CN118365492A - Knowledge point missing and gap-filling analysis method and system based on big data - Google Patents
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Abstract
Description
技术领域Technical Field
本发明属于大数据及计算机技术领域,具体涉及一种基于大数据的知识点查缺补漏分析方法及系统。The present invention belongs to the field of big data and computer technology, and specifically relates to a knowledge point deficiency checking and gap-filling analysis method and system based on big data.
背景技术Background technique
目前的大数据在线教育软件平台操作过程如下:首先,将学生的试卷扫描到在线平台中;然后,由该平台对试卷中客观题进行批分,主观题则由老师批分。最后,由平台将学生试卷中各小题的得分以及试卷总分统计出来。The current big data online education software platform operates as follows: First, the student's test paper is scanned into the online platform; then, the platform grades the objective questions in the test paper, and the subjective questions are graded by the teacher. Finally, the platform calculates the scores of each question in the student's test paper and the total score of the test paper.
该平台能初步对数据做以下分析:1、以学生试卷总分看,根据所有学生试卷总分,获知某一学生的位次;2、以学生试卷的某一小题得分看,根据所有学生某小题得分,获知某一学生该小题的位次;3、获知试卷的总分最高分和试卷中各小题的最高分;4、试卷中各小题的平均分等。The platform can make the following preliminary analyses on the data: 1. Based on the total score of the student's test paper, the ranking of a certain student can be known according to the total score of all students' test papers; 2. Based on the score of a certain question in the student's test paper, the ranking of a certain student in that question can be known according to the score of all students' questions; 3. The highest total score of the test paper and the highest score of each question in the test paper; 4. The average score of each question in the test paper, etc.
学生可以使用该平台,通过对比本人在各小题的得分、全体学生的各小题均分和全体学生的各小题最高分,初步了解到自己在哪些科目、哪些题目上存在着知识短板。Students can use this platform to compare their own scores on each question, the average score of all students on each question, and the highest score of all students on each question, to gain a preliminary understanding of which subjects and questions they have knowledge gaps in.
对于单个学生而言,有一些短板是容易提升的,还有一些短板是较难提升的。For individual students, some shortcomings are easy to improve, while others are more difficult to improve.
学生依靠上述方式了解的知识短板是比较原始的,并不能精确地指导学生找到容易提升的知识短板,不利于学生快速对知识点补短。The knowledge gaps that students understand through the above method are relatively primitive, and cannot accurately guide students to find knowledge gaps that are easy to improve, which is not conducive to students quickly making up for their knowledge gaps.
发明内容Summary of the invention
本发明的基于大数据的知识点查缺补漏分析方法及系统,其能够辅助老师或者学生快速地找出易提升的知识点,从而能辅助老师、学生进行快速补短。The knowledge point deficiency checking and gap-filling analysis method and system based on big data of the present invention can assist teachers or students to quickly find knowledge points that are easy to improve, thereby assisting teachers and students to quickly fill in the gaps.
为了实现上述目的,本发明的基于大数据的知识点查缺补漏分析方法,包括以下步骤:In order to achieve the above-mentioned purpose, the knowledge point deficiency checking and gap filling analysis method based on big data of the present invention comprises the following steps:
S1:获取目标区域全体学生某次考试某科目试卷的总得分;获取全体学生该科目试卷中各小题的得分;根据某位学生本次考试该科目试卷的总得分,计算该学生本次考试该科目试卷中各小题的预期分;所述目标区域全体学生是指采用相同试卷进行考试的所有学生;S1: Obtain the total score of a certain subject test paper of all students in the target area in a certain exam; obtain the score of each question in the test paper of the subject of all students; calculate the expected score of each question in the test paper of the subject of a certain student according to the total score of the test paper of the subject of this exam; the said all students in the target area refer to all students who take the exam with the same test paper;
S2:计算出步骤S1中该学生本次考试该科目试卷中各小题得分的预期偏差值;S2: Calculate the expected deviation value of the score of each question in the test paper of the subject of the student in step S1;
S3:计算本次考试之前的Z次考试中,该学生该科目Z次考试得分位次的加权平均值;S3: Calculate the weighted average of the student's Z test scores for the subject in the Z tests before the current test;
S4:计算步骤S1中该学生该科目试卷中各小题的优先补短系数;S4: Calculate the priority compensation coefficient of each question in the subject test paper of the student in step S1;
S5:将优先补短系数大的小题反馈给步骤S1中的该学生及其对应教师。S5: Feedback the small questions with large short-completion coefficients to the student and the corresponding teacher in step S1.
