I have created a 'SiteObject' which includes the following fields:
data class SiteObject(
//Site entry fields (10 fields)
var siteReference: String = "",
var siteAddress: String = "",
var sitePhoneNumber: String = "",
var siteEmail: String = "",
var invoiceAddress: String = "",
var invoicePhoneNumber: String = "",
var invoiceEmail: String = "",
var website: String = "",
var companyNumber: String = "",
var vatNumber: String = "",
)
I want to filter an ArrayList<SiteObject> (call it allSites) by checking if any of the fields of the objects within the list match those in a specific <SiteObject> (call it currentSite).
So for example, I know how to filter looking at one field:
fun checkIfExistingSite(currentSite: SiteObject) : ArrayList<SiteObject> {
var matchingSites = ArrayList<SiteObject>()
allSites.value?.filter { site ->
site.siteReference.contains(currentSite.siteReference)}?.let { matchingSites.addAll(it)
}
return matchingSites
}
But I am looking for an elegant way to create a list where I compare the matching fields in each of the objects in allSites with the corresponding fields in currentSite..
This will give me a list of sites that may be the same (allowing for differences in the way user inputs data) which I can present to the user to check.
Use equals property of Data Class:
val matchingSites: List<SiteObject> = allSites
.filterNotNull()
.filter { it.equals(currentSite) }
If you are looking for a more loose equlity criteria than the full match of all fields values, I would suggest usage of reflection (note that this approach could have performance penalties):
val memberProperties = SiteObject::class.memberProperties
val minMatchingProperties = 9 //or whatever number that makes sense in you case
val matchingItems = allSites.filter {
memberProperties.atLeast(minMatchingProperties) { property -> property.get(it) == property.get(currentSite) }
}
fun <E> Iterable<E>.atLeast(n: Int, predicate: (E) -> Boolean): Boolean {
val size = count()
return when {
n == 1 -> this.any(predicate)
n == size -> this.all(predicate)
n > size - n + 1 -> this.atLeast(size - n + 1) { !predicate.invoke(it) }
else -> {
var count = 0
for (element in this) {
if (predicate.invoke(element)) count++
if (count >= n) return true
}
return false
}
}
}
you could specify all the fields by which you want to match the currentSite inside the filter predicate:
fun checkIfExistingSite(currentSite: SiteObject) =
allSites.filter {
it.siteAddress == currentSite.siteAddress
|| it.sitePhoneNumber == currentSite.sitePhoneNumber
|| it.siteReference == currentSite.siteReference
}
Long but fast solution because of short circuiting.
If the list is nullable you can transform it to a non nullable list like:
allSites?filter{...}.orEmpty()
// or imho better
allSites.orEmpty().filter{...}
Related
The problem I'm working on accepts a number string and will output the product of the odd or even numbers in the string. While the product of purely number string is working fine, my code should also accept strings that is alphanumeric (ex: 67shdg8092) and output the product. I'm quite confused on how I should code the alphanumeric strings, because the code I have done uses toInt().
Here's my code:
fun myProd(Odd: Boolean, vararg data: Char): Int {
var bool = isOdd
var EvenProd = 1
var OddProd = 1
for (a in data)
{
val intVal = a.toString().toInt()
if (intVal == 0)
{
continue
}
if (intVal % 2 == 0)
{
EvenProd *= intVal
}
else
{
OddProd *= intVal
}
}
if(bool == true) return OddProd
else return EvenProd
}
Use toIntOrNull instead of toInt. It only converts numeric string
val intVal = a.toString().toIntOrNull()
if (intVal == null || intVal == 0) {
continue
}
Starting from Kotlin 1.6 you can also use a.digitToIntOrNull().
P.S. Your method could be also rewritten in functional style
fun myProd(isOdd: Boolean, input: String): Int {
return input.asSequence()
.mapNotNull { it.toString().toIntOrNull() } // parse to numeric, ignore non-numeric
.filter { it > 0 } // avoid multiplying by zero
.filter { if (isOdd) it % 2 != 0 else it % 2 == 0 } // pick either odd or even numbers
.fold(1) { prod, i -> prod * i } // accumulate with initial 1
}
I got 2 lists with x objects inside , for example:
data class Model(
var token: String = "",
var id: String = "",
var name: String = "",
var image: Int = 0,
)
array is initialized and filled, the other list has x objects also that contains the objects of the first list but with different values in their properties!
what I want to do is to change the properties of the first array by the second one if they got the same object.name
var arr1 = ArrayList<Model>() // locale
var arr2 = ArrayList<Model>() // from db
the first array I got for example
[Model(name = "David", token = "" , image = 0)]
the second array I got
[Model(name = "David", token = "1asd5asdd851", image = 1)]
How do I make the first array take the missing token?
I tried with .filter{} and with .map{}. groupBy {} for hours because Name is the only properties that are the same but I'm more and more confused.
