set-difference | 易学教程

MySQL: difference of two result sets

阅读更多关于 MySQL: difference of two result sets

问题 How can I get the set difference of two result sets? Say I have a result set (just one column in each): result1: 'a' 'b' 'c' result2: 'b' 'c' I want to minus what is in result1 by result2: result1 - result2 such that it equals: difference of result1 - result2: 'a' 回答1: 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

Difference and intersection of two arrays containing objects

阅读更多关于 Difference and intersection of two arrays containing objects

I have two arrays list1 and list2 which have objects with some properties; userId is the Id or unique property: list1 = [ { userId: 1234, userName: 'XYZ' }, { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1237, userName: 'WXYZ' }, { userId: 1238, userName: 'LMNO' } ] list2 = [ { userId: 1235, userName: 'ABC' }, { userId: 1236, userName: 'IJKL' }, { userId: 1252, userName: 'AAAA' } ] I'm looking for an easy way to execute the following three operations: list1 operation list2 should return the intersection of elements: [ { userId: 1235, userName: 'ABC' }, {

Find the set difference between two large arrays (matrices) in Python

阅读更多关于 Find the set difference between two large arrays (matrices) in Python

问题 I have two large 2-d arrays and I'd like to find their set difference taking their rows as elements. In Matlab, the code for this would be setdiff(A,B,'rows') . The arrays are large enough that the obvious looping methods I could think of take too long. 回答1: This should work, but is currently broken in 1.6.1 due to an unavailable mergesort for the view being created. It works in the pre-release 1.7.0 version. This should be the fastest way possible, since the views don't have to copy any

Difference and intersection of two arrays containing objects

阅读更多关于 Difference and intersection of two arrays containing objects

问题 I have two arrays list1 and list2 which have objects with some properties; userId is the Id or unique property: list1 = [ { userId: 1234, userName: \'XYZ\' }, { userId: 1235, userName: \'ABC\' }, { userId: 1236, userName: \'IJKL\' }, { userId: 1237, userName: \'WXYZ\' }, { userId: 1238, userName: \'LMNO\' } ] list2 = [ { userId: 1235, userName: \'ABC\' }, { userId: 1236, userName: \'IJKL\' }, { userId: 1252, userName: \'AAAA\' } ] I\'m looking for an easy way to execute the following three

Exclude characters from a character class

阅读更多关于 Exclude characters from a character class

问题 Is there a simple way to match all characters in a class except a certain set of them? For example if in a lanaguage where I can use \\w to match the set of all unicode word characters, is there a way to just exclude a character like an underscore \"_\" from that match? Only idea that came to mind was to use negative lookahead/behind around each character but that seems more complex than necessary when I effectively just want to match a character against a positive match AND negative match.

What is the fastest or most elegant way to compute a set difference using Javascript arrays?

阅读更多关于 What is the fastest or most elegant way to compute a set difference using Javascript arrays?

问题 Let A and B be two sets. I\'m looking for really fast or elegant ways to compute the set difference ( A - B or A \\B , depending on your preference) between them. The two sets are stored and manipulated as Javascript arrays, as the title says. Notes: Gecko-specific tricks are okay I\'d prefer sticking to native functions (but I am open to a lightweight library if it\'s way faster) I\'ve seen, but not tested, JS.Set (see previous point) Edit: I noticed a comment about sets containing duplicate

Get difference between two lists

阅读更多关于 Get difference between two lists

问题 I have two lists in Python, like these: temp1 = [\'One\', \'Two\', \'Three\', \'Four\'] temp2 = [\'One\', \'Two\'] I need to create a third list with items from the first list which aren\'t present in the second one. From the example I have to get: temp3 = [\'Three\', \'Four\'] Are there any fast ways without cycles and checking? 回答1: In [5]: list(set(temp1) - set(temp2)) Out[5]: ['Four', 'Three'] Beware that In [5]: set([1, 2]) - set([2, 3]) Out[5]: set([1]) where you might expect/want it to