Delta

KroneckerDelta in matlab

由 点点圈 提交于 2020-01-11 10:05:55
问题 This link shows that there is a kronecker delta function in matlab. However: >> help kroneckerDelta kroneckerDelta not found I am using R2011b, so maybe this wasn't programmed into the toolkit yet? EDIT:: It works in MuPad, just not in matlab... . 回答1: I don't see it in my R2012b so perhaps not. Unless you need the symbolic math, you could always write your own. Something as simple as function d = kronDel(j,k) if j == k d = 1; else d = 0; end 回答2: The Kronecker delta returns 1 if j==k... So

GitLab 使用git push 出现RPC failed; HTTP 500 curl 22 The requested URL returned error: 500

由 北城以北 提交于 2020-01-07 07:43:10
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> 今天前端人员提交代码时,出现了如下奇怪错误: $ git push origin Counting objects: 4, done. Delta compression using up to 24 threads. Compressing objects: 100% (4/4), done. Writing objects: 100% (4/4), 5.59 MiB | 16.73 MiB/s, done. Total 4 (delta 1), reused 0 (delta 0) error: RPC failed; HTTP 500 curl 22 The requested URL returned error: 500 Internal Server Error fatal: The remote end hung up unexpectedly fatal: The remote end hung up unexpectedly Everything up-to-date 一看,6MB不到的文件居然提交失败,之前20M的文件都提交过. 网上搜了下,大多是说使用 git config --global http.postBuffer 524288000 将本地http的缓存加大到500MB,但试了下

(Delta time) Getting 60 updates a second in java

由 旧时模样 提交于 2020-01-04 04:27:04
问题 I've seen this code several times. long lastTime = System.nanoTime(); final double ticks = 60D; double ns = 1000000000 / ticks; double delta = 0; The code above takes the System time and stores it to lastTime . The 60 ticks should equate to the number of times to update per second. while(running){ long now = System.nanoTime(); delta += (now - lastTime) / ns; lastTime = now; if(delta >= 1){ tick(); delta--; } It takes now and subtracts lastTime , then converts it to nanoseconds/60. Is there

how to set iOS 6/7 Deltas programmatically

由 天大地大妈咪最大 提交于 2020-01-01 08:04:57
问题 I was developing a UISplitView app by using Xcode 4.6 when I left iOS6 I had design: Now I migrate to new Xcode5 and now I have this design: UINavigationBar overlaps completelly my UISearchBar... Leo Natan told me about using a iOS 6/7 Deltas but since I'm creating and adding my UISplitViewControllers programmatically, this may doesn't work I need to set the iOS 6/7 programmatically but I don't know how, any help I'll appreciate 回答1: In iOS 7 there are now extended edges, and that's why

Getting the difference (delta) between two lists of dictionaries

由 安稳与你 提交于 2019-12-28 11:53:46
问题 I have the following Python data structures: data1 = [{'name': u'String 1'}, {'name': u'String 2'}] data2 = [{'name': u'String 1'}, {'name': u'String 2'}, {'name': u'String 3'}] I'm looking for the best way to get the delta between the two lists. Is there anything in Python that's as convenient as the JavaScript Underscore.js (_.difference) library? 回答1: Use itertools.filterfalse: import itertools r = list(itertools.filterfalse(lambda x: x in data1, data2)) + list(itertools.filterfalse(lambda

如何实现高抛平抛发射？从抛物线说起！Cocos Creator！

由 房东的猫 提交于 2019-12-27 08:18:07
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> > 游戏中经常会遇到抛物线轨迹问题，为此研究如何运用数学物理知识，实现高抛平抛效果。文章底部获取完整代码！ 效果预览： 先确认已知条件： 起点发射速度大小 V 重力加速 G 起始点与经过点 需要求出： 发射角度 a 对于抛物线运动，可以分两个方向去看。水平方向，匀速直线运动。垂直方向匀加速运动。所以可以得出以下式子： 在把 t 和 v_y 带入最后一个式子，化简整理后，可以得到一个关于 tan a 的一元二次方程。 再根据一元二次方程通解公式，可以解出角度的 tan 值。 接着使用反三角函数，求出角度的值。需要注意的是，反三角函数 arctan 的值域是 (-PI/2, PI/2) 。这个是第一、四象限的值，在二、三象限的时候要加 180 度(PI) 。也就是说，经过点在发射点左侧时，角度要加180度(PI) 。 角度大的正好是高抛效果，而角度小的是平抛效果。 接着看看代码吧。先通过两点的坐标确定水平位移 s 和垂直位移 h 。接着根据上面化简的一元二次方程求出 tan 值。最后求出角度。 const s = location.x - START_POS.x; const h = location.y - START_POS.y; // a*t^2 + b*t + c = 0 const a = G * s /

