Your question is still ambiguous. Your example doesn’t have the first field Team_Name introduced in your header. So the index here is probably off by one, but I think you get the concept: #read lines of text file and split into words lines = [line.split() for line in open(“test.txt”, “r”)] #sort lines for different columns, … Read more
What you have works. Just don’t forget to assign the sorted collection to a variable (sort_by and reverse do not change the collection). Bonus: here’s arguably a nicer version (one pass, instead of two) a.sort_by{ |v| -v[:a_value] } solved Multisort using Ruby [closed]
Since this is C and not C++, you’ll need to create your own sort program since I think qsort() can be O(n^2) worst case. Since you will be using your own sort program, you could just sort the array and the indices at the same time, treating them as pairs of elements. If there was … Read more
You can use interface java.lang.Comparable in java Example (pseudo code) assuming object.distance(player) returns an Integer value class Distance implements Comparable<Distance>{ /** * Compare a given Distance with this object. */ public int compareTo(Distance o) { return this.distance(player).compareTo(o.distance(o.player)); } } now you can sort your list like Collections.sort(YourListOfDistance) here some reference When should a class be … Read more
First of all a, e and g are variables and not necessarily strings. What you mean is probably “a”, “e” and “g”. Your question is essentially: How do I sort the values in a String[] so that numbers are sorted numerically and prioritized before letters (or words?) which are sorted alphabetically? Your question seems incomplete, … Read more
This operation is just doing the “merge” step of a Merge Sort and can be done in O(l1+l2) time. The general premise is to iterate over both (already sorted) lists at the same time, but only advance the list with the lowest head value, while using the advanced value in the resulting output. The operation … Read more
You need two counters Absolute Counter (always 1, 2, 3, 4, 5, etc). Rank Counter – it counts from 1 but if the scores are same, it does not update. As soon as scores are different, it updates with the absolute counter. Sample Code $counter = 1; // init absolute counter $rank = 1; // … Read more
Ok.I don’t know if your question is ‘legal’ for this website but I try to explain to you: Each bold Title (Algorithm1,Algorithm2, ecc.) represents a sequence of steps of 4 differents sorting algorithms. Each line represents the sequence that needs to be sorted. In Algorithms 1 for example at step 0 (first line) the number … Read more
Here is a simple, but not most efficient way. you can achieve that by two step: convert list to set, to remove the duplicate number. use heap to find nlargest and nsmallest in set def difference(list1): set1 = set(list1) return heapq.nlargest(2, set1)[1] – heapq.nsmallest(2, set1)[1] Here is a one pass way, more efficient way, use … Read more
After eliminating useless code, renaming variables and formatting it: private static void Sort(int[] array) { for (int j = 1; j < array.length; j++) { int value = array[j]; int index = j-1; while ( (index >= 0) && (array[index] < value) ) { array[index + 1] = array[index]; index–; } array[index + 1] = … Read more
This is not going to be the fastest data structure & algorithm for your particular purpose I guess, but it may be fast enough. Test it yourself. Note that a std::forward_list or even a std::vector might be faster depending on the actual scenario (-> constant factors in big-O-notation). tmyklebu mentioned another approach in the comments: … Read more
Hello you have to sort your object. Just use .sort of the array for that. Here is a sample: var obj = { “foo”: “bar”, “baz”: [ { “order”: 2, “fruit”: “banana” }, { “order”: 1, “fruit”: “apple” }, { “order”: 3, “fruit”: “peach” }, ] } // get property var arr = obj[“baz”]; // … Read more
Your solution is has O(n^2) time complexity – those nested loops will be very slow for large lists. However, with the aid of a HashMap, an O(n) solution is possible. Map<String, NameAndValue> map = new HashMap<>(); for (NameAndValue x : arrayB) map.put(x.getName(), x); for (int i = 0; i < arrayA.size(); i++) arrayB.set(i, map.get(arrayA.get(i).getName())); This … Read more
Here is my version where I filter on a range based on first value of the array and the coef variable, based on the result I slice away those elements already counted and filter again on the smaller array in a loop. This solution requires the input array to be sorted in ascending order func … Read more
Use objects for the grouping. var categories = {}; function incrementBy(category, value) { if (!categories[category]) { categories[category] = 0; } categories[category] += value; } var datasetarr = []; var pos; for(var i = 0; i < arr.length; i++){ console.log(‘unsorted ‘ + arr[i].type + ‘ ‘ + arr[i].quantity); incrementBy(arr[i].type, arr[i].quantity) } for(var category in categories){ console.log(‘sorted … Read more