Create a list of years that appear on both the top_critic_years_more_than_four_games table and the top_user_years_more_than_four_games table. -- Select the year results that appear on both tables

时间: 2024-01-14 15:03:55 浏览: 65
RAR

sem.rar_The Count

Sure, here's the SQL query to create a list of years that appear on both the top_critic_years_more_than_four_games table and the top_user_years_more_than_four_games table: ``` SELECT year FROM top_critic_years_more_than_four_games WHERE year IN ( SELECT year FROM top_user_years_more_than_four_games ); ``` This query uses a subquery to select the years that appear in both tables, and then returns the list of years that meet this condition. Let me know if you have any other questions!
阅读全文

相关推荐

rar

attempt_9_ideal_test.rar_Demand-response_There And Now_Tricity_d

In regulation there are quieter periods and more active ones. Present times appear to demand a particularly active regulatory response. Energy—and elec- tricity in particular—has recently become a ...
zip

CenterMDIWnd_demo.zip_Frame MDI_child

This article shows you how, and we also talk about the ways of keeping child windows centered, even if the user moves or resizes the main window, or shows or hides the toolbar(s) and status bar.

list_num = [] list_num_count = 0 dict_num ={} #从文件读入,文件第一行为集合中元素的个数,以后每一行为一个元素 list_num_count = int(open('input.txt','r').readline()) for line_num, line in enumerate(open("input.txt",'r')): if line_num > 0: list_num += line.split() #将读到的元素加入的字典中 for item in list_num: if dict_num.has_key(item): dict_num[item] += 1 else: dict_num.setdefault(item,1) pass #找到出现次数最多的那个数,找到重数 dict_sort_by_top = {} top_value = 0 for valus in dict_num.itervalues(): if valus> top_value: top_value = valus pass #根据重数找到众数...这是因为考虑到可能有多个元素有相同多的重数 the_pop_num = 0 the_pop_num_count = 0 for keys,values in dict_num.iteritems(): if values == top_value: print 'the pop num is %s,and the appear num is %s' % (keys,values) the_pop_num = keys the_pop_num_count = values #输出到文件,第一行为从数,第二行为重数 write_line = '%s\n%s' %(the_pop_num, the_pop_num_count) open("output.txt",'w').write(write_line)

5. 遍历 dict_num 字典，找到出现次数最多的数字（众数），将其赋值给 the_pop_num 变量，并将其出现次数赋值给 the_pop_num_count 变量。由于可能有多个数字出现次数相同，因此这里还需要考虑到可能有多个众数的...

(Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

Here's a Python program ... Then, it identifies the factors that appear an odd number of times in the list, and multiplies them together to obtain n. Finally, the program prints out the value of n.

Java (Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.)

System.out.println("The smallest integer n such that m * n is a perfect square is: " + n); } } Explanation: 1. We first prompt the user to enter an integer m using the Scanner class. 2. We...

A random number of Rabbit images ranging from 1 to 10 are displayed for each operand and the user is expected to enter the values of the two operands and the result of adding the two operands, in the given text fields. When the user clicks on the button ‘Check!’, one of two things can happen: Case 1: all three input values are correct i) the text changes to ‘"Correct! Have another go?"’. ii) the number of Rabbit images displayed for each of the two operands changes. See Figure 2 for an example. iii) the three text fields are reset (i.e. they are made empty). 2/5 Case 2: at least one of the input values entered is incorrect i) the text changes to ‘Wrong! Try again!’. ii) the number of Rabbit images displayed does NOT change. iii) the text fields do NOT change.Implement SumItUp as a Java application. You application must satisfy ALL the specific requirements given below: a) The title of the top-level container must be ‘Welcome to SumItUp!’. b) The initial text should be ‘Enter two operands, result and click on 'Check!'’. See Figure 1. c) There should be no more than 4 Rabbit images per row. See Hint 1. d) The text fields should be wide enough to display at least TWO characters. e) The button ‘Check!’ must not resize when the GUI is resized. See Hint 2 and Figure 3. f) The ‘plus sign’ icon should appear vertically centered between the two sets of Rabbit images and must not resize when the GUI is resized. See Hint 2 and Figure 3. g) When first launched and whenever a correct answer is given, the number of displayed Rabbit images for each operand should change to any number between 1 and 10 (inclusive). See Hint 3 and Hint 4. Note: It is possible for the next number(s) to be the same as the current number(s). h) Nothing should happen if the user clicks the ‘Check!’ button while at least one of the text fields are empty, i.e. no errors should be thrown in this case. Note: You can assume that only a numeric value will be entered into the text fields.

promptLabel = new JLabel("Enter two operands, result and click on 'Check!'"); contentPane.add(promptLabel, c); // Add the first operand label and rabbit images c.gridx = 0; c.gridy = 1; c....

