sometime based on some condition it may want to jump (or move forward) a few steps inside the for loop,
how to do it is kolin?
a simplified use case:
val datArray = arrayOf(1, 2, 3......)
/**
* start from the index to process some data, return how many data has been
consumed
*/
fun processData_1(startIndex: Int) : Int {
// process dataArray starting from the index of startIndex
// return the data it has processed
}
fun processData_2(startIndex: Int) : Int {
// process dataArray starting from the index of startIndex
// return the data it has processed
}
in Java it could be:
for (int i=0; i<datArray.lenght-1; i++) {
int processed = processData_1(i);
i += processed; // jump a few steps for those have been processed, then start 2nd process
if (i<datArray.lenght-1) {
processed = processData_2(i);
i += processed;
}
}
How to do it in kotlin?
for(i in array.indices){
val processed = processData(i);
// todo
}
With while:
var i = 0
while (i < datArray.length - 1) {
var processed = processData_1(i)
i += processed // jump a few steps for those have been processed, then start 2nd process
if (i < datArray.length - 1) {
processed = processData_2(i)
i += processed
}
i++
}
You can do that with continue as stated in the Kotlin docs here: https://kotlinlang.org/docs/reference/returns.html
Example:
val names = arrayOf("james", "john", "jim", "jacob", "johan")
for (name in names) {
if(name.length <= 4) continue
println(name)
}
This would only print names longer than 4 characters (as it skips names with a length of 4 and below)
Edit: this only skips one iteration at a time. So if you want to skip multiple, you could store the process state somewhere else and check the status for each iteration.
Related
I am working on a compiler and one aspect currently is how to wait for interpolated variable names to be resolved. So I am wondering how to take a nested interpolated variable string and build some sort of simple data model/schema for unwrapping the evaluated string so to speak. Let me demonstrate.
Say we have a string like this:
foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}
That has 1, 2, and 3 levels of nested interpolations in it. So essentially it should resolve something like this:
wait for x, y, one, two, and c to resolve.
when both x and y resolve, then resolve a{x}-{y} immediately.
when both one and two resolve, resolve baz{one}-{two}.
when a{x}-{y}, baz{one}-{two}, and c all resolve, then finally resolve the whole expression.
I am shaky on my understanding of the logic flow for handling something like this, wondering if you could help solidify/clarify the general algorithm (high level pseudocode or something like that). Mainly just looking for how I would structure the data model and algorithm so as to progressively evaluate when the pieces are ready.
I'm starting out trying and it's not clear what to do next:
{
dependencies: [
{
path: [x]
},
{
path: [y]
}
],
parent: {
dependency: a{x}-{y} // interpolated term
parent: {
dependencies: [
{
}
]
}
}
}
Some sort of tree is probably necessary, but I am having trouble figuring out what it might look like, wondering if you could shed some light on that with some pseudocode (or JavaScript even).
watch the leaf nodes at first
then, when the children of a node are completed, propagate upward to resolving the next parent node. This would mean once x and y are done, it could resolve a{x}-{y}, but then wait until the other nodes are ready before doing the final top-level evaluation.
You can just simulate it by sending "events" to the system theoretically, like:
ready('y')
ready('c')
ready('x')
ready('a{x}-{y}')
function ready(variable) {
if ()
}
...actually that may not work, not sure how to handle the interpolated nodes in a hacky way like that. But even a high level description of how to solve this would be helpful.
export type SiteDependencyObserverParentType = {
observer: SiteDependencyObserverType
remaining: number
}
export type SiteDependencyObserverType = {
children: Array<SiteDependencyObserverType>
node: LinkNodeType
parent?: SiteDependencyObserverParentType
path: Array<string>
}
(What I'm currently thinking, some TypeScript)
Here is an approach in JavaScript:
Parse the input string to create a Node instance for each {} term, and create parent-child dependencies between the nodes.
