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  • Git subtree not properly using .gitignore when doing a partial clone

    - by D W
    I am a graduate student with many scripts, bibliography data in bibtex, thesis draft in latex, presentations in open office, posters in scribus, and figures and result data. I would like to put everything in one project under version control. Then when I need to work on a portion such as the bibliography data, I would like to check that subdirectory out, modify it as necessary and merge it back.I would like the ability to check out one version to my home computer, and a different one to my work computer and make changes to each independently and eventually merge them back. I would also like to be able to check out a piece of code from this big project and import it with versioning into a separate project. If I may changes I'd like to be able to merge them back to the original project. Based on my understanding git subtree can do this. http://github.com/apenwarr/git-subtree There is an example that is along the lines of what I'm trying to do at: http://psionides.jogger.pl/2010/02/04/sharing-code-between-projects-with-git-subtree/ Say the trunk of my project contained the directories: (bib bin cfg data fig src todo). When I use git subtree split -P bib -b export git checkout export I get a the bib directory, plus all files that should have been ignored or considered binary based on .gitignore such as the src directory and everything in it that ends in a tilde or the ./data directory. dwickrama@DWwork:~/research/trunk$ ls * -r biblography.bib JabRef src: script1.sh~ README~ script2.sh~ script3.sh~ script4.R~ script5.awk~ script5.py~ cfg: cfgFile1.ini~ cfgFile2.ini~ cfgFile3.ini~ bin: bigBinaryPackage1 bigBinaryPackage2 dwickrama@DWwork:~/research/trunk$ My .gitignore file is as follows: *.doc diff=word *.tex diff=tex *.bib diff=bibtex *.py diff=python *.eps binary *.jpg binary *.png binary ./bin/* binary *~ How do I prevent this?

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  • Git subtree workflow

    - by Cedric
    In my current project I'm using an open source forum (https://github.com/vanillaforums/Garden.git). I was planning on doing something like this : git remote add vanilla_remote https://github.com/vanillaforums/Garden.git git checkout -b vanilla vanilla_remote/master git checkout master git read-tree --prefix=vanilla -u vanilla This way I can make change into the vanilla folder (like changing config) and commit it to my master branch and I can also switch into my vanilla branch to fetch updates. My problem is when I try to merge the branch together git checkout vanilla git pull git checkout master git merge --squash -s subtree --no-commit vanilla The problem is that the "update commit" goes on top of my commits and "overwrite" my change. I would rather like to have my commits replay on top of the update. Is there a simple way to do that? I'm not very good in git so maybe this is the wrong approach. Also, I really don't want to mix my history with the vanilla history.

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  • Creating subtree from tree which is represented in xml - python

    - by Jay
    Hi I have an XML (in the form of tree), I require to create sub-tree out of it. For ex: <a> <b> <c>Hello</c> <d> <e>Hi</e> </a> Subtree would be <root> <a> <b> <c>Hello</c> </b> </a> <a> <d> <e>Hi</e> </d> </a> </root> What is the best XML library in python to do it? Any algorithm that already does this would also be helpful. Note: the XML doc won't be that big, it will easily fit in memory.

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  • Using git subtree to clone a subdirectory of a project with versioning history then merge it back af

    - by D W
    I am a graduate student with many scripts, bibliography data in bibtex, thesis draft in latex, presentations in open office, posters in scribus, and figures and result data. I would like to put everything in one project under version control. Then when I need to work on a portion such as the bibliography data, I would like to check that subdirectory out, modify it as necessary and merge it back.I would like the ability to check out one version to my home computer, and a different one to my work computer and make changes to each independently and eventually merge them back. I would also like to be able to check out a piece of code from this big project and import it with versioning into a separate project. If I may changes I'd like to be able to merge them back to the original project. Based on my understanding git subtree can do this. http://github.com/apenwarr/git-subtree There is an example that is along the lines of what I'm trying to do at: http://psionides.jogger.pl/2010/02/04/sharing-code-between-projects-with-git-subtree/ This code is from that site: git clone git://git2.kernel.org/pub/scm/git/git.git newtree=$(git subtree split --prefix=gitweb --annotate='(split) ' \ 0a8f4f0^.. --onto=1130ef3 --rejoin) git branch latest_gitweb $newtree gitk latest_gitweb Say the trunk of my project contained the directories: (bib bin cfg data fig src todo). How would I use git-subtree to split off the bib (bibliography) directory with versioning? When I use git-subtree split --prefix=bib I get 884842f6f4e9896e2e4e9402ee0ef762cd617257 as output, but I don't know where to go from there.

