R: Problem with calculating mean accuracy with strsplit function - r

Question

I am trying to calculate the accuracy of participants' response (columns EQ_R and MEM_R) based on the correct response (columns EQ_C and MEM_C).
dput(example)
structure(list(TRIAL = c("1", "2", "3", "4", "5", "6", "7", "8",
"9", "10", "11", "12", "13", "14", "15"), EQ_C = c("0101", "1010",
"1010", "00111", "01011", "01101", "100011", "010101", "001101",
"0110011", "1101001", "1100101", "11100001", "11001010", "11001010"
), EQ_R = c("0101", "0010", "1010", "00111", "01011", "01101",
"10101", "11010", "001101", "0100011", "1101001", "0100101",
"11110001", "11001010", "11001010"), MEM_C = c("ZLHK", "RZKX",
"DGWL", "BCJSP", "WRKTJ", "CHBXS", "HNDCWX", "SWVNDT", "WLDGPB",
"DSHRKBV", "HCXLZWB", "HDNBVZC", "BCRHKVDM", "RVTBWKFS", "NWHVZFLD"
), MEM_R = c("ZLHK", "RZKX", "DGWL", "BCJSP", "WRKLTJ", "CHBXS",
"HNDCWX", "SWVDTN", "WLDGPB", "DSHRKBV", "HCXLZWB", "HDNBVZC",
"BCRHKVDM", "RVTBWKFS", "NWHVZFLD"), EQ_SUM = c(NA, NA, NA, NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA), MEM_SUM = c(NA,
NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA, NA)), row.names
= c(NA,
15L), class = "data.frame")
I added a new column for the "sum"/accuracy scores that need to be calculated for the binary data (EQ) and letters (MEM).
OSPAN["EQ_SUM"] <- NA
OSPAN["MEM_SUM"]<- NA
I then tried to calculate the accuracy with strsplit, but I receive error notifications.
mean(strsplit(OSPAN$MEM_C, "") == strsplit(OSPAN$MEM_R, ""))
Error in strsplit(OSPAN$MEM_C, "") == strsplit(OSPAN$MEM_R, "") : comparison of these types is not implemented
In addition:
Warning messages:
1: In strsplit(OSPAN$MEM_R, "") : input string 342 is invalid UTF-8
2: In strsplit(OSPAN$MEM_R, "") : input string 580 is invalid UTF-8
My question is:
How do I match/calculate the accuracy or congruence between predictor (C) and actual (R) values into the sum columns?
For instance, in row #1, EQ_SUM would be 1 (or 100%), whereas it would be 0.75 or 75% in #2, as the participant chose the wrong answer (0 instead of 1). Thus, partial credit scores are given, and it is not a matter of absolute match/congruence.
Thank you in advance.

Answer 1

One possibility could be using the RecordLinkage library:
with(df, levenshteinSim(EQ_C, EQ_R))
[1] 1.0000000 0.7500000 1.0000000 1.0000000 1.0000000 1.0000000 0.6666667 0.6666667
[9] 1.0000000 0.8571429 1.0000000 0.8571429 0.8750000 1.0000000 1.0000000
It calculates the similarity between the two strings using the Levenshtein distance.

Answer 2

I'm sure there is a most efficient way, however, you can compare list by list and add it to your data frame.
for (i in 1:nrow(OSPAN)){
OSPAN$EQ_SUM[i] <- sum(strsplit(OSPAN$EQ_C, "", useBytes = TRUE)[[i]] == strsplit(OSPAN$EQ_R, "", useBytes = TRUE)[[i]])/length(strsplit(OSPAN$EQ_C, "")[[i]])
OSPAN$MEM_SUM[i] <- sum(strsplit(OSPAN$MEM_C, "", useBytes = TRUE)[[i]] == strsplit(OSPAN$MEM_R, "", useBytes = TRUE)[[i]])/length(strsplit(OSPAN$MEM_C, "")[[i]])
}
On the other hand, there are cases with different length, what do we do with them?