多分类变量的 R 频率 table

Question

我已经从 SPSS .SAV 文件中将访谈数据导入为 data.frame，现在我正在尝试根据问题编号和访谈地点创建一个频率 table。这是一个例子 data.frame:

loc<-c("city1","city2","city1","city2","city1","city1","city2","city2","city1","city2")
q1<-c("YES","YES","NO","MAYBE","NO","NO","YES","NO","MAYBE","MAYBE")
q2<-c("YES","NO","MAYBE","YES","NO","MAYBE","MAYBE","YES","YES","NO")
q3<-c("NO","NO","NO","NO","YES","YES","MAYBE","MAYBE","NO","MAYBE")
df<-data.frame(loc,q1,q2,q3)

df
     loc    q1    q2    q3
1  city1   YES   YES    NO
2  city2   YES    NO    NO
3  city1    NO MAYBE    NO
4  city2 MAYBE   YES    NO
5  city1    NO    NO   YES
6  city1    NO MAYBE   YES
7  city2   YES MAYBE MAYBE
8  city2    NO   YES MAYBE
9  city1 MAYBE   YES    NO
10 city2 MAYBE    NO MAYBE

现在我想根据问题编号"q1","q2","q3"和位置"city1","city"计算每个答案选项"YES","NO","MAYBE"出现的次数。结果 data.frame 应如下所示：

   loc quest  answ freq
1  city1    q1   YES    1
2  city1    q1    NO    3
3  city1    q1 MAYBE    1
4  city2    q1   YES    2
5  city2    q1    NO    1
6  city2    q1 MAYBE    2
7  city1    q2   YES    2
8  city1    q2    NO    1
9  city1    q2 MAYBE    2
10 city2    q2   YES    2
11 city2    q2    NO    2
12 city2    q2 MAYBE    1
13 city1    q3   YES    2
14 city1    q3    NO    3
15 city1    q3 MAYBE    0
16 city2    q3   YES    0
17 city2    q3    NO    2
18 city2    q3 MAYBE    3

到目前为止，我已经玩过 plyr 包中的 count()、ddply() 和 summarise()，但没有成功。我当前的解决方案真的很老套，涉及将 df 拆分为 loc，创建频率 table 和 as.data.frame(summary(df_city1))，从摘要字符串中检索频率并合并摘要 data.frame 的 city1 和 city2 重新组合在一起。我想必须有一个 easier/more 优雅的解决方案。

Answer 1

我们将数据集从 'wide' 转换为 'long'（gather 这样做），然后 group_by) 'loc','quest', 'answ'，并使用 tally 获取计数。但是，如果我们正在寻找在数据集中未找到的计数为 0 的组合，那么我们可能需要连接一个具有三列的所有 unique 组合的数据集（complete 和unique 这样做）。

library(dplyr)
library(tidyr)
dfN <- gather(df, quest, answ, q1:q3) %>%
                   complete(loc, quest, answ) %>%
                   unique()

res <- gather(df, quest, answ, q1:q3) %>%
               group_by(loc, quest, answ) %>%
               tally() %>%
               left_join(dfN, .) %>%
               mutate(n = ifelse(is.na(n), 0, n))
res
#     loc quest  answ     n
#   (fctr) (chr) (chr) (dbl)
#1   city1    q1 MAYBE     1
#2   city1    q1    NO     3
#3   city1    q1   YES     1
#4   city1    q2 MAYBE     2
#5   city1    q2    NO     1
#6   city1    q2   YES     2
#7   city1    q3 MAYBE     0
#8   city1    q3    NO     3
#9   city1    q3   YES     2
#10  city2    q1 MAYBE     2
#11  city2    q1    NO     1
#12  city2    q1   YES     2
#13  city2    q2 MAYBE     1
#14  city2    q2    NO     2
#15  city2    q2   YES     2
#16  city2    q3 MAYBE     3
#17  city2    q3    NO     2
#18  city2    q3   YES     0

多分类变量的 R 频率 table

R Frequency table of multiple categorical variable

r

summary

frequency

plyr

dplyr