进一步地,步骤S1中,具体包括以下步骤:Furthermore, step S1 specifically includes the following steps:
S1.1: 获取目标区域全体学生某次考试某科目试卷的总得分;全体学生某次考试某科目试卷中各小题的得分所组成的矩阵为:S1.1: Get the total score of a certain subject in a certain exam for all students in the target area; the matrix composed of the scores of each question in a certain subject in a certain exam for all students is:
; ;
矩阵A中每行代表着对应学生本次考试该科目试卷中各小题的得分;共m个学生,本次考试该科目试卷中共n个小题;A为全体学生本次考试该科目试卷中各小题的得分矩阵;Each row in the matrix A represents the score of each question in the test paper of this subject for the corresponding student in this exam; there are m students in total, and there are n questions in the test paper of this subject in this exam; A is the score matrix of each question in the test paper of this subject for all students in this exam;
S1.2: 将全体学生本次考试该科目试卷中各小题的得分从大到小进行排序;排序后的矩阵为:S1.2: Sort the scores of all students in this subject in this exam from large to small; the matrix after sorting is:
; ;
S1.3:根据该学生本次考试该科目试卷总得分,计算该学生本次考试该科目试卷中各小题的预期分;S1.3: Calculate the expected score of each question in the subject paper of the student in this exam based on the student's total score of the subject paper in this exam;
设:set up:
; ;
矩阵S中每行的数值S k ,k=1、2、……m;设该学生本次考试该科目试卷总得分为E 总,设该学生本次考试该科目试卷期望总得分为E k ,设E k 等于E 总;如果矩阵S中存在与E 总相同的同位次预期得分S k ,则,该学生本次考试该科目试卷中第i小题的预期得分的公式为:The value of each row in the matrix S is Sk , k = 1, 2, ... m; let the total score of the student's test paper for this subject be Etotal , let the expected total score of the student's test paper for this subject be Ek , let Ek equal Etotal ; if there is an expected score Sk with the same rank as Etotal in the matrix S , then the formula for the expected score of the i- th question in the test paper of this subject for this student is:
(1); (1);
设E k 为该学生本次考试该科目试卷的期望总得分; 为该学生本次考试该科目试卷中第i小题的预期得分;为该学生的同位次预期得分;Let E k be the expected total score of the student in this subject paper of this exam; is the expected score of the student for question i in the subject test paper of this exam; The expected score for the student in the same position;
如果该学生本次考试该科目试卷总得分E 总,矩阵S中不存在与之相同的同位次预期得分,则,该学生本次考试该科目试卷中第i小题的预期得分的公式为:If the total score of the student's test paper for this subject is Etotal , and there is no same expected score of the same rank in the matrix S , then the formula for the expected score of the i- th question in the test paper of this subject for this student is:
(2); (2);
其中,S t+1 为小于E 总,且最接近E 总的第t+1行的期望总得分;S t 为大于E 总,且最接近E 总的第t行的期望总得分;u ti 为矩阵U中第t行第i小题的得分;u (t+1)i 为矩阵U中第t+1行第i小题的得分;E k 为该学生本次考试该科目试卷的期望总得分,为该学生本次考试该科目试卷中第i小题的预期得分;为该学生的同位次预期得分。Among them, S t+1 is the expected total score of the t+1th row that is less than E total and closest to E total ; S t is the expected total score of the tth row that is greater than E total and closest to E total ; u ti is the score of the ith question in the tth row of the matrix U ; u (t+1)i is the score of the ith question in the t+1th row of the matrix U ; E k is the expected total score of the student in this exam for this subject. is the expected score of the student for question i in the subject test paper of this exam; The expected score for the student in the same ranking.
