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  • Cladogram, tree of life, cladistics, taxonomy in JS or canvas?

    - by boblet
    Good people - I need some help to find a way to create an interactive cladogram or phylogenetic tree (yes, I have read all related posts, and do not find what I am looking for). The thing is, I need the nodes to be name-able. An example would be something like this Most scripts I find are either applets, flash, or simply do not show the node classification, ie it would skip "feliformia" in this example. This is useless to me, as I would then end up with carnivore - anonymous node - anonymous node - anonymous node - tiger, and that is not good. This tree will in theory cover all life, so it could get rather large, and get links and names in english and latin from database. So: no flash, no applets. It must be horizontal, no supertrees (circular). I have gone through this http://bioinfo.unice.fr/biodiv/Tree_editors.html but most of them seems to be either old, not displaying sub-node levels, applets, or way too complex. I imagine this would be a delightful job for canvas/jQuery..? And chances are, someone got there before me? Any pointers much appreciated. Note: if anyone out there would like to do something like this as a project, I will be happy to help, even though it would not benefit me for this project.This type of taxonomy is not as simple as it may seem, and I would be happy see this happen.

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  • How can you get the call tree with python profilers?

    - by Oliver
    I used to use a nice Apple profiler that is built into the System Monitor application. As long as your C++ code was compiled with debug information, you could sample your running application and it would print out an indented tree telling you what percent of the parent function's time was spent in this function (and the body vs. other function calls). For instance, if main called function_1 and function_2, function_2 calls function_3, and then main calls function_3: main (100%, 1% in function body): function_1 (9%, 9% in function body): function_2 (90%, 85% in function body): function_3 (100%, 100% in function body) function_3 (1%, 1% in function body) I would see this and think, "Something is taking a long time in the code in the body of function_2. If I want my program to be faster, that's where I should start." Does anyone know how I can most easily get this exact profiling output for a python program? I've seen people say to do this: import cProfile, pstats prof = cProfile.Profile() prof = prof.runctx("real_main(argv)", globals(), locals()) stats = pstats.Stats(prof) stats.sort_stats("time") # Or cumulative stats.print_stats(80) # 80 = how many to print but it's quite messy compared to that elegant call tree. Please let me know if you can easily do this, it would help quite a bit. Cheers!

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  • A data structure based on the R-Tree: creating new child nodes when a node is full, but what if I ha

    - by Tom
    I realize my title is not very clear, but I am having trouble thinking of a better one. If anyone wants to correct it, please do. I'm developing a data structure for my 2 dimensional game with an infinite universe. The data structure is based on a simple (!) node/leaf system, like the R-Tree. This is the basic concept: you set howmany childs you want a node (a container) to have maximum. If you want to add a leaf, but the node the leaf should be in is full, then it will create a new set of nodes within this node and move all current leafs to their new (more exact) node. This way, very populated areas will have a lot more subdivisions than a very big but rarely visited area. This works for normal objects. The only problem arises when I have more than maxChildsPerNode objects with the exact same X,Y location: because the node is full, it will create more exact subnodes, but the old leafs will all be put in the exact same node again because they have the exact same position -- resulting in an infinite loop of creating more nodes and more nodes. So, what should I do when I want to add more leafs than maxChildsPerNode with the exact same position to my tree? PS. if I failed to explain my problem, please tell me, so I can try to improve the explanation.

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  • IIS7 Binary Stream Error

    - by WDuffy
    I'm struggling to understand the problem I'm seeing so please accept my apology if the question is vague. I'm running a classic asp app in IIS7 and everything seems to work fine except for one issue that has me stumped. Basically, files can be downloaded from the server which is done using the sendBinary method of Persits ASP Upload component. This component works fine for uploading etc it's just the downloading I have a problem. The strange thing is, I cannot have a pure asp page that serves the binary file. Everything works fine in II6, but in II7 there is a strange problem. For example this does not work. <% SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> However, if I put anything in front of the asp code the file is served fine. serve this <% SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> I can also write something to the response stream first and it works fine. <% Response.Write("server this") SET objUpload = server.createObject("Persits.Upload") objUpload.sendBinary "D:\sites\file.pdf", true, "application/octet-binary", true SET objUpload = NOTHING %> Does anyone have any ideas of what could be causing this or has anyone ran into a similar situation? I'm sure it has something to do with the setup in IIS 7.