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  • binary search tree recursive subtree in java

    - by Art Peterson
    Can anyone point me to a code example (java preferably) or psuedocode that uses recursion to return a subtree that contains all nodes with keys between fromKey and toKey. So if I was to call Tree.subtree(5,10) it should return all nodes in the BST that have keys between 5 and 10 inclusive - but I can't use loops or helper methods...only recursive calls to the subtree method, which takes fromKey and toKey as parameters. Thanks!

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  • Git Subtree. Why can't I branch from a subtree rather than the root?

    - by dugla
    I am struggling trying to make sense of using the Git subtree strategy. My intent was to pull some disparate repos together into a little family of toy repos under an umbrella repo. I'm using the subtree strategy detailed here: http://help.github.com/subtree-merge I am pulling my hair out trying to convince Git that I want to create a branch from one of these subtrees NOT from the root. When I cd into a subtree, create the branch, and then cd back to the root, running git branch from the root clearly indicates the branch was created at the root. Sigh. I love git/github but it is maddening getting this seemingly routine task to work properly. Could someone please enlighten me?

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  • Finding the largest subtree in a BST

    - by rakeshr
    Given a binary tree, I want to find out the largest subtree which is a BST in it. Naive approach: I have a naive approach in mind where I visit every node of the tree and pass this node to a isBST function. I will also keep track of the number of nodes in a sub-tree if it is a BST. Is there a better approach than this ?

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  • Keep only a subtree of the DOM tree

    - by Randomblue
    On BBC articles, such as this one, there is a DOM element with class story-body, deep in the DOM chain. I want to hide all DOM element "outside" of this (unique) DOM element. The problem is that I can't just do $('*').hide(); $('.story-body'); because I need to make sure to keep the parents, grand-parents, etc. of story-body. I also can't do $('*').hide(); var current = $('.story-body').show(); while(current = current.parent()) { current.show(); } because that would simply show everything. Any suggestions?

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  • Attributes in XML subtree that belong to the parent

    - by Bart van Heukelom
    Say I have this XML <doc:document> <objects> <circle radius="10" doc:colour="red" /> <circle radius="20" doc:colour="blue" /> </objects> </doc:document> And this is how it is parsed (pseudo code): // class DocumentParser public Document parse(Element edoc) { doc = new Document(); doc.objects = ObjectsParser.parse(edoc.getChild("objects")); for ( ...?... ) { doc.objectColours.put(object, colour); } return doc; } ObjectsParser is responsible for parsing the objects bit, but is not and should not be aware of the existence of documents. However, in Document colours are associated with objects by use of a Map. What kind of pattern would you recommend to give the colour settings back to DocumentParser.parse from ObjectsParser.parse so it can associate it with the objects they belong to in a map? The alternative would be something like this: <doc:document> <objects> <circle id="1938" radius="10" /> <circle id="6398" radius="20" /> </objects> <doc:objectViewSettings> <doc:objectViewSetting object="1938" colour="red" /> <doc:objectViewSetting object="6398" colour="blue" /> </doc:objectViewSettings> </doc:document> Ugly!

