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  • Programming Language Family Tree?

    - by user134353
    As a man interested in programming, I must ask if there is a cataloged hierarchy of languages. I'd like to learn to actually understand what's happening- that is to say, I don't want to use a compiler until I understand what a compiler does and how to make my own. I really do want to start from total scratch. I'm told that means "machine code"? I don't know. What I do know is that "C++" is not the start. I'm not interested in learning that until I can actually break software down to its very base and see how the pieces go together.

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  • Case Insensitive Ternary Search Tree

    - by Yan Cheng CHEOK
    I had been using Ternary Search Tree for a while, as the data structure to implement a auto complete drop down combo box. Which means, when user type "fo", the drop down combo box will display foo food football The problem is, my current used of Ternary Search Tree is case sensitive. My implementation is as follow. It had been used by real world for around 1++ yeas. Hence, I consider it as quite reliable. My Ternary Search Tree code However, I am looking for a case insensitive Ternary Search Tree, which means, when I type "fo", the drop down combo box will show me foO Food fooTBall Here are some key interface for TST, where I hope the new case insentive TST may have similar interface too. /** * Stores value in the TernarySearchTree. The value may be retrieved using key. * @param key A string that indexes the object to be stored. * @param value The object to be stored in the tree. */ public void put(String key, E value) { getOrCreateNode(key).data = value; } /** * Retrieve the object indexed by key. * @param key A String index. * @return Object The object retrieved from the TernarySearchTree. */ public E get(String key) { TSTNode<E> node = getNode(key); if(node==null) return null; return node.data; } An example of usage is as follow. TSTSearchEngine is using TernarySearchTree as the core backbone. Example usage of Ternary Search Tree // There is stock named microsoft and MICROChip inside stocks ArrayList. TSTSearchEngine<Stock> engine = TSTSearchEngine<Stock>(stocks); // I wish it would return microsoft and MICROCHIP. Currently, it just return microsoft. List<Stock> results = engine.searchAll("micro");

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  • Flex Tree with infinite parents and children

    - by Tempname
    I am working on a tree component and I am having a bit of the issue with populating the data-provider for this tree. The data that I get back from my database is a simple array of value objects. Each value object has 2 properties. ObjectID and ParentID. For parents the ParentID is null and for children the ParentID is the ObjectID of the parent. Any help with this is greatly appreciated. Essentially the tree should look something like this: Parent1 Child1 Child1 Child2 Child1 Child2 Parent2 Child1 Child2 Child3 Child1 This is the current code that I am testing with: public function setDataProvider(data:Array):void { var tree:Array = new Array(); for(var i:Number = 0; i < data.length; i++) { // do the top level array if(!data[i].parentID) { tree.push(data[i], getChildren(data[i].objectID, data)); } } function getChildren(objectID:Number, data:Array):Array { var childArr:Array = new Array(); for(var k:Number = 0; k < data.length; k++) { if(data[k].parentID == objectID) { childArr.push(data[k]); //getChildren(data[k].objectID, data); } } return childArr; } trace(ObjectUtil.toString(tree)); } Here is a cross section of my data: ObjectID ParentID 1 NULL 10 NULL 8 NULL 6 NULL 4 6 3 6 9 6 2 6 11 7 7 8 5 8

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  • How do I output the preorder traversal of a tree given the inorder and postorder tranversal?

    - by user342580
    Given the code for outputing the postorder traversal of a tree when I have the preorder and the inorder traversal in an interger array. How do I similarily get the preorder with the inorder and postorder array given? void postorder( int preorder[], int prestart, int inorder[], int inostart, int length) { if(length==0) return; //terminating condition int i; for(i=inostart; i<inostart+length; i++) if(preorder[prestart]==inorder[i])//break when found root in inorder array break; postorder(preorder, prestart+1, inorder, inostart, i-inostart); postorder(preorder, prestart+i-inostart+1, inorder, i+1, length-i+inostart-1); cout<<preorder[prestart]<<" "; } Here is the prototype for preorder() void preorder( int inorderorder[], int inostart, int postorder[], int poststart, int length)

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  • How can I save a directory tree to an array in PHP?

