Use count() twice; one time in default mode and one time in recursive mode. If the values match, the array is not multidimensional, as a multidimensional array would have a higher recursive count. if (count($array) == count($array, COUNT_RECURSIVE)) { echo ‘array is not multidimensional’; } else { echo ‘array is multidimensional’; } This option second … Read more
If you just need to test basic equality, this can be done with the basic JDK without modifying the input lists in the one line !Collections.disjoint(list1, list2); If you need to test a specific property, that’s harder. I would recommend, by default, list1.stream() .map(Object1::getProperty) .anyMatch( list2.stream() .map(Object2::getProperty) .collect(toSet()) ::contains) …which collects the distinct values in … Read more
You could use commons lang’s ArrayUtils. array = ArrayUtils.removeElement(array, element) commons.apache.org library:Javadocs
Using linq: YourList.OrderBy(x => rnd.Next()).Take(5)
AND (both classes) var list = document.getElementsByClassName(“class1 class2”); var list = document.querySelectorAll(“.class1.class2”); OR (at least one class) var list = document.querySelectorAll(“.class1,.class2”); XOR (one class but not the other) var list = document.querySelectorAll(“.class1:not(.class2),.class2:not(.class1)”); NAND (not both classes) var list = document.querySelectorAll(“:not(.class1),:not(.class2)”); NOR (not any of the two classes) var list = document.querySelectorAll(“:not(.class1):not(.class2)”);
My way is .fadein, .fadeout .fadein, .fadeout …… $(“#someElement”).fadeOut(100).fadeIn(100).fadeOut(100).fadeIn(100); function go1() { $(“#demo1″).fadeOut(100).fadeIn(100).fadeOut(100).fadeIn(100)} function go2() { $(‘#demo2’).delay(100).fadeOut().fadeIn(‘slow’) } #demo1, #demo2 { text-align: center; font-family: Helvetica; background: IndianRed; height: 50px; line-height: 50px; width: 150px; } <script src=”https://cdnjs.cloudflare.com/ajax/libs/jquery/3.3.1/jquery.min.js”></script> <button onclick=”go1()”>Click Me</button> <div id=’demo1′>My Element</div> <br> <button onclick=”go2()”>Click Me</button> (from comment) <div id=’demo2′>My Element</div>
You can use operator.itemgetter: from operator import itemgetter a = [-2, 1, 5, 3, 8, 5, 6] b = [1, 2, 5] print(itemgetter(*b)(a)) # Result: (1, 5, 5) Or you can use numpy: import numpy as np a = np.array([-2, 1, 5, 3, 8, 5, 6]) b = [1, 2, 5] print(list(a[b])) # Result: [1, … Read more
You can use an object to hold the results: const arr = [5, 5, 5, 2, 2, 2, 2, 2, 9, 4]; const counts = {}; for (const num of arr) { counts[num] = counts[num] ? counts[num] + 1 : 1; } console.log(counts[5], counts[2], counts[9], counts[4]); So, now your counts object can tell you what … Read more
Getting the element type the jQuery way: var elementType = $(this).prev().prop(‘nodeName’); doing the same without jQuery var elementType = this.previousSibling.nodeName; Checking for specific element type: var is_element_input = $(this).prev().is(“input”); //true or false
var eElement; // some E DOM instance var newFirstElement; //element which should be first in E eElement.insertBefore(newFirstElement, eElement.firstChild);