Question

我有20名工人，每人完成100项任务。我已经为每项任务生成了真实的答案，这是5个答案中的1个

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))

我的dataSet包含列workerID，taskID，truth。现在我需要生成另一个向量，我模拟工作人员将根据某个概率回答的内容。例如，如果我的任务1的真相，工人1是＆＃34;肝脏＆＃34;，我希望工人1回答＆＃34;肝脏＆＃34;对于任务1的概率很高。同样，对于所有2000个任务的五个答案中的每一个，我都希望工人得到答案。为此，我使用以下for和if循环。

for (i in nrow(dataSet)){
if (dataSet$truth[i] == "liver")
{
df <- (rep(sample(answers, no.tasks, prob = c(0.9, 0.02, 0.02, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "blood")
{ 
df <-  (rep(sample(answers, no.tasks, prob = c(0.02, 0.9, 0.02, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "lung")
{
df <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.9, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "brain")
{
df <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.9, 0.02), no.workers)))
} else if (dataSet$truth[i] == "heart")
{
df <-  (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.02, 0.9), no.workers)))
} else {
df <- (rep(sample(answers, no.tasks, prob = c(0.2, 0.2, 0.2, 0.2, 0.2), no.workers)))
}
}

但是，由于我的任务1的真相是大脑，输出向量df有很多答案，它们是大脑＆＃34;。有人可以暗示这里出了什么问题吗？

Answer 1

考虑使用包含1,000个元素的基础字符向量的列表进行初始化。

df <- vector("list", 2000) 

for (i in 1:nrow(dataSet)){
if (dataSet$truth[i] == "liver")
{
df[[i]] <-(rep(sample(answers, no.tasks, prob = c(0.9, 0.02, 0.02, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "blood")
{ 
df[[i]] <-(rep(sample(answers, no.tasks, prob = c(0.02, 0.9, 0.02, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "lung")
{
df[[i]] <-(rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.9, 0.02, 0.02), no.workers)))
} else if (dataSet$truth[i] == "brain")
{
df[[i]] <-(rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.9, 0.02), no.workers)))
} else if (dataSet$truth[i] == "heart")
{
df[[i]] <-(rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.02, 0.9), no.workers)))
} 
}

或者，您可以使用lapply()输出与输入相同长度的列表向量（即 dataSet 的行），不需要初始化：

df2 <- lapply(seq_len(nrow(dataSet)), function(i){
  if (dataSet$truth[i] == "liver")
  {
  temp <- (rep(sample(answers, no.tasks, prob = c(0.9, 0.02, 0.02, 0.02, 0.02), no.workers)))
  } else if (dataSet$truth[i] == "blood")
  { 
  temp <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.9, 0.02, 0.02, 0.02), no.workers)))
  } else if (dataSet$truth[i] == "lung")
  {
  temp <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.9, 0.02, 0.02), no.workers)))
  } else if (dataSet$truth[i] == "brain")
  {
  temp <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.9, 0.02), no.workers)))
  } else if (dataSet$truth[i] == "heart")
  {
  temp <- (rep(sample(answers, no.tasks, prob = c(0.02, 0.02, 0.02, 0.02, 0.9), no.workers)))
  } 
  return(temp)
})

更好的是，您可以通过匹配 answers 向量中的当前if来修剪嵌套的dataSet$truth语句，然后用<替换概率向量中的相应索引EM> 0.9 ：

df3 <- lapply(seq_len(nrow(dataSet)), function(i){
  probs <- c(0.02, 0.02, 0.02, 0.02, 0.2)      
  probs[match(dataSet$truth[i], answers)] <- 0.9

  temp <- (rep(sample(answers, no.tasks, prob = probs, no.workers)))
})

R嵌套多个if循环以生成新的向量

1 个答案: