elements

How can I get jQuery to return the native DOM elements it encapsulates? When you find elements with jQuery, you can get them with the "get" function: var regularElement = $('#myElementId').get(0); Inside a ".each()" function, the "this" pointer refers to a "real" element: $('input.special').each(function() { var type = this.type; this.value = "exploding balloon"; // etc }) Using jQuery doesn't make Javascript "different." It is Javascript, and the DOM is still the DOM. $('myTag').get(0) returns the HTML element. jQuery uses the Sizzle Selector Engine *. You can use it on its own too. *

i have a question about how to do a common programming task in linq. lets say we have do different collections or arrays. What i would like to do is match elements between arrays and if there is a match then do something with that element. eg: string[] collection1 = new string[] { "1", "7", "4" }; string[] collection2 = new string[] { "6", "1", "7" }; foreach (string str1 in collection1) { foreach (string str2 in collection2) { if (str1 == str2) { // DO SOMETHING EXCITING/// } } } This can obviously be accomplished using the code above but what i am wondering if there is a fast and neat way

var elements = document.getElementsByClassName("item");console.log(elements)elements[0].classList.add("active"); 来源： https://www.cnblogs.com/guyuedashu/p/11533613.html

13:50:24 2019-09-14 继续把未看完的看完 排序算法 定理:任意$N$个不同元素组成的序列平均具有$N(N-1)/4$个逆序对 定理:任何仅以交换相邻两元素来排序的算法，其平均时间复杂度为$Ω(N^2)$ 这样子 冒泡排序 插入排序 的最坏情况都是 N^2 要使排序算法变高效 得使用间隔元素来交换 在一次排序中 交换多个逆序对的元素 希尔排序 定义增量序列$D_M>D_{M-1}>...>D_1=1$ $对每个D_k进行“D_k-间隔”排序（k=M,M-1,...1)$ $注意：“D_k-间隔"有序的序列，在执行”D_{k-1}间隔“排序后，仍然是“D_k-间隔"有序的$ 如果增量元素不互质，则小增量可能根本不起作用 比如 $N=N/2$ 最坏情况 时候 $T=\theta(N^{2})$ 那么就会有其他的增量序列: 　　Hibbard增量序列 $D_k=2^k-1$ --相邻元素互质 最坏情况$T=\theta(N^{3/2})$ 　　猜想 Hibbard增量的 平均复杂度 $T_{avg}=O(N^{5/4})$ 　　Sedgewick增量序列 $ \{1,5,19,41,109,.....\}$ 　　$9*4^i-9*2^i或4^i-3*2^i+1$猜想$T_{avg}=O(N^{7/6}) T_{worst}=O(N^{4/3})$ 冒泡排序 插入排序

I've got the following problem. I want to zoom-in a Scatterplot and then select all the displayed elements. It would be sufficient to somehow get the displayed area in the zoomed-in Scatterplot. From the range of this area i could determine which elements are displayed in the area and which are not. \edit: Found the Solution (Implementing AxisChangeListener Interface) import java.awt.Color; import java.awt.Dimension; import org.jfree.chart.ChartPanel; import org.jfree.chart.event.AxisChangeEvent; import org.jfree.chart.event.AxisChangeListener; import org.jfree.chart.plot.PlotOrientation;

I want to return odd numbers of an array yet Eclipse doesn't seem to accept my return array[i]; code. I think it requires returning a whole array since I set an array as a parameter to my method. As I said before, I need to pass an array and get a specific element of that array in return. Even if I make that array static, how do I return a single element? Edit : Alright then, here it is: public class newClass{ public static void main(String[] args) { int [] newArray= new int [4]; int [] array = {4,5,6,7}; newArray[0] = array[0]+array[1]+array[2]+array[3]; newArray[1] = array[0]*array[1]*array

I have this array: Array ( [0] => Array ( [0] => b [1] => d [2] => **c** [3] =>a [4] => ) [1] => Array ( [0] => **c** [1] => a [2] => d [3] => [4] => ) [2] => Array ( [0] => b [1] => d [2] => a [3] => [4] => ) [3] => Array ( [0] => **c** [1] => d [2] => a [3] =>b [4] => ) ) and need to delete (unset?) all elements where value is "c" so that one ends up with: Array ( [0] => Array ( **[0] => b [1] => d [2] => a [3] => [4] =>** ) [1] => Array ( **[0] => a [1] => d [2] => [3] =>** ) [2] => Array ( [0] => b [1] => d [2] => a [3] => [4] => ) [3] => Array ( **[0] => d [1] => a [2] =>b [3] =>** ) )