How to count different aswers in R - r

Question

I am analyzing a questionnaire and I've written the code below to count how many answers there are to each question. The questions are in columns and the answer is coded as a number, where 1=a, 2=b.
The main objective is to count how many times an answer was chosen, ignoring pattern to summarize the information.
DS is the data frame, containing questions Q_092 to Q_096. I have the code to change column names, but it expects a fixed number of columns.
Is there a prettier way to do it?
conta_respostas <- function (arr_resp) {
arr_resp[(is.na(arr_resp))]<-99
arr_result = c(
sum(arr_resp[(arr_resp=="1")])/1,
sum(arr_resp[(arr_resp=="2")])/2,
sum(arr_resp[(arr_resp=="3")])/3,
sum(arr_resp[(arr_resp=="4")])/4,
sum(arr_resp[(arr_resp=="5")])/5,
sum(arr_resp[(arr_resp=="6")])/6,
sum(arr_resp[(arr_resp=="7")])/7,
sum(arr_resp[(arr_resp=="8")])/8,
sum(arr_resp[(arr_resp=="9")])/9,
sum(arr_resp[(arr_resp=="10")])/10,
sum(arr_resp[(arr_resp=="99")])/99
)
}
adply(DS, 2, conta_respostas)
X1 V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
1 Q_092 431 1987 5053 1388 0 0 0 0 0 0 36
2 Q_093 281 1489 5728 1336 0 0 0 0 0 0 61
3 Q_094 594 3380 4365 519 0 0 0 0 0 0 37
4 Q_095 89 216 5042 3511 0 0 0 0 0 0 37
5 Q_096 213 1764 5384 1511 0 0 0 0 0 0 23

Answer 1

what it sounds like your data looks like:
DS <- data.frame(
'Q_092' = c(1, 3, 4, 5, 2, 99, 10),
'Q_093' = c(2, 5, 6, 2, 99, 1, 1),
'Q_094' = c(3, 5, 6, 2, 4, 7, 8),
'Q_095' = c(10, 5, 5, 6, 7, 8, 6),
'Q_096' = c(1, 3, 4, 5, 2, 99, 10)
)
DS
Q_092 Q_093 Q_094 Q_095 Q_096
1 1 2 3 10 1
2 3 5 5 5 3
3 4 6 6 5 4
4 5 2 2 6 5
5 2 99 4 7 2
6 99 1 7 8 99
7 10 1 8 6 10
Recreating your code:
library(plyr)
conta_respostas <- function (arr_resp) {
arr_resp[(is.na(arr_resp))]<-99
arr_result = c(
sum(arr_resp[(arr_resp=="1")])/1,
sum(arr_resp[(arr_resp=="2")])/2,
sum(arr_resp[(arr_resp=="3")])/3,
sum(arr_resp[(arr_resp=="4")])/4,
sum(arr_resp[(arr_resp=="5")])/5,
sum(arr_resp[(arr_resp=="6")])/6,
sum(arr_resp[(arr_resp=="7")])/7,
sum(arr_resp[(arr_resp=="8")])/8,
sum(arr_resp[(arr_resp=="9")])/9,
sum(arr_resp[(arr_resp=="10")])/10,
sum(arr_resp[(arr_resp=="99")])/99
)
}
adply(DS, 2, conta_respostas)
X1 V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 V11
1 Q_092 1 1 1 1 1 0 0 0 0 1 1
2 Q_093 2 2 0 0 1 1 0 0 0 0 1
3 Q_094 0 1 1 1 1 1 1 1 0 0 0
4 Q_095 0 0 0 0 2 2 1 1 0 1 0
5 Q_096 1 1 1 1 1 0 0 0 0 1 1
Without having to write that function, you can do something like this:
t(apply(DS, 2, function(x) table(factor(x, levels = c('1', '2', '3', '4', '5',
'6', '7', '8', '9', '10',
'99')))))
This will do the following:
transform your data into factors with the levels as input in levels =. Having your data as a factor will allow you to avoid the levels where no respondents chose that response to be left out.
This creates a table for each variable with a cell for each of the factor levels.
This function is applied over the five variable columns that you are interested in.
Finally, the output from the apply() function is transposed to match the output from your original output:
1 2 3 4 5 6 7 8 9 10 99
Q_092 1 1 1 1 1 0 0 0 0 1 1
Q_093 2 2 0 0 1 1 0 0 0 0 1
Q_094 0 1 1 1 1 1 1 1 0 0 0
Q_095 0 0 0 0 2 2 1 1 0 1 0
Q_096 1 1 1 1 1 0 0 0 0 1 1

Answer 2

One option would be to use the apply() function with FUN=table. The only issue here is that your tables may be of different lengths, thus the final result may not be combined row-by-row.