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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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  • 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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  • 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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  • 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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  • Searchable list of objects in Java

    - by Christian
    I want to create a large (~300,000 entries) List of self defined objects of the class Drug. Every Drug has an ID and I want to be able to search the Drugs in logarithmic time via that ID. What kind of List do I have to use? How do I declare that it should be searchable via the ID?

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  • Python filter/remove URLs from a list

    - by Eef
    Hi. I have a text file of URLs, about 14000. Below is a couple of examples: http://www.domainname.com/pagename?CONTENT_ITEM_ID=100&param2=123 http://www.domainname.com/images?IMAGE_ID=10 http://www.domainname.com/pagename?CONTENT_ITEM_ID=101&param2=123 http://www.domainname.com/images?IMAGE_ID=11 http://www.domainname.com/pagename?CONTENT_ITEM_ID=102&param2=123 I have loaded the text file into a Python list and I am trying to get all the URLs with CONTENT_ITEM_ID separated off into a list of their own. What would be the best way to do this in Python? Cheers

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  • List<> own comparer

    - by netmajor
    I have a List where element is: struct element { double priority; int value; } How can I implement my own comparer which allow me sort List by priority ? I try with SortredList... but it don't allow douplicated keys :( Big thanks for help!

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  • filtering directly and indirectly connected things from list

    - by Andreas Romeyke
    Hello, if you have a function "test a b" which returns true if a and b are connected directly and if you have a given unordered list of things, what would be an elegant and fast solution to filter all connected things from given list? Example: let test a b = let diff = a - b in diff == 0 ;; let lst = [4;1;7;3;8;9;2;0] ;; filter_connected 2 lst ;; - [4;1;3;2;0] Any hints?

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  • sorting a doubly linked list with merge sort.

    - by user329820
    Hi I have found this code in the internet and it was for arrays ,I want to change it for doubly linked list(instead of index we should use pointer) would you please help me that how can i change merge method(I have changed sort method by myself) also this is not my home work ,I love working with linked list!! public class MergeSort { private DoublyLinkedList LocalDoublyLinkedList; public MergeSort(DoublyLinkedList list) { LocalDoublyLinkedList = list; } public void sort() { if (LocalDoublyLinkedList.size() <= 1) { return; } DoublyLinkedList listOne = new DoublyLinkedList(); DoublyLinkedList listTwo = new DoublyLinkedList(); for (int x = 0; x < (LocalDoublyLinkedList.size() / 2); x++) { listOne.add(x, LocalDoublyLinkedList.getValue(x)); } for (int x = (LocalDoublyLinkedList.size() / 2) + 1; x < LocalDoublyLinkedList.size`(); x++) {` listTwo.add(x, LocalDoublyLinkedList.getValue(x)); } //Split the DoublyLinkedList again MergeSort sort1 = new MergeSort(listOne); MergeSort sort2 = new MergeSort(listTwo); sort1.sort(); sort2.sort(); merge(listOne, listTwo); } private void merge(DoublyLinkedList a, DoublyLinkedList b) { int x = 0; int y = 0; int z = 0; while (x < first.length && y < second.length) { if (first[x] < second[y]) { a[z] = first[x]; x++; } else { a[z] = second[y]; y++; } z++; } //copy remaining elements to the tail of a[]; for (int i = x; i < first.length; i++) { a[z] = first[i]; z++; } for (int i = y; i < second.length; i++) { a[z] = second[i]; z++; } } }

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  • How to publish a list of integers?

    - by Mason Wheeler
    I want to make a component that includes a list of integers as one of its serialized properties. I know I can't declare a TList<integer> as a published property, because it doesn't descend from TPersistent. I've read that you can define "fake" published properties if you override DefineProperties, but I'm not quite sure how that works, especially when it comes to creating a fake property that's a list, not a single value. Can someone point me in the right direction?

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