Collect the leaf Nodes of this tree as the tree is being constructed: group these leaf nodes by their identifier. Note that the same identifier could occur multiple times in the input string, leading to multiple Nodes. If a variable x is resolved, then all Nodes with that name (the group) will be resolved.
Each node has a resolve method to set its final value
Each node has a notify method that any of its child nodes can call in order to notify it that the child has been resolved with a value. This may (or may not yet) lead to a cascading call of resolve.
In a demo, a timer is set up that at every tick will resolve a randomly picked variable to some number
I think that in your example, foo, and a might be functions that need to be called, but I didn't elaborate on that, and just considered them as literal text that does not need further treatment. It should not be difficult to extend the algorithm with such function-calling features.
class Node {
constructor(parent) {
this.source = ""; // The slice of the input string that maps to this node
this.texts = []; // Literal text that's not part of interpolation
this.children = []; // Node instances corresponding to interpolation
this.parent = parent; // Link to parent that should get notified when this node resolves
this.value = undefined; // Not yet resolved
}
isResolved() {
return this.value !== undefined;
}
resolve(value) {
if (this.isResolved()) return; // A node is not allowed to resolve twice: ignore
console.log(`Resolving "${this.source}" to "${value}"`);
this.value = value;
if (this.parent) this.parent.notify();
}
notify() {
// Check if all dependencies have been resolved
let value = "";
for (let i = 0; i < this.children.length; i++) {
const child = this.children[i];
if (!child.isResolved()) { // Not ready yet
console.log(`"${this.source}" is getting notified, but not all dependecies are ready yet`);
return;
}
value += this.texts[i] + child.value;
}
console.log(`"${this.source}" is getting notified, and all dependecies are ready:`);
this.resolve(value + this.texts.at(-1));
}
}
function makeTree(s) {
const leaves = {}; // nodes keyed by atomic names (like "x" "y" in the example)
const tokens = s.split(/([{}])/);
let i = 0; // Index in s
function dfs(parent=null) {
const node = new Node(parent);
const start = i;
while (tokens.length) {
const token = tokens.shift();
i += token.length;
if (token == "}") break;
if (token == "{") {
node.children.push(dfs(node));
} else {
node.texts.push(token);
}
}
node.source = s.slice(start, i - (tokens.length ? 1 : 0));
if (node.children.length == 0) { // It's a leaf
const label = node.texts[0];
leaves[label] ??= []; // Define as empty array if not yet defined
leaves[label].push(node);
}
return node;
}
dfs();
return leaves;
}
// ------------------- DEMO --------------------
let s = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
const leaves = makeTree(s);
// Create a random order in which to resolve the atomic variables:
function shuffle(array) {
for (var i = array.length - 1; i > 0; i--) {
var j = Math.floor(Math.random() * (i + 1));
[array[j], array[i]] = [array[i], array[j]];
}
return array;
}
const names = shuffle(Object.keys(leaves));
// Use a timer to resolve the variables one by one in the given random order
let index = 0;
function resolveRandomVariable() {
if (index >= names.length) return; // all done
console.log("\n---------------- timer tick --------------");
const name = names[index++];
console.log(`Variable ${name} gets a value: "${index}". Calling resolve() on the connected node instance(s):`);
for (const node of leaves[name]) node.resolve(index);
setTimeout(resolveRandomVariable, 1000);
}
setTimeout(resolveRandomVariable, 1000);
your idea of building a dependency tree it's really likeable.
Anyway I tryed to find a solution as simplest possible.
Even if it already works, there are many optimizations possible, take this just as proof of concept.
The background idea it's produce a List of Strings which you can read in order where each element it's what you need to solve progressively. Each element might be mandatory to solve something that come next in the List, hence for the overall expression. Once you solved all the chunks you have all pieces to solve your original expression.
