我有100个任务和20个人来完成这些任务。我随机地为每个人分配7个任务（以便稍后计算评估者之间的协议）。所以我有一个dataID，其中包含personID，taskID。每项任务有5种可能的答案。我通过
模拟了“真实”答案（来自5个可能的答案） truth <- sample(answers, no.tasks, replace = FALSE, prob = c(0.2, 0.2, 0.2, 0.2, 0.2))
并将其添加到我的dataSet，它现在有三列：personID，taskID，truth;
我还说50％的人表现不佳（穷人），10％的任务是艰巨的任务（hardtasks）。我假设每个概率：性能差，性能好，任务艰巨，任务简单
ppoor <- 0.7
pgood <- 0.99
phard <- 0.2
peasy <- 0.8
现在，我需要根据她在任务中的好坏来模拟这个人的答案，以及该任务是容易还是困难。对于我正在做的一个这样的组合：
for(i in 1:length(dataSet$taskID)) { ifelse(dataSet$personID[i] %in% poorperson && dataSet[dataSet$personID == i,]$taskID %in% hardtasks, probs[i] <- ppoor * phard, NA)}
但是，我没有达到我的期望，即我希望得到每个任务和工人组合的'probs'。我在这里做错了什么？
答案 0 :(得分：1)
我无法理解您的代码，但我以一种希望开始讨论您需要向前发展的方式重新编写代码。如果您有任何问题，请告诉我们！
# Initialize
no.tasks <- 100
no.workers <- 20
tasksperworker <- no.tasks/no.workers # 5 each ( why did you have 7?)
# ANSWERS for Tasks
answers <- c("liver", "blood", "lung", "brain", "heart")
(truth <- sample(answers, no.tasks, replace = TRUE, prob = c(0.2, 0.2, 0.2, 0.2, 0.2)))
# TASKS
prop_hardtasks <- .10
hardtasks <- sample(1:no.tasks, prop_hardtasks * no.tasks)
easytasks <- setdiff(1:no.tasks, hardtasks)
phard <- 0.2
peasy <- 0.8
(task_Difficulty <- ifelse(1:no.tasks %in% easytasks, peasy, phard))
# WORKERS
prop_poorworkers <- .50
poorworkers <- sample(1:no.workers, prop_poorworkers * no.workers)
goodworkers <- setdiff(1:no.workers, poorworkers)
ppoor <- 0.7
pgood <- 0.99
(worker_Ability <- ifelse(1:no.workers %in% goodworkers, pgood, ppoor))
# The dataset
# One step data creation
dataSet <- data.frame("workerID" = rep(1:no.workers, each = tasksperworker),
"taskID" = 1:no.tasks,
"truth" = truth,
"taskDifficulty" = factor(task_Difficulty, labels = c("hard","easy")),
"workerAbility" = factor(rep(worker_Ability, each = tasksperworker), labels = c("poor","good")),
"probCorrect" = task_Difficulty * worker_Ability)
# I am coding out the old method as I believe it samples twice which is not necessary
# (assignmentMatrix <- replicate(no.workers, sample(1:no.tasks, tasksperworker, replace=FALSE)))
#(assignEach <- reshape::melt.matrix(assignmentMatrix))
#(dataSet <- cbind.data.frame("workerID" = assignEach[,2],
# "taskID" = assignEach[,3],
# "truth" = truth[assignEach[,2]],
# "taskDifficulty" = factor(task_Difficulty, labels = c("hard",'easy')),
# "workerAbility" = factor(rep(worker_Ability,each = tasksperworker), labels = c("poor","good")),
# "probCorrect" = task_Difficulty * worker_Ability))
编辑：
鉴于有5种可能性，每项任务都有一个事实（我们随机生成），我们希望保存工人得到正确答案的概率，同时为剩余的概率分配其他可能的答案。此向量可能会针对每个任务/工作人员进行更改。 sample(..., prob)参数必须始终等于1.因此，我继续执行以下操作：
