Here’s a useful trick 🙂 function randomWithProbability() { var notRandomNumbers = [1, 1, 1, 1, 2, 2, 2, 3, 3, 4]; var idx = Math.floor(Math.random() * notRandomNumbers.length); return notRandomNumbers[idx]; }
There are probably a lot of different (and good) answers, but in my humble opinion, the common characteristics of probabilistic data structures is that they provide you with approximate, not precise answer. How many items are here? About 1523425 with probability of 99% Update: Quick search produced link to decent article on the issue: Probabilistic … Read more
If you know in advance that you’re going to want to multiple samples without overlaps, easiest is to do random.shuffle() on list(range(100)) (Python 3 – can skip the list() in Python 2), then peel off slices as needed. s = list(range(100)) random.shuffle(s) first_sample = s[-10:] del s[-10:] second_sample = s[-10:] del s[-10:] # etc Else … Read more
Here are some links to articles, which could help as starting points: Poker Logic in C# and Fast, Texas Holdem Hand Evaluation and Analysis “This code snippet will let you calculate poker probabilities the hard way, using C# and .NET.” The theoretical fundamentals are given in this Wikipedia article about Poker Probabilities and in this … Read more
Equal hash means equal file, unless someone malicious is messing around with your files and injecting collisions. (this could be the case if they are downloading stuff from the internet) If that is the case go for a SHA2 based function. There are no accidental MD5 collisions, 1,47×10-29 is a really really really small number. … Read more
Per the SVC documentation, it looks like you need to change how you construct the SVC: model = SVC(probability=True) and then use the predict_proba method: class_probabilities = model.predict_proba(sub_main)
random.random() returns a uniformly distributed pseudo-random floating point number in the range [0, 1). This number is less than a given number p in the range [0,1) with probability p. Thus: def flip(p): return ‘H’ if random.random() < p else ‘T’ Some experiments: >>> N = 100 >>> flips = [flip(0.2) for i in xrange(N)] … Read more
If you call f(x) twice in a row, the following outcomes are possible (assuming that successive calls to f(x) are independent, identically distributed trials): 00 (probability 1/4 * 1/4) 01 (probability 1/4 * 3/4) 10 (probability 3/4 * 1/4) 11 (probability 3/4 * 3/4) 01 and 10 occur with equal probability. So iterate until you … Read more
You can easily achieve this with numpy. It has a choice function which accepts the parameter of probabilities. np.random.choice( [‘pooh’, ‘rabbit’, ‘piglet’, ‘Christopher’], 5, p=[0.5, 0.1, 0.1, 0.3] )