It's written in Java, I hope it's understandable.
import java.util.ArrayList;
import java.util.Collections;
import java.util.List;
import java.util.Objects;
public class StackOverflow {
public static void main(String[] args) {
String exp = "foo{a{x}-{y}}-{baz{one}-{two}}-foo{c}";
List<String> chunks = expToChunks(exp);
//it just reverse the order of the list
Collections.reverse(chunks);
System.out.println(chunks);
//output -> [c, two, one, baz{one}-{two}, y, x, a{x}-{y}]
}
public static List<String> expToChunks(String exp) {
List<String> chunks = new ArrayList<>();
//this first piece just find the first inner open parenthesys and its relative close parenthesys
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
//this if put an end to recursive calls
if(begin > 0 && begin < exp.length() && end > 0) {
//add the chunk to the final list
String substring = exp.substring(begin, end);
chunks.add(substring);
//remove from the starting expression the already considered chunk
String newExp = exp.replace("{" + substring + "}", "");
//recursive call for inner element on the chunk found
chunks.addAll(Objects.requireNonNull(expToChunks(substring)));
//calculate other chunks on the remained expression
chunks.addAll(Objects.requireNonNull(expToChunks(newExp)));
}
return chunks;
}
}
Some details on the code:
The following piece find the begin and the end index of the first outer chunk of expression. The background idea is: in a valid expression the number of open parenthesys must be equal to the number of closing parenthesys. The count of open(+1) and close(-1) parenthesys can't ever be negative.
So using that simple loop once I find the count of parenthesys to be 0, I also found the first chunk of the expression.
int begin = exp.indexOf("{") + 1;
int numberOfParenthesys = 1;
int end = -1;
for(int i = begin; i < exp.length(); i++) {
char c = exp.charAt(i);
if (c == '{') numberOfParenthesys ++;
if (c == '}') numberOfParenthesys --;
if (numberOfParenthesys == 0) {
end = i;
break;
}
}
The if condition provide validation on the begin and end indexes and stop the recursive call when no more chunks can be found on the remained expression.
if(begin > 0 && begin < exp.length() && end > 0) {
...
}
I have a script that changes the layer name of my illustrator file to "Test 1, Test 2, etc..." All I want to accomplish is to add leading zero to single digits. "Test 01, Test 02 ... Test 10, Test 11, etc..."
var doc = app.activeDocument;
idLayers("Test "); // Rename visible layers
// Hidden layers will be skipped and not counted
function idLayers(prefix){
var counter = 1;
for(i=0;doc.layers.length>i;i++){
var currentLayer = doc.layers[i];
// if layer is visible...
if (currentLayer.visible) {
currentLayer.name= prefix + counter;
counter++;
}
}
}
I found the following that would help but I'm not sure where to add it to the above code.
function pad(n) {
return (n < 10) ? ("0" + n) : n;
}
Total noob here so any help would be greatly appreciated. Thank you in advance!
You simply need to add the function that you already found at the end of your script (or at the beginning, it does not really matter) and then call it in the line, where the layer is named. So the whole script would look like this:
var doc = app.activeDocument;
idLayers("Test "); // Rename visible layers
// Hidden layers will be skipped and not counted
function idLayers(prefix){
var counter = 1;
for(i=0;doc.layers.length>i;i++){
var currentLayer = doc.layers[i];
// if layer is visible...
if (currentLayer.visible) {
currentLayer.name= prefix + pad(counter);
counter++;
}
}
}
function pad(n) {
return (n < 10) ? ("0" + n) : n;
}
I am doing an edit distance with the user input. I am storing my values in array. then the edit distance will compare the user input with my array of strings. I am doing a loop that if the edit distance is more than 2 it will display invalid else valid.
The only problem I've got is that although the program is working out fine, the output is the result of all the '28' strings that I have in my array. I would like to display only invalid or valid once.
Test is my array of strings and user is - String user - the user input.
void testingLD()
{
for (int i=0; i<test.length; i++)
{
if(getLevenshteinDistance(test[i],user) > 2)
{
println ("Invalid re-input");
}
else
{
println ("Valid");
}
}
}
You have your print line functions inside your loop so they get printed once per iteration.
