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  • What is the most efficient/elegant way to parse a flat table into a tree?

    - by Tomalak
    Assume you have a flat table that stores an ordered tree hierarchy: Id Name ParentId Order 1 'Node 1' 0 10 2 'Node 1.1' 1 10 3 'Node 2' 0 20 4 'Node 1.1.1' 2 10 5 'Node 2.1' 3 10 6 'Node 1.2' 1 20 What minimalistic approach would you use to output that to HTML (or text, for that matter) as a correctly ordered, correctly intended tree? Assume further you only have basic data structures (arrays and hashmaps), no fancy objects with parent/children references, no ORM, no framework, just your two hands. The table is represented as a result set, which can be accessed randomly. Pseudo code or plain English is okay, this is purely a conceptional question. Bonus question: Is there a fundamentally better way to store a tree structure like this in a RDBMS? EDITS AND ADDITIONS To answer one commenter's (Mark Bessey's) question: A root node is not necessary, because it is never going to be displayed anyway. ParentId = 0 is the convention to express "these are top level". The Order column defines how nodes with the same parent are going to be sorted. The "result set" I spoke of can be pictured as an array of hashmaps (to stay in that terminology). For my example was meant to be already there. Some answers go the extra mile and construct it first, but thats okay. The tree can be arbitrarily deep. Each node can have N children. I did not exactly have a "millions of entries" tree in mind, though. Don't mistake my choice of node naming ('Node 1.1.1') for something to rely on. The nodes could equally well be called 'Frank' or 'Bob', no naming structure is implied, this was merely to make it readable. I have posted my own solution so you guys can pull it to pieces.

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  • What data is actually stored in a B-tree database in CouchDB?

    - by Andrey Vlasovskikh
    I'm wondering what is actually stored in a CouchDB database B-tree? The CouchDB: The Definitive Guide tells that a database B-tree is used for append-only operations and that a database is stored in a single B-tree (besides per-view B-trees). So I guess the data items that are appended to the database file are revisions of documents, not the whole documents: +---------|### ... | | +------|###|------+ ... ---+ | | | | +------+ +------+ +------+ +------+ | doc1 | | doc2 | | doc1 | ... | doc1 | | rev1 | | rev1 | | rev2 | | rev7 | +------+ +------+ +------+ +------+ Is it true? If it is true, then how the current revision of a document is determined based on such a B-tree? Doesn't it mean, that CouchDB needs a separate "view" database for indexing current revisions of documents to preserve O(log n) access? Wouldn't it lead to race conditions while building such an index? (as far as I know, CouchDB uses no write locks).

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  • How to paste text and variables into a logical expression in R?

    - by Jasper
    I want to paste variables in the logical expression that I am using to subset data, but the subset function does not see them as column names when pasted (either with ot without quotes). I have a dataframe with columns named col1, col2 etc. I want to subset for the rows in which colx < 0.05 This DOES work: subsetdata<-subset(dataframe, col1<0.05) subsetdata<-subset(dataframe, col2<0.05) This does NOT work: for (k in 1:2){ subsetdata<-subset(dataframe, paste("col",k,sep="")<0.05) } for (k in 1:2){ subsetdata<-subset(dataframe, noquote(paste("col",k,sep=""))<0.05) } I can't find the answer; any suggestions?

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  • Help me understand Inorder Traversal without using recursion

    - by vito
    I am able to understand preorder traversal without using recursion, but I'm having a hard time with inorder traversal. I just don't seem to get it, perhaps, because I haven't understood the inner working of recursion. This is what I've tried so far: def traverseInorder(node): lifo = Lifo() lifo.push(node) while True: if node is None: break if node.left is not None: lifo.push(node.left) node = node.left continue prev = node while True: if node is None: break print node.value prev = node node = lifo.pop() node = prev if node.right is not None: lifo.push(node.right) node = node.right else: break The inner while-loop just doesn't feel right. Also, some of the elements are getting printed twice; may be I can solve this by checking if that node has been printed before, but that requires another variable, which, again, doesn't feel right. Where am I going wrong? I haven't tried postorder traversal, but I guess it's similar and I will face the same conceptual blockage there, too. Thanks for your time! P.S.: Definitions of Lifo and Node: class Node: def __init__(self, value, left=None, right=None): self.value = value self.left = left self.right = right class Lifo: def __init__(self): self.lifo = () def push(self, data): self.lifo = (data, self.lifo) def pop(self): if len(self.lifo) == 0: return None ret, self.lifo = self.lifo return ret