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  • Compiling vs using pre-built binaries performance?

    - by Nick Rosencrantz
    Will performance be better (quicker) if I manually compile the source for a software component for the actual machine that it will be used on, compared to if the source was compiled on another platform perhaps for many different architectures? I got some good results compiling source that I downloaded and I wonder whether this was due to compiling it instead of downloading a pre-compiled binary which is often the case with software updates.

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  • How portable are Binaries compiled in Ubuntu?

    - by hiobs
    The title says it all, actually. But allow me to specify the question: Assuming I were to compile an application that uses libffi, libGL, dlfcn, and SDL, would said binary run on other Linux distributions with same architecture, etc? The reason I ask is because of the directory /usr/lib/i386-linux-gnu - I might be wrong, but I assume this directory is something rather Ubuntu-specific, no? So, how portable are binaries compiled on Ubuntu really?

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  • How to find the insertion point in an array using binary search?

    - by ????
    The basic idea of binary search in an array is simple, but it might return an "approximate" index if the search fails to find the exact item. (we might sometimes get back an index for which the value is larger or smaller than the searched value). For looking for the exact insertion point, it seems that after we got the approximate location, we might need to "scan" to left or right for the exact insertion location, so that, say, in Ruby, we can do arr.insert(exact_index, value) I have the following solution, but the handling for the part when begin_index >= end_index is a bit messy. I wonder if a more elegant solution can be used? (this solution doesn't care to scan for multiple matches if an exact match is found, so the index returned for an exact match may point to any index that correspond to the value... but I think if they are all integers, we can always search for a - 1 after we know an exact match is found, to find the left boundary, or search for a + 1 for the right boundary.) My solution: DEBUGGING = true def binary_search_helper(arr, a, begin_index, end_index) middle_index = (begin_index + end_index) / 2 puts "a = #{a}, arr[middle_index] = #{arr[middle_index]}, " + "begin_index = #{begin_index}, end_index = #{end_index}, " + "middle_index = #{middle_index}" if DEBUGGING if arr[middle_index] == a return middle_index elsif begin_index >= end_index index = [begin_index, end_index].min return index if a < arr[index] && index >= 0 #careful because -1 means end of array index = [begin_index, end_index].max return index if a < arr[index] && index >= 0 return index + 1 elsif a > arr[middle_index] return binary_search_helper(arr, a, middle_index + 1, end_index) else return binary_search_helper(arr, a, begin_index, middle_index - 1) end end # for [1,3,5,7,9], searching for 6 will return index for 7 for insertion # if exact match is found, then return that index def binary_search(arr, a) puts "\nSearching for #{a} in #{arr}" if DEBUGGING return 0 if arr.empty? result = binary_search_helper(arr, a, 0, arr.length - 1) puts "the result is #{result}, the index for value #{arr[result].inspect}" if DEBUGGING return result end arr = [1,3,5,7,9] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 6 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1,3,5,7,9,11] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = -60 arr.insert(binary_search(arr, b), b) p arr arr = [1] b = 60 arr.insert(binary_search(arr, b), b) p arr arr = [] b = 60 arr.insert(binary_search(arr, b), b) p arr and result: Searching for 6 in [1, 3, 5, 7, 9] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 6, arr[middle_index] = 5, begin_index = 3, end_index = 2, middle_index = 2 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9] Searching for 6 in [1, 3, 5, 7, 9, 11] a = 6, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 6, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 6, arr[middle_index] = 7, begin_index = 3, end_index = 3, middle_index = 3 the result is 3, the index for value 7 [1, 3, 5, 6, 7, 9, 11] Searching for 60 in [1, 3, 5, 7, 9] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = 60, arr[middle_index] = 7, begin_index = 3, end_index = 4, middle_index = 3 a = 60, arr[middle_index] = 9, begin_index = 4, end_index = 4, middle_index = 4 the result is 5, the index for value nil [1, 3, 5, 7, 9, 60] Searching for 60 in [1, 3, 5, 7, 9, 11] a = 60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = 60, arr[middle_index] = 9, begin_index = 3, end_index = 5, middle_index = 4 a = 60, arr[middle_index] = 11, begin_index = 5, end_index = 5, middle_index = 5 the result is 6, the index for value nil [1, 3, 5, 7, 9, 11, 60] Searching for -60 in [1, 3, 5, 7, 9] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 4, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 9, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9] Searching for -60 in [1, 3, 5, 7, 9, 11] a = -60, arr[middle_index] = 5, begin_index = 0, end_index = 5, middle_index = 2 a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 1, middle_index = 0 a = -60, arr[middle_index] = 11, begin_index = 0, end_index = -1, middle_index = -1 the result is 0, the index for value 1 [-60, 1, 3, 5, 7, 9, 11] Searching for -60 in [1] a = -60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 0, the index for value 1 [-60, 1] Searching for 60 in [1] a = 60, arr[middle_index] = 1, begin_index = 0, end_index = 0, middle_index = 0 the result is 1, the index for value nil [1, 60] Searching for 60 in [] [60]

