If A and B are sets A – B is everything in A that’s not in B. >>> A = {1,2,3} >>> B = {1,4,5} >>> >>> A – B {2, 3} >>> B – A {4, 5} A.symmetric_difference(B) are all the elements that are in exactly one set, i.e. the union of A – … Read more
To perform result1 – result2, you can join result1 with result2, and only output items that exist in result1. For example: SELECT DISTINCT result1.column FROM result1 LEFT JOIN result2 ON result1.column = result2.column WHERE result2.column IS NULL In set theory terms this is a set difference of result1 and result2, i.e. elements in result1 but … Read more
To get elements which are in temp1 but not in temp2 (assuming uniqueness of the elements in each list): In [5]: list(set(temp1) – set(temp2)) Out[5]: [‘Four’, ‘Three’] Beware that it is asymmetric : In [5]: set([1, 2]) – set([2, 3]) Out[5]: set([1]) where you might expect/want it to equal set([1, 3]). If you do want set([1, 3]) … Read more
You could define three functions inBoth, inFirstOnly, and inSecondOnly which all take two lists as arguments, and return a list as can be understood from the function name. The main logic could be put in a common function operation that all three rely on. Here are a few implementations for that operation to choose from, … Read more
Yes there is, it is in <algorithm> and is called: std::set_difference. The usage is: #include <algorithm> #include <set> #include <iterator> // … std::set<int> s1, s2; // Fill in s1 and s2 with values std::set<int> result; std::set_difference(s1.begin(), s1.end(), s2.begin(), s2.end(), std::inserter(result, result.end())); In the end, the set result will contain the s1-s2.
The comm command does that.
set.difference, set.union… can take any iterable as the second arg while both need to be sets to use -, there is no difference in the output. Operation Equivalent Result s.difference(t) s – t new set with elements in s but not in t With .difference you can do things like: s1 = set([1,2,3]) print(s1.difference(*[[3],[4],[5]])) {1, … Read more
I don’t know if this is most effective, but perhaps the shortest: var A = [1, 2, 3, 4]; var B = [1, 3, 4, 7]; var diff = A.filter(function(x) { return B.indexOf(x) < 0; }); console.log(diff); // [2] Updated to ES6: const A = [1, 2, 3, 4]; const B = [1, 3, 4, … Read more
To get elements which are in temp1 but not in temp2 : In [5]: list(set(temp1) – set(temp2)) Out[5]: [‘Four’, ‘Three’] Beware that it is asymmetric : In [5]: set([1, 2]) – set([2, 3]) Out[5]: set([1]) where you might expect/want it to equal set([1, 3]). If you do want set([1, 3]) as your answer, you can use set([1, … Read more