Intersection of two circles Written by Paul Bourke The following note describes how to find the intersection point(s) between two circles on a plane, the following notation is used. The aim is to find the two points P3 = (x3, y3) if they exist. First calculate the distance d between the center of the circles. … Read more
You need to first transform data1, in your case by calling ToString() on each element. Use this if you want to return strings. List<int> data1 = new List<int> {1,2,3,4,5}; List<string> data2 = new List<string>{“6″,”3”}; var newData = data1.Select(i => i.ToString()).Intersect(data2); Use this if you want to return integers. List<int> data1 = new List<int> {1,2,3,4,5}; List<string> … Read more
You can use retainAll method: columnsOld.retainAll (columnsNew);
Place both series in Python’s set container then use the set intersection method: s1.intersection(s2) and then transform back to list if needed. Just noticed pandas in the tag. Can translate back to that: pd.Series(list(set(s1).intersection(set(s2)))) From comments I have changed this to a more Pythonic expression, which is shorter and easier to read: Series(list(set(s1) & set(s2))) … Read more
set.intersection(*map(set,d))
A little known fact is that you don’t need to construct sets to do this: Python 3 d1 = {‘a’: 1, ‘b’: 2} d2 = {‘b’: 2, ‘c’: 3} print(d1.keys() & d2.keys()) # {‘b’} Python 2 In Python 2, we replace keys with viewkeys. The same applies to values (viewvalues) and items(viewitems). In [78]: d1 … Read more
Yes there are differences which may or may not be relevant in your scenario. Perhaps the most significant is the difference in how members with the same property key are handled when present in both types. Consider: interface NumberToStringConverter { convert: (value: number) => string; } interface BidirectionalStringNumberConverter extends NumberToStringConverter { convert: (value: string) => … Read more
Here’s a plain implementation without using any third-party library. Main advantage over retainAll, removeAll and addAll is that these methods don’t modify the original lists input to the methods. public class Test { public static void main(String… args) throws Exception { List<String> list1 = new ArrayList<String>(Arrays.asList(“A”, “B”, “C”)); List<String> list2 = new ArrayList<String>(Arrays.asList(“B”, “C”, “D”, … Read more
Short answer: use not set(a).isdisjoint(b), it’s generally the fastest. There are four common ways to test if two lists a and b share any items. The first option is to convert both to sets and check their intersection, as such: bool(set(a) & set(b)) Because sets are stored using a hash table in Python, searching them … Read more
I assume X and Y are arrays? If so, there’s a very simple way to do this: x = [1, 1, 2, 4] y = [1, 2, 2, 2] # intersection x & y # => [1, 2] # union x | y # => [1, 2, 4] # difference x – y # => … Read more