Search Results

Search found 1765 results on 71 pages for 'z sorting'.

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

Page 35/71 | < Previous Page | 31 32 33 34 35 36 37 38 39 40 41 42  | Next Page >

  • Two-pass multi way merge sort?

    - by Nimesh
    If I have a relation (SQL) that does not fit in memory and I want to sort the relation using TPMMS (Two-pass multi-way merge sort method). How would I divide the table in sub tables (and how many) that can fit in memory and than merge them? Let's say I am using C#.

    Read the article

  • How to sort & Group in Android?

    - by crickpatel0024
    I have ArrayList and I want to sort and group all data by header in Android. How it is possible in Android? please help me.below me from owner And set header Me And Joe Manager From owner And set Header in listview. How to do that in Android? My code in below:: public class Request extends Activity { private String assosiatetoken; private ArrayList<All_Request_data_dto> list = new ArrayList<All_Request_data_dto>(); ListView lv; Button back; private Spinner spndata; String[] reqspinner = { "Request Date", "Last Update", "Type", "Owner", "State" }; ArrayAdapter<String> adapter; @Override protected void onCreate(Bundle savedInstanceState) { super.onCreate(savedInstanceState); setContentView(R.layout.request); assosiatetoken = MyApplication.getToken(); new doinbackground(this).execute(); back = (Button) findViewById(R.id.button1); spndata = (Spinner) findViewById(R.id.list_all_quize_req); adapter = new ArrayAdapter<String>(this, android.R.layout.simple_spinner_item, reqspinner); spndata.setAdapter(adapter); lv = (ListView) findViewById(R.id.listrequestdata); lv.setOnItemClickListener(new OnItemClickListener() { public void onItemClick(AdapterView<?> a, View v, int position, long id) { Intent edit = new Intent(Request.this, Request_webview.class); // edit.putExtra("Cat_url", url_link); startActivity(edit); } }); spndata.setOnItemSelectedListener(new OnItemSelectedListener() { public void onItemSelected(AdapterView<?> arg0, View arg1, int position, long arg3) { switch (position) { case 0: list = DBAdpter.requestUserData(assosiatetoken); Collections.sort(list, byDate1); // Collections.reverse(list); for (int i = 0; i < list.size(); i++) { if (list.get(i).submitDate != null) { lv.setAdapter(new MyListAdapter( getApplicationContext(), list)); } } break; case 1: list = DBAdpter.requestUserData(assosiatetoken); Collections.sort(list, byDate); for (int i = 0; i < list.size(); i++) { if (list.get(i).lastModifiedDate != null) { lv.setAdapter(new MyListAdapter( getApplicationContext(), list)); } } break; case 2: list = DBAdpter.requestUserData(assosiatetoken); Collections.sort(list, byDate3); // Collections.reverse(list); for (int i = 0; i < list.size(); i++) { if (list.get(i).state != null) { lv.setAdapter(new MyListAdapter( getApplicationContext(), list)); } } break; case 3: list = DBAdpter.requestUserData(assosiatetoken); for (int i = 0; i < list.size(); i++) { lv.setAdapter(new MyListAdapter( getApplicationContext(), list)); } break; default: break; } } public void onNothingSelected(AdapterView<?> arg0) { } }); back.setOnClickListener(new View.OnClickListener() { @Override public void onClick(View v) { finish(); } }); } static final Comparator<All_Request_data_dto> byDate = new Comparator<All_Request_data_dto>() { SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy hh:mm:ss a"); public int compare(All_Request_data_dto ord1, All_Request_data_dto ord2) { java.util.Date d1 = null; java.util.Date d2 = null; try { d1 = sdf.parse(ord1.lastModifiedDate); d2 = sdf.parse(ord2.lastModifiedDate); } catch (java.text.ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } return (d1.getTime() > d2.getTime() ? -1 : 1); // descending // return (d1.getTime() > d2.getTime() ? 1 : -1); //ascending } }; static final Comparator<All_Request_data_dto> byDate1 = new Comparator<All_Request_data_dto>() { SimpleDateFormat sdf = new SimpleDateFormat("MM/dd/yyyy hh:mm:ss a"); public int compare(All_Request_data_dto ord1, All_Request_data_dto ord2) { java.util.Date d1 = null; java.util.Date d2 = null; try { d1 = sdf.parse(ord1.submitDate); d2 = sdf.parse(ord2.submitDate); } catch (java.text.ParseException e) { // TODO Auto-generated catch block e.printStackTrace(); } return (d1.getTime() > d2.getTime() ? -1 : 1); // descending // return (d1.getTime() > d2.getTime() ? 1 : -1); //ascending } }; static final Comparator<All_Request_data_dto> byDate3 = new Comparator<All_Request_data_dto>() { public int compare(All_Request_data_dto ord1, All_Request_data_dto ord2) { String d1 = null; String d2 = null; d1 = ord1.state; d2 = ord2.state; return d1.compareToIgnoreCase(d2); } }; class doinbackground extends AsyncTask<Void, Void, Void> { ProgressDialog pd; private Context ctx; public doinbackground(Context c) { ctx = c; } @Override protected void onPreExecute() { super.onPreExecute(); pd = new ProgressDialog(ctx); pd.setMessage("Loading..."); pd.show(); } @Override protected Void doInBackground(Void... Params) { return null; } @Override protected void onPostExecute(Void result) { super.onPostExecute(result); pd.cancel(); } } public class MyListAdapter extends BaseAdapter { private ArrayList<All_Request_data_dto> list; public MyListAdapter(Context mContext, ArrayList<All_Request_data_dto> list) { this.list = list; } public int getCount() { return list.size(); } public All_Request_data_dto getItem(int position) { return list.get(position); } public long getItemId(int position) { return position; } public View getView(int position, View convertView, ViewGroup parent) { // if (convertView == null) { LayoutInflater inflator = (LayoutInflater) getSystemService(LAYOUT_INFLATER_SERVICE); convertView = inflator.inflate(R.layout.custom_request_data, null); TextView req_id = (TextView) convertView.findViewById(R.id.req_txt); TextView date = (TextView) convertView.findViewById(R.id.date_txt); TextView owner = (TextView) convertView .findViewById(R.id.owner_txt); TextView state = (TextView) convertView .findViewById(R.id.state_txt); req_id.setText(list.get(position).requestId + " - " + list.get(position).title); date.setText(list.get(position).lastModifiedDate + " - " + list.get(position).submitDate); owner.setText(list.get(position).owner); state.setText(list.get(position).state); // } return convertView; } } }

    Read the article

  • Alternatives to sign() in sqlite for custom order by

    - by Pentium10
    I have a string column which contains some numeric fields, but a lot are 0, empty string or null. The rest are numbers having different range, all positive. I tried to create a custom order by. The order would be done by two fields. First I would like to order the fields that have this number 0 and then sort by name. So something this would work: select * from table order by sign(referenceid) desc, name asc; But Sqlite lacks the sign() -1/0/1 function, and I am on Android and I can't create user defined functions. What other options I have to get this sort done.

    Read the article

  • Combining entries, filtering of Python dictionaries

    - by matt
    I have two large lists that are filled with dictionaries. I need to combine the entries if a value from dict2==dict1 and place the newly combined matches somewhere else. I'm having trouble explaining it. List one contains: {'keyword':value, 'keyword2':value2} List two: {'keyword2':value2, 'keyword3':value3} I want a new list with dictionaries including keyword1, keyword2, and keyword3 if both lists share the same 'keyword2' value. What's the best way to do this? When I try, I only come up with tons of nested for loops. Thanks

    Read the article

  • How to iterate through an XDocument's Nodes

    - by Ed R
    I am trying to iterate through my xml document's nodes to get the value for <username>Ed</username> in each node. I am using Linq to sort the XDocument first, then attempting to loop through the nodes. I can't seem to find the correct foreach loop to achieve this. Any help is appreciated. var doc = XDocument.Load("files\\config.xml"); var newDoc = new XDocument(new XElement("Config", from p in doc.Element("Config").Elements("Profile") orderby int.Parse(p.Element("order").Value) select p)); foreach (XElement xe in newDoc.Nodes()) { MessageBox.Show(xe.Element("username").Value); }

    Read the article

  • linked list sort function only loops once

    - by Tristan Pearce
    i have a singly linked list that i am trying to sort from least to greatest by price. here is what i have so far struct part { char* name; float price; int quantity; struct part *next; }; typedef struct part partType; partType *sort_price(partType **item) { partType *temp1 = *item; partType *temp2 = (*item)->next; if ( *item == NULL || (*item)->next == NULL ) return *item; else { while ( temp2 != NULL && temp2->next != NULL ){ if (temp2->price > temp2->next->price){ temp1->next = temp2->next; temp2->next = temp2->next->next; temp1->next->next = temp2; } temp1 = temp2; temp2 = temp2->next; } } return *item; } the list is already populated but when i call the sort function it only swaps the first two nodes that satisfy the condition in the if statement. I dont understand why it doesnt do the check again after the two temp pointers are incremented.

    Read the article

  • Python sort 2-D list by time string

    - by Mark Kennedy
    How do I sort a multi dimensional list like this based on a time string? The sublists can be of different sizes (i.e. 4 and 5, here) I want to sort by comparing the first time string in each sublist (sublist[-4]) x = (['1513', '08:19PM', '10:21PM', 1, 4], ['1290', '09:45PM', '11:43PM', 1, 4], ['0690', '07:25AM', '09:19AM', 1, 4], ['0201', '08:50AM', '10:50AM', 1, 4], ['1166', '04:35PM', '06:36PM', 1, 4], ['0845', '05:40PM', '07:44PM', 1, 4], ['1267', '07:05PM', '09:07PM', 1, 4], ['1513', '08:19PM', '10:21PM', 1, 4], ['1290', '09:45PM', '11:43PM', 1, 4], ['8772', '0159', '12:33PM', '02:43PM', 1, 5], ['0888', '0570', '09:42PM', '12:20AM', 1, 5], ['2086', '2231', '04:10PM', '06:20PM', 1, 5]) The sorted result would be sortedX = (['0690', '07:25AM', '09:19AM', 1, 4], ['0201', '08:50AM', '10:50AM', 1, 4], ['1166', '04:35PM', '06:36PM', 1, 4], ['0845', '05:40PM', '07:44PM', 1, 4], ['1267', '07:05PM', '09:07PM', 1, 4], ['1513', '08:19PM', '10:21PM', 1, 4], ['1513', '08:19PM', '10:21PM', 1, 4], ['1290', '09:45PM', '11:43PM', 1, 4], ['1290', '09:45PM', '11:43PM', 1, 4], ['8772', '0159', '12:33PM', '02:43PM', 1, 5], ['2086', '2231', '04:10PM', '06:20PM', 1, 5], ['0888', '0570', '09:42PM', '12:20AM', 1, 5]) I tried the following: sortedX = sorted(x, key=lambda k : k[-4]) #k[-4] is the first time string and it works but it doesn't respect the sublist size ordering

    Read the article

  • 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.

    Read the article

  • How can I do a multi level parent-child sort using Linq?

    - by Tenacious T
    How can I do a multi-level parent-child sort using Linq if I have a table structure like the one below: [Table: Sections] Id Seq Name ParentSectionId 1 1 TOP NULL 2 1 AAAA 1 3 2 SSSS 1 4 3 DDDD 1 5 1 SectionA1 2 6 2 SectionA2 2 7 1 SectionS1 3 8 3 ASummary 2 Expected sort result: TOP AAAA SectionA1 SectionA2 ASummary SSSS SectionS1 DDDD

    Read the article

  • 2 compareTo method overriden in the same class definition, how could I force to use the second?

