CODEWITHSUNDEEP
r
ron
ron

Reputation: 1

Comparing two lists and displaying the non-matching elements

I have two lists list_a and list_b. The structure of the lists is as follows

> list_b[1]
 [[1]]
 [1] ORANGE    COLUMNS        BLACK       BLUE    TRIGGER       
 [6] GREEN     DRAUGHT

 > list_a[1]
 [[1]]
 [1] RED       SPARROW         HAWK      TRIGGER     BLUE       
 [6] ORANGE     COLUMNS        TIGER      CAMEROON   BULLSEYE

> length(list_a)
[1] 1012
> length(list_b)
[1] 1

As we can see, list_b has only 1 list item which has several strings. I want to compare each element of list_b[1] with all the items of list_a i.e compare list_b[1] with list_a[1];list_b[1] with list_a[2];... and so on. The sequence in which the strings appear in list_b[1] and list_a[1:1012] is not important to me. For each match I would like to view the number of elements in list_b[1] that did not match and also the elements that did not match. Hence the comparison between list_b[1] and list_a[1] should indicate that the elements in list_b that did not match are BLACK,GREEN and DRAUGHT and therefore the count of non-matching elements is 3. I tried this using intersect() but was not successful. Please help.

Upvotes: 0

Views: 2948

Answers (1)

Erin
Erin

Reputation: 386

list_b = list(c('o', 'c', 'bk', 'bl', 't', 'g', 'd'))
list_a = list(c('r', 's', 'h', 't', 'bl', 'o', 'c', 'ti', 'ca', 'bu'),
              c('r', 's', 'h', 't', 'bl', 'o', 'c', 'ti', 'ca', 'bu'))
lapply(list_a, function(a) {
  b = list_b[[1]]
  non.matching = setdiff(b, a)
  return(c(length(non.matching), non.matching))
})

or

lapply(list_a, function(a) {
      b = list_b[[1]]
      non.matching = setdiff(b, a)
     return(list(num = length(non.matching), elements = non.matching))
   })

Upvotes: 2

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