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  • Properly maintain sorted state of Array/Set

    - by Jeff
    I'm trying to get data out of my MOC and then create some new objects based on those objects, and put it all back together, while keeping my sort state. The securities come out of the MOC in proper order. And everything seems to be fine until I do the assignment to the game at the bottom from setWithArray. The documentation says that setWithArray removed the duplicate objects, if there are any. I'm wonder if that's messing up my data, but I don't see a good alternative. The data is ultimately being pulled out into a UITableView. When I add items to the game manually, then they stay sorted, so I don't think the breaking of the sort is beyond the scope of what I've included here. NSError *error; NSArray *allTheSecurities = [managedObjectContext executeFetchRequest:request error:&error]; if (allTheSecurities == nil) { // Handle the error. } [request release]; /**/ NSLog( @"Enumerate..." ); NSEnumerator *enumerator = [allTheSecurities objectEnumerator]; id anObject; NSMutableArray *portfolioStocks = [[NSMutableArray alloc] init]; while (anObject = [enumerator nextObject]) { NSLog( @"Iteration... %@", [anObject name] ); NSLog( @"Build a stock..." ); PortfolioStocks *this_stock = (PortfolioStocks *)[NSEntityDescription insertNewObjectForEntityForName:@"PortfolioStocks" inManagedObjectContext:context]; NSLog( @"Set a value..." ); [this_stock setSecurity:(Security *)anObject]; [this_stock setQuantity:[NSNumber numberWithInt:0]]; NSLog( @"Add to portfolioStocks..." ); [portfolioStocks addObject:this_stock]; } //Sorted properly up to here! NSLog( @"Add to portfolio..." ); [game setPortfolio:[NSSet setWithArray:portfolioStocks]]; // <-- This is where it's not sorted anymore.

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  • How to randomize a sorted list?

    - by Faken
    Here's a strange question for you guys, I have a nice sorted list that I wish to randomize. How would i go about doing that? In my application, i have a function that returns a list of points that describe the outline of a discretized object. Due to the way the problem is solved, the function returns a nice ordered list. i have a second boundary described in math and want to determine if the two objects intersect each other. I simply itterate over the points and determine if any one point is inside the mathematical boundary. The method works well but i want to increase speed by randomizing the point data. Since it is likely that that my mathematical boundary will be overlapped by a series of points that are right beside each other, i think it would make sense to check a randomized list rather than iterating over a nice sorted one (as it only takes a single hit to declare an intersection). So, any ideas on how i would go about randomizing an ordered list?

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  • Best data-structure to use for two ended sorted list

    - by fmark
    I need a collection data-structure that can do the following: Be sorted Allow me to quickly pop values off the front and back of the list Remain sorted after I insert a new value Allow a user-specified comparison function, as I will be storing tuples and want to sort on a particular value Thread-safety is not required Optionally allow efficient haskey() lookups (I'm happy to maintain a separate hash-table for this though) My thoughts at this stage are that I need a priority queue and a hash table, although I don't know if I can quickly pop values off both ends of a priority queue. I'm interested in performance for a moderate number of items (I would estimate less than 200,000). Another possibility is simply maintaining an OrderedDictionary and doing an insertion sort it every-time I add more data to it. Furthermore, are there any particular implementations in Python. I would really like to avoid writing this code myself.

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  • Seaching for an element in a circular sorted array

    - by guirgis
    I wanted to share this with you, i had this problem in a google interview. we want to search for a given element in a circular sorted array in complexity not greater than O(Log n). ex: search for 13 in {5,9,13,1,3}. My idea was to convert the circular array into a regular sorted array then do a binary search on the resulting array, but my problem was the algorithm i came up was stupid that it takes O(n) in the worst case: for(i = 1; i < a.length; i++){ if (a[i] < a[i-1]){ minIndex = i; break; } } then the corresponding index of ith element will be determined from the following relation: (i + minInex - 1) % a.length it is clear that my conversion (from circular to regular) algorithm may take O(n), so we need a better one, any suggestions?