出一道与下面代码类似的题Write a function named frequency_of_elements that takes a list and returns the frequency of elements of this list in a dictionary, where the keys are elements, and values are the frequencies of corresponding elements. For example, if the input is ['a',2,'a','b',2], the function should return {'a': 2, 2: 2, 'b': 1}. The order of the keys should be the same as the order of when the elements first appear in the list.

This will calculate the frequency of elements in the given list and return a dictionary with the keys as elements and values as their corresponding frequencies. The order of the keys in the output ...

Timed out waiting for process (android) to appear on Nexus_10_API_19 [emulator-5554].

这个错误通常是由于模拟器启动过程中出现问题而导致的。以下是一些可能的解决方法： 1. 重启Android Studio和模拟器，然后再次尝试。 2. 确保你的电脑具有足够的内存和处理器速度以运行模拟器。...

用C++编写程序，实现以下问题2、题目ID Codes（POJ1146） Time Limit: 1000MS Memory Limit: 10000K 描述： It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入： Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出： Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

以下是使用C++编写的程序，可以解决该问题： c++ #include #include #include using namespace std; int main() ... while(cin >> str && str[0] !... if(next_permutation(str, str + strlen(str))) ...

Java(Algebra: perfect square) Write a program that prompts the user to enter an integer m and find the smallest integer n such that m * n is a perfect square. (Hint: Store all smallest factors of m into an array list. n is the product of the factors that appear an odd number of times in the array list. For example, consider m = 90, store the factors 2, 3, 3, and 5 in an array list. 2 and 5 appear an odd number of times in the array list. Thus, n is 10.) Here are some sample runs: Enter an integer m: 1500 The smallest number n for m * n to be a perfect square is 15 m * n is 22500

- After we have all the factors in the array list, we need to find the product of the factors that appear an odd number of times. We use a nested loop to count the occurrences of each factor in the ...

Write a SQL query to find the number of instances of each name in the following table, where the number of instances is greater than 1. Names that appear only once (such as “d”, “e” and “f”) should not be included in the result. NameList name a a b b b c c d e f gg gg

Sure! Here is the SQL query to find the number of instances of each name in the given table,... The result will show the names that appear more than once and the number of times they appear in the table.

用c++解决pipeline system consists of N transfer station, some of which are connected by pipelines. For each of M pipelines the numbers of stations A[i] and B[i], which are connected by this pipeline, and its profitability C[i] are known. A profitability of a pipeline is an amount of dollars, which will be daily yielded in taxes by transferring the gas through this pipeline. Each two stations are connected by not more than one pipeline. The system was built by Soviet engineers, who knew exactly, that the gas was transferred from Ukrainian gas fields to Siberia and not the reverse. That is why the pipelines are unidirectional, i.e. each pipeline allows gas transfer from the station number A[i] to the station number B[i] only. More over, if it is possible to transfer the gas from the station X to the station Y (perhaps, through some intermediate stations), then the reverse transfer from Y to X is impossible. It is known that the gas arrives to the starting station number S and should be dispatched to the buyers on the final station number F. The President ordered the Government to find a route (i.e. a linear sequence of stations which are connected by pipelines) to transfer the gas from the starting to the final station. A profitability of this route should be maximal. A profitability of a route is a total profitability of its pipelines. Unfortunately, the President did not consider that some pipelines ceased to exist long ago, and, as a result, the gas transfer between the starting and the final stations may appear to be impossible... Input The first line contains the integer numbers N (2 ≤ N ≤ 500) and M (0 ≤ M ≤ 124750). Each of the next M lines contains the integer numbers A[i], B[i] (1 ≤ A[i], B[i] ≤ N) and C[i] (1 ≤ C[i] ≤ 10000) for the corresponding pipeline. The last line contains the integer numbers S and F (1 ≤ S, F ≤ N; S ≠ F). Output If the desired route exists, you should output its profitability. Otherwise you should output "No solution".

int u = pq.top().second; pq.pop(); for (auto p : g[u]) { int v = p.first; int w = p.second; if (dist[v] [u] + w) { dist[v] = dist[u] + w; pq.push({dist[v], v}); } } } if (dist[f] == -INF...

