比较列中的文本以创建匹配标志

Question

有一个包含对调查的回答的数据集。尝试将项目回答与主答案进行比较。所有列都是完整的文本值，我想创建一个将文本与主行匹配的标志，然后创建一个标志来对项目响应进行二分法。我看过整数的代码，但没有看过文本的代码。为了简单起见，我有大约25列，示例：

具备以下条件：

Chart 1 user    Assistive device needed Assistance needed with 1 or more ADLs   Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None
Chart 1 user    Assistive device needed Assistance needed with 1 or more ADLs   Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None
Chart 2 user    Assistive device needed Independent Oriented    No capable caregiver availability   One hour or less if at least one of the disciplines None    None
Chart 2 user    Assistive device needed Independent Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None
Chart 2 user    Assistive device needed Assistance needed with 1 or more ADLs   Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None
Chart 1 expert  Assistive device needed Assistance needed with 1 or more ADLs   Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None
Chart 2 expert  Assistive device needed Assistance needed with 1 or more ADLs   Oriented    < 24 hours/day  One hour or less if at least one of the disciplines None    None

我想像这样输出一个数据框：

Chart 1 user    1   1   1   1   1   1   1
Chart 1 user    1   1   1   1   1   1   1
Chart 2 user    1   0   1   0   1   1   1
Chart 2 user    1   0   1   1   1   1   1
Chart 2 user    1   1   1   1   1   1   1

当前代码：

amb_flg <- ifelse(c[,3] == "Assistive device needed",1,0)
adl_flg <- ifelse(c[,4] == "Assistance needed with 1 or more ADLs",1,0) 
cog_flg <- ifelse(c[,5] == "Oriented",1,0)
cgv_flg <- ifelse(c[,6] == "< 24 hours/day",1,0)
tim_flg <- ifelse(c[,7] == "One hour or less if at least one of the disciplines",1,0)

上面的问题是，我必须硬编码每个图表的答案(1对2)。为了提高效率，我想创建一个函数来生成一个标志，以确定该值是否与主值相对应。实验代码：

 ## Create agreement score
f.a <- function (a) {
    y <- subset(c, c$user == 'expert')
    z <- subset(c, c$user != 'expert')
    df <- apply(z, 2, function(b){
        for (i in y)
          ifelse(i %in% z, 1, 0)
    })
    return(as.data.frame(df))
}
df <- f.a(c)

Answer 1

你的意思是这些方面的东西吗？如果你想在算法上识别主答案，那就复杂得多。

> df <- data.frame(Name = c("Jim", "John", "Amy", "Master"),
+                  Item1 = c("Does agree", "Does not agree", "Does agree", "Does not agree"))
> df
    Name          Item1
1    Jim     Does agree
2   John Does not agree
3    Amy     Does agree
4 Master Does not agree

> df$Item1_flag <- ifelse(df$Item1 == "Does not agree", "expert", "not expert")
> df$Item1_count <- ifelse(df$Item1 == "Does not agree", "1", "0")
> df
    Name          Item1 Item1_flag Item1_count
1    Jim     Does agree not expert           0
2   John Does not agree     expert           1
3    Amy     Does agree not expert           0
4 Master Does not agree     expert           1