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  • Find next lower item in a sorted list

    - by Sebastian
    Hey guys, let's say I have a sorted list of Floats. Now I'd like to get the index of the next lower item of a given value. The usual for-loop aprroach has a complexity of O(n). Since the list is sorted there must be a way to get the index with O(log n). My O(n) approach: index=0 for i,value in enumerate(mylist): if value>compareValue: index=i-1 Is there a datatype for solving that problem in O(log n)? best regards Sebastian

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  • How to fetch managed objects sorted by calculated value

    - by Marcin Zbijowski
    Hello, I'm working on the app that uses CoreData. There is location entity that holds latitude and longitude values. I'd like to fetch those entities sorted by distance to the user's location. I tried to set sort descriptor to distance formula sqrt ((x1 - x2)^2 + (y1 - y2)^2) but it fails with exception "... keypath ... not found in entity". NSString *distanceFormula = [NSString stringWithFormat:@"sqrt(((latitude - %f) * (latitude - %f)) + ((longitude - %f) * (longitude - %f)))", location.coordinate.latitude, location.coordinate.latitude, location.coordinate.longitude, location.coordinate.longitude]; NSSortDescriptor *sortDescriptor = [[NSSortDescriptor alloc] initWithKey:distanceFormula ascending:YES]; [fetchRequest setSortDescriptors:[NSArray arrayWithObject:sortDescriptor]]; NSError *error; NSArray *result = [[self managedObjectContext] executeFetchRequest:fetchRequest error:&error]; I'd like to fetch already sorted objects rather then fetch them all and then sort in the code. Any tips appreciated.

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  • Inserting a number into a sorted array!

    - by Jay
    I would like to write a piece of code for inserting a number into a sorted array at the appropriate position (i.e. the array should still remain sorted after insertion) My data structure doesn't allow duplicates. I am planning to do something like this: 1. Find the right index where I should be putting this element using binary search 2. Create space for this element, by moving all the elements from that index down. 3. Put this element there. Is there any other better way?

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  • Merging k sorted linked lists - analysis

    - by Kotti
    Hi! I am thinking about different solutions for one problem. Assume we have K sorted linked lists and we are merging them into one. All these lists together have N elements. The well known solution is to use priority queue and pop / push first elements from every lists and I can understand why it takes O(N log K) time. But let's take a look at another approach. Suppose we have some MERGE_LISTS(LIST1, LIST2) procedure, that merges two sorted lists and it would take O(T1 + T2) time, where T1 and T2 stand for LIST1 and LIST2 sizes. What we do now generally means pairing these lists and merging them pair-by-pair (if the number is odd, last list, for example, could be ignored at first steps). This generally means we have to make the following "tree" of merge operations: N1, N2, N3... stand for LIST1, LIST2, LIST3 sizes O(N1 + N2) + O(N3 + N4) + O(N5 + N6) + ... O(N1 + N2 + N3 + N4) + O(N5 + N6 + N7 + N8) + ... O(N1 + N2 + N3 + N4 + .... + NK) It looks obvious that there will be log(K) of these rows, each of them implementing O(N) operations, so time for MERGE(LIST1, LIST2, ... , LISTK) operation would actually equal O(N log K). My friend told me (two days ago) it would take O(K N) time. So, the question is - did I f%ck up somewhere or is he actually wrong about this? And if I am right, why doesn't this 'divide&conquer' approach can't be used instead of priority queue approach?

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  • Efficient way to maintain a sorted list of access counts in Python

    - by David
    Let's say I have a list of objects. (All together now: "I have a list of objects.") In the web application I'm writing, each time a request comes in, I pick out up to one of these objects according to unspecified criteria and use it to handle the request. Basically like this: def handle_request(req): for h in handlers: if h.handles(req): return h return None Assuming the order of the objects in the list is unimportant, I can cut down on unnecessary iterations by keeping the list sorted such that the most frequently used (or perhaps most recently used) objects are at the front. I know this isn't something to be concerned about - it'll make only a miniscule, undetectable difference in the app's execution time - but debugging the rest of the code is driving me crazy and I need a distraction :) so I'm asking out of curiosity: what is the most efficient way to maintain the list in sorted order, descending, by the number of times each handler is chosen? The obvious solution is to make handlers a list of (count, handler) pairs, and each time a handler is chosen, increment the count and resort the list. def handle_request(req): for h in handlers[:]: if h[1].handles(req): h[0] += 1 handlers.sort(reverse=True) return h[1] return None But since there's only ever going to be at most one element out of order, and I know which one it is, it seems like some sort of optimization should be possible. Is there something in the standard library, perhaps, that is especially well-suited to this task? Or some other data structure? (Even if it's not implemented in Python) Or should/could I be doing something completely different?

