes the characters altogether, does not offer a replacement. Please note that it is not about being alphanumerical, there are ALLOWED non-alphanumerical characters and there are not allowed alphanumerical characters, referenceStr happens to be arbitrary.
var referenceStr = "abcdefg"
var inputStr = "abcqwyzt"
inputStr = inputStr.filter{it in referenceStr}
This yields:
"abc"
But I need:
"abc00000"
I also considered replace but it looks more like when you have a complete reference list of characters that are NOT allowed. My case is the other way around.
Given:
val referenceStr = "abcd][efg"
val replacementChar = '0'
val inputStr = "abcqwyzt[]"
You can do this with a regex [^<referenceStr>], where <referenceStr> should be replaced with referenceStr:
val result = inputStr.replace("[^${Regex.escape(referenceStr)}]".toRegex(), replacementChar.toString())
println(result)
Note that Regex.escape is used to make sure that the characters in referenceStr are all interpreted literally.
Alternatively, use map:
val result = inputStr.map {
if (it !in referenceStr) replacementChar else it
}.joinToString(separator = "")
In the lambda decide whether the current char "it" should be transformed to replacementChar, or itself. map creates a List<Char>, so you need to use joinToString to make the result a String again.
There is a data class as fruits.
data class Fruits(
val code: String, //Unique
val name: String
)
The base list indexed items with boolean variable is as below.
val indexList: MutableList<Boolean> = MutableList(baseFruitList.size) { false }
Now the Favourite Indexed list is as below
val favList: MutableList<Boolean> = MutableList(favFruitList.size) { true}
I want a combined full list which basically has the fav item indicated as true.
Ex:
baseFruitList = {[FT1,apple],[FT2,grapes],[FT3,banana],[FT4,mango],[FT5,pears]}
favList = {[FT2,grapes],[FT4,mango]}
The final index list should have
finalIndexed = {false,true,false,true,false}
How can we achieve in Kotlin, without iterating through each element.
You can do
val finalIndexed = baseFruitList.map { it in favList }
assuming, like #Tenfour04 is asking, that name is guaranteed to be a specific value (including matching case) for a specific code (since that combination is how a data class matches another, e.g. for checking if it's in another list)
If you can't guarantee that, this is safer:
val finalIndexed = baseFruitList.map { fruit ->
favList.any { fav.code == fruit.code }
}
but here you have to iterate over all the favs (at least until you find a match) looking to see if one has the code.
But really, if code is the unique identifier here, why not just store those in your favList?
favList = listOf("FT2", "FT4") // or a Set would be more efficient, and more correct!
val finalIndexed = baseFruitList.map { it.code in favList }
I don't know what you mean about "without iterating through each element" - if you mean without an explicit indexed for loop, then you can use these simple functions like I have here. But there's always some amount of iteration involved. Sets are always an option to help you minimise that
In a function i have this:
val sunRise = SunEquation(2459622)
binding.timeDisplay.setText("$sunRise.n")
The SunEquation-Class looks like this:
class SunEquation(var jDate: Int,) {
val jYear = 2451545
val ttOffset = .0008
var n = jDate - jYear + ttOffset
}
the button- text that appears is:
com.example.soluna.SunEquation#6d1a94b.n
i would expect a double-value
You have to add curly brackets around the value you want to inject into the String, like this:
binding.timeDisplay.setText("${sunRise.n}")
The shorthand syntax without brackets only works for a single variable, but not
for access to a nested field or other more complex expressions.
In your case, this results in the object itself being injected into the String, which is resembled by com.example.soluna.SunEquation#6d1a94b based on the result of the corresponding toString() call, which defaults to the class name and the reference id of the object. Followed by the literal String .n.
Alternatively, you could extract the value into a val beforehand and reference that.
val customN = sunRise.n
binding.timeDisplay.setText("$customN")
I would like to swap a string from "abcde" to "bcdea". So I wrote my code as below in Kotlin
var prevResult = "abcde"
var tmp = prevResult[0]
for (i in 0..prevResult.length - 2) {
prevResult[i] = prevResult[i+1] // Error on preveResult[i]
}
prevResult[prevResult.length-1] = tmp // Error on preveResult[prevResult.lengt-1]
It errors out as stated above comment line. What did I do wrong? How could I fix this and get what I want?
Strings in Kotlin just like in Java are immutable, so there is no string.set(index, value) (which is what string[index] = value is equivalent to).
To build a string from pieces you could use a StringBuilder, construct a CharSequence and use joinToString, operate on a plain array (char[]) or do result = result + nextCharacter (creates a new String each time -- this is the most expensive way).