进一步地,步骤S2中,Further, in step S2,
; ;
其中,d i 为该学生本次考试该科目试卷中,第i小题实际得分与第i小题的预期得分之间的偏离值;为该学生本次考试该科目试卷中,第i小题的预期得分;a i 为该学生本次考试该科目试卷中,第i小题的实际得分。Among them, d i is the deviation between the actual score of question i and the expected score of question i in the test paper of this subject of the student in this exam; is the expected score of the i- th question in the test paper of this subject for this student in this examination; a i is the actual score of the i- th question in the test paper of this subject for this student in this examination.
进一步地,步骤S3中,设定本次考试之前,该学生该科目第j次考试得分在全体学生中的位次为p j, ;该学生第j次考试该科目得分位次的加权系数为f j ;Furthermore, in step S3, the ranking of the student's score in the jth test of the subject among all students before the current test is set to p j , ; The weighted coefficient of the student's score ranking in the subject in the jth examination is f j ;
则,本次考试之前的z次考试中,该学生该科目考试得分的位次加权平均值w为:Then, the weighted average w of the student's scores in this subject in the z exams before this exam is:
。 .
进一步地,设该学生本次考试该科目试卷中第i小题的实际得分在全体学生中的位次为h i ;Furthermore, let the actual score of the student on the i - th question in the test paper of this subject be ranked among all students as h i ;
优先补短系数公式为:The priority short-filling coefficient formula is:
(3); (3);
式中, 为该学生本次考试该科目试卷中第i小题的优先补短系数。In the formula , It is the priority short-compensation coefficient for the i- th question in the test paper of this subject for this student in this exam.
基于大数据的知识点查缺补漏分析系统,包括:The knowledge point checking and gap-filling analysis system based on big data includes:
题库,用于存储各个学科、各种知识点的题目,题目信息包括文字和图片,并通过自然语言处理技术抽取出题目的关键词;The question bank is used to store questions of various subjects and knowledge points. The question information includes text and pictures, and the keywords of the questions are extracted through natural language processing technology;
优先补短系数计算模块,能够根据学生的本次考试该科目试卷中各小题得分、本次考试该科目试卷中各小题的位次、以及本次考试该科目试卷之前,Z次该科目试卷中的位次加权平均值,计算出本次考试该科目试卷中各小题的优先补短系数;The priority-compensation coefficient calculation module can calculate the priority-compensation coefficient of each question in the test paper of this subject according to the student's score of each question in the test paper of this subject, the ranking of each question in the test paper of this subject, and the weighted average of the ranking in the Z test papers of this subject before the test paper of this subject;
习题识别模块,能通过文字和图片识别本次考试该科目试卷中各小题的学科、知识点;能通过自然语言处理技术抽取出关键词、通过反向图片搜索引擎在题库中找到相似的图片,从而关联出各小题对应的学科和知识点;The question recognition module can identify the subjects and knowledge points of each question in the test paper of this subject through text and pictures; it can extract keywords through natural language processing technology and find similar pictures in the question bank through reverse image search engine, so as to associate the subjects and knowledge points corresponding to each question;
分析模块,分别与优先补短系数计算模块和习题识别模块连接;分别获得本次考试该科目试卷中各小题的学科、知识点,以及能获得本次考试该科目试卷中各小题的优先补短系数,能够判断并分析出需要补短的学科和知识点;The analysis module is connected with the priority compensation coefficient calculation module and the question identification module respectively; the subject and knowledge point of each question in the test paper of the subject of this exam are obtained respectively, and the priority compensation coefficient of each question in the test paper of the subject of this exam is obtained, and the subject and knowledge point that need to be compensated are judged and analyzed;
相似习题推荐模块,与分析模块连接,能够接收分析模块的指令,能够从题库中找出对应知识点的相似习题;The similar exercise recommendation module is connected to the analysis module, can receive instructions from the analysis module, and can find similar exercises corresponding to the knowledge points from the question bank;
显示模块,能够显示相似习题推荐模块找出的对应知识点的习题。The display module can display the exercises of corresponding knowledge points found by the similar exercise recommendation module.
