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• Count number of given object in a list with LINQ

  - by Aaginor

  I have a list which can contain multiple occurrences of the same object. Now I need to count, how often the given object is contained in this list. int count = 0; foreach (IMyObject item in myList) if (item == object2Count) count++; I am sure that this can be done nicer with LINQ, but LINQ is still a mystery to me. My first question is: How would I count the objects via LINQ and the second question: Will this LINQ version be slower or faster? I am using a ObservableCollection and the number of items in the list is usally rather small ... usally not more then 20. Thanks in advance, Frank

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• PHP list of specific files in a directory

  - by Jessica

  The following code will list all the file in a directy <?php if ($handle = opendir('.')) { while (false !== ($file = readdir($handle))) { if ($file != "." && $file != "..") { $thelist .= '<LI><a href="'.$file.'">'.$file.'</a>'; } } closedir($handle); } ?> <P>List of files:</p> <UL> <P><?=$thelist?></p> </UL> While this is very simple code it does it job. I'm now looking for a way to list ONLY file that have .xml (or .XML) at the end, how do I do that? Code is appreciate, thanks! jess

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• Sorting a meta-list by first element of children lists in Python

  - by thismachinechills

  I have a list, root, of lists, root[child0], root[child1], etc. I want to sort the children of the root list by the first value in the child list, root[child0][0], which is an int. Example: import random children = 10 root = [[random.randint(0, children), "some value"] for child in children] I want to sort root from greatest to least by the first element of each of it's children. I've taken a look at some previous entries that used sorted() and a lamda function I'm entirely unfamiliar with, so I'm unsure of how to apply that to my problem. Appreciate any direction that can by given Thanks

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• Thunderbird: 'reply to sent mail' use-case [closed]

  - by BoD

  I seem to have this use-case quite often: Send a mail to A, with B and C in cc Realize you forgot to say something in the mail. You now want to send a 'reply mail' (including your original mail as quotation) to A, with B and C in cc. To do that: go to the 'Sent mails' folder, and press the 'Reply all' button Problem: Now you have yourself in the 'To:' field, and A, B and C in the cc fields. Question: Is there a better way (maybe an extension?) to do step 4, and correctly have A in the To: field, and B and C in cc.

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• How to know if a std::list has been modified

  - by Nick

  I have this class: class C { public: C* parent; std::list<C> children; }; I can use this class in this way for example: C root; C child; root.children.push_back(child); // or other method of std::list (es: push_front, insert, ...) // Here child.parent is root // How can I set the parent of child? I want to do this work internally to my class without losing the functionality of std::list, is it possible?

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• Windows Virtual Machine ping reply fails the first few times

  - by user28471

  I am running a windows Web Server 2008 virtual machine. When I try to ping it from another windows machine the ping reply comes straight back with 100% success. But when I try to ping it from one of the Linux boxes, the ping reply fails with partial or 100% loss the first couple of time and succeeds the third time onwards. I know it is quite strange but can any one think anything that could be causing it. Could anything in the Windows Domain Security or Firewall be "taking time" to allow responding to the ping request? In fact even the firewall is turned off on the virtual machine.

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• Building a linked list with LINQ

  - by FreshCode

  What is the fastest way to order an unordered list of elements by predecessor (or parent) element index using LINQ? Each element has a unique ID and the ID of that element's predecessor (or parent) element, from which a linked list can be built to represent an ordered state. Example ID | Predecessor's ID --------|-------------------- 20 | 81 81 | NULL 65 | 12 12 | 20 120 | 65 The sorted order is {81, 20, 12, 65, 120}. An (ordered) linked list can easily be assembled iteratively from these elements, but can it be done in fewer LINQ statements? Edit: I should have specified that IDs are not necessarily sequential. I chose 1 to 5 for simplicity. See updated element indices which are random.

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• How to display List<> using ListView?

  - by eriX

  When I use ListView to display a array of my object, I can use the following code: MyObject[] myObject; ... ArrayAdapter<MyObject> itemList = new ArrayAdapter<MyObject>(this, R.layout.list, myObject); setListAdapter(itemList); In case that the input is a list: List<MyObject> myobject; How can I assign it to ListAdapter? Please advise, Thx!

