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我正在处理调查数据。它询问了几个关于受访者对几个项目的满意度的矩阵问题。下面是一个例子。

q1：您对 A 项的满意程度如何？“非常满意” “有点满意” “有点不满意” “非常不满意”

q2：您对 B 项的满意程度如何？“非常满意” “有点满意” “有点不满意” “非常不满意”

q3：您对C项的满意程度如何？“非常满意” “有点满意” “有点不满意” “非常不满意”

q4：您对 D 项的满意程度如何？“非常满意” “有点满意” “有点不满意” “非常不满意”

数据如下：

df <- data.frame(q1 = c("Very satisfied", "Somewhat satisfied", "Very satisfied", "Very satisfied", "Somewhat satisfied", "Very dissatisfied", "Very satisfied", "Very dissatisfied", "Very dissatisfied", "Somewhat dissatisfied"),
                 q2 = c("Somewhat satisfied", "Very satisfied", "Somewhat satisfied", "Very satisfied", "Very satisfied", "Somewhat dissatisfied", "Somewhat dissatisfied", "Somewhat dissatisfied", "Very dissatisfied", "Very dissatisfied"),
                 q3 = c("Very satisfied", "Somewhat satisfied", "Very satisfied", "Very satisfied", "Somewhat satisfied", "Very dissatisfied", "Very satisfied", "Very dissatisfied", "Very dissatisfied", "Somewhat dissatisfied"),
                 q4 = c("Somewhat satisfied", "Very satisfied", "Somewhat satisfied", "Very satisfied", "Very satisfied", "Somewhat dissatisfied", "Somewhat dissatisfied", "Somewhat dissatisfied", "Very dissatisfied", "Very dissatisfied"))

                      q1                    q2                    q3                    q4
1         Very satisfied    Somewhat satisfied        Very satisfied    Somewhat satisfied
2     Somewhat satisfied        Very satisfied    Somewhat satisfied        Very satisfied
3         Very satisfied    Somewhat satisfied        Very satisfied    Somewhat satisfied
4         Very satisfied        Very satisfied        Very satisfied        Very satisfied
5     Somewhat satisfied        Very satisfied    Somewhat satisfied        Very satisfied
6      Very dissatisfied Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied
7         Very satisfied Somewhat dissatisfied        Very satisfied Somewhat dissatisfied
8      Very dissatisfied Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied
9      Very dissatisfied     Very dissatisfied     Very dissatisfied     Very dissatisfied
10 Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied     Very dissatisfied

我应该通过以下模式找出所有这些观察结果：

情况1

如果q1 =“非常满意” 和 q2 =“有点满意” 并且 q3 =“非常满意” 和 q4 =“有点满意”

案例2

或q1 =“非常满意” 和 q2 =“有点不满意” 和 q3 =“非常满意” 和 q4 =“有点不满意”

案例3

或q1 =“非常满意” 和 q2 =“非常不满意” 和 q3 =“非常满意” 和 q4 =“非常不满意”

I can find this pattern using below command. However, since I have to do this for several matrices and number of questions in each matrix varies, I wonder if anyone knows an easy way of doing this.


df %>%
  mutate(case1 = ifelse((q1 %in% "Very satisfied" & q2 %in% "Somewhat satisfied" & q3 %in% "Very satisfied" & q4 %in% "Somewhat satisfied"), TRUE, FALSE),
         case2 = ifelse((q1 %in% "Very satisfied" & q2 %in% "Somewhat dissatisfied" & q3 %in% "Very satisfied" & q4 %in% "Somewhat dissatisfied"), TRUE, FALSE),
         case3 = ifelse((q1 %in% "Very satisfied" & q2 %in% "Very dissatisfied" & q3 %in% "Very satisfied" & q4 %in% "Very dissatisfied"), TRUE, FALSE),
         zigzag = ifelse((case1 %in% TRUE | case2 %in% TRUE | case3 %in% TRUE), 1, 0)
         )


                      q1                    q2                    q3                    q4 case1 case2 case3 zigzag
1         Very satisfied    Somewhat satisfied        Very satisfied    Somewhat satisfied  TRUE FALSE FALSE      1
2     Somewhat satisfied        Very satisfied    Somewhat satisfied        Very satisfied FALSE FALSE FALSE      0
3         Very satisfied    Somewhat satisfied        Very satisfied    Somewhat satisfied  TRUE FALSE FALSE      1
4         Very satisfied        Very satisfied        Very satisfied        Very satisfied FALSE FALSE FALSE      0
5     Somewhat satisfied        Very satisfied    Somewhat satisfied        Very satisfied FALSE FALSE FALSE      0
6      Very dissatisfied Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied FALSE FALSE FALSE      0
7         Very satisfied Somewhat dissatisfied        Very satisfied Somewhat dissatisfied FALSE  TRUE FALSE      1
8      Very dissatisfied Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied FALSE FALSE FALSE      0
9      Very dissatisfied     Very dissatisfied     Very dissatisfied     Very dissatisfied FALSE FALSE FALSE      0
10 Somewhat dissatisfied     Very dissatisfied Somewhat dissatisfied     Very dissatisfied FALSE FALSE FALSE      0

** 先感谢您！**

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1 回答 1

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对于您提供的示例，因为 q1 和 q3 的条件在案例测试中都是相同的，您可以通过以下方式获得您的 zigzag 结果：

df[with(df, q1 == "Very satisfied" & 
            q2 == q4 & 
            q3 == "Very satisfied" & 
     q4 %in% c( "Very dissatisfied", "Somewhat dissatisfied", "Somewhat satisfied") ), ]

r2evans 已经指出了使用ifelse. 如果您想要结果的数值，zigzag您可以更紧凑地使用：

zigzag = as.numeric( case1 | case2 | case3 ) # since 1 == TRUE
于 2020-02-01T23:16:16.057 回答