7-3 Score Processing 分数 10 作者 翁恺 单位 浙江大学 Write a program to process students score data. The input of your program has lines of text, in one of the two formats: Student's name and student id, as <student id>, <name>, and Score for one student of one course, as <student id>, <course name>, <marks>. Example of the two formats are: 3190101234, Zhang San 3190101111, Linear Algebra, 89.5 Comma is used as the seperator of each field, and will never be in any of the fields. Notice that there are more than one word for name of the person and name of the course. To make your code easier, the score can be treated as double. The number of the students and the number of the courses are not known at the beginning. The number of lines are not known at the beginning either. The lines of different format appear in no order. One student may not get enrolled in every course. Your program should read every line in and print out a table of summary in .csv format. The first line of the output is the table head, consists fields like this: student id, name, <course name 1>, <course name 2>, ..., average where the course names are all the courses read, in alphabet order. There should be one space after each comma. Then each line of the output is data for one student, in the ascended order of their student id, with score of each course, like: 3190101234, Zhang San, 85.0, , 89.5, , , 87.3 For the course that hasn't been enrolled, leave a blank before the comma, and should not get included in the average. The average has one decimal place. There should be one space after each comma. And the last line of the output is a summary line for average score of every course, like: , , 76.2, 87.4, , , 76.8 All the number output, including the averages have one decimal place. Input Format As described in the text above. Output Format As described in the text above. The standard output is generated by a program compiled by gcc, that the round of the first decimal place is in the "gcc way". Sample Input 3180111435, Operating System, 34.5 3180111430, Linear Algebra, 80 3180111435, Jessie Zhao 3180111430, Zhiwen Yang 3180111430, Computer Architecture, 46.5 3180111434, Linear Algebra, 61.5 3180111434, Anna Teng Sample Output student id, name, Computer Architecture, Linear Algebra, Operating System, average 3180111430, Zhiwen Yang, 46.5, 80.0, , 63.2 3180111434, Anna Teng, , 61.5, , 61.5 3180111435, Jessie Zhao, , , 34.5, 34.5 , , 46.5, 70.8, 34.

很好，这是一道关于学生成绩数据处理的编程题。下面是我的解答： C++ #include #include #include #include #include using namespace std;...这个程序的主要思路是使用四个字典来记录学生姓名、学生成绩、...

To create the screen using Pygame, you will need to make use of the display.set_mode() function. Also, you will have to make use of the init() and the quit() methods to initialize and uninitialize everything at the start and the end of the code. The update() method is used to update any changes made to the screen. Besides, without using a loop,the screen will appear then immediately close. So you should make use of a game loop using the while loop before the player actually quit the game. Now, you have a screen to play your Snake Game, but how to quit the games? To quit the game, you should specify that your screen should exit when you hit that close button. To do that, Pygame provides an event called “QUIT”. Use these requests to fill this note. import pygame pygame.init() dis=pygame.display.set_mode((400,300)) pygame.display.update() pygame.display.set_caption('Snake game') game_over=False while not game_over: for event in pygame.event.get(): if event.type==pygame.QUIT: game_over=True pygame.quit() quit()

以下是已经填充完成的笔记： import pygame # 初始化 Pygame pygame.init() # 创建一个屏幕 dis = pygame.display.set_mode((400, 300)) pygame.display.set_caption('Snake game') # 游戏循环 ...quit()

解释如下代码：def detect_peaks(image): """ Takes an image and detect the peaks usingthe local maximum filter. Returns a boolean mask of the peaks (i.e. 1 when the pixel's value is the neighborhood maximum, 0 otherwise) """ # define an 8-connected neighborhood neighborhood = generate_binary_structure(2,2) #apply the local maximum filter; all pixel of maximal value #in their neighborhood are set to 1 local_max = maximum_filter(image, footprint=neighborhood)==image #local_max is a mask that contains the peaks we are #looking for, but also the background. #In order to isolate the peaks we must remove the background from the mask. #we create the mask of the background background = (image==0) #a little technicality: we must erode the background in order to #successfully subtract it form local_max, otherwise a line will #appear along the background border (artifact of the local maximum filter) eroded_background = binary_erosion(background, structure=neighborhood, border_value=1) #we obtain the final mask, containing only peaks, #by removing the background from the local_max mask (xor operation) detected_peaks = local_max ^ eroded_background return detected_peaks

这段代码是一个用于检测图像中峰值的函数。函数接受一个图像作为输入，并使用局部最大过滤器来检测峰值。这里使用了一个8-邻域的结构来定义峰值的邻域。局部最大过滤器会将所有像素中邻域内最大值的像素设为1，其余...

