Reputation: 721
I've imported interview data from a SPSS .SAV file as a data.frame
and now I'm trying to create a frequency table based on the question number and interview location. Here's an example 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
Now I would like to count the number of occurances for each answer option "YES","NO","MAYBE"
according to the question number "q1","q2","q3"
and the location "city1","city"
. The resulting data.frame
should look like this:
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
So far I've played with count()
,ddply()
and summarise()
from the plyr
package with no luck. My current solution is really hacky and involves splitting df
by loc
, creating a frequency table with as.data.frame(summary(df_city1))
, retrieving the frequency from the summary string and merging the summary data.frame
s of city1
and city2
back together. I guess there has to be an easier/more elegant solution.
Upvotes: 1
Views: 3387
Reputation: 887118
We convert the dataset from 'wide' to 'long' (gather
does that), then group_by
) 'loc','quest', 'answ', and use tally
to get the count. But, if we are looking for combinations that are not found in the dataset to have a count of 0, then we may need to join with a dataset having all the unique
combinations of three columns (complete
and unique
does that).
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
Upvotes: 2