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R多个分类变量的频率表

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我已经将SPSS .SAV文件中的访谈数据导入 data.frame ，现在我'm trying to create a frequency table based on the question number and interview location. Here'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" 的出现次数 . 结果 data.frame 应该如下所示：

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，我已经使用 plyr 包中的 count() ， ddply() 和 summarise() 而没有运气 . 我当前的解决方案非常hacky并且涉及将 df 拆分为 loc ，使用 as.data.frame(summary(df_city1)) 创建频率表，从摘要字符串中检索频率并将 city1 和 city2 的摘要 data.frame 合并在一起 . 我想必须有一个更容易/更优雅的解决方案 .

r dplyr plyr frequency summary

1 回答

  • 2

    我们将数据集从'wide'转换为'long'（ gather 执行此操作），然后将 group_by ）'loc'，'quest'，'answ'转换为 tally 并使用 tally 来获取计数 . 但是，如果我们正在寻找数据集中未找到的组合计数为0，那么我们可能需要连接具有三列所有 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
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