You can use Linq’s GroupBy in a modified version which groups only if the two items are adjacent, then it’s easy as: var result = classes .GroupAdjacent(c => c.Value) .Select(g => new { SequenceNumFrom = g.Min(c => c.SequenceNumber), SequenceNumTo = g.Max(c => c.SequenceNumber), Value = g.Key }); foreach (var x in result) Console.WriteLine(“SequenceNumFrom:{0} SequenceNumTo:{1} Value:{2}”, … Read more
Well, the brute force answer is: subList = [theList[n:n+N] for n in range(0, len(theList), N)] where N is the group size (3 in your case): >>> theList = list(range(10)) >>> N = 3 >>> subList = [theList[n:n+N] for n in range(0, len(theList), N)] >>> subList [[0, 1, 2], [3, 4, 5], [6, 7, 8], [9]] … Read more
values = set(map(lambda x:x[1], mylist)) newlist = [[y[0] for y in mylist if y[1]==x] for x in values]
I based my answer on the title of your post only, as I don’t know C# and didn’t understand the given query. But in MySQL I suggest you try subselects. First get a set of primary keys of interesting columns then select data from those rows: SELECT somecolumn, anothercolumn FROM sometable WHERE id IN ( … Read more
Everything in the g element is positioned relative to the current transform matrix. To move the content, just put the transformation in the g element: <g transform=”translate(20,2.5) rotate(10)”> <rect x=”0″ y=”0″ width=”60″ height=”10″/> </g> Links: Example from the SVG 1.1 spec
Use function array_reduce() to combine the items having the same city: $input = array( array(‘city’ => ‘NewYork’, ‘cash’ => ‘1000’), array(‘city’ => ‘Philadelphia’, ‘cash’ => ‘2300’), array(‘city’ => ‘NewYork’, ‘cash’ => ‘2000’), ); $output = array_reduce( // Process the input list $input, // Add each $item from $input to $carry (partial results) function (array $carry, … Read more
The documentation of accumarray says: Note If the subscripts in subs are not sorted, fun should not depend on the order of the values in its input data. And your subs is not sorted (at least not in ascending order). If you rewrite the code so that subs is sorted and vals is also rearranged … Read more
You basically need a Stream that is made out of a Pair (I choose AbstractMap.SimpleEntry here) that has the left part as a Hobby and right as the Student (could be the other way around, does not matter). Later just group those based on Hobby (that is a String in your case). data.stream() .flatMap(student -> … Read more
Here’s one approach. First group into lists and then process the lists into the values you actually want: import static java.util.Comparator.comparingLong; import static java.util.stream.Collectors.groupingBy; import static java.util.stream.Collectors.toMap; Map<Route,Integer> routeCounts = routes.stream() .collect(groupingBy(x -> x)) .values().stream() .collect(toMap( lst -> lst.stream().max(comparingLong(Route::getLastUpdated)).get(), List::size ));
To get objects to work with many of LINQ’s operators, such as GroupBy or Distinct, you must either implement GetHashCode & Equals, or you must provide a custom comparer. In your case, with a property as a list you probably need a comparer, unless you made the list read only. Try this comparer: public class … Read more