Use Comparable and Comparator as shown below, you can also visit https://www.journaldev.com/780/comparable-and-comparator-in-java-example for further details. import java.util.Comparator; class Employee implements Comparable<Employee> { private int id; private String name; private int age; private long salary; public int getId() { return id; } public String getName() { return name; } public int getAge() { return age; } … Read more
Here it’s done with std::sort from the header algorithm #include <iostream> #include <string> #include <algorithm> #include <vector> int main(){ std::vector<std::string> nums{ “922003001020293839297830207344987344973074734”, “766352786207892397340783784078348747606208602”, “182823068326283756515117829362376823572395775” }; std::cout << “unsorted: ” << std::endl; for (auto i : nums){ std::cout << i << std::endl; } std::sort(nums.begin(), nums.end()); //sort it std::cout << “\nsorted: ” << std::endl; for (auto i … Read more
You can use OrderBy followed by ThenBy. var result = MyCollection.OrderBy(x => x.CategoryCounter) .ThenBy(y => y.GroupCounter) .ThenBy(z => z.QuestionCounter); 1 solved Sort list on 3 values [closed]
You can cast pointer to the first row of a two-dimensional array to pointer to int and sort the array as a one-dimensional array. Here you are #include <iostream> #include <iomanip> #include <cstdlib> #include <ctime> void bubble_sort( int *a, size_t n ) { for ( size_t last /* = n */; not ( n < … Read more
You need give an order (actually a equivalence relation) to be able to sort. The order on characters is usually a,b,c,… and the order usually given on words such as ‘one’ is called lexicographic order. However you want to sort by the meaning behind, its semantic: integers. Your computer doesn’t know that you want this, … Read more
array_count_values will output the count as like Array ( [keyword1] => 10 [keyword2] => 3 [keyword3] => 6 [keyword4] => 5 [keyword5] => 1 ) But For your desired output you need to use foreach Demo $arra = Array ( 0 => “keyword1”, 1 => “keyword1”, 2 => “keyword1”, 3 => “keyword1”, 4 => “keyword1”, … Read more
I could not post it to comment section. Please see below: List<Integer> integers = Arrays.asList(1, 2, 3, 4); int i = 0; StringBuilder sb = new StringBuilder(“[“); for (int value : integers) { sb.append((i == 0) ? “” : (i % 2 == 0 ? “,” : “:”)).append(value); i++; } sb.append(“]”); System.out.println(sb.toString()); Output is: [1:2,3:4] … Read more
from itertools import combinations, permutations perm=[] index = [9,7,6,5,2,10,8,4,3,1] perm.append(index) M = 3 slicer = [x for x in combinations(range(1, len(index)), M – 1)] slicer = [(0,) + x + (len(index),) for x in slicer] result = [tuple(p[s[i]:s[i + 1]] for i in range(len(s) – 1)) for s in slicer for p in perm] 1 … Read more
data_tab.sort(function(a,b){return a[0]-b[0];}); solved Array sort in javascript [closed]
You just: std::sort(v.begin(), v.end()); std::pair is lexicographically compared. On the other hand if you want to sort them with respect the second element of the std::pair then you would have to defind a custom comparator in the following manner: std::sort(v.begin(), v.end(), [](std::pair<int, std::string> const &p1, std::pair<int, std::string> const &p2) { return (p1.second == p2.second)? p1.first … Read more
after you’ve sorted the director list…. just take one director as an key value in outer loop and then start comparing each and every movie name in inner loop… How can I sort a List alphabetically? this would be helpful… do some searching next time….it is easy to find such solutions… 🙂 😉 3 solved … Read more
Simplest will be bubble sort. item* sort(item *start){ item *node1,*node2; int temp; for(node1 = start; node1!=NULL;node1=node1->next){ for(node2 = start; node2!=NULL;node2=node2->next){ if(node2->draw_number > node1->draw_number){ temp = node1->draw_number; node1->draw_number = node2->draw_number; node2->draw_number = temp; } } } return start; } 1 solved Assorting linked list in c [duplicate]
You can use pattern matching to determine which type of Array you’re dealing with. def arrTest[A](a :Array[A]) :Boolean = a match { case sa :Array[String] => sa.length < 2 || sa.sliding(2).forall(x => x(0) < x(1)) case ia :Array[Int] => ia.length > 2 && ia(0) == 1 && ia(1) == 1 && ia.sliding(3).forall(x => x(0) + … Read more
You can use first character to sort and consider if first character is number it is small, let myObj = [{ “60+”: 0.1413972485314015, “18-29”: 0.0832178903621611, “40-49”: 0.1033361204013377, “30-39”: 0.0835906328864075, “Under 18”: 0.1326368677036551, “50-59”: 0.1224973366151133 }]; const ordered = {}; Object.keys(myObj[0]).sort( function(a, b) { if (isNaN(Number(a.charAt(0)))) return -1; if (isNaN(Number(b.charAt(0)))) return 1; return a.charAt(0) – b.charAt(0); … Read more
I think I found why. That is because there are some nan in dpd.txt. And nan is unable to compare: float(‘nan’) > 1 # False while float(‘nan’) < 1 # False So this totally breaks comparison. If you change your key compare function to: def _key(id_): import math result = -dpd[index_map[id_]], id_.lower() if math.isnan(result[0]): result … Read more