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  • reiserfsck --rebuild-tree failed: Not enough allocable blocks

    - by mojo
    I have a reiserfs volume that required a --rebuild-tree, but is currently failing to complete when I pass it --rebuild-tree. Here is the output that I receive when running it: reiserfsck 3.6.19 (2003 www.namesys.com) # reiserfsck --rebuild-tree started at Mon Oct 26 13:22:16 2009 # Pass 0: # Pass 0 The whole partition (7864320 blocks) is to be scanned Skipping 8450 blocks (super block, journal, bitmaps) 7855870 blocks will be read 0%....20%....40%....60%....80%....100% left 0, 9408 /sec 287884 directory entries were hashed with "r5" hash. "r5" hash is selected Flushing..finished Read blocks (but not data blocks) 7855870 Leaves among those 6105606 Objectids found 287892 Pass 1 (will try to insert 6105606 leaves): # Pass 1 Looking for allocable blocks .. finished 0%....20%....40%....60%....80%....Not enough allocable blocks, checking bitmap...there are 1 allocable blocks, btw out of disk space Aborted I can't mount it, and I can't fsck it. I've tried extending the volume, but that hasn't helped either.

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  • reiserfsck --rebuild-tree failed: Not enough allocable blocks

    - by mojo
    I have a reiserfs volume that required a --rebuild-tree, but is currently failing to complete when I pass it --rebuild-tree. Here is the output that I receive when running it: reiserfsck 3.6.19 (2003 www.namesys.com) # reiserfsck --rebuild-tree started at Mon Oct 26 13:22:16 2009 # Pass 0: # Pass 0 The whole partition (7864320 blocks) is to be scanned Skipping 8450 blocks (super block, journal, bitmaps) 7855870 blocks will be read 0%....20%....40%....60%....80%....100% left 0, 9408 /sec 287884 directory entries were hashed with "r5" hash. "r5" hash is selected Flushing..finished Read blocks (but not data blocks) 7855870 Leaves among those 6105606 Objectids found 287892 Pass 1 (will try to insert 6105606 leaves): # Pass 1 Looking for allocable blocks .. finished 0%....20%....40%....60%....80%....Not enough allocable blocks, checking bitmap...there are 1 allocable blocks, btw out of disk space Aborted I can't mount it, and I can't fsck it. I've tried extending the volume, but that hasn't helped either.

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  • How can I force a tree itemrenderer to redraw during a drag and drop operation?

