R 如何操作汇总表结果
我有一些数据如下所示:R 如何操作汇总表结果,r,R,我有一些数据如下所示: 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"
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summary,我根据可能的响应得到每种类型响应的计数,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,我想操纵这些数字以获得(经常+总是)/总体响应。虽然摘要输出是字符输出——我可以在冒号上拆分,但肯定有更好的方法
鉴于上述数据集,我如何计算每个问题的经常+总回答百分比?这可以使用
apply
和表来完成(假设d
是您的数据框):
请注意,只有在每列中始终至少有一个“始终”和一个“经常”时,上述方法才有效。以下内容稍微不够简洁,但即使某列中缺少“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)
})
这可以使用apply
和table
来完成(假设d
是您的数据帧):
请注意,只有在每列中始终至少有一个“始终”和一个“经常”时,上述方法才有效。以下内容稍微不够简洁,但即使某列中缺少“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)
})