Referential Integrity • Ensures that a value that appears in one relation for a given set of attributes also appears for a certain set of attributes in another relation. • Example: If “Perryridge” is a branch name appearing in one of the tuples in the account relation, then there exists a tuple in the branch relation for branch “Perryridge”. • Primary and candidate keys and foreign keys can be specified as part of the SQL create table statement: • The primary key clause lists attributes that comprise the primary key. • The unique key clause lists attributes that comprise a candidate key. • The foreign key clause lists the attributes that comprise the foreign key and the name of the relation referenced by the foreign key. By default, a foreign key references the primary key attributes of the referenced table详细讲解以上

The foreign key clause lists the attributes that make up the foreign key and the name of the referenced table. By default, a foreign key references the primary key attributes of the referenced table. ...

We consider a contingency table for a pair of binary variables A and B. • We use ~A(~B) to indicate that A(B) is absent from a transaction. • Each entry fij in this table denotes a frequency count. For example, • f11 is the number of times A and B appear together in the same transaction. • f01 is the number of transactions that contain B but not A. B ~B A f11 f10 f1+ ~A f01 f00 f0+ f+1 f+0 N翻译

我们考虑一个二元变量A和B的列联表。 我们使用 ~A(~B) 表示某个交易中缺少 A(B)。 该表中的每一个条目 fij 表示一个频率计数。例如， - f11 表示 A 和 B 同时出现在一个交易中的次数。 - f01 表示包含 B 但不包含...
zip

Pytorch版代码幻灯片.zip

Jupyter-Notebook
zip

Jupyter_Chat甄嬛是利用甄嬛传剧本中所有关于甄嬛的台词和语句基于ChatGLM2进行LoRA微调得到的模仿甄.zip

Jupyter-Notebook

最新推荐

recommend-type

高清艺术文字图标资源，PNG和ICO格式免费下载

资源摘要信息:"艺术文字图标下载" 1. 资源类型及格式：本资源为艺术文字图标下载，包含的图标格式有PNG和ICO两种。PNG格式的图标具有高度的透明度以及较好的压缩率，常用于网络图形设计，支持24位颜色和8位alpha透明度，是一种无损压缩的位图图形格式。ICO格式则是Windows操作系统中常见的图标文件格式，可以包含不同大小和颜色深度的图标，通常用于桌面图标和程序的快捷方式。 2. 图标尺寸：所下载的图标尺寸为128x128像素，这是一个标准的图标尺寸，适用于多种应用场景，包括网页设计、软件界面、图标库等。在设计上，128x128像素提供了足够的面积来展现细节，而大尺寸图标也可以方便地进行缩放以适应不同分辨率的显示需求。 3. 下载数量及内容：资源提供了12张艺术文字图标。这些图标可以用于个人项目或商业用途，具体使用时需查看艺术家或资源提供方的版权声明及使用许可。在设计上，艺术文字图标融合了艺术与文字的元素，通常具有一定的艺术风格和创意，使得图标不仅具备标识功能，同时也具有观赏价值。 4. 设计风格与用途：艺术文字图标往往具有独特的设计风格，可能包括手绘风格、抽象艺术风格、像素艺术风格等。它们可以用于各种项目中，如网站设计、移动应用、图标集、软件界面等。艺术文字图标集可以在视觉上增加内容的吸引力，为用户提供直观且富有美感的视觉体验。 5. 使用指南与版权说明：在使用这些艺术文字图标时，用户应当仔细阅读下载页面上的版权声明及使用指南，了解是否允许修改图标、是否可以用于商业用途等。一些资源提供方可能要求在使用图标时保留作者信息或者在产品中适当展示图标来源。未经允许使用图标可能会引起版权纠纷。 6. 压缩文件的提取：下载得到的资源为压缩文件，文件名称为“8068”，意味着用户需要将文件解压缩以获取里面的PNG和ICO格式图标。解压缩工具常见的有WinRAR、7-Zip等，用户可以使用这些工具来提取文件。 7. 具体应用场景：艺术文字图标下载可以广泛应用于网页设计中的按钮、信息图、广告、社交媒体图像等；在应用程序中可以作为启动图标、功能按钮、导航元素等。由于它们的尺寸较大且具有艺术性，因此也可以用于打印材料如宣传册、海报、名片等。 通过上述对艺术文字图标下载资源的详细解析，我们可以看到，这些图标不仅是简单的图形文件，它们集合了设计美学和实用功能，能够为各种数字产品和视觉传达带来创新和美感。在使用这些资源时，应遵循相应的版权规则，确保合法使用，同时也要注重在设计时根据项目需求对图标进行适当调整和优化，以获得最佳的视觉效果。
recommend-type