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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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  • New Sample Demonstrating the Traversing of Tree Bindings

    - by Duncan Mills
    A technique that I seem to use a fair amount, particularly in the construction of dynamic UIs is the use of a ADF Tree Binding to encode a multi-level master-detail relationship which is then expressed in the UI in some kind of looping form – usually a series of nested af:iterators, rather than the conventional tree or treetable. This technique exploits two features of the treebinding. First the fact that an treebinding can return both a collectionModel as well as a treeModel, this collectionModel can be used directly by an iterator. Secondly that the “rows” returned by the collectionModel themselves contain an attribute called .children. This attribute in turn gives access to a collection of all the children of that node which can also be iterated over. Putting this together you can represent the data encoded into a tree binding in all sorts of ways. As an example I’ve put together a very simple sample based on the HT schema and uploaded it to the ADF Sample project. It produces this UI: The important code is shown here for a Region -> Country -> Location Hierachy: <af:iterator id="i1" value="#{bindings.AllRegions.collectionModel}" var="rgn"> <af:showDetailHeader text="#{rgn.RegionName}" disclosed="true" id="sdh1"> <af:iterator id="i2" value="#{rgn.children}" var="cnty">     <af:showDetailHeader text="#{cnty.CountryName}" disclosed="true" id="sdh2">       <af:iterator id="i3" value="#{cnty.children}" var="loc">         <af:panelList id="pl1">         <af:outputText value="#{loc.City}" id="ot3"/>           </af:panelList>         </af:iterator>       </af:showDetailHeader>     </af:iterator>   </af:showDetailHeader> </af:iterator>  You can download the entire sample from here:

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  • Failed to load viewstate.The control tree into which viewstate is being loaded...etc

    - by alaa9jo
    Two days ago,a colleague of mine tried to publish an asp.net website (which is built in VS2008 using framework 3.5) to our server,he configured everything in IIS (he made sure that the selected asp.net version is 2.0) and launched the website..at first it was working great but when he tried to click on a specific treeview...BOOM..: "Failed to load viewstate. The control tree into which viewstate is being loaded must match the control tree that was used to save viewstate during the previous request. For example, when adding controls dynamically, the controls added during a post-back must match the type and position of the controls added during the initial request." In that page there were these control: a TreeView and a Placeholder,when the user selects any node then it's controls will be created dynamically into that placeholder..for the first time it's working fine but when (s)he select another node then that issue appears. He called me to help him with this issue,for me this is the first time I see such an issue,scratch my head then I decided to eliminate the possibilities of this issue one by one,at the development machine it's working perfectly,he published the website at the local IIS and again..it's working perfectly,I took a copy of the website and published it into my laptop but no issues at all,so this is means that it's not an issue in the code. So there is something missing/wrong in our server [it has Windows Server 2003],we went to the server and checked on the web-config and the configurations on IIS...nothing wrong so far,so I decided to check if the framework 3.5 is installed or not and the answer: it wasn't installed Of course he assumed that it was installed and there was nothing to tell if it wasn't from the "ASP.Net version" in IIS because frameworks 3.0 and 3.5 will not be listed there [2.0 will be listed there instead],the only way to check if it was installed or not is to search for the framework in this path:[WINDOWS Folder]\Microsoft.NET\Framework or check if it was installed in Add or remove programs. The obvious solution for his case: We installed Framework 3.5 SP1 into our server,did a restart to the machine and it worked ! If anyone faced the same issue and solved it using the same solution or with a different one please post it here to share experience.

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  • How exactly is an Abstract Syntax Tree created?

    - by Howcan
    I think I understand the goal of an AST, and I've build a couple of tree structures before, but never an AST. I'm mostly confused because the nodes are text and not number, so I can't think of a nice way to input a token/string as I'm parsing some code. For example, when I looked at diagrams of AST's, the variable and its value were leaf nodes to an equal sign. This makes perfect sense to me, but how would I go about implementing this? I guess I can do it case by case, so that when I stumble upon an "=" I use that as a node, and add the value parsed before the "=" as the leaf. It just seems wrong, because I'd probably have to make cases for tons and tons of things, depending on the syntax. And then I came upon another problem, how is the tree traversed? Do I go all the way down the height, and go back up a node when I hit the bottom, and do the same for it's neighbor? I've seen tons of diagrams on ASTs, but I couldn't find a fairly simple example of one in code, which would probably help.

