![]() ![]() ![]() Now notice that I needed to add a pop to the variable since now it keeps the elements after the call to the function.Īlso I needed to use the nonlocal term so the function will know the cur variable. Yield from permutations_with_replacement_rec(n, m) Yield from permutations_with_replacement_rec(n_rec - 1, m_rec) This is how it turned out: def permutations_with_replacement(n: int, m: int) -> Iterator]:ĭef permutations_with_replacement_rec(n_rec: int, m_rec: int) -> Iterator]: I created a sub function that is recursive taken out the cur variable. So you said you do not like the way cur is implemented in your code, I thought of a way to take it out of the function declaration. Since it's a generator it should be: Iterator]. I added type annotation to the return value of the function.I do not think n and m should be in main.Is there any better way to do that?įirst of all, this is a very nicely written code.īefore writing about the cur implementation, a few notes: I don't like the way I've implemented cur in my code. The arguments to the constructor are the elements of the permutation’s word representation, i.e., the images of the integers 1 through some n under the permutation. There's a better way to do this if we used itertools.product, but there's no fun in that! from itertools import productįor i in product(list(range(1, m + 1)), repeat=n): A Permutation object represents a permutation of finitely many positive integers, i.e., a bijective function from some integer range 1, n to itself. If x is an integer, randomly permute np.arange(x).If x is an array, make a copy and shuffle the elements randomly. Yield from permutations_with_replacement(n - 1, m, cur + )įor i in permutations_with_replacement(n, m): permutation (x, axis 0) Randomly permute a sequence, or return a permuted range. So, I've written a function to do just that: from sys import setrecursionlimitĭef permutations_with_replacement(n: int, m: int, cur=None): I have looked at the random library but couldn't find something that fits my need.Many a times, I've had the need to use a permutations with replacement function. The permutations command from itertools generates all the possible permutations which in my case are too many to handle. I could not figure out or find a way to do this. (Or find a trade off between memory usage and the looping, so maybe do 5 loops for 4000 shuffles at the time). Therefore I though about trying to generate a 2D array of shape (len(ys),20000) with essentially 20000 shuffled ys array, and then access them at ones and calculate the 20000 differences (diff in the code). I am looking to speed up the permutation test code. Permutation importance is one of the most reliable ways to see the important. The two loops (within (lon,lat) and for 20000 shuffles) adds up. Think of those Python one-liners that are dreadful in terms of readability. Ideally I want to run a permutation test with 20000 permutations. I run it using eratools: for ii in om_iterable(zip(*dataset.variable())):ĭataset.perm_test = exact_mc_perm_test1(iis, ii.values, 1000.,selected_position) the dataset has (lon,lat,time) coordinates, I need to run it for each (lon,lat) position (along the time dimension) I have to run it over a 3D array stored in an xarray. I took this SO answer and modify it for my specific test. # difference now between fixed all time series and shuffled subsamplevaluesĭiff_shuffled = np.abs(np.mean(ys) - mean_ys) Two such features Ive discovered recently are the permutations and combinations functions of Pythons itertools module. We will be using the combinations () and permutations () methods under the itertools module of Python. We will look at sets of characters and numbers. ![]() My code: def exact_mc_perm_test(ys, nmc,boolean_selection):ĭiff = np.abs(np.mean(ys) - mean_ys) In this tutorial, we will learn how to get the permutations and combinations of a group of elements in Python. This method is defined in itertool package. However I have to repeat it many times, and it is taking a too long time. Python provides one utility method to find out the permutation of a iterable. I developed, thanks to many SO questions, a decent permutation test routine. ![]() I have browsed SO extensively and I have found many questions about generating all possible permutations, but none regarding generating a specific number of permutations. ![]()
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