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  • Sorted queue with dropping out elements

    - by ffriend
    I have a list of jobs and queue of workers waiting for these jobs. All the jobs are the same, but workers are different and sorted by their ability to perform the job. That is, first person can do this job best of all, second does it just a little bit worse and so on. Job is always assigned to the person with the highest skills from those who are free at that moment. When person is assigned a job, he drops out of the queue for some time. But when he is done, he gets back to his position. So, for example, at some moment in time worker queue looks like: [x, x, .83, x, .7, .63, .55, .54, .48, ...] where x's stand for missing workers and numbers show skill level of left workers. When there's a new job, it is assigned to 3rd worker as the one with highest skill of available workers. So next moment queue looks like: [x, x, x, x, .7, .63, .55, .54, .48, ...] Let's say, that at this moment worker #2 finishes his job and gets back to the list: [x, .91, x, x, .7, .63, .55, .54, .48, ...] I hope the process is completely clear now. My question is what algorithm and data structure to use to implement quick search and deletion of worker and insertion back to his position. For the moment the best approach I can see is to use Fibonacci heap that have amortized O(log n) for deleting minimal element (assigning job and deleting worker from queue) and O(1) for inserting him back, which is pretty good. But is there even better algorithm / data structure that possibly take into account the fact that elements are already sorted and only drop of the queue from time to time?

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  • How to list rpm packages/subpackages sorted by total size

    - by smci
    Looking for an easy way to postprocess rpm -q output so it reports the total size of all subpackages matching a regexp, e.g. see the aspell* example below. (Short of scripting it with Python/PERL/awk, which is the next step) (Motivation: I'm trying to remove a few Gb of unnecessary packages from a CentOS install, so I'm trying to track down things that are a) large b) unnecessary and c) not dependencies of anything useful like gnome. Ultimately I want to pipe the ouput through sort -n to what the space hogs are, before doing rpm -e) My reporting command looks like [1]: cat unwanted | xargs rpm -q --qf '%9.{size} %{name}\n' > unwanted.size and here's just one example where I'd like to see rpm's total for all aspell* subpackages: root# rpm -q --qf '%9.{size} %{name}\n' `rpm -qa | grep aspell` 1040974 aspell 16417158 aspell-es 4862676 aspell-sv 4334067 aspell-en 23329116 aspell-fr 13075210 aspell-de 39342410 aspell-it 8655094 aspell-ca 62267635 aspell-cs 16714477 aspell-da 17579484 aspell-el 10625591 aspell-no 60719347 aspell-pl 12907088 aspell-pt 8007946 aspell-nl 9425163 aspell-cy Three extra nice-to-have things: list the dependencies/depending packages of each group (so I can figure out the uninstall order) Also, if you could group them by package group, that would be totally neat. Human-readable size units like 'M'/'G' (like ls -h does). Can be done with regexp and rounding on the size field. Footnote: I'm surprised up2date and yum don't add this sort of intelligence. Ideally you would want to see a tree of group-package-subpackage, with rolled-up sizes. Footnote 2: I see yum erase aspell* does actually produce this summary - but not in a query command. [1] where unwanted.txt is a textfile of unnecessary packages obtained by diffing the output of: yum list installed | sed -e 's/\..*//g' > installed.txt diff --suppress-common-lines centos4_minimal.txt installed.txt | grep '>' and centos4_minimal.txt came from the Google doc given by that helpful blogger.

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  • Winamp question: I want to search through my songs by which playlists they're sorted into

    - by Daddy Warbox
    I'm trying to think of a way to do this. I sort my songs into a variety of playlists corresponding to different 'moods' I might have, but some songs fit for more than one kind of mood (e.g. a techno song might be 'stylish' and 'surreal', or something to that effect). I also give them different star ratings for a general sort of opinion about them. I want to be able to filter my media library by moods I want or don't want, as well as by star rating. Anyone have a good way to do something like this? Alternatively, is there a way to 'tag' my songs by mood and then search up by those tags? because then I could just dump the queried results into playlists as needed. Thanks in advance.

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  • JQuery tablesorter - Keeping grouped subheaders together, but still sorted

    - by hfidgen
    Hiya, I'm not really a Javascript programmer, so I'm struggling with this! I'm using the tablesorter plugin along with the Tablegroup plugin, which work very nicely to group the table rows by a parent, and then sort the parents. My problem is though, that I'd also like the child rows to be sorted whilst within the parent group I've done my best to get this working but I'm afraid I've hit a wall. Can anyone suggest a new starter for 10? The example below is working fine - There are 2x groups here: Nordics (Norway and Denmark) DACH (Germany and Austria) If I click on the header row, the groups are sorted, but the child rows within the group are not sorted. <script type="text/javascript"> $(document).ready(function () { $(".tablesorter") .tablesorter({ // set default sort column sortList: [[0,0]], // don't sort by first column headers: {0: {sorter: false}} , onRenderHeader: function (){ this.wrapInner("<span></span>"); } , debug: true }) }); </script> <table id="results-header" class="grid tablesorter table-header" cellpadding="0" cellspacing="0" border="0"> <thead> <tr class="title"> <th class="countries">&nbsp;</th> <th>% market share</th> <th>% increase in mkt share</th> <th>Target achieved</th> <th>% targets</th> <th>% sales inc. M-o-M</th> <th>% sales inc. M-o-M for country</th> <th>% training</th> </tr> </thead> <tbody> <tr id="Nord" class="collapsible parent parent-even collapsed"> <td class="countries">Nordics</td> <td>39.5</td> <td>49</td> <td>69.8</td> <td>51.8</td> <td>43</td> <td>42.5</td> <td>38</td> </tr> <tr id="row-Norway" class="expand-child child child-Nord"> <td class="countries">Norway</td> <td>6</td> <td>45</td> <td>101</td> <td>10</td> <td>20</td> <td>40</td> <td>30</td> </tr> <tr id="row-Denmark" class="expand-child child child-Nord"> <td class="countries">Denmark</td> <td>10</td> <td>20</td> <td>3</td> <td>40</td> <td>50</td> <td>25</td> <td>8</td> </tr> <tr id="DACH" class="collapsible parent parent-odd collapsed"> <td class="countries">DACH</td> <td>77</td> <td>61</td> <td>43</td> <td>98</td> <td>65</td> <td>92.5</td> <td>59.5</td> </tr> <tr id="row-Germany" class="expand-child child child-DACH"> <td class="countries">Germany</td> <td>56</td> <td>24</td> <td>84</td> <td>98</td> <td>32</td> <td>87</td> <td>21</td> </tr> <tr id="row-Austria" class="expand-child child child-DACH"> <td class="countries">Austria</td> <td>98</td> <td>98</td> <td>2</td> <td>98</td> <td>98</td> <td>98</td> <td>98</td> </tr> </tbody> </table>

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  • Insert a Row with a sorted DataGridView

    - by Ruben Trancoso
    My DataGridView is bound to the same BindingSource as the Insert Form, and the Name column is sorted. After insert is done, the groupBindingSource.Current is not returning the new inserted DataRowView but the last row in the sort order what makes the Update do nothing. FormGroup formGroup = new FormGroup(); formGroup .Source = groupBindingSource; formGroup .setMode(FormGroup.Mode.Insert); if (formGroup .ShowDialog() == DialogResult.OK) { DataRowView drv = (DataRowView)groupBindingSource.Current; grupoTableAdapter.Update(drv.Row); }

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  • Sorted directory size display in Unix

