Question

I would like to compare two lists (two rows of a data frame) and count how many differences I have between the two lists.

for instance:

list1=a,b,c,a
list2=a,a,d,d

two elements of list 1 are in list 2

I am able to do that with a loop and sum but it is very inefficient. Is there any function to do that in R?

I have checked setdiff and the compare package but did not find anything that helps.

Thanks for your ideas,

Vincent

My function looks like:

        NRebalancing=function(NamePresent)
        {
          Nbexchange=NamePresent[,2]
          Nbexchange=NamePresent[1,2]=0

          for (i in 2:nrow(NamePresent))
          {
            print(i)
            compteur=0
            NameNeeded=NamePresent[i,]
            NameNeeded=unique(NameNeeded)
            NameNeeded=na.omit(NameNeeded)
            for(j in 2:length(NameNeeded))
              #j=1 correspond a une date
            {
              compteur = compteur+(abs(sum(NamePresent[i,]==as.character(NameNeeded[j]))-sum(NamePresent[i-1,]==as.character(NameNeeded[j]))))
            }
          Nbexchange[i]=compteur  
          }

          return(Nbexchange)
        }

Answer 1

One main point: your list isn’t an R list – that’s something a bit special. You are using vectors:

R> is.vector(l1)
[1] TRUE
R> is.list(l1)
[1] FALSE

don’t call variables list1 if they are vectors.

---

Since you have a vector there are lots of possibilities open.

1. The %in% operator

   R> l1 = c("a", "b", "c", "d")
   R> l2 = c("a", "a", "d", "d")
   R> l1[l1 %in% l2]
    [1] "a" "d"
   

2. Or use is.element

   R> l1[is.element(l1, l2)]
    [1] "a" "d"
   

3. There is also unique:

   R> unique(l2)
    [1] "a" "d"
   

   Following your comment to @mrdwab, you can count the number of occurances using a combination of sapply and unique

   sapply(unique(l1), function(i) sum(i==l2))
   

   i==l2 checks for membership, sum counts the number of times TRUE appears and sapply is basically just a for loop over unique(l1)

   R> sapply(unique(l1), function(i) sum(i==l2))
   a b c d 
   2 0 0 2
   

4. A very nice suggestion from @mrdwab is to use table and colSums:

   R> table(l1, l2)
     l2 l1  
      a d
    a 1 0
    b 1 0
    c 0 1
    d 0 1
   R> colSums(table(l1, l2))
    a d 
    2 2