4.65761066e-03 + 9.95342389e-01 = 1 9.75851270e-01 + 2.41487300e-02 = 1 9.99983374e-01 + 1.66258341e-05 = 1 The first column is the probability that the entry has the -1 label and the second column is the probability that the entry has the +1 label. Note that classes are ordered as they are in self.classes_. If you would … Read more
Duplicate of Expected collisions for perfect 32bit crc The answer referenced this article: http://arstechnica.com/civis/viewtopic.php?f=20&t=149670 Found the image below from: http://preshing.com/20110504/hash-collision-probabilities
You’d use Random to generate a random number, then test it against a literal to match the probability you’re trying to achieve. So given: boolean val = new Random().nextInt(25)==0; val will have a 1/25 probability of being true (since nextInt() has an even probability of returning any number starting at 0 and up to, but … Read more
One way is checking that the return value of Random.Next(100) is less than your desired probability. I can’t speak to the true ‘randomness’ of this method though. Proper example, using desired probability of 20%: Random gen = new Random(); int prob = gen.Next(100); return prob < 20;
Yes, there does seem to be 3rd libraries (none in Java Math). Two that have come up are: http://opsresearch.com/app/ http://www.iro.umontreal.ca/~simardr/ssj/indexe.html but, it is actually not that difficult to write your own methods to calculate mean, median, mode and range. MEAN public static double mean(double[] m) { double sum = 0; for (int i = 0; … Read more
Use : norm = [float(i)/sum(raw) for i in raw] to normalize against the sum to ensure that the sum is always 1.0 (or as close to as possible). use norm = [float(i)/max(raw) for i in raw] to normalize against the maximum
Yours is a pretty good way already and works well with any range. Just thinking: another possibility is to get rid of the fractions by multiplying with a constant multiplier, and then build an array with the size of this multiplier. Multiplying by 10 you get P(1) = 2 P(2) = 3 P(3) = 5 … Read more
The multivariate normal is now available on SciPy 0.14.0.dev-16fc0af: from scipy.stats import multivariate_normal var = multivariate_normal(mean=[0,0], cov=[[1,0],[0,1]]) var.pdf([1,0])
Are the 160 bit hash values generated by SHA-1 large enough to ensure the fingerprint of every block is unique? Assuming random hash values with a uniform distribution, a collection of n different data blocks and a hash function that generates b bits, the probability p that there will be one or more collisions is … Read more
Well I had a little fun with this one. The first thing I thought of when I first read the problem was group theory (the symmetric group Sn, in particular). The for loop simply builds a permutation σ in Sn by composing transpositions (i.e. swaps) on each iteration. My math is not all that spectacular … Read more