有益效果:本方法通过大数据分析学生对于同一份试卷中各小题解答分数的差异,依据学生的总得分以及各小题的位次,计算出优先补短系数,依据优先补短系数找出学生的知识缺陷并进行针对性弥补,有助于快速解决或改善这些知识缺陷。通过找出最容易提升的知识缺陷并加以改进,避免了面面俱到的覆盖性学习,从而提高了学习的针对性和有效性。可以有效减轻学生的过重学业负担。Beneficial effects: This method uses big data to analyze the differences in students' answers to each question in the same test paper, and calculates the priority compensation coefficient based on the student's total score and the ranking of each question. Based on the priority compensation coefficient, the student's knowledge deficiencies are found and targeted compensation is made, which helps to quickly solve or improve these knowledge deficiencies. By finding the knowledge deficiencies that are easiest to improve and improving them, comprehensive and comprehensive learning is avoided, thereby improving the pertinence and effectiveness of learning. It can effectively reduce the excessive academic burden on students.
附图说明BRIEF DESCRIPTION OF THE DRAWINGS
图1是本发明基于大数据的知识点查缺补漏分析方法的流程图;FIG1 is a flow chart of a method for analyzing knowledge points based on big data;
图2是本发明基于大数据的知识点查缺补漏分析的模块图。FIG. 2 is a module diagram of the knowledge point deficiency and gap-filling analysis based on big data of the present invention.
具体实施方式Detailed ways
为使本发明实施例的目的、技术方案和优点更加清楚,下面将结合附图对本发明的技术方案进行清楚、完整地描述,显然,所描述的实施例是本发明一部分实施例,而不是全部的实施例。基于本发明中的实施例,本领域普通技术人员在没有做出创造性劳动前提下所获得的所有其他实施例,都属于本发明保护的范围。In order to make the purpose, technical solution and advantages of the embodiments of the present invention clearer, the technical solution of the present invention will be clearly and completely described below in conjunction with the accompanying drawings. Obviously, the described embodiments are part of the embodiments of the present invention, not all of the embodiments. Based on the embodiments of the present invention, all other embodiments obtained by ordinary technicians in this field without creative work are within the scope of protection of the present invention.
实施例1:如图1所示,基于大数据的知识点查缺补漏分析方法,包括以下步骤:Embodiment 1: As shown in FIG1 , a knowledge point deficiency checking and gap filling analysis method based on big data includes the following steps:
S1:获取目标区域全体学生某次考试某科目试卷的总得分;获取全体学生该科目试卷中各小题的得分;根据某位学生本次考试该科目试卷的总得分,计算该学生本次考试该科目试卷中各小题的预期分;目标区域全体学生是指采用相同试卷进行考试的所有学生;目标区域如:某一区或者县统一考试的同一张试卷。S1: Get the total score of a certain subject in a certain exam for all students in the target area; get the score of each question in the subject in the exam for all students; calculate the expected score of each question in the subject in the exam for a student based on the total score of the subject in the exam; all students in the target area refer to all students who take the same exam paper; the target area is, for example, the same exam paper in a unified exam in a district or county.
