R语言 CDM包 Data-sim()函数中文帮助文档(中英文对照)

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Data-sim(CDM)
Data-sim()所属R语言包：CDM

                                         Artificial Data: DINA and DINO
                                         人工数据：DINA和DINO

                                         译者：生物统计家园网 机器人LoveR

描述----------Description----------

Artificial data: dichotomously coded fictitious answers of 400 respondents  to 9 items assuming 3 underlying attributes.  
人工数据：二分法编码的9个项目，假设3个基本属性虚构的400多名受访者的答案。


用法----------Usage----------


  data(sim.dina)
  data(sim.dino)
  data(sim.qmatrix)



格式----------Format----------

The sim.dina and sim.dino data sets include dichotomous  answers of N=400 respondents to J=9 items, thus they are  400 times 9 data matrices.  For both data sets K=3 attributes are assumed to underlie the process of responding, stored  in sim.qmatrix.  
sim.dina和sim.dino的数据集，包括二分的的答案N=400回应J=9项目的，因此，他们是400倍9数据矩阵。对于这两个数据集K=3属性被假定为基础的应诉过程中，存储在sim.qmatrix。

The sim.dina data set is simulated according to the DINA condensation rule, whereas the sim.dino data set is simulated according to the  DINO condensation rule.  The slipping errors for the items 1 to 9 in both  data sets are 0.20, 0.20, 0.20, 0.20, 0.00, 0.50, 0.50, 0.10, 0.03  and the guessing errors are 0.10, 0.125, 0.15, 0.175, 0.2, 0.225,    0.25, 0.275, 0.3.  The attributes are assumed to be mastered with expected  probabilities of -0.4 , 0.2, 0.6, respectively.  The correlation of  the attributes is 0.3 for attributes 1 and 2, 0.4 for  attributes 1 and 3 and 0.1 for attributes 2 and 3.  
sim.dina数据集是根据的DINA冷凝规则的模拟，而sim.dino数据集是根据的DINO冷凝规则的模拟。滑倒错误项目两组数据0.20, 0.20, 0.20, 0.20, 0.00, 0.50, 0.50, 0.10, 0.03猜谜错误0.10, 0.125, 0.15, 0.175, 0.2, 0.225,    0.25, 0.275, 0.3被假定为是分别掌握与预期概率-0.4 , 0.2, 0.6的属性。的属性的相关性是0.3属性1和2，0.4属性1和3和0.1的属性2和3。


（作者）----------Author(s)----------



Alexander Robitzsch, Thomas Kiefer, Ann Cathrice George, Ali Uenlue

Maintainer: Alexander Robitzsch <a.robitzsch@bifie.at>,
Thomas Kiefer <t.kiefer@bifie.at>,
Ann Cathrice George <george@statistik.tu-dortmund.de>,
Ali Uenlue <ali.uenlue@tum.de>




参考文献----------References----------

Measurement: Theory, Methods, and Applications. New York: The Guilford Press.

实例----------Examples----------


##[＃]
## (2) examples based on dataset sim.dina[＃（2）的例子基于数据集sim.dina的]
##[＃]

# DINA Model[DINA模型]
d1 <- din(sim.dina, q.matr = sim.qmatrix, rule = "DINA",
  conv.crit = 0.01, maxit = 500, progress = TRUE)
summary(d1)

# Mixed DINA and DINO Model[混合DINA和DINO型号]
d1b <- din(sim.dina, q.matr = sim.qmatrix, rule =
  c(rep("DINA", 7), rep("DINO", 2)), conv.crit = 0.01,
  maxit = 500, progress = FALSE)
summary(d1b)

# DINO Model[DINO型号]
d2 <- din(sim.dina, q.matr = sim.qmatrix, rule = "DINO",
  conv.crit = 0.01, maxit = 500, progress = FALSE)
summary(d2)

# Comparison of DINA and DINO estimates[的DINA和DINO估计的比较]
lapply(list("guessing" = rbind("DINA" = d1$guess[,1],
  "DINO" = d2$guess[,1]), "slipping" = rbind("DINA" =
  d1$slip[,1], "DINO" = d2$slip[,1])), round, 2)