    - by Greg
    I'm trying to take a directory with the structure: top folder1 file1 folder2 file1 file2 And save it into an array like: array ( 'folder1' => array('file1'), 'folder2' => array('file1', 'file2') ) This way, I can easily resuse the tree throughout my site. I've been playing around with this code but it's still not doing what I want: private function get_tree() { $uploads = __RELPATH__ . DS . 'public' . DS . 'uploads'; $iterator = new RecursiveIteratorIterator(new RecursiveDirectoryIterator($uploads), RecursiveIteratorIterator::SELF_FIRST); $output = array(); foreach($iterator as $file) { $relativePath = str_replace($uploads . DS, '', $file); if ($file->isDir()) { if (!in_array($relativePath, $output)) $output[$relativePath] = array(); } } return $output; }

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  • How can I do block-oriented disk I/O with Java? Or similar for a B+ tree

    - by Sanoj
    I would like to implement an B+ tree in Java and try to optimize it for disk based I/O. Is there an API for accessing individual disk blocks from Java? or is there an API that can do similar block-oriented access that fits my purpose? I would like to create something like Tokyo Cabinet in 100% Java. Is there anyone that knows what Java only databases like JavaDB is using in the back-end for this? I know that there are probably other languages than Java that can do this better, but I do this in a learning purpose only.

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  • HTML Markup in einem APEX Tree - ganz einfach per Plugin!

    - by carstenczarski
    Die APEX Tree Region kennt sicherlich jeder APEX-Entwickler. Und vielfach besteht der Bedarf, das Aussehen des APEX Tree mit Hilfe von HTML Markup zu beeinflussen. Leider ist es seit APEX 4.0 nicht mehr möglich, eigenes HTML-Markup in einen APEX-Tree aufzunehmen - aus Sicherheitsgründen (Schutz vor Cross-Site-Scripting) werden alle HTML Sonderzeichen maskiert. Wenn kein XSS-Risiko besteht (die vom Tree dargestellten Inhalte basieren nicht auf Benutzereingaben und werden komplett vom Entwickler bestimmt), kann dies mit wenigen Zeilen JavaScript und jQuery-Code erreicht werden. Damit es noch einfacher wird,  haben wir die Funktionalität für Sie in einem APEX-Plugin gekapselt. Und so funktioniert es: APEX Plugin "HTML Markup for APEX Tree Region" herunterladenhttp://apex-plugin.com/oracle-apex-plugins/dynamic-action-plugin/html-markup-for-apex-tree_174.html APEX Plugin in die Anwendung importieren APEX Tree Region erzeugen und eigene Ersetzungen für HTML-Sonderzeichen verwenden, also bspw."[" für "<", "]" für ">" und "§" für "&". Eine neue dynamische Aktion erzeugen, die beim Laden der Seite ausgeführt wird und mit Hilfe des Plugins die Ersetzungen im Tree durch die "richtigen" HTML-Sonderzeichen ersetzt. Fertig. Wie das Plugin wirkt, können Sie sich auf einer Demo-Seite ansehen.

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  • Tool to compute SHA256 Tree Hash

    - by Benjamin
    I've started using AWS Glacier, and noticed that it hashes the files using an algorithm called SHA-256 Tree Hash. To my surprise, this algorithm is different from SHA-256, so I can't use the tools I'm used to, to compare hashes and verify file integrity. Do you know a Windows tool, if possible integrated in the context menu, to compute the SHA-256 Tree Hash of a file? I'd also accept a Linux command-line tool, as a second choice :-)