Here's how you could do this with StringBuilder:
var prevResult = "abcde"
var tmp = prevResult[0]
var builder = StringBuilder()
for (i in 0..prevResult.length - 2) {
builder.append(prevResult[i+1])
}
builder.append(tmp) // Don't really need tmp, use prevResult[0] instead.
var result = builder.toString()
However, a much simpler way to achieve your goal ("bcdea" from "abcde") is to just "move" one character:
var result = prevResult.substring(1) + prevResult[0]
or using the Sequence methods:
var result = prevResult.drop(1) + prevResult.take(1)
You can use drop(1) and first() (or take(1)) to do it in one line:
val str = "abcde"
val r1 = str.drop(1) + str.first()
val r2 = str.drop(1) + str.take(1)
As to your code, Kotlin String is immutable and you cannot modify its characters. To achieve what you want, you can convert a String to CharArray, modify it and then make a new String of it:
val r1 = str.toCharArray().let {
for (i in 0..it.lastIndex - 1)
it[i] = it[i+1]
it[it.lastIndex] = str[0] // str is unchanged
String(it)
}
(let is used for conciseness to avoid creating more variables)
Also, you can write a more general version of this operation as an extension function for String:
fun String.rotate(n: Int) = drop(n % length) + take(n % length)
Usage:
val str = "abcde"
val r1 = str.rotate(1)
Simpler solution: Just use toMutableList() to create a MutableList of Char and then join it all together with joinToString.
Example:
Given a String input, we want to exchange characters at positions posA and posB:
val chars = input.toMutableList()
val temp = chars[posA]
chars[posA] = chars[posB]
chars[posB] = temp
return chars.joinToString(separator = "")
Since Strings are immutable, you will have to copy the source string into an array, make changes to the array, then create a new string from the modified array. Look into:
getChars() to copy the string chars into an array.
Perform your algorithm on that array, making changes to it as needed.
Convert the modified array back into a String with String(char[]).
"first part" &&&& fun _ ->
let ident
"second part" &&&& fun _ ->
ident ....
I need to use variable "ident".
I just need to pass value of variable from first part of test to second one...
I want to ask you if there is any easy way how to define and use global variable or even if you have better (and easy) idea of doing that
Keep in mind, please, that I am a beginner, so I would prefer easier ones.
Global variables will often make your code difficult to work with - particularly if they are mutable.
Instead, consider returning the values you need to keep track of as composite values. An easy data type to start with would be a tuple:
let ``first part`` id =
let someOtherValue = "Foo"
someOtherValue, id + 1
This function takes an int (the current ID) as input, and returns string * int (a tuple where the first element is a string, and the second element and int) as output.
You can call it like this:
> let other, newId = ``first part`` 42;;
val other : string = "Foo"
val newId : int = 43
Notice that you can use pattern matching to immediately destructure the values into two named symbols: other and newId.
Your second function could also take an ID as input:
let ``second part`` id otherArgument =
// use id here, if you need it
"Bar"
You can call it like this, with the newId value from above:
> let result = ``second part`` newId "Baz";;
val result : string = "Bar"
If you find yourself doing this a lot, you can define a record for the purpose:
type Identifiable<'a> = { Id : int; Value : 'a }
Now you can begin to define higher-order functions to deal with such a type, such as e.g. a map function:
module Identifiable =
let map f x = { Id = x.Id; Value = f x.Value }
// Other functions go here...
This is a function that maps the Value of an Identifiable from one value to another, but preserves the identity.
Here's a simple example of using it:
> let original = { Id = 42; Value = "1337" };;
val original : Identifiable<string> = {Id = 42;
Value = "1337";}
> let result' = original |> Identifiable.map System.Int32.Parse;;
val result' : Identifiable<int> = {Id = 42;
Value = 1337;}
As you can see, it preserves the value 42, but changes the Value from a string to an int.
You can still change the ID explicitly, if you want to do that:
> let result'' = { result' with Id = 7 };;
val result'' : Identifiable<int> = {Id = 7;
Value = 1337;}
Since this was getting out of hand for comments this is how I would do it for an example
let mutable t = 0
let first =
t <- 1 + 1
//other stuff
let second =
//can use t here and it will have a value of 2
In some cases you have to use a ref:
let t = ref 0
let first =
t := 1 + 1
//other stuff
let second =
//can use t here and it will have a value of 2 -
// you use "!t" to get the value
If you define ident at the top of your file like this :
let ident = "foo"
// rest of your code using ident
ident are global and you can use in the next part of your file.
EDIT :
If ident wil change in the next part of your code, use this :
let ident = ref "foo"