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• Sort a list numerically in Python

  - by Matthew

  So I have this list, we'll call it listA. I'm trying to get the [3] item in each list e.g. ['5.01','5.88','2.10','9.45','17.58','2.76'] in sorted order. So the end result would start the entire list over again with Santa at the top. Does that make any sense? [['John Doe', u'25.78', u'20.77', '5.01'], ['Jane Doe', u'21.08', u'15.20', '5.88'], ['James Bond', u'20.57', u'18.47', '2.10'], ['Michael Jordan', u'28.50', u'19.05', '9.45'], ['Santa', u'31.13', u'13.55', '17.58'], ['Easter Bunny', u'17.20', u'14.44', '2.76']]

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• Remove duplicates in a String [] list f#

  - by Jose Maria de la Torre

  I need some noob help... I have a String[] list (a list that each element is a Array of strings) I wish to remove all duplicates in the list. But to see if they are duplicates I need to see the fist element in the array. what I have tried is let NotDuplicated = Duplicated.[0] |> Seq.distinct let NotDuplicated = Duplicated |> Seq.distinctBy id but nothing is working... can you guys please help me.. thanks!!

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• Request to server x Reply from server y

  - by klaasio

  I need some advice from you guys: I'm dealing with a custom loadbalancer/software for which we will use 2 main servers and about 8 slave servers. In short: User sends request to main server, main server will receive and handle the requests, sends a request to a slave server and slave server should send data DIRECTLY to the "user". User - Main server Main server - Slave server Slave server - User The reason for which data should be send directly to the user and not through the main server is because of bandwidth and low budget. Now I have the following idea's: -IPinIP, but that is not possible in Layer7 (so far i know there some expensive routers for that) -IP Spoof, using C/C++ we will make it look like the reply came from main server. But I was thinking, perhaps the reply "slave server - User" could just come from a different IP without causing issues in the firewall from the user or his anti-virus. I don't know so well about "home" firewalls/routers and/or anti-virus software. I guess the user machine wouldn't handle it well?

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• Scolling list with jQuery

  - by Chris

  My javascript is not up to scratch at the moment and I am stumped with this! I need to create an animated list with javascript like this one - http://www.fiveminuteargument.com/blog/scrolling-list. What I want is to take a list like so <ul> <li>Item 1</li> <li>Item 2</li> <li>Item 3</li> <li>Item 4</li> <li>Item 5</li> <li>Item 6</li> </ul> And display two at once, then display them in a loop, 2 at a time. Even pseudo code would help to get me started.

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• Find an element in List of List, c#

  - by Sandy

  Hi, I have a list of lists as below: List<List <T> > userList Class T { string uniqueidentifier, string param2, int param2} I have a uniqueidentifier and i need to find the element T in the list that has the same 'uniqueidentifier' value. I can do it using two 'foreach' loops. This does not seem to be nice way doing things. I guess there should be some inbuilt method like 'Find' that does the same thing and is highly optimized.

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• Order a List (C#) by many fields?

  - by Esabe

  Hi everyone, I want to order a List of objects in C# by many fields, not just by one. For example, let's suppose I have a class called X with two Attributes, A and B, and I have the following objects, in that order: object1 = A = "a", B = "h" object2 = A = "a", B = "c" object3 = A = "b", B = "x" object4 = A = "b", B = "b" and I want to order the list by A attribute first, and when they are equals, by B element, so the order would be: "a" "c" "a" "h" "b" "b" "b" "x" As far as I know, the OrderBy method order by one parameter. Question: How can I order a C# List by more than one field? Thank you very much

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• displaying a listbox using Html.List, and I have a List<User> collection

  - by Blankman

  Is a List enumerable? Html.List("mylistbox", Model.UserList); doesnt' seem to be working, where UserList is a List collection.