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  • Binary Search Tree in C

    - by heapzero
    Hi, I'm a Python guy. Learning C language and I've been trying to implement Binary Search Tree in C. I wrote down the code, and I've been trying from few hours but, not able to get the output as expected. Please help! Please correct me. #include<stdlib.h> #include<stdio.h> typedef int ElementType; typedef struct TreeNode { ElementType element; struct TreeNode *left, *right; } TreeNode; TreeNode *createTree(){ //Create the root of tree TreeNode *tempNode; tempNode = malloc(sizeof(TreeNode)); tempNode->element = 0; tempNode->left = NULL; tempNode->right = NULL; return tempNode; } TreeNode *createNode(ElementType X){ //Create a new leaf node and return the pointer TreeNode *tempNode; tempNode = malloc(sizeof(TreeNode)); tempNode->element = X; tempNode->left = NULL; tempNode->right = NULL; return tempNode; } TreeNode *insertElement(TreeNode *node, ElementType X){ //insert element to Tree if(node==NULL){ return createNode(X); } else{ if(X < node->element){ node->left = insertElement(node->left, X); } else if(X > node->element){ node->right = insertElement(node->right, X); } else if(X == node->element){ printf("Oops! the element is already present in the tree."); } } } TreeNode *displayTree(TreeNode *node){ //display the full tree if(node==NULL){ return; } displayTree(node->left); printf("| %d ", node->element); displayTree(node->right); } main(){ //pointer to root of tree #2 TreeNode *TreePtr; TreeNode *TreeRoot; TreeNode *TreeChild; //Create the root of tree TreePtr = createTree(); TreeRoot = TreePtr; TreeRoot->element = 32; printf("%d\n",TreeRoot->element); insertElement(TreeRoot, 8); TreeChild = TreeRoot->left; printf("%d\n",TreeChild->element); insertElement(TreeRoot, 2); insertElement(TreeRoot, 7); insertElement(TreeRoot, 42); insertElement(TreeRoot, 28); insertElement(TreeRoot, 1); insertElement(TreeRoot, 4); insertElement(TreeRoot, 5); // the output is not as expected :( displayTree(TreeRoot); }

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  • Algorithm to Render a Horizontal Binary-ish Tree in Text/ASCII form

    - by Justin L.
    It's a pretty normal binary tree, except for the fact that one of the nodes may be empty. I'd like to find a way to output it in a horizontal way (that is, the root node is on the left and expands to the right). I've had some experience expanding trees vertically (root node at the top, expanding downwards), but I'm not sure where to start, in this case. Preferably, it would follow these couple of rules: If a node has only one child, it can be skipped as redundant (an "end node", with no children, is always displayed) All nodes of the same depth must be aligned vertically; all nodes must be to the right of all less-deep nodes and to the left of all deeper nodes. Nodes have a string representation which includes their depth. Each "end node" has its own unique line; that is, the number of lines is the number of end nodes in the tree, and when an end node is on a line, there may be nothing else on that line after that end node. As a consequence of the last rule, the root node should be in either the top left or the bottom left corner; top left is preferred. For example, this is a valid tree, with six end nodes (node is represented by a name, and its depth): [a0]------------[b3]------[c5]------[d8] \ \ \----------[e9] \ \----[f5] \--[g1]--------[h4]------[i6] \ \--------------------[j10] \-[k3] Which represents the horizontal, explicit binary tree: 0 a / \ 1 g * / \ \ 2 * * * / \ \ 3 k * b / / \ 4 h * * / \ \ \ 5 * * f c / \ / \ 6 * i * * / / \ 7 * * * / / \ 8 * * d / / 9 * e / 10 j (branches folded for compactness; * representing redundant, one-child nodes; note that *'s are actual nodes, storing one child each, just with names omitted here for presentation sake) (also, to clarify, I'd like to generate the first, horizontal tree; not this vertical tree) I say language-agnostic because I'm just looking for an algorithm; I say ruby because I'm eventually going to have to implement it in ruby anyway. Assume that each Node data structure stores only its id, a left node, and a right node. A master Tree class keeps tracks of all nodes and has adequate algorithms to find: A node's nth ancestor A node's nth descendant The generation of a node The lowest common ancestor of two given nodes Anyone have any ideas of where I could start? Should I go for the recursive approach? Iterative?