# Initialize a matrix where each row contains the probability vector we will use to sample.
truthProb <- matrix(NA, nrow = no.tasks, ncol = length(answers), dimnames = list(1:no.tasks, answers))
# run a for loop to populate it
for(i in 1:no.tasks){
# Because Answer always changes,
# Find location of truth amongst answer vector using which
# And place the probCorrect value into that spot
truthProb[i, which(answers %in% dataSet$truth[i])] <- dataSet$probCorrect[i]
# I would assign equal remaining probabilities to other incorrect answers.
truthProb[i, -which(answers %in% dataSet$truth[i])] <- (1 - dataSet$probCorrect[i]) / (length(answers) - 1)
}
rowSums(truthProb) # Should sum to 1
# Add their answers here by using apply to say, given a probability from each row (task), draw the single answer
dataSet$results <- apply(truthProb, 1, function(x) sample(answers, 1, replace = F, prob = x))
tail(truthProb) # see the probabilities for each task
tail(dataSet) # can compare the last bit
# how did we do? table showing results at each probCorrect (combination of difficulty and ability)
table(dataSet$truth, dataSet$results, dataSet$probCorrect)
# double check again that this idea works as intended...
dataSet$truth[1]
truthProb[1,]
sum(1 * (dataSet$truth[1] == replicate(100, sample(answers, 1, replace = F, prob = truthProb[1,])))) /100
编辑所有功能版本：
# Simulate a Function!!
sim <- function(answers, no.tasks, no.workers, prop_hardtasks, prop_poorworkers, prob_hardeasy, prob_poorgood){
# Initialize
tasksperworker <- no.tasks/no.workers
# ANSWERS for Tasks
truth <- sample(answers, no.tasks, replace = TRUE) # assumes equal probability of each answer
# TASKS
hardtasks <- sample(1:no.tasks, prop_hardtasks * no.tasks)
easytasks <- setdiff(1:no.tasks, hardtasks)
phard <- prob_hardeasy[1]
peasy <- prob_hardeasy[2]
(task_Difficulty <- ifelse(1:no.tasks %in% easytasks, peasy, phard))
# WORKERS
poorworkers <- sample(1:no.workers, prop_poorworkers * no.workers)
goodworkers <- setdiff(1:no.workers, poorworkers)
ppoor <- prob_poorgood[1]
pgood <- prob_poorgood[2]
(worker_Ability <- ifelse(1:no.workers %in% goodworkers, pgood, ppoor))
# One step data creation
dataSet <- data.frame("workerID" = rep(1:no.workers, each = tasksperworker),
"taskID" = 1:no.tasks, "truth" = truth, "taskDifficulty" = factor(task_Difficulty, labels = c("hard","easy")),
"workerAbility" = factor(rep(worker_Ability, each = tasksperworker), labels = c("poor","good")),
"probCorrect" = task_Difficulty * rep(worker_Ability, each = tasksperworker))
# SIMULATE ANSWER
truthProb <- matrix(NA, nrow = no.tasks, ncol = length(answers), dimnames = list(1:no.tasks, answers))
for(i in 1:no.tasks){
truthProb[i, which(answers %in% dataSet$truth[i])] <- dataSet$probCorrect[i]
truthProb[i, -which(answers %in% dataSet$truth[i])] <- (1 - dataSet$probCorrect[i]) / (length(answers) - 1)
}
dataSet$results <- apply(truthProb, 1, function(x) sample(answers, 1, replace = F, prob = x))
# Return
return(dataSet)
}
dat <- sim(answers = LETTERS[1:5], # c("liver", "blood", "lung", "brain", "heart")
no.tasks = 100,
prop_hardtasks = 10/100,
prob_hardeasy = c(.2, .8),
no.workers = 20,
prop_poorworkers = 10/20,
prob_poorgood = c(.77, .99))
head(dat)
table(dat$truth, dat$results, dat$probCorrect)