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  • Is there a program that will show a tree of the differences in two file trees?

    - by Huckle
    In windows I manually back up from time to time by formatting my external drive and copying the contents of my data partition over. Inevitably there is a difference in the number and size of the files copied because of system files, etc. Is there a program that would diff two directories recursively and compile the differences into a nice GUI tree that I could peruse (preferably filter) to ensure that everything I want made it over to the drive? It should only show files that are not in both directories. (Also, please ignore the inadequacy of my backup solution)

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  • d3 tree - parents having same children

    - by Larry Anderson
    I've been transitioning my code from JIT to D3, and working with the tree layout. I've replicated http://mbostock.github.com/d3/talk/20111018/tree.html with my tree data, but I wanted to do a little more. In my case I will need to create child nodes that merge back to form a parent at a lower level, which I realize is more of a directed graph structure, but would like the tree to accomodate (i.e. notice that common id's between child nodes should merge). So basically a tree that divides like normal on the way from parents to children, but then also has the ability to bring those children nodes together to be parents (sort of an incestual relationship or something :)). Asks something similar - How to layout a non-tree hierarchy with D3 It sounds like I might be able to use hierarchical edge bundling in conjunction with the tree hierarchy layout, but I haven't seen that done. I might be a little off with that though.

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  • DrScheme versus mzscheme: treatment of definitions

    - by speciousfool
    One long term project I have is working through all the exercises of SICP. I noticed something a bit odd with the most recent exercise. I am testing a Huffman encoding tree. When I execute the following code in DrScheme I get the expected result: (a d a b b c a) However, if I execute this same code in mzscheme by calling (load "2.67.scm") or by running mzscheme -f 2.67.scm, it reports: symbols: expected symbols as arguments, given: (leaf D 1) My question is: why? Is it because mzscheme and drscheme use different rules for loading program definitions? The program code is below. ;; Define an encoding tree and a sample message ;; Use the decode procedure to decode the message, and give the result. (define (make-leaf symbol weight) (list 'leaf symbol weight)) (define (leaf? object) (eq? (car object) 'leaf)) (define (symbol-leaf x) (cadr x)) (define (weight-leaf x) (caddr x)) (define (make-code-tree left right) (list left right (append (symbols left) (symbols right)) (+ (weight left) (weight right)))) (define (left-branch tree) (car tree)) (define (right-branch tree) (cadr tree)) (define (symbols tree) (if (leaf? tree) (list (symbol-leaf tree)) (caddr tree))) (define (weight tree) (if (leaf? tree) (weight-leaf tree) (cadddr tree))) (define (decode bits tree) (define (decode-1 bits current-branch) (if (null? bits) '() (let ((next-branch (choose-branch (car bits) current-branch))) (if (leaf? next-branch) (cons (symbol-leaf next-branch) (decode-1 (cdr bits) tree)) (decode-1 (cdr bits) next-branch))))) (decode-1 bits tree)) (define (choose-branch bit branch) (cond ((= bit 0) (left-branch branch)) ((= bit 1) (right-branch branch)) (else (error "bad bit -- CHOOSE-BRANCH" bit)))) (define (test s-exp) (display s-exp) (newline)) (define sample-tree (make-code-tree (make-leaf 'A 4) (make-code-tree (make-leaf 'B 2) (make-code-tree (make-leaf 'D 1) (make-leaf 'C 1))))) (define sample-message '(0 1 1 0 0 1 0 1 0 1 1 1 0)) (test (decode sample-message sample-tree))

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  • Cloud Computing - Multiple Physical Computers, One Logical Computer

    - by Koobz
    I know that you can set up multiple virtual machines per physical computer. I'm wondering if it's possible to make multiple physical computers behave as one logical unit? Fundamentally the way I imagine it working is that you can throw 10 computers into a facility one day. You've got one client that requires the equivalent of two computers worth, and 100 others that eat up the remaining 8. As demands change you're just reallocating logical resources, maybe the 2 computer client now requires a third physical system. You just add it to the cloud, and don't worry about sharding the database, or migrating data over to a new server. Can it work this way? If yes, why would anyone ever do things like partition their database servers anymore? Just add more computing resources. You scale horizontally with the hardware, but your server appears to scale vertically. There's no need to modify your application's infrastructure to support multiple databases etc.