# Comparison of the information criteria[信息标准的比较]
c("DINA"=d1$AIC, "MIXED"=d1b$AIC, "DINO"=d2$AIC)

# following estimates:[下面的估计：]
d1$coef            # guessing and slipping parameter[猜测和滑动参数]
d1$guess           # guessing parameter[猜测参数]
d1$slip            # slipping parameter[打滑参数]
d1$skill.patt      # probabilities for skills[对技能的概率]
d1$attribute.patt  # skill classes with probabilities[技能类的概率]
d1$subj.pattern    # pattern per subject[每科模式]

# posterior probabilities for every response pattern[每一个响应模式的后验概率]
d1$posterior      

##[＃]
## (3) examples based on dataset sim.dino[＃（3）的例子基于数据集sim.dino的]
##[＃]

# DINO Estimation[DINO估计。]
d3 <- din(sim.dino, q.matr = sim.qmatrix, rule = "DINO",
  conv.crit = 0.005, progress = FALSE)

# Mixed DINA and DINO Model[混合DINA和DINO型号]
d3b <- din(sim.dino, q.matr = sim.qmatrix, rule =
  c(rep("DINA", 4), rep("DINO", 5)), conv.crit = 0.001,
  progress = FALSE)
                        
# DINA Estimation[DINA估计。]
d4 <- din(sim.dino, q.matr = sim.qmatrix, rule = "DINA",
  conv.crit = 0.005, progress = FALSE)
            
# Comparison of DINA and DINO estimates[的DINA和DINO估计的比较]
lapply(list("guessing" = rbind("DINO" = d3$guess[,1],
  "DINA" = d4$guess[,1]), "slipping" = rbind("DINO" =
  d3$slip[,1], "DINA" = d4$slip[,1])), round, 2)

# Comparison of the information criteria[信息标准的比较]
c("DINO"=d3$AIC, "MIXED"=d3b$AIC, "DINA"=d4$AIC)

##[＃]
## (4) example estimation with weights based on dataset sim.dina[＃（4）例如估计配重块根据数据集sim.dina]
##[＃]

# Here, a weighted maximum likelihood estimation is used [在这里，加权最大似然估计]
# This could be useful for survey data.[这可能是有用的调查数据。]

# i.e. first 200 persons have weight 2, the other have weight 1[即第一个200人的重量为2，其他权重为1]
(weights <- c(rep(2, 200), rep(1, 200)))

d5 <- din(sim.dina, sim.qmatrix, rule = "DINA", conv.crit =
  0.005, weights = weights, progress = FALSE)
        
# Comparison of the information criteria[信息标准的比较]
c("DINA"=d1$AIC, "WEIGHTS"=d5$AIC)


##[＃]
## (5) example estimation within a Balanced Incomplete [＃（5）如在一个平衡的不完全估计]
##     Block (BIB) Design generated on dataset sim.dina[＃块（BIB）设计生成的数据集sim.dina]
##[＃]

# generate BIB data[产生BIB数据]

# The next example shows that the din function[下面的例子演示的喧嚣功能]
# works for (relatively arbitrary) missing value pattern[（比较乱）的作品缺失值模式]

# Here, a missing by design is generated in the dataset dinadat.bib[在这里，一个失踪的设计中产生的数据集dinadat.bib]
sim.dina.bib <- sim.dina
sim.dina.bib[1:100, 1:3] <- NA
sim.dina.bib[101:300, 4:8] <- NA
sim.dina.bib[301:400, c(1,2,9)] <- NA


d6 <- din(sim.dina.bib, sim.qmatrix, rule = "DINA",
  conv.crit = 0.0005, weights = weights, maxit=200)

d7 <- din(sim.dina.bib, sim.qmatrix, rule = "DINO",
  conv.crit = 0.005, weights = weights)

# Comparison of DINA and DINO estimates[的DINA和DINO估计的比较]
lapply(list("guessing" = rbind("DINA" = d6$guess[,1],
  "DINO" = d7$guess[,1]), "slipping" = rbind("DINA" =
  d6$slip[,1], "DINO" = d7$slip[,1])), round, 2)

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