You can try this – $products = array( array (“08:10”, “10:30”, “13:15”), array (“GSÖ2B2U”, “VSH2B2U”, “FOR2B2U”), array (“GUS”, “GJG”, “GRL”) ); $new= array(); for ($i = 0; $i < count($products); $i++) { $new[] = array_column($products, $i); } foreach($new as $vals) { echo implode(‘ ‘, $vals) . ‘<br>’; } Output 08:10 GSÖ2B2U GUS<br> 10:30 VSH2B2U GJG<br> … Read more
You might consider learning how to implement breadth-first searches and depth-first searches in order to accomplish your objective. The following example shows how both of these search strategies can be easily handled in one function. A modular approach should make the code simple to change. #! /usr/bin/env python3 from collections import deque from operator import … Read more
Something like var items = new Array(‘0,364,2113,0′,’0,0,0,0’); var finalitems = []; for (var i in items) { vals = items[i].split(‘,’); finalitems.push(vals); } ? 2 solved Multidimensional Array in JavaScript?
Code should look something like this, if you were intending on using strings as your keys. $this_array= array( ‘string_name’ => ‘string’, ‘string_array’ => array( ‘string_key’ => ‘string_val’ ) ); Yes, you will access it by: $this_array[‘string_array’][‘string_key’]; 0 solved PHP Multidimensional Array Access
Such amazing complexity for a simple problem. You don’t need arrays to do this, not even multi-dimensional arrays. int main() { int count = 0; for (int i = 1; i <= 100; ++i) { if (is_prime(i)) { cout << i << ‘ ‘; ++count; if (count == 5) { cout << endl; count = … Read more
Maybe you could use one of the containers, e.g. vector. you could push_back all the moves into this vector until home is reached. The size of this vector will be the number of moves. You can setup a counter array[400] for counting the number of moves into same coordinates. For each move, – `push_back` the … Read more
// Three-dimensional array. int[, ,] array3D = new int[,,] { { { 1, 2, 3 }, { 4, 5, 6 } }, { { 7, 8, 9 }, { 10, 11, 12 } } }; Which was taken from here, which I found using Google and searching with this string, “multidimensional arrays in C#”. solved … Read more
From that, I want to get both the first and last non-null elements from the first and last rows each. 2D arrays are a pain in the ass, but they behave like one long array if you iterate them var w = array.GetLength(1); var h = array.GetLength(0); var a = array.Cast<int?>(); var ff = a.First(e … Read more
I hope I’ve understood correctly what you wanted to ask. First of all as it was already mentioned you can’t store integers in String array. Secondly, in Java two-dimensional array is actually the array of arrays. So when you declare int[][] nums = int[4][3] you create an int[] array nums that has four elemets and … Read more
How about: foreach(array(458752, 131586, 65793) as $x) { $j = intval($x / 65536); $x = $x % 65536; $i = intval($x / 256); $k = $x % 256; echo “x=$x, i=$i, j=$j, k=$k\n”; } output: x=0, i=0, j=7, k=0 x=514, i=2, j=2, k=2 x=257, i=1, j=1, k=1 2 solved PHP value=arraykey + arraykey + arraykey … Read more
You can use a TreeMap to keep the items sorted by key and just increment a counter for each occurrence. import java.util.Map; import java.util.Map.Entry; import java.util.TreeMap; public class Counter { public static void main(String[] args) { int[] arr = { 2, 5, 6, 1, 2, 1, 5, 3, 6, 1, 2, 1, 5 }; Map<Integer, … Read more
using combination of array_map, array_column and array_reverse : $myArrays = array( array(1, 2, 3, 4), array(5, 6, 7, 8), array(9, 10, 11, 12), array(13, 14, 15, 16) ); $index = 0; $myArrays = array_map(function ($v) use (&$index, $myArrays) { if (($index % 2) != 0) { echo implode(“\n”, array_reverse(array_column($myArrays, $index))) . “\n”; } else { … Read more
Have you tried looking at ListViewItem and then populating a list view? If not then you can easily use a list of arrays, or even populate it in a foreach loop like bellow? foreach( var v in Muvees ) {ListViewItem movieToAdd = new ListViewItem(); movieToAdd.Text = v.movieid; movieToAdd.SubItems.Add(v.rating); movieToAdd.SubItems.Add(v.movieName); listOfMovies.Items.Add( movieToAdd ).BackColor = Color.Green;} //add … Read more
As usual there are many ways to do the task. For example you can use the following solution #include <iostream> int main() { const size_t M = 5; const size_t N = 10; int a[M][N]; for ( size_t i = 0; i < M; i++ ) { for ( size_t j = 0; j < … Read more
I solved it, thank’s for all, I really appreciate your answers #define n 3 int *p = &vec[0]; int count1 = 0; for (i = 0; i < n; i++) { for (j = 0; j < n; j++) { if (*p <= vec[2]) { if (*p == mat[i][j]) { p++; count1++; i=0; j=0; } … Read more