删除跨列的值包含4个以上唯一字符中的2个的行

Question

Hopefully the wording of the title makes sense. I have a data frame that consists of values: "A", "B", "C", "D", "", "A/B". I want to identify which rows contain only 2 of "A", "B", "C", or "D". The frequency of each of these letters within the row does not matter. I just want to know if more than 2 of those 4 letters exists in the row.

Here is a sample data frame:

    df.sample = as.data.frame(rbind(c("A","B","A","A/B","B","B","B","B","","B"),c("A","B","C","A","B","","","B","","B"),c("A","B","D","D","B","B","B","B","","B"),c("A","B","A","A","B","B","B","B","B","B")))
    df.sample

      V1 V2 V3  V4 V5 V6 V7 V8 V9 V10
    1  A  B  A A/B  B  B  B  B      B
    2  A  B  C   A  B        B      B
    3  A  B  D   D  B  B  B  B      B
    4  A  B  A   A  B  B  B  B  B   B

I want to apply a function to each row that determines how many of each of the 4 letters ("A","B","C",or "D") exist, not the frequency of each, but essentially just a 0 or 1 value for "A", "B", "C", and "D". If the sum of those 4 values is > 3, then I want to assign the index of that row to a new vector which will be used to remove those rows from the data frame.

    myfun (x){
      #which rows contain > 2 different letters of A, B, C, or D.
      #The number of times each letter occurs in a given row does not matter. 
      #What matters is if each row contains more than 2 of the 4 letters. Each row should only contain 2 of them. The combination does not matter.

      out = which(something > 2)
    }

    row.indexes = apply(df.sample,1,function(x) myfun(x)) #Return a vector of row indexes that contain more than 2 of the 4 letters.

    new.df.sample = df.sample[-row.indexes,] #create new data frame excluding rows containing more than 2 of the 4 letters.

In the df.sample above, rows 2 and 3 contain more than 2 of those 4 letters and thus should be indexed for removal. After running the df.sample through the function and removing rows in row.indexes, my new.df.sample data frame should look like this:

      V1 V2 V3  V4 V5 V6 V7 V8 V9 V10
    1  A  B  A A/B  B  B  B  B      B
    4  A  B  A   A  B  B  B  B  B   B

I have tried to think of this as a logical statement for each of the 4 letters which then assigns a 0 or 1 to each letter, sums them up, and then identifies which ones sum to > 2. For instance, I thought perhaps I could try 'grep()' and convert that to a logical for each letter, which was then converted to a 0 or 1 and summed. That seems too lengthy and didn't work with the way I tried it. Any ideas?

Answer 1

Here's a function for this task. The function returns a logical value. TRUE indicates rows with more than two different strings:

myfun <- function(x) {
  sp <- unlist(strsplit(x, "/"))
  length(unique(sp[sp %in% c("A", "B", "C", "D")])) > 2
}

row.indexes <- apply(df.sample, 1, myfun)
# [1] FALSE  TRUE  TRUE FALSE

new.df.sample <- df.sample[!row.indexes, ] # negate the index with '!'

#   V1 V2 V3  V4 V5 V6 V7 V8 V9 V10
# 1  A  B  A A/B  B  B  B  B      B
# 4  A  B  A   A  B  B  B  B  B   B