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  • Finding the most frequent subtrees in a collection of (parse) trees

    - by peter.murray.rust
    I have a collection of trees whose nodes are labelled (but not uniquely). Specifically the trees are from a collection of parsed sentences (see http://en.wikipedia.org/wiki/Treebank). I wish to extract the most common subtrees from the collection - performance is not (yet) an issue. I'd be grateful for algorithms (ideally Java) or pointers to tools which do this for treebanks. Note that order of child nodes is important. EDIT @mjv. We are working in a limited domain (chemistry) which has a stylised language so the varirty of the trees is not huge - probably similar to children's readers. Simple tree for "the cat sat on the mat". <sentence> <nounPhrase> <article/> <noun/> </nounPhrase> <verbPhrase> <verb/> <prepositionPhrase> <preposition/> <nounPhrase> <article/> <noun/> </nounPhrase> </prepositionPhrase> </verbPhrase> </sentence> Here the sentence contains two identical part-of-speech subtrees (the actual tokens "cat". "mat" are not important in matching). So the algorithm would need to detect this. Note that not all nounPhrases are identical - "the big black cat" could be: <nounPhrase> <article/> <adjective/> <adjective/> <noun/> </nounPhrase> The length of sentences will be longer - between 15 to 30 nodes. I would expect to get useful results from 1000 trees. If this does not take more than a day or so that's acceptable. Obviously the shorter the tree the more frequent, so nounPhrase will be very common. EDIT If this is to be solved by flattening the tree then I think it would be related to Longest Common Substring, not Longest Common Sequence. But note that I don't necessarily just want the longest - I want a list of all those long enough to be "interesting" (criterion yet to be decided).

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  • is there a way to get a "subtree" from hclust ? (R)

    - by Tal Galili
    Hello all, I wish to create a "subtree" from an hclust object. For example, let's say I have the following object: a <- list() # initialize empty object a$merge <- matrix(c(-1, -2, -3, -4, 1, 2, -5,-6, 3,4), nc=2, byrow=TRUE ) a$height <- c(1, 1.5, 3,4,4.5) # define merge heights a$order <- 1:6 # order of leaves(trivial if hand-entered) a$labels <- 1:6# LETTERS[1:4] # labels of leaves class(a) <- "hclust" # make it an hclust object plot(a) # look at the result Now I wish the extract from it the following subtree: a <- list() # initialize empty object a$merge <- matrix(c(-1, -2, -3, -4, 1, 2 ), nc=2, byrow=TRUE ) a$height <- c(1, 1.5, 3) # define merge heights a$order <- 1:4 # order of leaves(trivial if hand-entered) a$labels <- 1:4# LETTERS[1:4] # labels of leaves class(a) <- "hclust" # make it an hclust object plot(a) # look at the result How could I access it? (I know that cutree could get me the objects of the sub tree, but not create an actual hclust object) Thanks for any help, Tal

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  • In org-mode, is there a way to show the number of undone todo items in a subtree when collapsed?

    - by Gerry Lufwansa
    For example, suppose I have this document: * category 1 ** TODO item 1.1 ** not a todo item ** DONE a done todo item ** TODO item 1.4 * category 2 ** not a todo item ** not a todo item * category 3 ** DONE done item * category 4 ** TODO item 4.1 *** TODO subitem 4.1.1 *** TODO subitem 4.1.2 When collapsed, I'd like to see something like: * category 1 (2)... * category 2 (0)... * category 3 (2)... * category 4 (3)...

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  • How to keep subtree removal (`rm -rf`) from starving other processes for Disk I/O?

    - by David Eyk
    We have a very large (multi-GB) Nginx cache directory for a busy site, which we occasionally need to clear all at once. I've solved this in the past by moving the cache folder to a new path, making a new cache folder at the old path, and then rm -rfing the old cache folder. Lately, however, when I need to clear the cache on a busy morning, the I/O from rm -rf is starving my server processes of disk access, as both Nginx and the server it fronts for are read-intensive. I can watch the load average climb while the CPUs sit idle and rm -rf takes 98-99% of Disk IO in iotop. I've tried ionice -c 3 when invoking rm, but it seems to have no appreciable effect on the observed behavior. Is there any way to tame rm -rf to share the disk more? Do I need to use a different technique that will take its cues from ionice? Update: The filesystem in question is an AWS EC2 instance store (the primary disk is EBS). The /etc/fstab entry looks like this: /dev/xvdb /mnt auto defaults,nobootwait,comment=cloudconfig 0 2

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  • C++ Unlocking a std::mutex before calling std::unique_lock wait