We can first group the second array by name using associateBy() and then iterate over first array and reassign properties:
val arr2ByName = arr2.associateBy { it.name }
arr1.forEach { item1 ->
arr2ByName[item1.name]?.let { item2 ->
item1.token = item2.token
item1.image = item2.image
}
}
Alternatively, if you don't need to modify items in arr1, but create another array and you can use items from both arr1 and arr2, then it will be much easier:
val arr3 = arr1.map { arr2ByName[it.name] ?: it }
One possible way would be to use fold() as follows:
fun main(args: Array<String>) {
val arr1 = listOf(Model(name = "David", token = "" , image = 0))
val arr2 = listOf(Model(name = "David", token = "1asd5asdd851", image = 1))
val mergedModels = arr2.fold(arr1) { localModels, dbModel ->
localModels.map { localModel ->
if (localModel.name == dbModel.name) localModel.copy(token = dbModel.token, image = dbModel.image)
else localModel
}
}
println(mergedModels)
}
If you want to reuse arr1 variable then you can do the following (but I would still use the previous option):
fun main(args: Array<String>) {
var arr1 = listOf(Model(name = "David", token = "" , image = 0))
val arr2 = listOf(Model(name = "David", token = "1asd5asdd851", image = 1))
arr1 = arr2.fold(arr1) { localModels, dbModel ->
localModels.map { localModel ->
if (localModel.name == dbModel.name) localModel.copy(token = dbModel.token, image = dbModel.image)
else localModel
}
}
println(arr1)
}
fun main () {
var integers = mutableListOf(0)
for (x in 1..9) {
integers.add(x)
}
//for or while could be used in this instance
var lowerCase = listOf("a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z")
var upperCase = listOf('A','B','C','D','E','F','G','H','I','J','K','L','M','N','O','P','Q','R','S','T','U','V','W','X','Y','Z')
println(integers)
println(lowerCase)
println(upperCase)
//Note that for the actual program, it is also vital that I use potential punctuation
val passwordGeneratorKey1 = Math.random()*999
val passwordGeneratorKey2 = passwordGeneratorKey1.toInt()
var passwordGeneratorL1 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorL2 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorL3 = lowerCase[(Math.random()*lowerCase.size).toInt()]
var passwordGeneratorU1 = upperCase[(Math.random()*upperCase.size).toInt()]
var passwordGeneratorU2 = upperCase[(Math.random()*upperCase.size).toInt()]
var passwordGeneratorU3 = upperCase[(Math.random()*upperCase.size).toInt()]
val password = passwordGeneratorKey2.toString()+passwordGeneratorL1+passwordGeneratorL2+passwordGeneratorL3+passwordGeneratorU1+passwordGeneratorU2+passwordGeneratorU3
println(password)
//No, this isn't random, but it's pretty close to it
//How do I now run through every possible combination of the lists //lowerCase, integers, and upperCase?
}
How do I run through every possible permutation to eventually solve for the randomly generated password? This is in Kotlin.
I think you should append all the lists together and then draw from it by random index, this way you ensure that position of numbers, lower cases and uppercases is random too. Also you don't need to write all the characters, you can use Range which generates them for you.
fun main() {
val allChars = mutableListOf<Any>().apply {
addAll(0..9) // creates range from 0 to 9 and adds it to a list
addAll('a'..'z') // creates range from a to z and adds it to a list
addAll('A'..'Z') // creates range from A to Z and adds it to a list
}
val passwordLength = 9
val password = StringBuilder().apply {
for (i in 0 until passwordLength) {
val randomCharIndex =
Random.nextInt(allChars.lastIndex) // generate random index from 0 to lastIndex of list
val randomChar = allChars[randomCharIndex] // select character from list
append(randomChar) // append char to password string builder
}
}.toString()
println(password)
}
Even shorter solution can be achieved using list methods
fun main() {
val password = mutableListOf<Any>()
.apply {
addAll(0..9) // creates range from 0 to 9 and adds it to a list
addAll('a'..'z') // creates range from a to z and adds it to a list
addAll('A'..'Z') // creates range from A to Z and adds it to a list
}
.shuffled() // shuffle the list
.take(9) // take first 9 elements from list
.joinToString("") // join them to string
println(password)
}
As others pointed out there are less painful ways to generate the initial password in the format of: 1 to 3 digits followed by 3 lowercase characters followed by 3 uppercase characters.
To brute force this password, you will need to consider all 3-permutations of "a..z" and all 3-permitations of "A..Z". In both cases the number of such 3-permutations is 15600 = 26! / (26-3)!. In worst case you will have to examine 1000 * 15600 * 15600 combination, half of this on the average.
