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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: After exporting a .dot file using scikit-learn's handy export_graphviz function. I am trying to render the dot file using Graphviz into a cell in my Jupyter Notebook: import graphviz from IPython.display import display with open("tree_1.dot") as f: dot_graph = f.read() display(graphviz.Source(dot_graph)) However the out[ ] is just an empty cell. I am using graphviz 0.5 (pip then conda installed), iPython 5.1 , and Python 3.5 The dot file looks correct here are the first characters: digraph Tree {\nnode [shape=box, style="filled", color=

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 am looking for a fast way to calculate the dot product of vectors with 3 or 4 components. I tried several things, but most examples online use an array of floats while our data structure is different. We use structs which are 16 byte aligned. Code excerpt (simplified): struct float3 { float x, y, z, w; // 4th component unused here } struct float4 { float x, y, z, w; } In previous tests (using SSE4 dot product intrinsic or FMA) I could not get a speedup, compared to using the following regular c++ code. float dot(const float3 a, const

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): 由 翻译 强力驱动 问题: I am running Python3.4 on Windows 7. I am trying to use the Python interface for graphviz. This is a script I intend to run: from graphviz import Digraph import pydotplus dot = Digraph ( comment = 'The Round Table' ) dot . node ( 'A' , 'King Arthur' ) dot . node ( 'B' , 'Sir Bedevere the Wise' ) dot . node ( 'L' , 'Sir Lancelot the Brave' ) dot . edges ([ 'AB' , 'AL' ]) dot . edge ( 'B' , 'L' , constraint = 'false' ) print ( dot . source ) dot . render ( 'test-output/round-table.gv' , view = True ) I get the following error at

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 am using html5 canvas element to draw a graph with dots denoting various points in here . I want to display different tool-tip on different points on mouse hover.the text to be displayed as tool-tip will be provided by the user. I tried but couldn't figure out how to add tool-tip to various points in the graph.The code I'm using for displaying dots is.. // Draw the dots c.fillStyle = '#333'; for (var i = 0; i What addition should I make in this code so that i am able to display user input as tool-tip? 回答1: You can display tooltips when

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): 由 翻译 强力驱动 问题: While in the process of debugging code written by a co-worker, I stumbled across the following that has me mystified: NSMutableArray * array = [ NSMutableArray array ]; NSUInteger arrayCount = array . count ; Why does this work? It also works for NSDictionary and other types, but nowhere in the documentation nor Cocoa headers can those @property definitions be found. Googling for "NSArray property" doesn't yield many useful results, so I'm reaching out to SO for what will surely be a very embarrassing question. 回答1: It works

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've seen this pseudo-random number generator for use in shaders referred to here and there around the web : float rand(vec2 co){ return fract(sin(dot(co.xy ,vec2(12.9898,78.233))) * 43758.5453); } It's variously called "canonical", or "a one-liner I found on the web somewhere". What's the origin of this function? Are the constant values as arbitrary as they seem or is there some art to their selection? Is there any discussion of the merits of this function? EDIT: The oldest reference to this function that I've come across is this archive

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 have two arrays A=[1,2,3] and B=[[1],[0],[1],[0]] . The question how to perform their tensor dot product in python. I am expecting to get: C=[[1,2,3], [0,0,0], [1,2,3], [0,0,0]] The function np.tensordot() returns an error concerning shapes of arrays. A little addition to this question. How to do such operation if matrix are totally different in shape, like: A=[[1,1,1,1], [1,1,1,1], [2,2,2,2], [3,3,3,3]] B=[2,1] C=[[[2,1],[2,1],[2,1],[2,1]], [[2,1],[2,1],[2,1],[2,1]], [[4,2],[4,2],[4,2],[4,2]], [[6,3],[6,3],[6,3],[6,3]]] 回答1: Try using

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 have hard time understanding the explanation from stroustrup for what difficulties one must have faced, if operator overloading for '.' was allowed. See this quote from Bjarne Stroustrup: Operator . (dot) could in principle be overloaded using the same technique as used for ->. However, doing so can lead to questions about whether an operation is meant for the object overloading . or an object referred to by . For example: class Y { public: void f(); // ... }; class X { // assume that you can overload . Y* p; Y& operator.() { return *p; }

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: For example we have collection {field: {subfield: 'name'}} {field: {subfield: 'phone'}} Can I find document without dot notation? Like this db.test.find({field: {subfield: /regex/}}) or maybe like this db.test.find({field: {$someOperator: {subfield: /regex/}}}) I just don't want to build dot notation like db.test.find({"field.subfield": /regex/}) 回答1: The problem is that: db.test.find({field: {$someOperator: {subfield: /regex/}}}) Is actually another way of querying in MongoDB which uses object euqality to search for subdocuments. So no, you

题目内容： 给定一组二维坐标，表示直角坐标系内的一个多边形的连续的顶点的坐标序列。计算能包围这个多边形的平行于坐标轴的最小矩形，输出它的左下角和右上角的坐标。 输入格式: 第一行是一个正整数n表示顶点的数量，第二行是n组整数，依次表示每个顶点坐标的x和y值。 输出格式： 四个整数，依次表示所计算的矩形的左下角的坐标的x、y值和右上角坐标的x、y值。输出最后带有回车换行。 输入样例： 5 1 1 1 4 3 7 4 4 4 1 输出样例： 1 1 4 7 1 #include <stdio.h> 2 3 typedef struct _dot { 4 int x; 5 int y; 6 } Dot; 7 8 int main( int argc, char const * argv[]) 9 { 10 Dot left_down, right_up; 11 left_down = right_up = (Dot){ 0 , 0 }; 12 13 Dot p; 14 int n; 15 scanf( " %d " , & n); 16 17 for ( int i = 0 ; i < n; i++ ) { 18 scanf( " %d %d " , &p.x, & p.y); 19 if ( left_down.x || left_down.y || right_up.x ||