    - by jayjaypg22
    I want to sort a list List<Blabla> donnees by a criterion on one of its field. My problem is that compareTo is already overriden for this Class. So I've got something like : Blabla { public int compareTo(Object other) { ... } public int compareTo(Blabla other) { ... } } In a business layer class I call : Business { method (){ Collections.sort(List<Blabla > donnees); } } But this call N°1 compareTo method with object parameter. How could I sort my list with the N°2 method?

    Read the article

  • How to fill a dataset after getting a gridview?

    - by user175084
    I have a gridview which has many columns.. the columns are got separately and displayed in a gridview. now i need to sort this gridview but i cannot do that.... i have found a way but i will need to get the gridview in a datatable or a dataset.... is there a a way to do this? DataSet ds= new DataSet(); ds = Gridview1.???? please help..

    Read the article

  • getting rid of repeated customer id's in mysql query

    - by bsandrabr
    I originally started by selecting customers from a group of customers and then for each customer querying the records for the past few days and presenting them in a table row. All working fine but I think I might have got too ambitious as I tried to pull in all the records at once having heard that mutiple queries are a big no no. here is the mysqlquery i came up with to pull in all the records at once SELECT morning, afternoon, date, date2, fname, lname, customers.customerid FROM customers LEFT OUTER JOIN attend ON ( customers.customerid = attend.customerid ) RIGHT OUTER JOIN noattend ON ( noattend.date2 = attend.date ) WHERE noattend.date2 BETWEEN '$date2' AND '$date3' AND DayOfWeek( date2 ) %7 >1 AND group ={$_GET['group']} ORDER BY lname ASC , fname ASC , date2 DESC tables are customer-customerid,fname,lname attend-customerid,morning,afternoon,date noattend-date2 (a table of all the days to fill in the blanks) Now the problem I have is how to start a new row in the table when the customer id changes My query above pulls in customer 1 morning 2 customer 1 morning 1 customer 2 morning 2 customer 2 morning 1 whereas I'm trying to get customer1 morning2 morning1 customer2 morning2 morning1 I dont know whether this is possible in the sql or more likely in the php

    Read the article

  • Need to sort 3 arrays by one key array

    - by jeff6461
    I am trying to get 3 arrays sorted by one key array in objective c for the iphone, here is a example to help out... Array 1 Array 2 Array 3 Array 4 1 15 21 7 3 12 8 9 6 7 8 0 2 3 4 8 When sorted i want this to look like Array 1 Array 2 Array 3 Array 4 1 15 21 7 2 3 4 8 3 12 8 9 6 7 8 0 So array 2,3,4 are moving with Array 1 when sorted. Currently i am using a bubble sort to do this but it lags so bad that it crashes by app. The code i am using to do this is int flag = 0; int i = 0; int temp = 0; do { flag=1; for(i = 0; i < distancenumber; i++) { if(distance[i] > distance[i+1]) { temp = distance[i]; distance[i]=distance[i + 1]; distance[i + 1]=temp; temp = FlowerarrayNumber[i]; FlowerarrayNumber[i] = FlowerarrayNumber[i+1]; FlowerarrayNumber[i + 1] = temp; temp = BeearrayNumber[i]; BeearrayNumber[i] = BeearrayNumber[i + 1]; BeearrayNumber[i + 1] = temp; flag=0; } } }while (flag==0); where distance number is the amount of elements in all of the arrays, distance is array 1 or my key array. and the other 2 are getting sorted. If anyone can help me get a merge sort(or something faster, it is running on a iPhone so it needs to be quick and light) to do this that would be great i cannot figure out how the recursion works in this method and so having a hard time to get the code to work. Any help would be greatly appreciated

    Read the article

  • Django: ordering numerical value with order_by

    - by h3
    I'm in a situation where I must output a quite large list of objects by a CharField used to store street addresses. My problem is, that obviously the data is ordered by ASCII codes since it's a Charfield, with the predictable results .. it sort the numbers like this; 1, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 2, 20, 21.... Now the obvious step would be to change the Charfield the proper field type (IntegerField let's say), however it cannot work since some address might have apartments .. like "128A". I really don't know how I can order this properly ..