Probably doable in a few hours with the code below:
import kotlin.random.Random
import kotlin.system.exitProcess
val lowercaseList = ('a'..'z').toList()
val uppercaseList = ('A'..'Z').toList()
val lowercase = lowercaseList.joinToString(separator = "")
val uppercase = uppercaseList.joinToString(separator = "")
fun genPassword(): String {
val lowercase = lowercaseList.shuffled().take(3)
val uppercase = uppercaseList.shuffled().take(3)
return (listOf(Random.nextInt(0, 1000)) + lowercase + uppercase).joinToString(separator = "")
}
/**
* Generate all K-sized permutations of str of length N. The number of such permutations is:
* N! / (N-K)!
*
* For example: perm(2, "abc") = [ab, ac, ba, bc, ca, cb]
*/
fun perm(k: Int, str: String): List<String> {
val nk = str.length - k
fun perm(str: String, accumulate: String): List<String> {
return when (str.length == nk) {
true -> listOf(accumulate)
false -> {
str.flatMapIndexed { i, c ->
perm(str.removeRange(i, i + 1), accumulate + c)
}
}
}
}
return perm(str, "")
}
fun main() {
val password = genPassword().also { println(it) }
val all3LowercasePermutations = perm(3, lowercase).also { println(it) }.also { println(it.size) }
val all3UppercasePermutations = perm(3, uppercase).also { println(it) }.also { println(it.size) }
for (i in 0..999) {
println("trying $i")
for (l in all3LowercasePermutations) {
for (u in all3UppercasePermutations) {
if ("$i$l$u" == password) {
println("found: $i$l$u")
exitProcess(0)
}
}
}
}
}
I need to get the index of the array containing the member fileName = "Andres"
data class File(var fileName: String, var _id : String? = null)
data class Files(val files: Array<File>)
val miObjetG = Gson().fromJson(response_files, Files::class.java)
var indice = miObjetG.files.filterIndexed { index, file -> file.fileName == "Andres"}
I think indexOfFirst is what you are looking for:
val index = miObjetG.files.indexOfFirst{ it.fileName == "Andres" }
The below code will look for "=" and then split them. If there's no "=", filter them away first
myPairStr.asSequence()
.filter { it.contains("=") }
.map { it.split("=") }
However seeing that we have both
.filter { it.contains("=") }
.map { it.split("=") }
Wonder if there's a single operation that could combine the operation instead of doing it separately?
You can use mapNotNull instead of map.
myPairStr.asSequence().mapNotNull { it.split("=").takeIf { it.size >= 2 } }
The takeIf function will return null if the size of the list returned by split method is 1 i.e. if = is not present in the string. And mapNotNull will take only non null values and put them in the list(which is finally returned).
In your case, this solution will work. In other scenarios, the implementation(to merge filter & map) may be different.
I see your point and under the hood split is also doing an indexOf-check to get the appropriate parts.
I do not know of any such function supporting both operations in a single one, even though such a function would basically just be similar to what we have already for the private fun split-implementation.
So if you really want both in one step (and require that functionality more often), you may want to implement your own splitOrNull-function, basically copying the current (private) split-implementation and adapting mainly 3 parts of it (the return type List<String>?, a condition if indexOf delivers a -1, we just return null; and some default values to make it easily usable (ignoreCase=false, limit=0); marked the changes with // added or // changed):
fun CharSequence.splitOrNull(delimiter: String, ignoreCase: Boolean = false, limit: Int = 0): List<String>? { // changed
require(limit >= 0, { "Limit must be non-negative, but was $limit." })
var currentOffset = 0
var nextIndex = indexOf(delimiter, currentOffset, ignoreCase)
if (nextIndex == -1 || limit == 1) {
if (currentOffset == 0 && nextIndex == -1) // added
return null // added
return listOf(this.toString())
}
val isLimited = limit > 0
val result = ArrayList<String>(if (isLimited) limit.coerceAtMost(10) else 10)
do {
result.add(substring(currentOffset, nextIndex))
currentOffset = nextIndex + delimiter.length
// Do not search for next occurrence if we're reaching limit
if (isLimited && result.size == limit - 1) break
nextIndex = indexOf(delimiter, currentOffset, ignoreCase)
} while (nextIndex != -1)
result.add(substring(currentOffset, length))
return result
}
Having such a function in place you can then summarize both, the contains/indexOf and the split, into one call:
myPairStr.asSequence()
.mapNotNull {
it.splitOrNull("=") // or: it.splitOrNull("=", limit = 2)
}
Otherwise your current approach is already good enough. A variation of it would just be to check the size of the split after splitting it (basically removing the need to write contains('=') and just checking the expected size, e.g.:
myPairStr.asSequence()
.map { it.split('=') }
.filter { it.size > 1 }
If you want to split a $key=$value-formats, where value actually could contain additional =, you may want to use the following instead:
myPairStr.asSequence()
.map { it.split('=', limit = 2) }
.filter { it.size > 1 }
// .associate { (key, value) -> key to value }