管理建模和仿真的文件

管理Boualem Benatallah引用此版本：布阿利姆·贝纳塔拉。管理建模和仿真。约瑟夫-傅立叶大学-格勒诺布尔第一大学，1996年。法语。NNT：电话：00345357HAL ID：电话：00345357https://theses.hal.science/tel-003453572008年12月9日提交HAL是一个多学科的开放存取档案馆，用于存放和传播科学研究论文，无论它们是否被公开。论文可以来自法国或国外的教学和研究机构，也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaire
recommend-type

DMA技术：绕过CPU实现高效数据传输

![DMA技术：绕过CPU实现高效数据传输](https://res.cloudinary.com/witspry/image/upload/witscad/public/content/courses/computer-architecture/dmac-functional-components.png) # 1. DMA技术概述 DMA（直接内存访问）技术是现代计算机架构中的关键组成部分，它允许外围设备直接与系统内存交换数据，而无需CPU的干预。这种方法极大地减少了CPU处理I/O操作的负担，并提高了数据传输效率。在本章中，我们将对DMA技术的基本概念、历史发展和应用领域进行概述，为读
recommend-type

SGM8701电压比较器如何在低功耗电池供电系统中实现高效率运作？

SGM8701电压比较器的超低功耗特性是其在电池供电系统中高效率运作的关键。其在1.4V电压下工作电流仅为300nA，这种低功耗水平极大地延长了电池的使用寿命，尤其适用于功耗敏感的物联网(IoT)设备，如远程传感器节点。SGM8701的低功耗设计得益于其优化的CMOS输入和内部电路，即使在电池供电的设备中也能提供持续且稳定的性能。 参考资源链接：[SGM8701：1.4V低功耗单通道电压比较器](https://wenku.csdn.net/doc/2g6edb5gf4?spm=1055.2569.3001.10343) 除此之外，SGM8701的宽电源电压范围支持从1.4V至5.5V的电
recommend-type