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  • Cloud Computing - Multiple Physical Computers, One Logical Computer

    - by bundini
    I know that you can set up multiple virtual machines per physical computer. I'm wondering if it's possible to make multiple physical computers behave as one logical unit? Fundamentally the way I imagine it working is that you can throw 10 computers into a facility one day. You've got one client that requires the equivalent of two computers worth, and 100 others that eat up the remaining 8. As demands change you're just reallocating logical resources, maybe the 2 computer client now requires a third physical system. You just add it to the cloud, and don't worry about sharding the database, or migrating data over to a new server. Can it work this way? If yes, why would anyone ever do things like hand partition their database servers anymore? Just add more computing resources. You scale horizontally with the hardware, but your server appears to scale vertically. There's no need to modify your application's supporting infrastructure to support multiple databases etc.

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  • Changing default logical filename in SQL 2005

    - by Andrew
    I have a issue about creating databases in SQL 2005. I want to be able to change the default logical filename for the mdf file. At the moment the log logical filename ends in _log by default. I want the data logical filename to automatically end with _data for consistency. Is there a way i can set this? Andrew

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  • User reduced LVM logical volume without resizing filesystem

    - by Matthew
    I received an email yesterday that one of our users was trying to make room for a heartbeat/clustering package which requires its own partition to act as a voting disk. To do this, he attempted to reduce the size of the root partition's logical volume, and then create a new logical volume for this purpose. However, he forgot to resize the filesystem first (or include the -r switch in the command). He also forgot to unmount the root partition by running this process from a rescue cd. The system is now refusing to boot into the OS with the following error: Either the superblock or the partition table is likely to be corrupt! Unexpected Inconsistency; run fsck manually. The system them drops the user into single user mode. Is it possible to rescue the filesystem, or is it hosed? Its running ext3.

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  • "reduce" or "apply" using logical functions in Clojure

    - by Alex B
    I cannot use logical functions on a range of booleans in Clojure (1.2). Neither of the following works due to logical functions being macros: (reduce and [... sequence of bools ...]) (apply or [... sequence of bools ...]) The error saying that I "can't take value of a macro: #'clojure.core/and". How to apply these logical functions (macros) without writing boilerplate code?

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  • How to recover logical volume deleted with lvremove

    - by John P
    I am on CentOS 5.5 and am running Xen. I have a large volume group that I create logical volumes on using lvcreate. Today I had a customer cancel her account, then change her mind about an hour later. Unfortunately I had already removed the LVM her Xen image resided on. (just using a standard lvremove ). There has been no other LVM activity on this disk since then (nothing else added or deleted). Is it possible to "undo" a lvremove, or recover logical volume? If so, how would I go about it?

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  • Is there a Generic USB TouchScreen Driver 12.04?