    - by swidnikk
    I have a tree control with a custom item renderer. The item renderer has different states that should be set while an item is being dragged over the item renderer. I understand from reading this post http://forums.adobe.com/message/2091088 that the 'right way' to do this is to override the 'getCurrentState' method and append some text. I do that. Now in my tree control I handle the drag over event and get a reference to the itemrenderer that is being dragged over and I set the boolean 'dragOver' property to true. Now I just need to force my itemRenderer to redraw. I can't figure that out. A workaround, is to just set the currentState of the itemRenderer. My question then, how can I force my itemRenderer to refresh? (and I've tried calling validateNow, invalideDisplayList/Properties/Size, to no avail) <?xml version="1.0" encoding="utf-8"?> <s:MXTreeItemRenderer xmlns:fx="http://ns.adobe.com/mxml/2009" xmlns:s="library://ns.adobe.com/flex/spark" xmlns:mx="library://ns.adobe.com/flex/mx"> <fx:Script> <![CDATA[ import mx.events.DragEvent; import mx.events.FlexEvent; import spark.layouts.supportClasses.DropLocation; public var dragOver:Boolean = false; override protected function getCurrentRendererState():String { var skinState:String = super.getCurrentRendererState(); if( dragOver ) skinState += "AndDragOver"; trace('getCurrentRendererState', skinState); return skinState; } ]]> </fx:Script> <s:states> <s:State name="normal" /> <s:State name="hovered" /> <s:State name="selected" /> <s:State name="normalAndDragOver" stateGroups="dragOverGroup" /> <s:State name="hoveredAndDragOver" stateGroups="dragOverGroup" /> <s:State name="selectedAndDragOver" stateGroups="dragOverGroup" /> </s:states> ...

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  • How to modify preorder tree traversal algorithm to handle nodes with multiple parents?

    - by poldo
    I've been searching for a while now and can't seem to find an alternative solution. I need the tree traversal algorithm in such a way that a node can have more than 1 parent, if it's possible (found a great article here: Storing Hierarchical Data in a Database). Are there any algorithms so that, starting from a root node, we can determine the sequence and dependencies of nodes (currently reading topological sorting)?

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  • merge sort recursion tree height

    - by Tony
    Hello! I am learning about recursion tree's and trying to figure out how the height of the tree is log b of n where n = 2 and one has 10 elements as input size. I am working with Merge sort. The number of times the split is done is the height of the tree as far as I understood, and the number of levels in the tree is height + 1. But if you take (for merge sort) log2 of 10 you get 1, where if you draw the tree you get at least 2 times that the recursion occurs. Where have I gone wrong? (I hope I am making sense here) NOTE: I am doing a self study, this is not homework!

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  • How to compile scheme into native binary files ?

    - by Joe
    I am very new to scheme. And now I am trying to compile some scheme code into binary file which will be loaded faster into interpreter. (The interpreter is a hybrid interpreter)Some one told me that I can compile the code into native binary file and then load it into interperter. And my question is: 1. What is the native binary file? 2. How can I compile the scheme code into a native binary file? 3. How can I load native bianry file into scheme interpreter? Thanks in advance. Joe Suggested that I want to compile below code into native binary file: (define test (lambda() (display "this is a test")) And then load the bianry file into interpreter and call the function "test".

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  • Tree item edit event listener in flex

    - by Biroka
    I want to catch itemEditEnd in a tree control, but i'm having a little trouble. I call the item editing a bit different: <mx:Tree id="tree" width="100%" height="100%" itemDoubleClick="treeProjects_itemDoubleClickHandler(event)" doubleClickEnabled="true" editable="true" itemEditBeginning="treeProjects_itemEditBeginingHandler(event)" itemEditEnd="treeProjects_itemEditEndHandler(event)" > but: protected function treeProjects_itemEditBeginingHandler(event:ListEvent):void { event.preventDefault(); } i do this, because I want to edit the field just when clicking rename in the context menu. And when I click rename in the context menu, this is what i'm doing (contextParentItem is the event catched on itemRollOver in the tree) : Tree( contextParentItem.target ).editedItemPosition = { columnIndex: 0, rowIndex: contextParentItem.rowIndex }; Thisway I can't catch the itemEditEnd event, but I want to handle some stuff, like: if the new entered value is an empty string, then I don't want to accept the new value if I hit ESC then, the old value should remain.