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• Hiding dropdown list when list is empty

  - by Amina

  I currently have a dropdown list that is populated on page load and sometimes the dropdown list doesn't contain anything and is empty. I would like the hide the dropdown list whenever there aren't any items in the list. I thought i could do this via javascript but i am not sure what i could be doing wrong because after adding the javascript it stills appear on the page. dropdown select: <select data-bind="options: endReason() ? endReason().SubReasons : [], value: subReasonId, optionsText: 'Text', optionsValue: 'Value', visible: isChecked" name="subReasons"> </select> This is my Javascript: <script type="text/javascript"> function OnClientDropDownOpening(sender, eventArgs) { var combo = $find("<%= subReasons %>"); items = combo.get_items(); if(items.get_count()==0) { eventArgs.set_cancel(true); } } </script>

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• Second largest number in list python

  - by Manu Lakaster

  So I have to find THE SECOND LARGEST NUMBER IN A LIST. I am doing it through simple loops.My approach is I am going to divide a list into two parts and then find the largest number into two parts and then compare two nuumbers. I will choose the smaller number from two of them. I can not use ready functions or different approaches. Basically, this is my code....But it does not run correctly....Help me please to fix it because I spent a lot of time on it :( Thanks....P.S. Can we use indices to "divide" a list ??? #!/usr/local/bin/python2.7 alist=[-45,0,3,10,90,5,-2,4,18,45,100,1,-266,706] largest=alist[0] h=len(alist)/2 m=len(alist)-h print(alist) for i in alist: if alist[h]>largest: largest=alist[h] i=i+1 print(largest)

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• Python finding repeating sequence in list of integers?

  - by tijko

  I have a list of lists and each list has a repeating sequence. I'm trying to count the length of repeated sequence of integers in the list: list_a = [111,0,3,1,111,0,3,1,111,0,3,1] list_b = [67,4,67,4,67,4,67,4,2,9,0] list_c = [1,2,3,4,5,6,7,8,9,0,1,2,3,4,5,6,7,8,9,0,23,18,10] Which would return: list_a count = 4 (for [111,0,3,1]) list_b count = 2 (for [67,4]) list_c count = 10 (for [1,2,3,4,5,6,7,8,9,0]) Any advice or tips would be welcome. I'm trying to work it out with re.compile right now but, its not quite right.

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• Python 2.7 creating a multidimensional list

  - by poop

  I don't know why I am having so much trouble creating a 3 dimensional list. I need the program to create an empty n by n list. So for n = 4: x = [[[],[],[],[]],[[],[],[],[]],[[],[],[],[]],[[],[],[],[]]] I've tried using: y = [n*[n*[]]] y = [[[]]* n for i in range(n)] Which both appear to be creating copies of a reference. I've also tried naieve application of the list builder with little success: y = [[[]* n for i in range(n)]* n for i in range(n)] y = [[[]* n for i in range(1)]* n for i in range(n)] I've also tried building up the array iteratively using loops, with no success. In my rapid flurry of attempts to not post something stupidly easy to SO, I came upon a solution: y = [] for i in range(0,n): y.append([[]*n for i in range(n)]) Is there an easier/ more intuitive way of doing this?

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• Move <option> to top of list with Javascript

