Xcode 11 • Swift 5.1 extension Collection { func choose(_ n: Int) -> ArraySlice<Element> { shuffled().prefix(n) } } Playground testing var alphabet = [“A”,”B”,”C”,”D”,”E”,”F”,”G”,”H”,”I”,”J”,”K”,”L”,”M”,”N”,”O”,”P”,”Q”,”R”,”S”,”T”,”U”,”V”,”W”,”X”,”Y”,”Z”] let shuffledAlphabet = alphabet.shuffled() // “O”, “X”, “L”, “D”, “N”, “K”, “R”, “E”, “S”, “Z”, “I”, “T”, “H”, “C”, “U”, “B”, “W”, “M”, “Q”, “Y”, “V”, “A”, “G”, “P”, “F”, “J”] … Read more
You need to pass pointers to a[0] and a[10], not the elements themselves: random_shuffle(&a[0], &a[10]); // end must be 10, not 9 In C++11, you can use std::begin and std::end: random_shuffle(std::begin(a), std::end(a));
In Scala you can use scala.util.Random: util.Random.shuffle((1 to 10).toSeq) //Vector(9, 6, 8, 7, 10, 1, 2, 5, 3, 4) util.Random.shuffle(List(‘A’, ‘B’, ‘C’, ‘D’, ‘E’, ‘F’)) //List(B, D, A, E, C, F) Your results may vary…
If you have a shuffled desk, into which you wish to shuffle a batch of new cards (and you know that none of the cards are duplicates), then I think the following is valid. ForEach card in batch: gap = random(deck.size() + 1) # choose a gap between cards, before first, or after last. deck.insertAt(gap,card) … Read more
TL;DR: It “breaks” on lists with over 2080 elements, but don’t worry too much 🙂 Complete answer: First of all, notice that “shuffling” a list can be understood (conceptually) as generating all possible permutations of the elements of the lists, and picking one of these permutations at random. Then, you must remember that all self-contained … Read more
You could use the non-inplace numpy.random.permutation with explicit non-zero indexing: >>> X = np.array([[2,3,1,0], [0,0,2,1]]) >>> for i in range(len(X)): … idx = np.nonzero(X[i]) … X[i][idx] = np.random.permutation(X[i][idx]) … >>> X array([[3, 2, 1, 0], [0, 0, 2, 1]])
Given the relationship demonstrated in the question, I’m going to assume the lists are the same length and that list1[i] corresponds to list2[i] for any index i. With that assumption in place, shuffling the lists is as simple as shuffling the indices: from random import shuffle # Given list1 and list2 list1_shuf = [] list2_shuf … Read more
Yes, you can look at the code; it basically does a Fisher-Yates shuffle. Here it is (thanks OpenJDK, and yay for open source :-P): public static void shuffle(List<?> list, Random rnd) { int size = list.size(); if (size < SHUFFLE_THRESHOLD || list instanceof RandomAccess) { for (int i=size; i>1; i–) swap(list, i-1, rnd.nextInt(i)); } else … Read more
Built in now: [1,2,3,4].shuffle => [2, 1, 3, 4] [1,2,3,4].shuffle => [1, 3, 2, 4]
To shuffle the rows of a matrix, you can use RANDPERM shuffledArray = orderedArray(randperm(size(orderedArray,1)),:); randperm will generate a list of N random values and sort them, returning the second output of sort as result.