Try this.
void testingLD()
{
boolean isValid = true; // assume true, check each and update
// begin loop
for (int i=0; i<test.length; i++)
{
if(getLevenshteinDistance(test[i],user) > 2)
{
isValid = false;
break; // don't need to test the rest of the loop if we already found a false
}
}
// end loop
if(isValid){
println("Valid");
}else{
println("Invalid re-input");
}
}
Similarly you could count the number of valid int validCount = 0; validCount++ and then display stats about how many were valid, the percentage etc. Or keep an array of the invalid strings and display those as the ones that fail etc!
Wrap up:
When you want to check an entire collection or array for some condition and output one answer make sure to have your output outside of the loop!
.proto file structure
message repetedMSG
{
required string data = 1;
}
message mainMSG
{
required repetedMSG_id = 1;
repeated repetedMSG rptMSG = 2;
}
I have one mainMSG and in it too many (suppose 10) repetedMSG are present.
Now i want to delete any particular repetedMSG (suppose 5th repetedMSG )from mainMSG. For this i tried 3 ways but none of them worked.
for (int j = 0; j<mainMSG->repetedMSG_size(); j++){
repetedMSG reptMsg = mainMsg->mutable_repetedMSG(j);
if (QString::fromStdString(reptMsg->data).compare("deleteMe") == 0){
*First tried way:-* reptMsg->Clear();
*Second tried Way:-* delete reptMsg;
*Third tried way:-* reptMsg->clear_formula_name();
break;
}
}
I get run-time error when i serialize the mainMSG for writing to a file i.e. when execute this line
mainMSG.SerializeToOstream (std::fstream output("C:/A/test1", std::ios::out | std::ios::trunc | std::ios::binary)) here i get run-time error
You can use RepeatedPtrField::DeleteSubrange() for this. However, be careful about using this in a loop -- people commonly write code like this which is O(n^2):
// BAD CODE! O(n^2)!
for (int i = 0; i < message.foo_size(); i++) {
if (should_filter(message.foo(i))) {
message.mutable_foo()->DeleteSubrange(i, 1);
--i;
}
}
Instead, if you plan to remove multiple elements, do something like this:
// Move all filtered elements to the end of the list.
int keep = 0; // number to keep
for (int i = 0; i < message.foo_size(); i++) {
if (should_filter(message.foo(i))) {
// Skip.
} else {
if (keep < i) {
message.mutable_foo()->SwapElements(i, keep)
}
++keep;
}
}
// Remove the filtered elements.
message.mutable_foo()->DeleteSubrange(keep, message.foo_size() - keep);
I just started a project for my Java2 class and I've come to a complete stop. I just can't get
my head around this method. Especially when the assignment does NOT let us use any other DATA STRUCTURE or shuffle methods from java at all.
So I have a Deck.class in which I've already created a linked list containing 52 nodes that hold 52 cards.
public class Deck {
private Node theDeck;
private int numCards;
public Deck ()
{
while(numCards < 52)
{
theDeck = new Node (new Card(numCards), theDeck);
numCards++;
}
}
public void shuffleDeck()
{
int rNum;
int count = 0;
Node current = theDeck;
Card tCard;
int range = 0;
while(count != 51)
{
// Store whatever is inside the current node in a temp variable
tCard = current.getItem();
// Generate a random number between 0 -51
rNum = (int)(Math.random()* 51);
// Send current on a loop a random amount of times
for (int i=0; i < rNum; i ++)
current = current.getNext(); ******<-- (Btw this is the line I'm getting my error, i sort of know why but idk how to stop it.)
