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
A solution without indexOf. var firstObject = { x: 0, y: 1, z: 2, a: 10, b: 20, e: 30 }, secondObject = { x: 0, y: 1, z: 2, a: 10, c: 20, d: 30 }; function intersection(o1, o2) { return Object.keys(o1).concat(Object.keys(o2)).sort().reduce(function (r, a, i, aa) { if (i && aa[i – 1] === … Read more
When you do set() you are creating an empty set. When you do set().intersection(…) you are intersecting this empty set with other stuff. The intersection of an empty set with any other collection of sets is empty. If you actually have a list of sets, you can get their intersection similar to how you did … Read more
Set.retainAll() is how you find the intersection of two sets. If you use HashSet, then converting your ArrayLists to Sets and using retainAll() in a loop over all of them is actually O(n).
The simplest approach is this: List<T> intersect = list1.stream() .filter(list2::contains) .collect(Collectors.toList());
How about an Enumerable.Intersect and StringComparer combo: // other options include StringComparer.CurrentCultureIgnoreCase // or StringComparer.InvariantCultureIgnoreCase var results = array1.Intersect(array2, StringComparer.OrdinalIgnoreCase);
// returns true if the line from (a,b)->(c,d) intersects with (p,q)->(r,s) function intersects(a,b,c,d,p,q,r,s) { var det, gamma, lambda; det = (c – a) * (s – q) – (r – p) * (d – b); if (det === 0) { return false; } else { lambda = ((s – q) * (r – a) + … 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
Just do: map.keySet().retainAll(set); As per the javadoc, the changes in the key set are reflected back in the map. … The set is backed by the map, so changes to the map are reflected in the set, and vice-versa. … Here’s a demo: var map = new HashMap<String, String>(); map.put(“1”, “one”); map.put(“2”, “two”); map.put(“3”, “three”); … Read more
/** * Finds the intersection point between * * the rectangle * with parallel sides to the x and y axes * * the half-line pointing towards (x,y) * originating from the middle of the rectangle * * Note: the function works given min[XY] <= max[XY], * even though minY may not be the “top” … Read more