    - by Srikanth M
    How do i display directory sizes in a sorted manner in Unix while listing only innermost sub-directories? For example, I don't want to something like this - 100270480 /a/b/BP/b/bat/qc3 100270416 /a/b/BP/b/bat/qc3/logs 99020464 /a/b/BP/b/bat/qc3/logs/i 99005456 /a/b/BP/b/bat/qc5 99005408 /a/b/BP/b/bat/qc5/logs 97726832 /a/b/BP/b/bat/qc5/logs/i I just want - 99020464 /a/b/BP/b/bat/qc3/logs/i 97726832 /a/b/BP/b/bat/qc5/logs/i

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  • Merging some sorted lists with unknown order sequence

    - by Gabriel
    I've some sorted lists with variable number of elements. I wold like to merge the lists into one big list which contains all other lists in same order, without duplicates. Example: 1. XS,M,L,XL 2. S,M,XXL 3. XXS,XS,S,L Result: XXS,XS,S,M,L,XL,XXL The function should notify, if there are elements which have ambiguous positions. Here, it would be XXL and I need to specify its position after XL.

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  • Binary search in rotated sorted list

    - by Algorist
    I am having a sorted list which is rotated and I would like to do a binary search on that list to find the minimum element. Lets suppose initial list is {1,2,3,4,5,6,7,8} rotated list can be like {5,6,7,8,1,2,3,4} Normal binary search doesn't work in this case. Any idea how to do this.

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  • Keeping DB Table sorted using multi-field formula (Microsoft SQL Server)

    - by user298167
    I have a JOB table, with two interesting columns: Creation Date Importance (high - 3, medium 2, low - 1). A JOB record's priority calculated like this: Priority = Importance * (time passed since creation) The problem is, every time I would like to pick 200 jobs with highest priority, and I don't want to resort the table. Is there a way to keep rows sorted? I was also thinking about having three tables one for High, Medium and Low and then sort those by Creation Date.

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  • data from few MySQL tables sorted by ASC

    - by Andrew
    In the dbase I 've few tables named as aaa_9xxx, aaa_9yyy, aaa_9zzz. I want to find all data with a specified DATE and show it with the TIME ASC. First, I must find a tables in the dbase: $STH_1a = $DBH->query("SELECT table_name FROM information_schema.tables WHERE table_name LIKE 'aaa\_9%' "); foreach($STH_1a as $row) { $table_name_s1[] = $row['table_name']; } Second, I must find a data wit a concrete date and show it with TIME ASC: foreach($table_name_s1 as $table_name_1) { $STH_1a2 = $DBH->query("SELECT * FROM `$table_name_1` WHERE date = '2011-11-11' ORDER BY time ASC "); while ($row = $STH_1a2->fetch(PDO::FETCH_ASSOC)) { echo " ".$table_name_1."-".$row['time']."-".$row['ei_name']." <br>"; } } .. but it shows the data sorted by tables name, then by TIME ASC. I must to have all this data (from all tables) sorted by TIME ASC. Thank You dev-null-dweller, Andrew Stubbs and Jaison Erick for your help. I test the Erick solution : foreach($STH_1a as $row) { $stmts[] = sprintf('SELECT * FROM %s WHERE date="%s"', $row['table_name'], '2011-11-11'); } $stmt = implode("\nUNION\n", $stmts); $stmt .= "\nORDER BY time ASC"; $STH_1a2 = $DBH->query($stmt); while ($row_1a2 = $STH_1a2->fetch(PDO::FETCH_ASSOC)) { echo " ".$row['table_name']."-".$row_1a2['time']."-".$row_1a2['ei_name']." <br>"; } it's working but I've problem with 'table_name' - it's always the LAST table name. //---------------------------------------------------------------------- end the ending solution with all fixes, thanks all for your help, :)) foreach($STH_1a as $row) { $stmts[] = sprintf("SELECT *, '%s' AS table_name FROM %s WHERE date='%s'", $row['table_name'], $row['table_name'], '2011-11- 11'); } $stmt = implode("\nUNION\n", $stmts); $stmt .= "\nORDER BY time ASC"; $STH_1a2 = $DBH->query($stmt); while ($row_1a2 = $STH_1a2->fetch(PDO::FETCH_ASSOC)) { echo " ".$row_1a2['table_name']."-".$row_1a2['time']."-".$row_1a2['ei_name']." <br>"; }

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