我已将SPSS .SAV文件中的采访数据导入为data.frame
,现在我正在尝试根据问题编号和访谈位置创建频率表。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"YES","NO","MAYBE"
和位置"q1","q2","q3"
计算每个答案选项"city1","city"
的出现次数。生成的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,我已经使用了来自count()
套餐的ddply()
,summarise()
和plyr
,但没有运气。我当前的解决方案非常简洁,需要将df
拆分为loc
,创建一个包含as.data.frame(summary(df_city1))
的频率表,从摘要字符串中检索频率并合并摘要data.frame
city1
和city2
重新组合在一起。我想必须有一个更容易/更优雅的解决方案。
答案 0 :(得分:2)
我们转换广泛的数据集&#39;长期&#39; (gather
执行此操作),然后group_by
)&#39;定位&#39;任务&#39;,&#39;回答&#39;,然后使用{{1}得到计数。但是,如果我们要查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tally
和unique
的所有complete
组合的数据集。 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