    - by Sant Kadog
    I have a multithreaded application (using std::thread) with a manager (class Tree) that executes some piece of code on different subtrees (embedded struct SubTree) in parallel. The basic idea is that each instance of SubTree has a deque that store objects. If the deque is empty, the thread waits until a new element is inserted in the deque or the termination criteria is reached. One subtree can generate objects and push them in the deque of another subtree. For convenience, all my std::mutex, std::locks and std::variable_condition are stored in a struct called "locks". The class Tree creates some threads that run the following method (first attempt) : void Tree::launch(SubTree & st, Locks & locks ) { /* some code */ std::lock_guard<std::mutex> deque_lock(locks.deque_mutex_[st.id_]) ; // lock the access to the deque of subtree st if (st.deque_.empty()) // check that the deque is still empty { // some threads are still running, wait for them to terminate std::unique_lock<std::mutex> wait_lock(locks.restart_mutex_[st.id_]) ; locks.restart_condition_[st.id_].wait(wait_lock) ; } /* some code */ } The problem is that "deque_lock" is still locked while the thread is waiting. Hence no object can be added in the deque of the current thread by a concurrent one. So I turned the lock_guard into a unique_lock and managed the lock/unlock manually : void launch(SubTree & st, Locks & locks ) { /* some code */ std::unique_lock<std::mutex> deque_lock(locks.deque_mutex_[st.id_]) ; // lock the access to the deque of subtree st if (st.deque_.empty()) // check that the deque is still empty { deque_lock.unlock() ; // unlock the access to the deque to enable the other threads to add objects // DATA RACE : nothing must happen to the unprotected deque here !!!!!! // some threads are still running, wait for them to terminate std::unique_lock<std::mutex> wait_lock(locks.restart_mutex_[st.id_]) ; locks.restart_condition_[st.id_].wait(wait_lock) ; } /* some code */ } The problem now, is that there is a data race, and I would like to make sure that the "wait" instruction is performed directly after the "deque_lock.unlock()" one. Would anyone know a way to create such a critical instruction sequence with the standard library ? Thanks in advance.

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  • How to create a Binary Tree from a General Tree?

    - by mno4k
    I have to solve the following constructor for a BinaryTree class in java: BinaryTree(GeneralTree<T> aTree) This method should create a BinaryTree (bt) from a General Tree (gt) as follows: Every Vertex from gt will be represented as a leaf in bt. If gt is a leaf, then bt will be a leaf with the same value as gt If gt is not a leaf, then bt will be constructed as an empty root, a left subTree (lt) and a right subTree (lr). Lt is a stric binary tree created from the oldest subtree of gt (the left-most subtree) and lr is a stric binary tree created from gt without its left-most subtree. The frist part is trivial enough, but the second one is giving me some trouble. I've gotten this far: public BinaryTree(GeneralTree<T> aTree){ if (aTree.isLeaf()){ root= new BinaryNode<T>(aTree.getRootData()); }else{ root= new BinaryNode<T>(null); // empty root LinkedList<GeneralTree<T>> childs = aTree.getChilds(); // Childs of the GT are implemented as a LinkedList of SubTrees child.begin(); //start iteration trough list BinaryTree<T> lt = new BinaryTree<T>(childs.element(0)); // first element = left-most child this.addLeftChild(lt); aTree.DeleteChild(hijos.elemento(0)); BinaryTree<T> lr = new BinaryTree<T>(aTree); this.addRightChild(lr); } } Is this the right way? If not, can you think of a better way to solve this? Thank you!

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  • Need algorithm to add Node in binary tree

    - by m.qayyum
    •if your new element is less or equal than the current node, you go to the left subtree, otherwise to the right subtree and continue traversing •if you arrived at a node, where you can not go any deeper, because there is no subtree, this is the place to insert your new element (5)Root (3)-------^--------(7) (2)---^----(5) ^-----(8) (5)--^ i want to add this last node with data 5...but i can't figure it out...I need a algorithm to do that or in java language

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  • How to automate org-refile for multiple todo