具体包括如下步骤:The specific steps include:
S1.1:获取目标区域全体学生某次考试某科目试卷的总得分;获取全体学生本次考试该科目试卷中各小题的得分;S1.1: Get the total score of a certain subject in a certain exam for all students in the target area; get the score of each question in the subject in this exam for all students;
设全体学生本次考试该科目试卷中各小题得分所组成的矩阵为:Suppose the matrix composed of the scores of each question in the test paper of this subject for all students is:
; ;
其中,矩阵A中每行代表着对应学生本次考试该科目试卷中各小题的得分;共m个学生,本次考试该科目试卷中n个小题;A为全体学生本次考试该科目试卷中各小题的得分矩阵;Each row in the matrix A represents the score of each question in the test paper of this subject for the corresponding student in this exam; there are m students in total and n questions in the test paper of this subject for this exam; A is the score matrix of each question in the test paper of this subject for all students in this exam;
S1.2:将全体学生本次考试该科目试卷中各小题的得分从大到小进行排序;S1.2: Sort the scores of all students in this exam from the highest to the lowest.
将矩阵A中各小题的得分,按照从大到小的方式进行排序;排序后的矩阵如下:Sort the scores of each question in matrix A from large to small; the sorted matrix is as follows:
; ;
S1.3:根据该学生的本次考试该科目试卷总得分,计算该学生本次考试该科目试卷中各小题的预期得分。S1.3: Based on the student’s total score for this subject in this exam, calculate the expected score for each question in this subject in this exam.
具体地,设:Specifically, suppose:
; ;
矩阵S中每行的数值S k ,k=1、2、……m;设该学生本次考试该科目试卷总得分为E 总,设该学生本次考试该科目试卷期望总得分为E k ,设E k 等于E 总;如果矩阵S中存在与E 总相同的同位次预期得分S k ,则,该学生本次考试该科目试卷中第i小题的预期得分的公式为:The value of each row in the matrix S is Sk , k = 1, 2, ... m; let the total score of the student's test paper for this subject be Etotal , let the expected total score of the student's test paper for this subject be Ek , let Ek equal Etotal ; if there is an expected score Sk with the same rank as Etotal in the matrix S , then the formula for the expected score of the i- th question in the test paper of this subject for this student is:
(1); (1);
设E k 为该学生本次考试该科目试卷的期望总得分; 为该学生本次考试该科目试卷中第i小题的预期得分;为该学生的同位次预期得分。Let E k be the expected total score of the student in this subject paper of this exam; is the expected score of the student for question i in the subject test paper of this exam; The expected score for the student in the same ranking.
如果该学生的本次考试该科目试卷的试卷总得分E 总,矩阵S中不存在与之相同的同位次预期得分,则,该学生本次考试该科目试卷中第i小题的预期得分的公式为:If the total score E of the student's test paper for this subject in this exam does not exist in the matrix S with the same expected score of the same rank, then the formula for the expected score of the i-th question in the test paper of this subject in this exam is:
(2); (2);
其中,S t+1 为小于E 总,且最接近E 总的第t+1行的期望总得分;S t 为大于E 总,且最接近E 总的第t行的期望总得分;u ti 为矩阵U中第t行第i小题的得分;u (t+1)i 为矩阵U中第t+1行第i小题的得分。E k 为该学生本次考试该科目试卷的期望总得分,为该学生本次考试该科目试卷中第i小题的预期得分;为该同学的同位次预期得分。Among them, S t+1 is the expected total score of the t+1th row that is less than E total and closest to E total ; S t is the expected total score of the tth row that is greater than E total and closest to E total ; u ti is the score of the ith question in the tth row of the matrix U ; u (t+1)i is the score of the ith question in the t+1th row of the matrix U. E k is the expected total score of the student in this exam. is the expected score of the student for question i in the subject test paper of this exam; The expected score for students in the same ranking.
S2:计算出步骤S1中该学生本次考试该科目试卷中各小题得分的预期偏差值。S2: Calculate the expected deviation value of the score of each question in the test paper of the subject of the student in step S1.
