Count number of non-blank columns and assign to individual respondents - r

Question

I am trying to get a total number of friends that will become the denominator in a later step.
example data:
set.seed(24) ## for sake of reproducibility
n <- 5
data <- data.frame(id=1:n,
Q1= c("same", "diff", NA, NA, NA),
Q2= c("diff", "diff", "same", "diff", NA),
Q3= c("same", "diff", NA ,NA, "diff"),
Q4= c("diff", "same", NA, NA, NA))
i first need to create a column that contains a numeric count of how many columns each participant responded to (either "same" or "diff", not counting NAs/blanks). I have tried the following
friendship <- total.friends <- rowSums(c(data$Q1, data$Q2, data$Q3, data$Q4)), != "")
friendship <- total.friends <-rowSums(!is.na(c(data$Q1, data$Q2, data$Q3, data$Q4)))
Neither is effective, likely because my data is not numeric. the first did count the cells but did not group by id as I require. is there any function i can use to count the populated cells? how can i edit this to count cells populated only with "diff" so that i can then start on the second step (making the proportion)?

Answer 1

You could
data2 <- apply(data[,-1],MARGIN=1,function(x){c <- length(x[!is.na(x)])})
result <- as.data.frame(cbind(data[,1],data2)) %>% setNames(c("id","number"))
And result will hold the amount of not NA each id has.
The data2 is basically a count of the number of not NAs for each id, it uses the apply function with margin 1 which basically takes each row of your dataframe and applies a function to that row. The function that is being applied is the c<-length(x[!is.na(x)] part. Which basically, the 'x[!is.na(x)]' filters away all the NA entries in each row so that it only has NOT NA entries of the row, then we apply the length() function to that result so it gives us how many entries where there after filtering the NAs.
The result of that apply will be a single column array, in which each row is the result of computing that procedure to each row, and considering you have a row for each id. It translates as computing that function to each id
Lastly, in the result line I simply add the id back to the previous step, for the sake of having in it well identified and not just one column of results.
Hope this works for you :)

Answer 2

Here's a regex solution with grep:
data$count <- apply(data, 1, function(x) length(grep("[a-z]", x, value = T)))
Here using length you count the number of times grep finds a lower-case letter in any row cell.
Result:
data
id Q1 Q2 Q3 Q4 count
1 1 same diff same diff 4
2 2 diff diff diff same 4
3 3 <NA> same <NA> <NA> 1
4 4 <NA> diff <NA> <NA> 1
5 5 <NA> <NA> diff <NA> 1

Answer 3

You can also accomplish this using c_across and rowwise from the dplyr library:
library(dplyr)
data %>%
dplyr::rowwise() %>%
dplyr::mutate(Total = sum(!is.na(c_across(Q1:Q4)))) %>%
dplyr::ungroup()
Note: alternatively you can use starts_with("Q") inside of c_across to do this across all columns that start with "Q" (shown below).
To count the number of a specific response you can do or compute other variables that depend on a newly created variable, like a proportion, in the mutate statement:
data %>%
dplyr::rowwise() %>%
dplyr::mutate(Total = sum(!is.na(c_across(starts_with("Q")))),
Diff = sum(c_across(starts_with("Q")) == "diff", na.rm = T),
Prop = Diff / Total) %>%
dplyr::ungroup()
id Q1 Q2 Q3 Q4 Total Diff Prop
<int> <chr> <chr> <chr> <chr> <int> <int> <dbl>
1 1 same diff same diff 4 2 0.5
2 2 diff diff diff same 4 3 0.75
3 3 NA same NA NA 1 0 0
4 4 NA diff NA NA 1 1 1
5 5 NA NA diff NA 1 1 1