[英]zigzag pattern in matrix of questions
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程度的矩阵问题。 下面是一个例子。
Q1:您对 A 项的满意程度如何? "非常满意" "比较满意" 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" "比较满意" 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" "比较满意" 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" "比较满意" 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
我应该用以下模式找出所有这些观察结果:
情况1
如果q1 = "非常满意" and q2 = "有点满意" and q3 = "非常满意" and q4 = "有点满意"
案例2
或q1 = "非常满意" and q2 = "有点不满意" and q3 = "非常满意" and 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! **
对于您提供的示例,因为 q1 和 q3 的条件在个案测试中都相同,您只需使用以下命令即可获得锯齿形结果:
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
的冗余。 如果您想要zigzag
结果的数值,您可以更紧凑地使用:
zigzag = as.numeric( case1 | case2 | case3 ) # since 1 == TRUE
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