  - by Adam

  I'm trying to create a button that will move the currently selected OPTION in a SELECT MULTIPLE list to the top of that list. I currently have OptionTransfer.js implemented, which is allowing me to move items up and down the list. I want to add a new function function MoveOptionTop(obj) { ... } Here is the source of OptionTransfer.js // =================================================================== // Author: Matt Kruse // WWW: http://www.mattkruse.com/ // // NOTICE: You may use this code for any purpose, commercial or // private, without any further permission from the author. You may // remove this notice from your final code if you wish, however it is // appreciated by the author if at least my web site address is kept. // // You may *NOT* re-distribute this code in any way except through its // use. That means, you can include it in your product, or your web // site, or any other form where the code is actually being used. You // may not put the plain javascript up on your site for download or // include it in your javascript libraries for download. // If you wish to share this code with others, please just point them // to the URL instead. // Please DO NOT link directly to my .js files from your site. Copy // the files to your server and use them there. Thank you. // =================================================================== /* SOURCE FILE: selectbox.js */ function hasOptions(obj){if(obj!=null && obj.options!=null){return true;}return false;} function selectUnselectMatchingOptions(obj,regex,which,only){if(window.RegExp){if(which == "select"){var selected1=true;var selected2=false;}else if(which == "unselect"){var selected1=false;var selected2=true;}else{return;}var re = new RegExp(regex);if(!hasOptions(obj)){return;}for(var i=0;i(b.text+"")){return 1;}return 0;});for(var i=0;i3){var regex = arguments[3];if(regex != ""){unSelectMatchingOptions(from,regex);}}if(!hasOptions(from)){return;}for(var i=0;i=0;i--){var o = from.options[i];if(o.selected){from.options[i] = null;}}if((arguments.length=0;i--){if(obj.options[i].selected){if(i !=(obj.options.length-1) && ! obj.options[i+1].selected){swapOptions(obj,i,i+1);obj.options[i+1].selected = true;}}}} function removeSelectedOptions(from){if(!hasOptions(from)){return;}for(var i=(from.options.length-1);i=0;i--){var o=from.options[i];if(o.selected){from.options[i] = null;}}from.selectedIndex = -1;} function removeAllOptions(from){if(!hasOptions(from)){return;}for(var i=(from.options.length-1);i=0;i--){from.options[i] = null;}from.selectedIndex = -1;} function addOption(obj,text,value,selected){if(obj!=null && obj.options!=null){obj.options[obj.options.length] = new Option(text, value, false, selected);}} /* SOURCE FILE: OptionTransfer.js */ function OT_transferLeft(){moveSelectedOptions(this.right,this.left,this.autoSort,this.staticOptionRegex);this.update();} function OT_transferRight(){moveSelectedOptions(this.left,this.right,this.autoSort,this.staticOptionRegex);this.update();} function OT_transferAllLeft(){moveAllOptions(this.right,this.left,this.autoSort,this.staticOptionRegex);this.update();} function OT_transferAllRight(){moveAllOptions(this.left,this.right,this.autoSort,this.staticOptionRegex);this.update();} function OT_saveRemovedLeftOptions(f){this.removedLeftField = f;} function OT_saveRemovedRightOptions(f){this.removedRightField = f;} function OT_saveAddedLeftOptions(f){this.addedLeftField = f;} function OT_saveAddedRightOptions(f){this.addedRightField = f;} function OT_saveNewLeftOptions(f){this.newLeftField = f;} function OT_saveNewRightOptions(f){this.newRightField = f;} function OT_update(){var removedLeft = new Object();var removedRight = new Object();var addedLeft = new Object();var addedRight = new Object();var newLeft = new Object();var newRight = new Object();for(var i=0;i0){str=str+delimiter;}str=str+val;}return str;} function OT_setDelimiter(val){this.delimiter=val;} function OT_setAutoSort(val){this.autoSort=val;} function OT_setStaticOptionRegex(val){this.staticOptionRegex=val;} function OT_init(theform){this.form = theform;if(!theform[this.left]){alert("OptionTransfer init(): Left select list does not exist in form!");return false;}if(!theform[this.right]){alert("OptionTransfer init(): Right select list does not exist in form!");return false;}this.left=theform[this.left];this.right=theform[this.right];for(var i=0;i

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• Search in a List<DataRow>?

  - by grady

  Hello, I have a List which I create from a DataTabe which only has one column in it. Lets say the column is called MyColumn. Each element in the list is an object array containing my columns, in this case, only one (MyColumn). Whats the most elegant way to check if that object array contains a certain value? Thanks

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• Skip List vs. Binary Tree

  - by Claudiu

  I recently came across the data structure known as a Skip list. They seem to have very similar behavior to a binary search tree... my question is - why would you ever want to use a skip list over a binary search tree?

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• Python to MATLAB: exporting list of strings using scipy.io

  - by user292461

  I am trying to export a list of text strings from Python to MATLAB using scipy.io. I would like to use scipy.io because my desired .mat file should include both numerical matrices (which I learned to do here) and text cell arrays. I tried: import scipy.io my_list = ['abc', 'def', 'ghi'] scipy.io.savemat('test.mat', mdict={'my_list': my_list) In MATLAB, I load test.mat and get a character array: my_list = adg beh cfi How do I make scipy.io export a list into a MATLAB cell array?

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• Python most common element in a list

  - by Richard

  What is an efficient way to find the most common element in a Python list? My list items may not be hashable so can't use a dictionary. Also in case of draws the item with the lowest index should be returned. Example: >>> most_common(['duck', 'duck', 'goose']) 'duck' >>> most_common(['goose', 'duck', 'duck', 'goose']) 'goose'

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• List - Strings - Textfiles

  - by b3y4z1d

  I've got a few questions concerning text files,list and strings. I wonder if it is possible to put in a code which reads the text in a textfile,and then using "string line;" or something else to define each new row of the text and turn all of them into one list. So I can sort the rows, remove a row or two or even all of them or search through the text for a specific row.

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