closest

Here's my HTML: <tr> <td class="show_r3c">click here</td> </tr> <tr class="r3c"> <td>This is what I'd like to display</td> </tr> And I have currently got this JQuery code, $(document).ready(function(){ $(".r3c").hide(); $('.show_r3c').click(function() { $(this).closest('.r3c').toggle(); return false; }); }); For some reason the closest() function isn't working, and it won't toggle the table row .r3c - I've tried using parent and other alternatives but can't get it working either :( Apologies for the silly question, and something similar to a problem I've had before. Just wondering, What's the

Ok, I have a projectile that has its position defined such that: a.x = initialX + initialDX * time; a.y = initialY + initialDY * time + 0.5 * gravtiy * time^2; I want to be able to predict which obstacles in my environment this projectile will collide with. I plan on checking the distance from A the closest point on the curve to the point P . I figure that at the point A the tangent to the curve will be perpendicular to the vector AP , and that the tangent to the curve at A will simply be the velocity V of the projectile at that point. AP dot V = 0 ap.x = initialX + initialDX * time - p.x; ap

I am trying to understand the closest pair algorithm. I understand about dividing the set in half. But I am having trouble understanding how to recursively compute the closest pair. I understand recursion, but do not understand how to compute the closest pair by recursion. If you have (1,2)(1,11)(7,8) how would recursion work on these? If you mean this algorithm you do the following: Sort points: (1,2) (1,11) (7,8) Build two subsets: (1,2) (1,11) and (7,8) Run the algorithm on (1,2) (1,11) and on (7,8) separately <= this is where the recursion comes. The result is dLmin = 9 and dRmin =

So lets say I have 10,000 points in A and 10,000 points in B and want to find out the closest point in A for every B point. Currently, I simply loop through every point in B and A to find which one is closest in distance. ie. B = [(.5, 1, 1), (1, .1, 1), (1, 1, .2)] A = [(1, 1, .3), (1, 0, 1), (.4, 1, 1)] C = {} for bp in B: closestDist = -1 for ap in A: dist = sum(((bp[0]-ap[0])**2, (bp[1]-ap[1])**2, (bp[2]-ap[2])**2)) if(closestDist > dist or closestDist == -1): C[bp] = ap closestDist = dist print C However, I am sure there is a faster way to do this... any ideas? I typically use a kd-tree

The Problem Imagine I am stood in an airport. Given a geographic coordinate pair, how can one efficiently determine which airport I am stood in? Inputs A coordinate pair (x,y) representing the location I am stood at. A set of coordinate pairs [(a1,b1), (a2,b2)...] where each coordinate pair represents one airport. Desired Output A coordinate pair (a,b) from the set of airport coordinate pairs representing the closest airport to the point (x,y) . Inefficient Solution Here is my inefficient attempt at solving this problem. It is clearly linear in the length of the set of airports. shortest

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I want to merge two dataframes on three columns: email, subject and timestamp. The timestamps between the dataframes differ and I therefore need to identify the closest matching timestamp for a group of email & subject. Below is a reproducible example using a function for closest match suggested for this question. import numpy as np import pandas as pd from pandas.io.parsers import StringIO def find_closest_date(timepoint, time_series, add_time_delta_column=True): # takes a pd.Timestamp() instance and a pd.Series with dates in it # calcs the

可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效，请关闭广告屏蔽插件后再试): 问题: I need help with a function that returns coordinate from a list of coordinates, that's closest to some point. For example: closest((9, 2), {(0, 0), (10, 0), (10, 10)}) returns (10, 0) . 回答1: class Point: def __init__(self,x,y): self.x = x self.y = y def closest(self,*points): return min(points,key=lambda x:abs(x-self)) def __sub__(self,other): return Point((self.x-other.x) , (self.y - other.y)) def __pow__(self,powTo): return Point(self.x**powTo,self.y**powTo) def __iter__(self): yield self.x yield self.y def __abs__(self): return sum(self*

描述 Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target. Given target value is a floating point. You are guaranteed to have only one unique value in the BST that is closest to the target. 是 样例 {1} 4.428571 1 . 我的第一想法就是先序遍历即可，但是这样做遗漏了最重要的一个条件这是一个二叉搜索树，如果目标比当前节点的值要大，那就没有必要遍历当前节点的左子树了。所以遍历的时候加上筛选条件即可。 /** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * @param root: the given BST