How to calculate delta latitude and longitude for MapView component in React-Native?

由 谁都会走 提交于 2019-12-21 09:24:08
问题 How can i calculate delta-latitude and delta-longitude values form latitude and longitude values for MapView component in React-Native. Thank You 回答1: If you have an array of coordinates and you want a region that will fit all these points, you can do something like this: const regionContainingPoints = points => { let minLat, maxLat, minLng, maxLng; // init first point (point => { minLat = point.latitude; maxLat = point.latitude; minLng = point.longitude; maxLng = point.longitude; })(points[0

反复横跳的瞄准线！从向量计算说起！基于射线检测的实现！Cocos Creator!

由 a 夏天 提交于 2019-12-16 19:30:48
【推荐】2019 Java 开发者跳槽指南.pdf(吐血整理) >>> > 最近有小伙伴问我瞄准线遇到各种形状该怎么处理？如何实现反复横跳的瞄准线？最近刚好在《Cocos Creator游戏开发实战》中看到物理系统有一个射线检测，于是，基于这个射线检测，写了一个反复横跳的瞄准线效果。一起往下看吧！文章底部获取完整项目！ 国际惯例，先上最终效果！ 在讲解之前我们需要一些向量的知识，简单的介绍一些吧！ 向量的加法， OA + AB = OB 向量的点乘，表示一个向量在另一个向量上的投影，是个标量，有正负之分。向量夹角小于 90度 为正数，等于 90度 为 零，大于 90度 为负数。 向量的叉乘，结果为向量，正好垂直于两个向量构成的平面(右手系)，也称为法向量。这里暂时没用到，顺便提一下。 接下来进入正题，已知入射向量(单位向量)，法向量(单位向量)，如何得出反射向量？ 我们将反射向量平移至入射向量起点，延长法向量与其相交，这个延长线的长度，刚好是 入射向量在法向量上的投影的相反数的两倍 。再根据投影和向量加法可以推出反射向量的计算公式。 清楚了么？不清楚也没关系，记得最后的公式就可以了，接下来进入 cocos creator 操作环节。 既然是物理系统中的碰撞检测，我们在编辑器里添加的是物理系统中的碰撞器，而不是引擎的碰撞器，不要选错了哦。 不动的刚体类型设为 static

How I get the new Field Value from a TClientDataset in Delphi?

由 萝らか妹 提交于 2019-12-13 02:09:42
问题 hi i have a problem with TClientDataset in Delphi. I want to get a Dataset with the changed Data. here is my code: procedure TForm2.btnUpdateClick(Sender: TObject); var I: Integer; counter : Integer; //for testing value : String; begin if not Self.DatasetArtikel.Active then begin ShowMessage('Nicht aktiv'); Exit; end; if Self.DatasetArtikel.ChangeCount = 0 then begin ShowMessage('Delta is empty'); Exit; end; counter := DatasetArtikel.ChangeCount; //DatasetArtikelUpdate.ClearFields; /

Change from baseline for repeated ids with missing baseline points

由 霸气de小男生 提交于 2019-12-12 18:29:45
问题 Change from baseline for repeated ids with missing baseline points A similar question has been asked and answered below: Change from baseline for repeated ids My question differs from the original question in that I have missing baseline values. I am including a small reproducible example below: df1 <- data.frame( probeID = c( rep("A", 19), rep("B",19), rep("C",19)), Subject_ID = c( rep( c( rep(1,5), rep(2,4), rep(3,5), rep(4,5)),3)), time = c(rep( c( c(1:5), c(2:5), rep( 1:5,2)),3))) df1
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