    - by lbjoum
    Is there a Generic USB TouchScreen Driver 12.04? Device 03eb:201c I've been looking for 4 days solid (not very skilled) and can't find a solution. I have a generic tablet: C97- Atom N2600 9.7" 2GB 32GB Bluetooth WiFi WebCam Ext.3G Windows 7 Tablet PC Using 12.04 and cannot find a driver. I installed android and the touchscreen works but still lots of other bugs. Oh well, stuck with Windows 7 and not happy about it. Will keep trying, but too much time wasted already. If you have a solution I would love to try it. ubuntu@ubuntu:~$ lsusb Bus 001 Device 001: ID 1d6b:0002 Linux Foundation 2.0 root hub Bus 002 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 003 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 004 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 005 Device 001: ID 1d6b:0001 Linux Foundation 1.1 root hub Bus 001 Device 002: ID 0cf2:6238 ENE Technology, Inc. Bus 001 Device 003: ID 1a40:0101 Terminus Technology Inc. 4-Port HUB Bus 001 Device 005: ID 05e1:0100 Syntek Semiconductor Co., Ltd 802.11g + Bluetooth Wireless Adapter Bus 001 Device 006: ID 090c:3731 Silicon Motion, Inc. - Taiwan (formerly Feiya Technology Corp.) Bus 003 Device 002: ID 03eb:201c Atmel Corp. at90usbkey sample firmware (HID mouse) (from Windows: HID\VID_03EB&PID_201C\6&5F38127&0&0000 USB\VID_03EB&PID_201C\5&193ADADC&1&2 ) Bus 001 Device 007: ID 0518:0001 EzKEY Corp. USB to PS2 Adaptor v1.09 Bus 001 Device 008: ID 192f:0916 Avago Technologies, Pte. ubuntu@ubuntu:~$ sudo lsusb -v Bus 003 Device 002: ID 03eb:201c Atmel Corp. at90usbkey sample firmware (HID mouse) Device Descriptor: bLength 18 bDescriptorType 1 bcdUSB 2.00 bDeviceClass 0 (Defined at Interface level) bDeviceSubClass 0 bDeviceProtocol 0 bMaxPacketSize0 32 idVendor 0x03eb Atmel Corp. idProduct 0x201c at90usbkey sample firmware (HID mouse) bcdDevice 45.a2 iManufacturer 1 CDT iProduct 2 9.75 iSerial 0 bNumConfigurations 1 Configuration Descriptor: bLength 9 bDescriptorType 2 wTotalLength 34 bNumInterfaces 1 bConfigurationValue 1 iConfiguration 0 bmAttributes 0x00 (Missing must-be-set bit!) (Bus Powered) MaxPower 100mA Interface Descriptor: bLength 9 bDescriptorType 4 bInterfaceNumber 0 bAlternateSetting 0 bNumEndpoints 1 bInterfaceClass 3 Human Interface Device bInterfaceSubClass 0 No Subclass bInterfaceProtocol 0 None iInterface 0 HID Device Descriptor: bLength 9 bDescriptorType 33 bcdHID 1.11 bCountryCode 0 Not supported bNumDescriptors 1 bDescriptorType 34 Report wDescriptorLength 177 Report Descriptors: ** UNAVAILABLE ** Endpoint Descriptor: bLength 7 bDescriptorType 5 bEndpointAddress 0x81 EP 1 IN bmAttributes 3 Transfer Type Interrupt Synch Type None Usage Type Data wMaxPacketSize 0x0020 1x 32 bytes bInterval 5 Device Status: 0x00fb Self Powered Remote Wakeup Enabled Debug Mode ubuntu@ubuntu:~$ sudo lshw ubuntu description: Notebook product: To be filled by O.E.M. (To be filled by O.E.M.) vendor: To be filled by O.E.M. version: To be filled by O.E.M. serial: To be filled by O.E.M. width: 32 bits capabilities: smbios-2.7 dmi-2.7 smp-1.4 smp configuration: boot=normal chassis=notebook cpus=2 family=To be filled by O.E.M. sku=To be filled by O.E.M. uuid=00020003-0004-0005-0006-000700080009 *-core description: Motherboard product: Tiger Hill vendor: INTEL Corporation physical id: 0 version: To be filled by O.E.M. serial: To be filled by O.E.M. slot: To be filled by O.E.M. *-firmware description: BIOS vendor: American Megatrends Inc. physical id: 0 version: 4.6.5 date: 08/24/2012 size: 64KiB capacity: 960KiB capabilities: pci upgrade shadowing cdboot bootselect socketedrom edd int13floppy1200 int13floppy720 int13floppy2880 int5printscreen int9keyboard int14serial int17printer acpi usb biosbootspecification *-cpu:0 description: CPU product: Intel(R) Atom(TM) CPU N2600 @ 1.60GHz vendor: Intel Corp. physical id: 4 bus info: cpu@0 version: 6.6.1 serial: 0003-0661-0000-0000-0000-0000 slot: CPU 1 size: 1600MHz capacity: 1600MHz width: 64 bits clock: 400MHz capabilities: x86-64 boot fpu fpu_exception wp vme de pse tsc msr pae mce cx8 apic sep mtrr pge mca cmov pat pse36 clflush dts acpi mmx fxsr sse sse2 ss ht tm pbe nx constant_tsc arch_perfmon pebs bts nonstop_tsc aperfmperf pni dtes64 monitor ds_cpl est tm2 ssse3 cx16 xtpr pdcm movbe lahf_lm arat configuration: cores=2 enabledcores=1 id=2 threads=2 *-cache:0 