具体地,公式为:Specifically, the formula is:
; ;
其中,d i 为该学生本次考试该科目试卷中,第i小题实际得分与第i小题的预期得分之间的偏离值;为该学生本次考试该科目试卷中,第i小题的预期得分;a i 为该学生本次考试该科目试卷中,第i小题的实际得分。Among them, d i is the deviation between the actual score of question i and the expected score of question i in the test paper of this subject of the student in this exam; is the expected score of the i- th question in the test paper of this subject for this student in this examination; a i is the actual score of the i- th question in the test paper of this subject for this student in this examination.
S3:计算本次考试之前的Z次考试中,该学生该科目Z次考试得分位次的加权平均值。S3: Calculate the weighted average of the student's scores and rankings in the Z examinations before this examination.
设定本次考试之前,该学生该科目第j次考试得分在全体学生中的位次为p j, ;设该学生第j次考试该科目得分位次的加权系数为f j ;Assume that before this exam, the ranking of the student's score in the jth exam of this subject among all students is p j , ; Let the weighted coefficient of the student's score ranking in the subject in the jth examination be f j ;
例如:该学生为初三学生,Z的选值为该学生初三学年中,本次考试之前该科目的所有考试次数。For example: the student is in the third year of junior high school, and the value of Z is the number of all the examinations for this subject before this examination in the third year of junior high school.
则,该学生本次之前,Z次考试中该学生该科目得分位次的加权平均值w为:Then, the weighted average w of the student's score ranking in this subject in the Z exams before this one is:
。 .
S4:计算S1中该学生本次考试该科目试卷中各小题的优先补短系数。S4: Calculate the priority short-answer coefficient for each question in the test paper of this subject for the student in S1.
设该学生本次考试该科目试卷中第i小题的实际得分在全体学生中的位次为h i ;Suppose the actual score of the student on the i- th question in the test paper of this subject ranks among all students as h i ;
具体地,优先补短系数公式为:Specifically, the priority short-filling coefficient formula is:
(3); (3);
为该学生本次考试该科目试卷中第i小题的优先补短系数。 It is the priority short-compensation coefficient for the i- th question in the test paper of this subject for this student in this exam.
S5:将优先补短系数大的小题反馈给S1中的该学生及其教师。S5: Feedback the small questions with large short-term coefficients to the student and his/her teacher in S1.
优先补短系数大的题目表明需要其教师快速为该学生补短。Prioritizing questions with a large coefficient of shortcoming indicates that the teacher needs to quickly make up for the student's shortcomings.
实施例2:为了更清楚地反映实施例1的方案,本实施例中采用实例进行说明。Embodiment 2: In order to more clearly reflect the scheme of embodiment 1, an example is used for illustration in this embodiment.
步骤S1.1中,假设共有六名学生,该六名学生某次考试某一科目试卷中各小题得分所组成的矩阵A为:In step S1.1, it is assumed that there are six students in total, and the matrix A composed of the scores of each question in a certain subject test paper of the six students in a certain exam is:
; ;
即,第一位学生试卷的各小题的依次得分为[8,17,26,40];该学生试卷中的实际得分为91分;That is, the scores of the first student's test paper for each question are [8, 17, 26, 40] in order; the actual score of the student's test paper is 91 points;
第二位学生试卷的各小题的依次得分为[10,15,29,39];该学生试卷中的实际得分为93分。The scores of the questions in the second student's test paper are [10, 15, 29, 39] respectively; the actual score of this student is 93 points.
……,以此类推,... and so on.
第六位学生试卷的各小题的依次得分为[1,19,30,38]。该学生试卷中的实际得分为88分。The scores of the sixth student's test paper for each question are [1, 19, 30, 38], and the actual score of the student is 88 points.