// So wherever current landed get that item stored in that node and store it in the first on
theDeck.setItem(current.getItem());
// Now make use of the temp variable at the beginning and store it where current landed
current.setItem(tCard);
// Send current back to the beginning of the deck
current = theDeck;
// I've created a counter for another loop i want to do
count++;
// Send current a "count" amount of times for a loop so that it doesn't shuffle the cards that have been already shuffled.
for(int i=0; i<count; i++)
current = current.getNext(); ****<-- Not to sure about this last loop because if i don't shuffle the cards that i've already shuffled it will not count as a legitimate shuffle? i think? ****Also this is where i sometimes get a nullpointerexception****
}
}
}
Now I get different kinds of errors
When I call on this method:
it will sometimes shuffle just 2 cards but at times it will shuffle 3 - 5 cards then give me a NullPointerException.
I've pointed out where it gives me this error with asterisks in my code above
at one point I got it to shuffle 13 cards but then everytime it did that it didn't quite shuffle them the right way. one card kept always repeating.
at another point I got all 52 cards to go through the while loop but again it repeated one card various times.
So I really need some input in what I'm doing wrong. Towards the end of my code I think my logic is completely wrong but I can't seem to figure out a way around it.
Seems pretty long-winded.
I'd go with something like the following:
public void shuffleDeck() {
for(int i=0; i<52; i++) {
int card = (int) (Math.random() * (52-i));
deck.addLast(deck.remove(card));
}
}
So each card just gets moved to the back of the deck in a random order.
If you are authorized to use a secondary data structure, one way is simply to compute a random number within the number of remaining cards, select that card, move it to the end of the secondary structure until empty, then replace your list with the secondary list.
My implementation shuffles a linked list using a divide-and-conquer algorithm
public class LinkedListShuffle
{
public static DataStructures.Linear.LinkedListNode<T> Shuffle<T>(DataStructures.Linear.LinkedListNode<T> firstNode) where T : IComparable<T>
{
if (firstNode == null)
throw new ArgumentNullException();
if (firstNode.Next == null)
return firstNode;
var middle = GetMiddle(firstNode);
var rightNode = middle.Next;
middle.Next = null;
var mergedResult = ShuffledMerge(Shuffle(firstNode), Shuffle(rightNode));
return mergedResult;
}
private static DataStructures.Linear.LinkedListNode<T> ShuffledMerge<T>(DataStructures.Linear.LinkedListNode<T> leftNode, DataStructures.Linear.LinkedListNode<T> rightNode) where T : IComparable<T>
{
var dummyHead = new DataStructures.Linear.LinkedListNode<T>();
DataStructures.Linear.LinkedListNode<T> curNode = dummyHead;
var rnd = new Random((int)DateTime.Now.Ticks);
while (leftNode != null || rightNode != null)
{
var rndRes = rnd.Next(0, 2);
if (rndRes == 0)
{
if (leftNode != null)
{
curNode.Next = leftNode;
leftNode = leftNode.Next;
}
else
{
curNode.Next = rightNode;
rightNode = rightNode.Next;
}
}
else
{
if (rightNode != null)
{
curNode.Next = rightNode;
rightNode = rightNode.Next;
}
else
{
curNode.Next = leftNode;
leftNode = leftNode.Next;
}
}
curNode = curNode.Next;
}
return dummyHead.Next;
}
private static DataStructures.Linear.LinkedListNode<T> GetMiddle<T>(DataStructures.Linear.LinkedListNode<T> firstNode) where T : IComparable<T>
{
if (firstNode.Next == null)
return firstNode;
DataStructures.Linear.LinkedListNode<T> fast, slow;
fast = slow = firstNode;
while (fast.Next != null && fast.Next.Next != null)
{
slow = slow.Next;
fast = fast.Next.Next;
}
return slow;
}
}
Just came across this and decided to post a more concise solution which allows you to specify how much shuffling you want to do.
For the purposes of the answer, you have a linked list containing PlayingCard objects;
LinkedList<PlayingCard> deck = new LinkedList<PlayingCard>();
And to shuffle them use something like this;
public void shuffle(Integer swaps) {
for (int i=0; i < swaps; i++) {
deck.add(deck.remove((int)(Math.random() * deck.size())));
}
}
The more swaps you do, the more randomised the list will be.