description: L1 cache physical id: 5 slot: L1-Cache size: 24KiB capacity: 24KiB capabilities: internal write-back unified *-cache:1 description: L2 cache physical id: 6 slot: L2-Cache size: 512KiB capacity: 512KiB capabilities: internal varies unified *-logicalcpu:0 description: Logical CPU physical id: 2.1 width: 64 bits capabilities: logical *-logicalcpu:1 description: Logical CPU physical id: 2.2 width: 64 bits capabilities: logical *-logicalcpu:2 description: Logical CPU physical id: 2.3 width: 64 bits capabilities: logical *-logicalcpu:3 description: Logical CPU physical id: 2.4 width: 64 bits capabilities: logical *-memory description: System Memory physical id: 28 slot: System board or motherboard size: 2GiB *-bank:0 description: SODIMM [empty] product: [Empty] vendor: [Empty] physical id: 0 serial: [Empty] slot: DIMM0 *-bank:1 description: SODIMM DDR3 Synchronous 800 MHz (1.2 ns) vendor: 69 physical id: 1 serial: 00000210 slot: DIMM1 size: 2GiB width: 64 bits clock: 800MHz (1.2ns) *-cpu:1 physical id: 1 bus info: cpu@1 version: 6.6.1 serial: 0003-0661-0000-0000-0000-0000 size: 1600MHz capabilities: ht configuration: id=2 *-logicalcpu:0 description: Logical CPU physical id: 2.1 capabilities: logical *-logicalcpu:1 description: Logical CPU physical id: 2.2 capabilities: logical *-logicalcpu:2 description: Logical CPU physical id: 2.3 capabilities: logical *-logicalcpu:3 description: Logical CPU physical id: 2.4 capabilities: logical *-pci description: Host bridge product: Atom Processor D2xxx/N2xxx DRAM Controller vendor: Intel Corporation physical id: 100 bus info: pci@0000:00:00.0 version: 03 width: 32 bits clock: 33MHz *-display UNCLAIMED description: VGA compatible controller product: Atom Processor D2xxx/N2xxx Integrated Graphics Controller vendor: Intel Corporation physical id: 2 bus info: pci@0000:00:02.0 version: 09 width: 32 bits clock: 33MHz capabilities: pm msi vga_controller bus_master cap_list configuration: latency=0 resources: memory:dfe00000-dfefffff ioport:f100(size=8) *-multimedia description: Audio device product: N10/ICH 7 Family High Definition Audio Controller vendor: Intel Corporation physical id: 1b bus info: pci@0000:00:1b.0 version: 02 width: 64 bits clock: 33MHz capabilities: pm msi pciexpress bus_master cap_list configuration: driver=snd_hda_intel latency=0 resources: irq:42 memory:dff00000-dff03fff *-pci:0 description: PCI bridge product: N10/ICH 7 Family PCI Express Port 1 vendor: Intel Corporation physical id: 1c bus info: pci@0000:00:1c.0 version: 02 width: 32 bits clock: 33MHz capabilities: pci pciexpress msi pm normal_decode bus_master cap_list configuration: driver=pcieport resources: irq:40 ioport:2000(size=4096) memory:80000000-801fffff ioport:80200000(size=2097152) *-usb:0 description: USB controller product: N10/ICH 7 Family USB UHCI Controller #1 vendor: Intel Corporation physical id: 1d bus info: pci@0000:00:1d.0 version: 02 width: 32 bits clock: 33MHz capabilities: uhci bus_master configuration: driver=uhci_hcd latency=0 resources: irq:23 ioport:f0a0(size=32) *-usb:1 description: USB controller product: N10/ICH 7 Family USB UHCI Controller #2 vendor: Intel Corporation physical id: 1d.1 bus info: pci@0000:00:1d.1 version: 02 width: 32 bits clock: 33MHz capabilities: uhci bus_master configuration: driver=uhci_hcd latency=0 resources: irq:19 ioport:f080(size=32) *-usb:2 description: USB controller product: N10/ICH 7 Family USB UHCI Controller #3 vendor: Intel Corporation physical id: 1d.2 bus info: pci@0000:00:1d.2 version: 02 width: 32 bits clock: 33MHz capabilities: uhci bus_master configuration: driver=uhci_hcd latency=0 resources: irq:18 ioport:f060(size=32) *-usb:3 description: USB controller product: N10/ICH 7 Family USB UHCI Controller #4 vendor: Intel Corporation physical id: 1d.3 bus info: pci@0000:00:1d.3 version: 02 width: 32 bits clock: 33MHz capabilities: uhci bus_master configuration: driver=uhci_hcd latency=0 resources: irq:16 ioport:f040(size=32) *-usb:4 description: USB controller product: N10/ICH 7 Family USB2 EHCI Controller vendor: Intel Corporation physical id: 1d.7 bus info: pci@0000:00:1d.7 version: 02 width: 32 bits clock: 33MHz capabilities: pm debug ehci bus_master cap_list configuration: driver=ehci_hcd latency=0 resources: irq:23 memory:dff05000-dff053ff *-pci:1 description: PCI bridge product: 82801 