步骤S1.2中,将全体学生本次考试该科目试卷中各小题的得分从大到小进行排序,排序后的矩阵U为:In step S1.2, the scores of all students in the test paper of this subject are sorted from large to small, and the sorted matrix U is:
; ;
步骤S1.3中,In step S1.3,
; ;
对于第一位实际得分为91分的学生,因为矩阵S中不存在与91分相同的同位次预期得分,所以采用实施例1中式(2)计算各小题的预期得分:则,第一位学生的各小题的预期得分分别为:7.5、17.5、28和38。For the first student whose actual score is 91, since there is no expected score of the same rank as 91 in the matrix S , the expected score of each question is calculated using formula (2) in Example 1: Then, the expected scores of the first student for each question are 7.5, 17.5, 28 and 38 respectively.
具体地:specifically:
以第一位学生的第一小题预期得分举例计算:Take the expected score of the first question of the first student as an example:
从矩阵U中知:u (t+1)i =7,u ti =8,From the matrix U , we know that: u (t+1)i =7, u ti =8,
从矩阵S中知:E 总=91,S t+1=89,S t =93。From the matrix S we know: Etotal = 91 , S t +1 = 89, S t = 93.
因此,结合公式(2)计算该学生第一小题的预期得分为:Therefore, combined with formula (2), the expected score of the student for the first question is calculated as:
7+(91-89)(8-7)/(93-89)=7+2/4=7.5。该学生其他小题的预期得分以此类推,不再赘述。计算后,该学生的每小题的预期得分为7.5、17.5、28和38。7+(91-89)(8-7)/(93-89)=7+2/4=7.5. The expected scores of the other questions of this student are similar and will not be repeated here. After calculation, the expected scores of each question of this student are 7.5, 17.5, 28 and 38.
对于第二位实际得分为93分的学生,因为矩阵S中第三行存在与93分相同的同位次预期得分,所以采用实施例1中式(1)计算各小题的预期得分,即,第二位学生的各小题的预期得分直接从矩阵U中第三行获得,分别为:8、18、29和38。如果矩阵S中存在相同的多个相同的同位次预期得分,可以任意选择一个同位次预期得分,让该同位次预期分中各小题的预期得分,代表该学生试卷中各小题的预期得分。For the second student whose actual score is 93, because there is an expected score of the same rank as 93 in the third row of matrix S , the expected score of each question is calculated using formula (1) in Example 1, that is, the expected score of each question of the second student is directly obtained from the third row of matrix U , which are 8, 18, 29 and 38 respectively. If there are multiple identical expected scores of the same rank in matrix S , you can arbitrarily select one expected score of the same rank, and let the expected score of each question in the expected score of the same rank represent the expected score of each question in the student's test paper.
……,以此类推,... and so on.
对于第六位实际得分为88的学生,因为矩阵S中不存在与88分相同的同位次预期得分,所以采用实施例1中式(2)计算各小题的预期得分:For the sixth student whose actual score is 88, since there is no expected score of 88 in the matrix S , the expected score of each question is calculated using formula (2) in Example 1:
则,第六位学生各小题的预期得分分别为:6.4、16.9、26.9和37.8。Then, the expected scores of the sixth student for each question are: 6.4, 16.9, 26.9 and 37.8 respectively.
步骤S2中,第一位学生各小题得分的预期偏差值为:-0.5、0.5、2和-2。In step S2, the expected deviation values of the first student's scores for each question are: -0.5, 0.5, 2 and -2.
第二位学生各小题得分的预期偏差值为:-2、3、0和-1。The expected deviations of the second student's scores on each question are: -2, 3, 0, and -1.
……,以此类推,... and so on.
第六位学生各小题得分的预期偏差值为:5.4、-2.1、-3.1和-0.2。The expected deviation values of the sixth student's scores for each question are: 5.4, -2.1, -3.1 and -0.2.