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  • C++ : Lack of Standardization at the Binary Level

    - by Nawaz
    Why ISO/ANSI didn't standardize C++ at the binary level? There are many portability issues with C++, which is only because of lack of it's standardization at the binary level. Don Box writes, (quoting from his book Essential COM, chapter COM As A Better C++) C++ and Portability Once the decision is made to distribute a C++ class as a DLL, one is faced with one of the fundamental weaknesses of C++, that is, lack of standardization at the binary level. Although the ISO/ANSI C++ Draft Working Paper attempts to codify which programs will compile and what the semantic effects of running them will be, it makes no attempt to standardize the binary runtime model of C++. The first time this problem will become evident is when a client tries to link against the FastString DLL's import library from a C++ developement environment other than the one used to build the FastString DLL. Are there more benefits Or loss of this lack of binary standardization?

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  • dojo 1.4: difference in tree html rendering

    - by Fell
    Hi, I have written an application using dojo 1.3 in which i have used the dijit tree. I am loading the tree with json data specified in the store which is in turn used by the tree model. In the 1.3 version the tree elements pick up the id directly from the json data.However in 1.4 the tree elements have their own id which is something like dijit__treenode_4. The id's that I have specified in json are unique and im not able to understand why these are not being used anymore. Please help me understand how this functionality has changed and how I can override the automatic id generation.. Thanks in advance, Fell

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  • Very long tree control inside a <frame>

    - by ryancerium
    I have an tree control inside of a frame. It's quite tall, around 2000 pixels. Right now, we use the frame's scroll bars, which is mostly good. Unfortunately, when you select an item near the bottom of the list, the page reloads and the view goes back to the top of the list. I tried calling the tree control's ScrollNodeIntoView() function, but since we're not using the tree's scroll bars, it just thinks that it's displaying the full 2000 pixels and the control doesn't have its own scroll bar. I'm not much of an HTML guru, so I have two ideas, neither of which I know how to do. 1) Tell the frame to not to let the tree render all 2000 pixels and instead stay inside the viewable area. 2) Tell the tree to not render all 2000 pixels and instead stay within the viewable area. Setting the CSS height property on the within the frame doesn't do anything. Thanks.

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  • Virtual String Tree - Display subnode when parent node is hidden

    - by daemon_x
    Is there a way to show subnode if the parent node is hidden in the Virtual String Tree ? I have some tasks in the tree structure and I wish to display only tasks which belongs to the current user as the list, but from all levels. What I've done is the function to display a list, which hides tree buttons, tree lines, sets the fixed indent and enable toShowHiddenNodes option. Then in this function I iterate through the whole tree (all levels) and hide nodes which doesn't belong to the current user IsVisible[Node] := False and show those which belongs him IsVisible[Node] := True, but the subnodes which should be displayed are invisible when their parent is hidden. VT.TreeOptions.PaintOptions - toShowButtons - toShowTreeLines + toFixedIndent + toShowHiddenNodes

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  • Last element not getting insert in Tree

    - by rdk1992
    So I was asked to make a Binary Tree in Haskell taking as input a list of Integers. Below is my code. My problem is that the last element of the list is not getting inserted in the Tree. For example [1,2,3,4] it only inserts to the tree until "3" and 4 is not inserted in the Tree. data ArbolBinario a = Node a (ArbolBinario a) (ArbolBinario a) | EmptyNode deriving(Show) insert(x) EmptyNode= insert(tail x) (Node (head x) EmptyNode EmptyNode) insert(x) (Node e izq der) |x == [] = EmptyNode --I added this line to fix the Prelude.Head Empty List error, after I added this line the last element started to be ignored and not inserted in the tree |head x == e = (Node e izq der) |head x < e = (Node e (insert x izq) der) |head x > e = (Node e izq (insert x der)) Any ideas on whats going on here? Help is much appreciated

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