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  • How can I convert this PHP script to Ruby? (build tree from tabbed string)

    - by Jon Sunrays
    I found this script below online, and I'm wondering how I can do the same thing with a Ruby on Rails setup. So, first off, I ran this command: rails g model Node node_id:integer title:string Given this set up, how can I make a tree from a tabbed string like the following? <?php // Make sure to have "Academia" be root node with nodeID of 1 $data = " Social sciences Anthropology Biological anthropology Forensic anthropology Gene-culture coevolution Human behavioral ecology Human evolution Medical anthropology Paleoanthropology Population genetics Primatology Anthropological linguistics Synchronic linguistics (or Descriptive linguistics) Diachronic linguistics (or Historical linguistics) Ethnolinguistics Sociolinguistics Cultural anthropology Anthropology of religion Economic anthropology Ethnography Ethnohistory Ethnology Ethnomusicology Folklore Mythology Political anthropology Psychological anthropology Archaeology ...(goes on for a long time) "; //echo "Checkpoint 2\n"; $lines = preg_split("/\n/", $data); $parentids = array(0 => null); $db = new PDO("host", 'username', 'pass'); $sql = 'INSERT INTO `TreeNode` SET ParentID = ?, Title = ?'; $stmt = $db->prepare($sql); foreach ($lines as $line) { if (!preg_match('/^([\s]*)(.*)$/', $line, $m)) { continue; } $spaces = strlen($m[1]); //$level = intval($spaces / 4); //assumes four spaces per indent $level = strlen($m[1]); // if data is tab indented $title = $m[2]; $parentid = ($level > 0 ? $parentids[$level - 1] : 1); //All "roots" are children of "Academia" which has an ID of "1"; $rv = $stmt->execute(array($parentid, $title)); $parentids[$level] = $db->lastInsertId(); echo "inserted $parentid - " . $parentid . " title: " . $title . "\n"; } ?>

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  • How do I force all Tree itemrenderers to refresh?

    - by Richard Haven
    I have item renderers in an mx.controls.Tree that I need to refresh on demand. I have code in the updateDisplayList that fires for only some of the visible nodes no matter what I do. I've tried triggering a change that they should all be listening for; I have tried clearing and resetting the dataProvider and the itemRenderer properties. private function forceCategoryTreeRefresh(event : Event = null) : void { trace("forceCategoryTreeRefresh"); var prevDataProvider : Object = CategoryTree.dataProvider; CategoryTree.dataProvider = null; CategoryTree.validateNow(); CategoryTree.dataProvider = prevDataProvider; var prevItemRenderer : IFactory = CategoryTree.itemRenderer; CategoryTree.itemRenderer = null; CategoryTree.itemRenderer = prevItemRenderer as IFactory; _categoriesChangeDispatcher.dispatchEvent(new Event(Event.CHANGE)); } The nodes refresh properly when I scroll them into view (e.g. the .data gets set), but I cannot force the ones that already exist to refresh or reset themselves. Any ideas?

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  • How to find sum of node's value for given depth in binary tree?

    - by masato-san
    I've been scratching my head for several hours for this... problem: Binary Tree (0) depth 0 / \ 10 20 depth 1 / \ / \ 30 40 50 60 depth 2 I am trying to write a function that takes depth as argument and return the sum of values of nodes of the given depth. For instance, if I pass 2, it should return 180 (i.e. 30+40+50+60) I decided to use breath first search and when I find the node with desired depth, sum up the value, but I just can't figure out how to find out the way which node is in what depth. But with this approach I feel like going to totally wrong direction. function level_order($root, $targetDepth) { $q = new Queue(); $q->enqueue($root); while(!$q->isEmpty) { //how to determin the depth of the node??? $node = $q->dequeue(); if($currentDepth == $targetDepth) { $sum = $node->value; } if($node->left != null) { $q->enqueue($node->left); } if($node->right != null) { $q->enqueue($node->right); } //need to reset this somehow $currentDepth ++; } }

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  • Approaches to create a nested tree structure of NSDictionaries?