Mobile PCI Bridge vendor: Intel Corporation physical id: 1e bus info: pci@0000:00:1e.0 version: e2 width: 32 bits clock: 33MHz capabilities: pci subtractive_decode bus_master cap_list *-isa description: ISA bridge product: NM10 Family LPC Controller vendor: Intel Corporation physical id: 1f bus info: pci@0000:00:1f.0 version: 02 width: 32 bits clock: 33MHz capabilities: isa bus_master cap_list configuration: latency=0 *-storage description: SATA controller product: N10/ICH7 Family SATA Controller [AHCI mode] vendor: Intel Corporation physical id: 1f.2 bus info: pci@0000:00:1f.2 logical name: scsi0 version: 02 width: 32 bits clock: 66MHz capabilities: storage msi pm ahci_1.0 bus_master cap_list emulated configuration: driver=ahci latency=0 resources: irq:41 ioport:f0f0(size=8) ioport:f0e0(size=4) ioport:f0d0(size=8) ioport:f0c0(size=4) ioport:f020(size=16) memory:dff04000-dff043ff *-disk description: ATA Disk product: BIWIN SSD physical id: 0.0.0 bus info: scsi@0:0.0.0 logical name: /dev/sda version: 1206 serial: 123403501060 size: 29GiB (32GB) capabilities: partitioned partitioned:dos configuration: ansiversion=5 signature=8fbe402b *-volume:0 description: Windows NTFS volume physical id: 1 bus info: scsi@0:0.0.0,1 logical name: /dev/sda1 version: 3.1 serial: 249bde5d-8246-9a40-88c7-2d5e3bcaf692 size: 19GiB capacity: 19GiB capabilities: primary bootable ntfs initialized configuration: clustersize=4096 created=2011-04-04 02:27:51 filesystem=ntfs state=clean *-volume:1 description: Windows NTFS volume physical id: 2 bus info: scsi@0:0.0.0,2 logical name: /dev/sda2 version: 3.1 serial: de12d40f-d5ca-8642-b306-acd9349fda1a size: 10231MiB capacity: 10GiB capabilities: primary ntfs initialized configuration: clustersize=4096 created=2011-04-04 01:52:26 filesystem=ntfs state=clean *-serial UNCLAIMED description: SMBus product: N10/ICH 7 Family SMBus Controller vendor: Intel Corporation physical id: 1f.3 bus info: pci@0000:00:1f.3 version: 02 width: 32 bits clock: 33MHz configuration: latency=0 resources: ioport:f000(size=32) *-scsi:0 physical id: 2 bus info: usb@1:1 logical name: scsi4 capabilities: emulated scsi-host configuration: driver=usb-storage *-disk description: SCSI Disk physical id: 0.0.0 bus info: scsi@4:0.0.0 logical name: /dev/sdb size: 29GiB (31GB) capabilities: partitioned partitioned:dos configuration: signature=00017463 *-volume description: Windows FAT volume vendor: mkdosfs physical id: 1 bus info: scsi@4:0.0.0,1 logical name: /dev/sdb1 logical name: /cdrom version: FAT32 serial: 129b-4f87 size: 29GiB capacity: 29GiB capabilities: primary bootable fat initialized configuration: FATs=2 filesystem=fat mount.fstype=vfat mount.options=rw,relatime,fmask=0022,dmask=0022,codepage=cp437,iocharset=iso8859-1,shortname=mixed,errors=remount-ro state=mounted *-scsi:1 physical id: 3 bus info: usb@1:3.1 logical name: scsi6 capabilities: emulated scsi-host configuration: driver=usb-storage *-disk description: SCSI Disk physical id: 0.0.0 bus info: scsi@6:0.0.0 logical name: /dev/sdc size: 7400MiB (7759MB) capabilities: partitioned partitioned:dos configuration: signature=c3072e18 *-volume description: Windows FAT volume vendor: mkdosfs physical id: 1 bus info: scsi@6:0.0.0,1 logical name: /dev/sdc1 logical name: /media/JOUM8G version: FAT32 serial: e676-9311 size: 7394MiB capacity: 7394MiB capabilities: primary bootable fat initialized configuration: FATs=2 filesystem=fat label=Android mount.fstype=vfat mount.options=rw,nosuid,nodev,relatime,uid=999,gid=999,fmask=0022,dmask=0077,codepage=cp437,iocharset=iso8859-1,shortname=mixed,showexec,utf8,flush,errors=remount-ro state=mounted ubuntu@ubuntu:~$ ubuntu@ubuntu:~$ xinput list ? Virtual core pointer id=2 [master pointer (3)] ? ? Virtual core XTEST pointer id=4 [slave pointer (2)] ? ? Plus More Enterprise LTD. USB-compliant keyboard id=10 [slave pointer (2)] ? ? USB Optical Mouse id=11 [slave pointer (2)] ? Virtual core keyboard id=3 [master keyboard (2)] ? Virtual core XTEST keyboard id=5 [slave keyboard (3)] ? Power Button id=6 [slave keyboard (3)] ? Power Button id=7 [slave keyboard (3)] ? Sleep Button id=8 [slave keyboard (3)] ? Plus More Enterprise LTD. USB-compliant keyboard id=9 [slave keyboard (3)] ? USB 2.0 Webcam - Front id=12 [slave keyboard (3)] ? AT Translated Set 2 keyboard id=13 [slave keyboard (3)] ubuntu@ubuntu:~$