步骤S3,w为该学生本次考试之前,该学生该科目考试的位次加权平均值。w从本次考试该科目试卷之前,Z次该科目试卷中已经算出来,视为定值。Step S3, w is the weighted average of the ranking of the student in the subject before the current exam. w has been calculated from the Z times of the subject test papers before the current exam and is regarded as a fixed value.
步骤S4中,因为式(3)中的w为定值,所以由式(3)可知,的大小主要由h i 和d i 决定。以第六位学生为例,其d 1 =5.4,随着该学生该题得分的位次h 1 越大,则越大,表明该学生的第一小题越有补短的需求。In step S4, since w in equation (3) is a constant, it can be seen from equation (3) that: The size of is mainly determined by h i and d i . Taking the sixth student as an example, his d 1 =5.4. As the ranking of the student's score on this question increases , The larger the value is, the more the student needs to improve on the first question.
实施例3:如图2所示,基于大数据的知识点查缺补漏分析系统,包括:Embodiment 3: As shown in FIG2 , a knowledge point deficiency checking and gap filling analysis system based on big data includes:
题库,用于存储各个学科、各种知识点的题目,题目信息包括文字和图片,并通过自然语言处理技术抽取出题目的关键词;The question bank is used to store questions of various subjects and knowledge points. The question information includes text and pictures, and the keywords of the questions are extracted through natural language processing technology;
优先补短系数计算模块,能够根据学生的本次考试该科目试卷中各小题得分、本次考试该科目试卷中各小题的位次、以及本次考试该科目试卷之前,Z次该科目试卷中的位次加权平均值,计算出本次考试该科目试卷中各小题的优先补短系数;The priority-compensation coefficient calculation module can calculate the priority-compensation coefficient of each question in the test paper of this subject according to the student's score of each question in the test paper of this subject, the ranking of each question in the test paper of this subject, and the weighted average of the ranking in the Z test papers of this subject before the test paper of this subject;
习题识别模块,能通过文字和图片识别本次考试该科目试卷中各小题的学科、知识点等;能通过自然语言处理技术抽取出关键词、通过反向图片搜索引擎在题库中找到相似的图片,从而关联出各小题对应的学科和知识点;The question recognition module can identify the subject and knowledge points of each question in the test paper of this subject through text and pictures; it can extract keywords through natural language processing technology and find similar pictures in the question bank through reverse image search engine, so as to associate the subject and knowledge points corresponding to each question;
分析模块,分别与优先补短系数计算模块和习题识别模块连接;分别获得本次考试该科目试卷中各小题的学科、知识点等,以及能获得本次考试该科目试卷中各小题的优先补短系数,能够判断并分析出需要补短的学科和知识点。The analysis module is connected with the priority-making-up coefficient calculation module and the exercise identification module respectively; it obtains the subject, knowledge point, etc. of each question in the test paper of this subject in this examination, and can obtain the priority-making-up coefficient of each question in the test paper of this subject in this examination, and can judge and analyze the subjects and knowledge points that need to be made up.
相似习题推荐模块,与分析模块连接,能够接收分析模块的指令,能够从题库中找出对应知识点的相似习题;The similar exercise recommendation module is connected to the analysis module, can receive instructions from the analysis module, and can find similar exercises corresponding to the knowledge points from the question bank;
显示模块,能够显示相似习题推荐模块找出的对应知识点的习题。The display module can display the exercises of corresponding knowledge points found by the similar exercise recommendation module.
以上述依据本发明的理想实施例为启示,通过上述的说明内容,相关工作人员完全可以在不偏离本项发明技术思想的范围内,进行多样的变更以及修改。本项发明的技术性范围并不局限于说明书上的内容,必须要根据权利要求范围来确定其技术性范围。Based on the above ideal embodiments of the present invention, the relevant staff can make various changes and modifications without departing from the technical concept of the present invention through the above description. The technical scope of the present invention is not limited to the content in the specification, and its technical scope must be determined according to the scope of the claims.
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