    - by lawlist
    I'm looking to automate org-refile so that it will find all of the matches and re-file them to a specific location (but not archive). I found a fully automated method of archiving multiple todo, and I am hopeful to find or create (with some help) something similar to this awesome function (but for a different heading / location other than archiving): https://github.com/tonyday567/jwiegley-dot-emacs/blob/master/dot-org.el (defun org-archive-done-tasks () (interactive) (save-excursion (goto-char (point-min)) (while (re-search-forward "\* \\(None\\|Someday\\) " nil t) (if (save-restriction (save-excursion (org-narrow-to-subtree) (search-forward ":LOGBOOK:" nil t))) (forward-line) (org-archive-subtree) (goto-char (line-beginning-position)))))) I also found this (written by aculich), which is a step in the right direction, but still requires repeating the function manually: http://stackoverflow.com/questions/7509463/how-to-move-a-subtree-to-another-subtree-in-org-mode-emacs ;; I also wanted a way for org-refile to refile easily to a subtree, so I wrote some code and generalized it so that it will set an arbitrary immediate target anywhere (not just in the same file). ;; Basic usage is to move somewhere in Tree B and type C-c C-x C-m to mark the target for refiling, then move to the entry in Tree A that you want to refile and type C-c C-w which will immediately refile into the target location you set in Tree B without prompting you, unless you called org-refile-immediate-target with a prefix arg C-u C-c C-x C-m. ;; Note that if you press C-c C-w in rapid succession to refile multiple entries it will preserve the order of your entries even if org-reverse-note-order is set to t, but you can turn it off to respect the setting of org-reverse-note-order with a double prefix arg C-u C-u C-c C-x C-m. (defvar org-refile-immediate nil "Refile immediately using `org-refile-immediate-target' instead of prompting.") (make-local-variable 'org-refile-immediate) (defvar org-refile-immediate-preserve-order t "If last command was also `org-refile' then preserve ordering.") (make-local-variable 'org-refile-immediate-preserve-order) (defvar org-refile-immediate-target nil) "Value uses the same format as an item in `org-refile-targets'." (make-local-variable 'org-refile-immediate-target) (defadvice org-refile (around org-immediate activate) (if (not org-refile-immediate) ad-do-it ;; if last command was `org-refile' then preserve ordering (let ((org-reverse-note-order (if (and org-refile-immediate-preserve-order (eq last-command 'org-refile)) nil org-reverse-note-order))) (ad-set-arg 2 (assoc org-refile-immediate-target (org-refile-get-targets))) (prog1 ad-do-it (setq this-command 'org-refile))))) (defadvice org-refile-cache-clear (after org-refile-history-clear activate) (setq org-refile-targets (default-value 'org-refile-targets)) (setq org-refile-immediate nil) (setq org-refile-immediate-target nil) (setq org-refile-history nil)) ;;;###autoload (defun org-refile-immediate-target (&optional arg) "Set current entry as `org-refile' target. Non-nil turns off `org-refile-immediate', otherwise `org-refile' will immediately refile without prompting for target using most recent entry in `org-refile-targets' that matches `org-refile-immediate-target' as the default." (interactive "P") (if (equal arg '(16)) (progn (setq org-refile-immediate-preserve-order (not org-refile-immediate-preserve-order)) (message "Order preserving is turned: %s" (if org-refile-immediate-preserve-order "on" "off"))) (setq org-refile-immediate (unless arg t)) (make-local-variable 'org-refile-targets) (let* ((components (org-heading-components)) (level (first components)) (heading (nth 4 components)) (string (substring-no-properties heading))) (add-to-list 'org-refile-targets (append (list (buffer-file-name)) (cons :regexp (format "^%s %s$" (make-string level ?*) string)))) (setq org-refile-immediate-target heading)))) (define-key org-mode-map "\C-c\C-x\C-m" 'org-refile-immediate-target) It sure would be helpful if aculich, or some other maven, could please create a variable similar to (setq org-archive-location "~/0.todo.org::* Archived Tasks") so users can specify the file and heading, which is already a part of the org-archive-subtree functionality. I'm doing a search and mark because I don't have the wherewithal to create something like org-archive-location for this setup. EDIT: One step closer -- almost home free . . . (defun lawlist-auto-refile () (interactive) (beginning-of-buffer) (re-search-forward "\* UNDATED") (org-refile-immediate-target) ;; cursor must be on a heading to work. (save-excursion (re-search-backward "\* UNDATED") ;; must be written in such a way so that sub-entries of * UNDATED are not searched; or else infinity loop. (while (re-search-backward "\* \\(None\\|Someday\\) " nil t) (org-refile) ) ) )