mui框架HTML5应用界面组件使用示例教程

资源摘要信息:"HTML5基本类模块V1.46例子（mui角标+按钮+信息框+进度条+表单演示）-易语言" 描述中的知识点: 1. HTML5基础知识：HTML5是最新一代的超文本标记语言，用于构建和呈现网页内容。它提供了丰富的功能，如本地存储、多媒体内容嵌入、离线应用支持等。HTML5的引入使得网页应用可以更加丰富和交互性更强。 2. mui框架：mui是一个轻量级的前端框架，主要用于开发移动应用。它基于HTML5和JavaScript构建，能够帮助开发者快速创建跨平台的移动应用界面。mui框架的使用可以使得开发者不必深入了解底层技术细节，就能够创建出美观且功能丰富的移动应用。 3. 角标+按钮+信息框+进度条+表单元素：在mui框架中，角标通常用于指示未读消息的数量，按钮用于触发事件或进行用户交互，信息框用于显示临时消息或确认对话框，进度条展示任务的完成进度，而表单则是收集用户输入信息的界面组件。这些都是Web开发中常见的界面元素，mui框架提供了一套易于使用和自定义的组件实现这些功能。 4. 易语言的使用：易语言是一种简化的编程语言，主要面向中文用户。它以中文作为编程语言关键字，降低了编程的学习门槛，使得编程更加亲民化。在这个例子中，易语言被用来演示mui框架的封装和使用，虽然描述中提到“如何封装成APP，那等我以后再说”，暗示了mui框架与移动应用打包的进一步知识，但当前内容聚焦于展示HTML5和mui框架结合使用来创建网页应用界面的实例。 5. 界面美化源码：文件的标签提到了“界面美化源码”，这说明文件中包含了用于美化界面的代码示例。这可能包括CSS样式表、JavaScript脚本或HTML结构的改进，目的是为了提高用户界面的吸引力和用户体验。 压缩包子文件的文件名称列表中的知识点: 1. mui表单演示.e：这部分文件可能包含了mui框架中的表单组件演示代码，展示了如何使用mui框架来构建和美化表单。表单通常包含输入字段、标签、按钮和其他控件，用于收集和提交用户数据。 2. mui角标+按钮+信息框演示.e：这部分文件可能展示了mui框架中如何实现角标、按钮和信息框组件，并进行相应的事件处理和样式定制。这些组件对于提升用户交互体验至关重要。 3. mui进度条演示.e：文件名表明该文件演示了mui框架中的进度条组件，该组件用于向用户展示操作或数据处理的进度。进度条组件可以增强用户对系统性能和响应时间的感知。 4. html5标准类1.46.ec：这个文件可能是核心的HTML5类库文件，其中包含了HTML5的基础结构和类定义。"1.46"表明这是特定版本的类库文件，而".ec"文件扩展名可能是易语言项目中的特定格式。 总结来说，这个资源摘要信息涉及到HTML5的前端开发、mui框架的界面元素实现和美化、易语言在Web开发中的应用，以及如何利用这些技术创建功能丰富的移动应用界面。通过这些文件和描述，可以学习到如何利用mui框架实现常见的Web界面元素，并通过易语言将这些界面元素封装成移动应用。
recommend-type

"互动学习：行动中的多样性与论文攻读经历"

多样性她- 事实上SCI NCES你的时间表ECOLEDO C Tora SC和NCESPOUR l’Ingén学习互动，互动学习以行动为中心的强化学习学会互动，互动学习，以行动为中心的强化学习计算机科学博士论文于2021年9月28日在Villeneuve d'Asq公开支持马修·瑟林评审团主席法布里斯·勒菲弗尔阿维尼翁大学教授论文指导奥利维尔·皮耶昆谷歌研究教授：智囊团论文联合主任菲利普·普雷教授，大学。里尔/CRISTAL/因里亚报告员奥利维耶·西格德索邦大学报告员卢多维奇·德诺耶教授，Facebook /索邦大学审查员越南圣迈IMT Atlantic高级讲师邀请弗洛里安·斯特鲁布博士，Deepmind对于那些及时看到自己错误的人...3谢谢你首先，我要感谢我的两位博士生导师Olivier和Philippe。奥利维尔，"站在巨人的肩膀上"这句话对你来说完全有意义了。从科学上讲，你知道在这篇论文的（许多）错误中，你是我可以依
recommend-type

【数据传输高速公路】：总线系统的深度解析

![计算机组成原理知识点](https://img-blog.csdnimg.cn/6ed523f010d14cbba57c19025a1d45f9.png) # 1. 总线系统概述 在计算机系统和电子设备中，总线系统扮演着至关重要的角色。它是一个共享的传输介质，用于在组件之间传递数据和控制信号。无论是单个芯片内部的互连，还是不同设备之间的通信，总线技术都是不可或缺的。为了实现高效率和良好的性能，总线系统必须具备高速传输能力、高效的数据处理能力和较高的可靠性。 本章节旨在为读者提供总线系统的初步了解，包括其定义、历史发展、以及它在现代计算机系统中的应用。我们将讨论总线系统的功能和它在不同层
recommend-type

如何结合PID算法调整PWM信号来优化电机速度控制？请提供实现这一过程的步骤和代码示例。

为了优化电机的速度控制，结合PID算法调整PWM信号是一种常见且有效的方法。这里提供一个具体的实现步骤和代码示例，帮助你深入理解这一过程。 参考资源链接：[Motor Control using PWM and PID](https://wenku.csdn.net/doc/6412b78bbe7fbd1778d4aacb?spm=1055.2569.3001.10343) 首先，确保你已经有了一个可以输出PWM波形的硬件接口，例如Arduino或者其他微控制器。接下来，你需要定义PID控制器的三个主要参数：比例（P）、积分（I）、微分（D），这些参数决定了控制器对误差的响应速度和方式。
recommend-type

