{"id":114621,"date":"2020-06-25T12:45:52","date_gmt":"2020-06-25T12:45:52","guid":{"rendered":"https:\/\/blog.finxter.com\/?p=10063"},"modified":"2020-06-25T12:45:52","modified_gmt":"2020-06-25T12:45:52","slug":"the-most-pythonic-way-to-check-if-two-unordered-lists-are-identical","status":"publish","type":"post","link":"https:\/\/sickgaming.net\/blog\/2020\/06\/25\/the-most-pythonic-way-to-check-if-two-unordered-lists-are-identical\/","title":{"rendered":"The Most Pythonic Way to Check if Two Unordered Lists Are Identical"},"content":{"rendered":"

To check if two unordered lists x<\/code> and y<\/code> are identical, compare the converted sets with set(x) == set(y)<\/code>. However, this loses all information about duplicated elements. To consider duplicates, compare the sorted lists with sorted(x) == sorted(y)<\/code>. Due to the efficient merge-sort-like implementation of the sorted()<\/code> function, this is quite fast for almost-sorted lists.<\/p>\n

\"\"<\/figure>\n

Problem<\/strong>: Given are two lists <\/a>x<\/code> and y<\/code>. You want to return True<\/code> if both lists contain the same elements, and otherwise False<\/code>. A variant of this problem is to ignore duplicates <\/a>(which makes this problem far simpler).<\/p>\n

Examples<\/strong>:<\/p>\n

x = [1, 2, 3, 4, 5]\ny = [1, 2, 3]\n# compare(x, y) --> False x = [1, 2, 3, 4, 5]\ny = [1, 2, 3, 5, 4]\n# compare(x, y) --> True x = [1, 2, 3, 4, 5]\ny = [1, 2, 3, 4, 5]\n# compare(x, y) --> True<\/pre>\n
Let’s discuss the most Pythonic ways of solving this problem. Here’s a quick interactive code overview:<\/p>\n