    - by d11wtq
    I'm parsing some input which produces a tree structure containing NSDictionary instances on the branches and NSString instance at the nodes. After parsing, the whole structure should be immutable. I feel like I'm jumping through hoops to create the structure and then make sure it's immutable when it's returned from my method. We can probably all relate to the input I'm parsing, since it's a query string from a URL. In a string like this: a=foo&b=bar&a=zip We expect a structure like this: NSDictionary { "a" => NSDictionary { 0 => "foo", 1 => "zip" }, "b" => "bar" } I'm keeping it just two-dimensional in this example for brevity, though in the real-world we sometimes see var[key1][key2]=value&var[key1][key3]=value2 type structures. The code hasn't evolved that far just yet. Currently I do this: - (NSDictionary *)parseQuery:(NSString *)queryString { NSMutableDictionary *params = [NSMutableDictionary dictionary]; NSArray *pairs = [queryString componentsSeparatedByString:@"&"]; for (NSString *pair in pairs) { NSRange eqRange = [pair rangeOfString:@"="]; NSString *key; id value; // If the parameter is a key without a specified value if (eqRange.location == NSNotFound) { key = [pair stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; value = @""; } else { // Else determine both key and value key = [[pair substringToIndex:eqRange.location] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; if ([pair length] > eqRange.location + 1) { value = [[pair substringFromIndex:eqRange.location + 1] stringByReplacingPercentEscapesUsingEncoding:NSASCIIStringEncoding]; } else { value = @""; } } // Parameter already exists, it must be a dictionary if (nil != [params objectForKey:key]) { id existingValue = [params objectForKey:key]; if (![existingValue isKindOfClass:[NSDictionary class]]) { value = [NSDictionary dictionaryWithObjectsAndKeys:existingValue, [NSNumber numberWithInt:0], value, [NSNumber numberWithInt:1], nil]; } else { // FIXME: There must be a more elegant way to build a nested dictionary where the end result is immutable? NSMutableDictionary *newValue = [NSMutableDictionary dictionaryWithDictionary:existingValue]; [newValue setObject:value forKey:[NSNumber numberWithInt:[newValue count]]]; value = [NSDictionary dictionaryWithDictionary:newValue]; } } [params setObject:value forKey:key]; } return [NSDictionary dictionaryWithDictionary:params]; } If you look at the bit where I've added FIXME it feels awfully clumsy, pulling out the existing dictionary, creating an immutable version of it, adding the new value, then creating an immutable dictionary from that to set back in place. Expensive and unnecessary? I'm not sure if there are any Cocoa-specific design patterns I can follow here?

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  • How to structure classes in the filesystem?

    - by da_b0uncer
    I have a few (view) classes. Table, Tree, PagingColumn, SelectionColumn, SparkLineColumn, TimeColumn. currently they're flat under app/view like this: app/view/Table app/view/Tree app/view/PagingColumn ... I thought about restructuring it, because the Trees and Tables use the columns, but there are some columns, which only work in a tree, some who work in trees and tables and in the future there are probably some who only work in tables, I don't know. My first idea was like this: app/view/Table app/view/Tree app/view/column/PagingColumn app/view/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn But since the SelectionColumn is explicitly for trees, I have the fear that future developers could get the idea of missuse them. But how to restructure it probably? Like this: app/view/table/panel/Table app/view/tree/panel/Tree app/view/tree/column/PagingColumn app/view/tree/column/SelectionColumn app/view/column/SparkLineColumn app/view/column/TimeColumn Or like this: app/view/Table app/view/Tree app/view/column/SparkLineColumn app/view/column/TimeColumn app/view/column/tree/PagingColumn app/view/column/tree/SelectionColumn

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  • Dojo slider: update value dynamically

    - by xApple
    I am building an application where the user browses certain objects by scrolling and zooming. To change the zoom level I have successfully implemented a "dijit.form.HorizontalSlider" object. Every time the user changes the position of the silder, I can catch the "onChange" call and do something with that. However, the user can also zoom-in by double clicking inside the view zone, at which point the slider should change position automatically to reflect the new zoom level. My question is the following: what function or method should I call in my javascript to update the position of a dojo silder ? Here is the code that creates the silder object: var zoomSlider = new dijit.form.HorizontalSlider({ name: "zoom_slider", id: "zoom_slider", value: 0, minimum: 0, maximum: 19, discreteValues: 20, intermediateChanges: false, style: "width: 160px;", onChange: function(value) { brwsr.view.zoomTo(value); } }, "zoom_slider"); navbox_silder.appendChild(zoomSlider.domNode);

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  • Ensuring dynamically added Dojo Dijits are registered in the DOM.