Vue.js开发利器：chrome-vue-devtools插件解析

资源摘要信息:"Vue.js Devtools 是一款专为Vue.js开发设计的浏览器扩展插件，可用于Chrome浏览器。这个插件是开发Vue.js应用时不可或缺的工具之一，它极大地提高了开发者的调试效率。Vue.js Devtools能够帮助开发者在Chrome浏览器中直接查看和操作Vue.js应用的组件树，观察组件的数据变化，以及检查路由和Vuex的状态。通过这种直观的调试方式，开发者可以更加深入地理解应用的行为，快速定位和解决问题。这个工具支持Vue.js的版本2和版本3，并且随着Vue.js的更新不断迭代，以适应新的特性和调试需求。" 知识点： 1. Vue.js Devtools定义： - Vue.js Devtools是用于调试Vue.js应用程序的浏览器扩展工具。 - 它是一个Chrome插件，但也存在其他浏览器（如Firefox）的版本。 2. 功能特性： - 组件树结构展示：Vue.js Devtools可以显示应用中所有的Vue组件，并以树状图的形式展现它们的层级和关系。 - 组件数据监控：开发者可以实时查看组件内的数据状态，包括prop、data、computed等。 - 事件监听：可以查看和触发组件上的事件。 - 路由调试：能够查看当前的路由状态，以及路由变化的历史记录。 - Vuex状态管理：如果使用Vuex进行状态管理，Vue.js Devtools可以帮助调试状态树，查看和修改state，以及跟踪mutations和actions。 3. 使用场景： - 在开发阶段进行调试，帮助开发者了解应用内部工作原理。 - 生产环境问题排查，通过复现问题时使用Vue.js Devtools快速定位问题所在。 - 教学和学习，作为学习Vue.js和理解组件驱动开发的辅助工具。 4. 安装和更新： - 通过Chrome网上应用店搜索并安装Vue.js Devtools。 - 插件会定期更新，以保持与Vue.js的兼容性和最新的特性支持。 5. 兼容性： - 通常支持主流的Vue.js版本，包括Vue.js 2.x和3.x。 - 适用于大多数现代浏览器。 6. 开发背景： - Vue.js Devtools由社区开发和维护，它不是Vue.js官方产品，但得到了广大Vue.js社区的认可和支持。 - 随着Vue.js版本的迭代，社区会不断优化和增加Vue.js Devtools的新功能，以满足开发者日益增长的调试需求。 7. 技术实现： - Vue.js Devtools利用浏览器提供的调试接口和Vue.js自身的调试能力，构建了一个用户友好的界面。 - 它通过Vue.js实例的$vm属性访问组件实例，从而读取和修改组件的数据和方法。 8. 社区支持： - 在使用过程中遇到问题可以参考社区论坛、GitHub仓库中的issue或文档。 - 社区活跃，经常会有新的开发者贡献代码或提供问题解决方案。 通过使用Vue.js Devtools，开发者可以更加高效地进行问题定位、性能优化和代码调试，是提升Vue.js应用开发和维护效率的强力工具。
recommend-type

关系数据表示学习

关系数据卢多维奇·多斯桑托斯引用此版本：卢多维奇·多斯桑托斯。关系数据的表示学习机器学习[cs.LG]。皮埃尔和玛丽·居里大学-巴黎第六大学，2017年。英语。NNT：2017PA066480。电话：01803188HAL ID：电话：01803188https://theses.hal.science/tel-01803188提交日期：2018年HAL是一个多学科的开放存取档案馆，用于存放和传播科学研究论文，无论它们是否被公开。论文可以来自法国或国外的教学和研究机构，也可以来自公共或私人研究中心。L’archive ouverte pluridisciplinaireUNIVERSITY PIERRE和 MARIE CURIE计算机科学、电信和电子学博士学院（巴黎）巴黎6号计算机科学实验室D八角形T HESIS关系数据表示学习作者：Ludovic DOS SAntos主管：Patrick GALLINARI联合主管：本杰明·P·伊沃瓦斯基为满足计算机科学博士学位的要求而提交的论文评审团成员：先生蒂埃里·A·退休记者先生尤尼斯·B·恩