    Read the article

  • SSRS - Sort table based on column value

    - by Ehsan
    I am trying to sort the following table: hYear hSale ------------------------------------ [year] =Count(Fields!sale.Value) The table only has one row group (year) and no column group. I'd like to: -initially sort the table based on the calculated value; is it possible? -add interactive sort to calculated column based on the value. I assume I should sort 'Detail rows', but what will be the sort expression?

    Read the article

  • Best tree/heap data structure for fixed set of nodes with changing values + need top 20 values?

    - by user350139
    I'm writing something like a game in C++ where I have a database table containing the current score for each user. I want to read that table into memory at the start of the game, quickly change each user's score while the game is being played in response to what each user does, and then when the game ends write the current scores back to the database. I also want to be able to find the 20 or so users with the highest scores. No users will be added or deleted during the short period when the game is being played. I haven't tried it yet, but updating the database might take too much time during the period when the game is being played. Fixed set of users (might be 10,000 to 50,000 users) Will map user IDs to their score and other user-specific information. User IDs will be auto_increment values. If the structure has a high memory overhead that's probably not an issue. If the program crashes during gameplay it can just be re-started. Quickly get a user's current score. Quickly add to a user's current score (and return their current score) Quickly get 20 users with highest score. No deletes. No inserts except when the structure is first created, and how long that takes isn't critical. Getting the top 20 users will only happen every five or ten seconds, but getting/adding will happen much more frequently. If not for the last, I could just create a memory block equal to sizeof(user) * max(user id) and put each user at user id * sizeof(user) for fast access. Should I do that plus some other structure for the Top 20 feature, or is there one structure that will handle all of this together?

    Read the article

  • matlab: simple matrix filtering - group size

    - by Art
    I have a huuuge matrix storing information about X and Y coordinates of multiple particle trajectories , which in simplified version looks like that: col 1- track number; col 2- frame number; col 2- coordinate X; col 3- coordinate Y for example: A = 1 1 5.14832 3.36128 1 2 5.02768 3.60944 1 3 4.85856 3.81616 1 4 5.17424 4.08384 2 1 2.02928 18.47536 2 2 2.064 18.5464 3 1 8.19648 5.31056 3 2 8.04848 5.33568 3 3 7.82016 5.29088 3 4 7.80464 5.31632 3 5 7.68256 5.4624 3 6 7.62592 5.572 Now I want to filter out trajectories shorter than lets say 2 and keep remaining stuff like (note renumbering of trajectories): B = 1 1 5.14832 3.36128 1 2 5.02768 3.60944 1 3 4.85856 3.81616 1 4 5.17424 4.08384 2 1 8.19648 5.31056 2 2 8.04848 5.33568 2 3 7.82016 5.29088 2 4 7.80464 5.31632 2 5 7.68256 5.4624 2 6 7.62592 5.572 How to do it efficiently? I can think about some ideas using for loop and vertcat, but its the slowest solution ever :/ Thanks!