    - by Pro777
    I need to dynamically create Dijits to add N rules to a particular form. Currently I am adding them with the following javascript. var value = new dijit.form.Select({ id: "value_" + counter, store: ValueStore, searchAttr: "description" }, "stateSelect" + counter); value.placeAt(form, "last"); value.startup(); This is creating the element correctly in the form, but it is not associated when the form actually posts. I know Dojo is recognizing the Dijit because the styling is correct. How can I make sure they are included in the DOM form?

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  • Pseudo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • gunit syntax for tree walker with a flat list of nodes

    - by Kaleb Pederson
    Here's a simple gunit test for a portion of my tree grammar which generates a flat list of nodes: objectOption walks objectOption: <<one:"value">> -> (one "value") Although you define a tree in ANTLR's rewrite syntax using a caret (i.e. ^(ROOT child...)), gunit matches trees without the caret, so the above represents a tree and it's not surprising that it fails: it's a flat list of nodes and not a tree. This results in a test failure: 1 failures found: test2 (objectOption walks objectOption, line17) - expected: (one \"value\") actual: one \"value\" Another option which seems intuitive is to leave off the parenthesis, like this: objectOption walks objectOption: <<one:"value">> -> one "value" But gunit doesn't like this syntax. It seems to result in a parse failure in the gunit grammar: line 17:20 no viable alternative at input 'one' line 17:24 missing ':' at 'value' line 0:-1 no viable alternative at input '<EOF>' java.lang.NullPointerException at org.antlr.gunit.OutputTest.getExpected(OutputTest.java:65) at org.antlr.gunit.gUnitExecutor.executeTests(gUnitExecutor.java:245) ... What is the correct way to match a flat tree?

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  • Parsing a given binary tree using python?

    - by kaushik
    Parse a binary tree,referring to given set of features,answering decision tree question at each node to decide left child or right child and find the path to leaf node according to answer given to the decision tree.. input wil be a set of feature which wil help in answering the question at each level to choose the left or right half and the output will be the leaf node.. i need help in implementing this can anyone suggest methods?? Please answer... thanks in advance..

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  • Psuedo LRU tree algorithm.

    - by patros
    A lot of descriptions of Pseudo LRU algorithms involve using a binary search tree, and setting flags to "point away" from the node you're searching for every time you access the tree. This leads to a reasonable approximation of LRU. However, it seems from the descriptions that all of the nodes deemed LRU would be leaf nodes. Is there a pseudo-LRU algorithm that deals with a static tree that will still perform reasonably well, while determining that non-leaf nodes are suitable LRU candidates?

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  • Output of gcc -fdump-tree-original

    - by Job
    If I dump the code generated by GCC for a virtual destructor (with -fdump-tree-original), I get something like this: ;; Function virtual Foo::~Foo() (null) ;; enabled by -tree-original { <<cleanup_point <<< Unknown tree: expr_stmt (void) (((struct Foo *) this)->_vptr.Foo = &_ZTV3Foo + 8) >>> >>; } <D.20148>:; if ((bool) (__in_chrg & 1)) { <<cleanup_point <<< Unknown tree: expr_stmt operator delete ((void *) this) >>> >>; } My question is: where is the code after "<D.20148>:;" located? It is outside of the destructor so when is this code executed?

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  • What is validating a binary search tree?

    - by dotnetdev
    I read on here of an exercise in interviews known as validating a binary search tree. How exactly does this work? What would one be looking for in validating a binary search tree? I have written a basic search tree, but never heard of this concept. Thanks

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