Use the retainAll() method of Set: Set<String> s1; Set<String> s2; s1.retainAll(s2); // s1 now contains only elements in both sets If you want to preserve the sets, create a new set to hold the intersection: Set<String> intersection = new HashSet<String>(s1); // use the copy constructor intersection.retainAll(s2); The javadoc of retainAll() says it’s exactly what you … Read more
You can use np.sign in combination with np.diff and np.argwhere to obtain the indices of points where the lines cross (in this case, the points are [ 0, 149, 331, 448, 664, 743]): import numpy as np import matplotlib.pyplot as plt x = np.arange(0, 1000) f = np.arange(0, 1000) g = np.sin(np.arange(0, 10, 0.01) * … Read more
The point A is always outside of the rectangle and the point B is always at the center of the rectangle Assuming the rectangle is axis-aligned, this makes things pretty simple: The slope of the line is s = (Ay – By)/(Ax – Bx). If -h/2 <= s * w/2 <= h/2 then the line … Read more
set.intersection(*map(set,d))
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
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
You don’t need to define intersection. It’s already a first-class part of set. >>> b1 = [1,2,3,4,5,9,11,15] >>> b2 = [4,5,6,7,8] >>> set(b1).intersection(b2) set([4, 5])
If order is not important and you don’t need to worry about duplicates then you can use set intersection: >>> a = [1,2,3,4,5] >>> b = [1,3,5,6] >>> list(set(a) & set(b)) [1, 3, 5]
Simplest code for array intersection in javascript
Removing duplicates in lists