Sort within the same array variable Sort functions bellow are exactly the same, the only difference how short and expressive they are: Full declaration: myArr.sort { (lhs: EntryStruct, rhs: EntryStruct) -> Bool in // you can have additional code here return lhs.deadline < rhs.deadline } Shortened closure declaration: myArr.sort { (lhs:EntryStruct, rhs:EntryStruct) in return lhs.deadline … Read more
map.toList() .sortedBy { (key, value) -> value } .toMap()
Instead of asort, use asorti(source, destination) which sorts the indices into a new array and you won’t have to copy the array. Then you can use the destination array as pointers into the source array. For your example, you would use it like this: n=asorti(chr_count, sorted) for (i=1; i<=n; i++) { print sorted[i] ” : … Read more
This code: keys = sorted(attributes.keys(), reverse=True) result = [] for key in keys: result.append(attributes[key]) Is basically the use case for which list comprehensions were invented: result = [attributes[key] for key in sorted(attributes.keys(), reverse=True)]
As alluded to in the comments by btilly and Amadan, this is due to how the Timsort sorting algorithm works. Detailed description of the algorithm is here. Timsort speeds up operation on partially sorted arrays by identifying runs of sorted elements. A run is either “ascending”, which means non-decreasing: a0 <= a1 <= a2 <= … Read more
You almost had it. You just need to add an order property to your a1 terms aggregations, like this: GET myindex/_search { “size”:0, “aggs”: { “a1”: { “terms”: { “field”: “FIELD1”, “size”:0, “order”: {“a2”: “desc”} <— add this }, “aggs”:{ “a2”:{ “sum”:{ “field”:”FIELD2.SUBFIELD” } } } } } }
You can use Select-Object -Last 1 at the end of that pipeline.
Another way of solving this is using Quickselect. This should give you a total average time complexity of O(n). Consider this: Find the kth largest number x using Quickselect (O(n)) Iterate through the array again (or just through the right-side partition) (O(n)) and save all elements ≥ x Return your saved elements (If there are … Read more
Array.Sort() chooses one of three sorting algorithm, depending on the size of the input: If the size is fewer than 16 elements, it uses an insertion sort algorithm. If the size exceeds 2 * log^N, where N is the range of the input array, it uses a Heap Sort algorithm. Otherwise, it uses a Quicksort … Read more
That isn’t exactly a Java problem. You need to look into an efficient algorithm for sorting data that isn’t completely read into memory. A few adaptations to Merge-Sort can achieve this. Take a look at this: http://en.wikipedia.org/wiki/Merge_sort and: http://en.wikipedia.org/wiki/External_sorting Basically the idea here is to break the file into smaller pieces, sort them (either with … Read more