Question

我有一些看起来像这样的数据：

          A      B
6     Often  Often
7    Always Always
8    Rarely Rarely
9 Sometimes  Often

structure(list(A = structure(c(5L, 6L, 3L, 4L), .Label = c("", 
"Almost Never", "Rarely", "Sometimes", "Often", "Always"), class = c("ordered", 
"factor")), B = structure(c(5L, 6L, 3L, 5L), .Label = c("", "Almost Never", 
"Rarely", "Sometimes", "Often", "Always"), class = c("ordered", 
"factor"))), .Names = c("A", "B"), row.names = 6:9, class = "data.frame")

使用摘要，我会根据可能的响应得到每种响应的计数，这正是我想要的：

            A                B    
             :0               :0  
 Almost Never:0   Almost Never:0  
 Rarely      :1   Rarely      :1  
 Sometimes   :1   Sometimes   :0  
 Often       :1   Often       :2  
 Always      :1   Always      :1

现在我想操纵这些数字来获得（通常+总是）/总回复。摘要输出是字符输出，但是 - 我可以在冒号上拆分，但必须有更好的方法。

如果给出上面的数据，我如何计算每个问题的常数+总回答的百分比？

Answer 1

可以使用apply和table完成此操作（假设d是您的数据框）：

apply(d, 2, function(col) {
    tab = table(col)
    (tab["Often"] + tab["Always"]) / sum(tab)
})

请注意，只有在每列中始终至少包含“始终”和“常用”时，上述操作才有效。以下内容略显简洁，但即使列中缺少“始终”或“经常”，也可以使用：

sapply(1:NCOL(d), function(i) {
        tab = table(d[, i])
        (tab["Often"] + tab["Always"]) / sum(tab)
})

如何操作汇总表结果

1 个答案: