np.random.permutation has two differences from np.random.shuffle: if passed an array, it will return a shuffled copy of the array; np.random.shuffle shuffles the array inplace if passed an integer, it will return a shuffled range i.e. np.random.shuffle(np.arange(n)) If x is an integer, randomly permute np.arange(x). If x is an array, make a copy and shuffle the … Read more
Given the R data.frame: > df1 a b c 1 1 1 0 2 1 0 0 3 0 1 0 4 0 0 0 Shuffle row-wise: > df2 <- df1[sample(nrow(df1)),] > df2 a b c 3 0 1 0 4 0 0 0 2 1 0 0 1 1 1 0 By default sample() … Read more
The itertools module has a useful method called permutations(). The documentation says: itertools.permutations(iterable[, r]) Return successive r length permutations of elements in the iterable. If r is not specified or is None, then r defaults to the length of the iterable and all possible full-length permutations are generated. Permutations are emitted in lexicographic sort order. … Read more
To describe a permutation of n elements, you see that for the position that the first element ends up at, you have n possibilities, so you can describe this with a number between 0 and n-1. For the position that the next element ends up at, you have n-1 remaining possibilities, so you can describe … Read more
Basically, for each item from left to right, all the permutations of the remaining items are generated (and each one is added with the current elements). This can be done recursively (or iteratively if you like pain) until the last item is reached at which point there is only one possible order. So with the … Read more
Let’s look at some permutations: 1 2 3 4 1 2 4 3 1 3 2 4 1 3 4 2 1 4 2 3 1 4 3 2 2 1 3 4 … How do we go from one permutation to the next? Firstly, let’s look at things a little differently. We can view … Read more
There are several ways to do this. Common methods use recursion, memoization, or dynamic programming. The basic idea is that you produce a list of all strings of length 1, then in each iteration, for all strings produced in the last iteration, add that string concatenated with each character in the string individually. (the variable … Read more
First of all: it smells like recursion of course! Since you also wanted to know the principle, I did my best to explain it human language. I think recursion is very easy most of the times. You only have to grasp two steps: The first step All the other steps (all with the same logic) … Read more
Little late, but like to add a slightly more elegant version here. Can be any array… function permutator(inputArr) { var results = []; function permute(arr, memo) { var cur, memo = memo || []; for (var i = 0; i < arr.length; i++) { cur = arr.splice(i, 1); if (arr.length === 0) { results.push(memo.concat(cur)); } … Read more
Selecting a random permutation requires simultaneously more and less randomness than what your question implies. Let me explain. The bad news: need more randomness. The fundamental flaw in your approach is that it’s trying to choose between ~2226 possibilities using 64 bits of entropy (the random seed). To fairly choose between ~2226 possibilities you’re going … Read more