    Read the article

  • Interview question: C program to sort a binary array in O(n)

    - by Zacky112
    I've comeup with the following program to do it, but it does not seem to work and goes into infinite loop. Its working is similar to quicksort. int main() { int arr[] = {1,1,0,1,0,0,0,1,0,1,0,1,0,1,0,1,0,1}; int N = 18; int *front, *last; front = arr; last = arr + N; while(front <= last) { while( (front < last) && (*front == 0) ) front++; while( (front < last) && (*last == 1) ) last--; if( front < last) { int temp = *front; *front = *last; *last = temp; front ++; last--; } } for(int i=0;i<N;i++) printf("%d ",arr[i]); return 0; }

    Read the article

  • On counting pairs of words that differ by one letter

    - by Quintofron
    Let us consider n words, each of length k. Those words consist of letters over an alphabet (whose cardinality is n) with defined order. The task is to derive an O(nk) algorithm to count the number of pairs of words that differ by one position (no matter which one exactly, as long as it's only a single position). For instance, in the following set of words (n = 5, k = 4): abcd, abdd, adcb, adcd, aecd there are 5 such pairs: (abcd, abdd), (abcd, adcd), (abcd, aecd), (adcb, adcd), (adcd, aecd). So far I've managed to find an algorithm that solves a slightly easier problem: counting the number of pairs of words that differ by one GIVEN position (i-th). In order to do this I swap the letter at the ith position with the last letter within each word, perform a Radix sort (ignoring the last position in each word - formerly the ith position), linearly detect words whose letters at the first 1 to k-1 positions are the same, eventually count the number of occurrences of each letter at the last (originally ith) position within each set of duplicates and calculate the desired pairs (the last part is simple). However, the algorithm above doesn't seem to be applicable to the main problem (under the O(nk) constraint) - at least not without some modifications. Any idea how to solve this?

    Read the article

  • Ordering z-indexes in an array

    - by Tom Gullen
    I have an array which looks something along the lines of resourceData[0][0] = "pic1.jpg"; resourceData[0][1] = 5; resourceData[1][0] = "pic2.jpg"; resourceData[1][1] = 2; resourceData[2][0] = "pic3.jpg"; resourceData[2][1] = 900; resourceData[3][0] = "pic4.jpg"; resourceData[3][1] = 1; The numeric represents the z-index of the image. Minimum z-index value is 1. Maximum (not really important) is 2000. I have all the rendering and setting z-indexes done fine. My question is, I want to have four functions: // Brings image to z front function bringToFront(resourceIndex) { // Set z-index to max + 1 resourceData[resourceIndex][1] = getBiggestZindex() + 1; // Change CSS property of image to bring to front $('#imgD' + resourceIndex).css("z-index", resourceData[resourceIndex][1]); } function bringUpOne(resourceIndex) { } function bringDownOne(resourceIndex) { } // Send to back z function sendToBack(resourceIndex) { } So given then index [3] (900 z): If we send it to the back, it will take the value 1, and [3] will have to go to 2, but that conflicts with [1] who has a 2 z-index so they need to go to three etc. Is there an easy programatical way of doing this because as soon as I start doing this it's going to get messy. It's important that the indexes of the array don't change. We can't sort the array unfortunately due to design. Update Thanks for answers, I'll post the functions here once they are written incase anyone comes across this in the future (note this code has zindex listed in [6]) // Send to back z function sendToBack(resourceIndex) { resourceData[resourceIndex][6] = 1; $('#imgD' + resourceIndex).css("z-index", 1); for (i = 0; i < resourceData.length; i++) { if (i != resourceIndex) { resourceData[i][6]++; $('#imgD' + i).css("z-index", resourceData[i][6]); } } }

    Read the article

  • What's the fastest way to search a very long list of words for a match in actionscript 3?

    - by Nuthman
    So I have a list of words (the entire English dictionary). For a word matching game, when a player moves a piece I need to check the entire dictionary to see if the the word that the player made exists in the dictionary. I need to do this as quickly as possible. simply iterating through the dictionary is way too slow. What is the quickest algorithm in AS3 to search a long list like this for a match, and what datatype should I use? (ie array, object, Dictionary etc)

    Read the article

facebook twitter digg google delicious technorati stumbleupon myspace wordpress linkedin gmail igoogle windows live tumblr viadeo yahoo buzz yahoo mail yahoo bookmarks favorites email print

< Previous Page | 31 32 33 34 35 36 37 38 39 40 41 42  | Next Page >