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  • Binary Search Tree - Postorder logic

    - by daveb
    I am looking at implementing code to work out binary search tree. Before I do this I was wanting to verify my input data in postorder and preorder. I am having trouble working out what the following numbers would be in postorder and preorder I have the following numbers 4, 3, 14 ,8 ,1, 15, 9, 5, 13, 10, 2, 7, 6, 12, 11, that I am intending to put into an empty binary tree in that order. The order I arrived at for the numbers in POSTORDER is 2, 1, 6, 3, 7, 11, 12, 10, 9, 8, 13, 15, 14, 4. Have I got this right? I was wondering if anyone here would be able to kindly verify if the postorder sequence I came up with is indeed the correct sequence for my input i.e doing left subtree, right subtree and then root. The order I got for pre order (Visit root, do left subtree, do right subtree) is 4, 3, 1, 2, 5, 6, 14 , 8, 7, 9, 10, 12, 11, 15, 13. I can't be certain I got this right. Very grateful for any verification. Many Thanks

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  • UNIX tool to dump a selection of HTML?

    - by jldugger
    I'm looking to monitor changes on websites and my current approach is being defeated by a rotating top banner. Is there a UNIX tool that takes a selection parameter (id attribute or XPath), reads HTML from stdin and prints to stdout the subtree based on the selection? For example, given an html document I want to filter out everything but the subtree of the element with id="content". Basically, I'm looking for the simplest HTML/XML equivalent to grep.

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  • Deletion procedure for a Binary Search Tree

    - by Metz
    Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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  • [BST] Deletion procedure

    - by Metz
    Hi all. Consider the deletion procedure on a BST, when the node to delete has two children. Let's say i always replace it with the node holding the minimum key in its right subtree. The question is: is this procedure commutative? That is, deleting x and then y has the same result than deleting first y and then x? I think the answer is no, but i can't find a counterexample, nor figure out any valid reasoning. Thank you. EDIT: Maybe i've got to be clearer. Consider the transplant(node x, node y) procedure: it replace x with y (and its subtree). So, if i want to delete a node (say x) which has two children i replace it with the node holding the minimum key in its right subtree: y = minimum(x.right) transplant(y, y.right) // extracts the minimum (it doesn't have left child) y.right = x.right y.left = x.left transplant(x,y) The question was how to prove the procedure above is not commutative.

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  • Configure Oracle Identity Manager AD/LDAP Authentication

    - by Arda Eralp
    Requirements (on AD side) LDAP connection user with the necessary rights in AD to do subtree searches on your users and groups container, respectively in the scope we configure below For LDAP in OIM to work, you need an AD Group called "oimusers", in which all users who shall be able to login to OIM need to be member. The group need to be named exactly "oimusers". Step 1: Login Weblogic Administration Console  Step 2: Create New Provider Authentication Provider Name: ADAuthenticationProvider Type: ActiveDirectoryAuthenticator Control Flag: SUFFICIENT   User scope configuration User Base DN: Container where your users are found Rest of the parameters stay default   Group scope configuration Group Base DN: Container where your groups are found Your "oimusers" group must be found in this container or in the subtree Rest of the parameters stay default  Step 3: Restart Admin Server Step 4: Check oimusers group Step 5: Re order providers Step 6: Restart Admin Server

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  • Worst Case number of rotations for BST to AVL algorithm?

    - by spacker_lechuck
    I have a basic algorithm below and I know that the worst case input BST is one that has degenerated to a linked list from inserts to only one side. How would I compute the worst case complexity in terms of number of rotations for this BST to AVL conversion algorithm? IF tree is right heavy { IF tree's right subtree is left heavy { Perform Double Left rotation } ELSE { Perform Single Left rotation } } ELSE IF tree is left heavy { IF tree's left subtree is right heavy { Perform Double Right rotation } ELSE { Perform Single Right rotation } }

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