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  • How to completely remove ldap and remove the directory tree

    - by rugbert
    so I followed this guide: https://help.ubuntu.com/11.04/serverguide/C/openldap-server.html to install and configure ldap but then I discoverd both phpLDAPadmin and Luma and have decided to rebuild my tree from scratch using one of those tools. However Im not sure how to completely remove LDAP now. I can remove it using apt-get, but if I attempt to reinstall it and login using phpLDAPadmin it seems that it's still looking for older authentication and gives me a credential error

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  • RecursiveIterator: used to explode tree structure, or only flatten?

    - by Stephen J. Fuhry
    There are tons of examples of using the RecursiveIterator to flatten a tree structure.. but what about using it to explode a tree structure? Is there an elegant way to use this, or some other SPL library to recursively build a tree (read: turn a flat array into array of arbitrary depth) given a table like this: SELECT id, parent_id, name FROM my_tree EDIT: You know how you can do this with Directories? $it = new RecursiveDirectoryIterator("/var/www/images"); foreach(new RecursiveIteratorIterator($it) as $file) { echo $file . PHP_EOL; } .. What if you could do something like this: $it = new RecursiveParentChildIterator($result_array); foreach(new RecursiveIteratorIterator($it) as $group) { echo $group->name . PHP_EOL; echo implode(PHP_EOL, $group->getChildren()) . PHP_EOL . PHP_EOL; } :END EDIT

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  • Python loop | "do-while" over a tree

    - by johannix
    Is there a more Pythonic way to put this loop together?: while True: children = tree.getChildren() if not children: break tree = children[0] UPDATE: I think this syntax is probably what I'm going to go with: while tree.getChildren(): tree = tree.getChildren()[0]

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  • traverse a binary decison tree using python?

    - by kaushik
    how to traverse a binary decision tree using python language. given a tree,i want know how can we travesre from root to required leaf the feature of the required leaf are given in an dictionary form assume and have to traverse from root to leaf answering the questions at each node with the details given in feature list.. the decision tree node has format ((question)(left tree)(right tree)) while traversing it should answer question at each node and an choose left or right and traverse till leaf?

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