For example: Generate a random value rand (range is 1-100). The probability of a small value appearing in Rand is high, but the probability of a large value is very low. Hope the great God gives a concrete formula. Random values are substituted with Rand for the line. Best can be implemented in code, such as PHP python, thank you!
Reply content:
Given a cumulative distribution function (cumulative distribution function, CDF), a uniformly distributed random number can be mapped to it as long as it can find its inverse function. This is called Inverse transform sampling
。 Random Sampling (Numpy.random)
Numpy.random has implemented a sample function from a well-defined distribution,
Like what
beta(a, b[, size])Draw samples from a Beta distribution.binomial(n, p[, size])Draw samples from a binomial distribution.chisquare(df[, size])Draw samples from a chi-square distribution.dirichlet(alpha[, size])Draw samples from the Dirichlet distribution.exponential([scale, size])Draw samples from an exponential distribution.f(dfnum, dfden[, size])Draw samples from an F distribution.gamma(shape[, scale, size])Draw samples from a Gamma distribution.geometric(p[, size])Draw samples from the geometric distribution.gumbel([loc, scale, size])Draw samples from a Gumbel distribution.
Assuming that the landlord takes a 1-100 integer, 100 of the probability is greater than 99 times, 99:98 big one times,
Then press 1 for the benchmark,
Sum sum= 1+2+ 4 + 2^99 =2^100-1
Then the random rand (1,sum), if between the 2^30-1 to 2^31 range, then is taken to 30. The simple formula is log (rand (1,sum))/log2 The ugly, there are mistakes to point out!
It happened that I met this problem and shared it with you.
Let's say we get the probability of each value, then we accumulate the probability of each value, and the result is uniformly placed on a number axis for each increment. Let's take 10 values to illustrate the problem, such as:
[I] represents the first value, the horizontal axis (0-0.1) represents the probability of [1], (0.5-0.75) represents the probability of [7], (0-0.75) represents [1] to [7] the probability of accumulation and.
Next, for example in MATLAB, use the rand () function to produce a random number with a range of 0-1 evenly distributed, the number generated in the false is 0.65, it falls in the interval (0.5-0.75), and this interval represents the probability of [7]. Thus, we decided to take [7] out as the result of this probability-based filter.
Code attached with MATLAB
% where the number of max_column values, prob represents each
The probability of the% value, Prob_array represents the cumulative probability and
%largest_cumulative_prob represents the cumulative maximum value of probability
For j=1:max_column-1
Prob_array (j+1) = Prob_array (j) +prob (j);
End
Largest_cumulative_prob=prob_array (Max_column);
%largest_cumulative_prob and Prob_array need to do
Percent normalization, not done in the code
Choice = rand () * LARGEST_CUMULATIVE_PROB;
low = 1;
High = Max_column;
while (High> (low+1))
Middle = (high+low)/2;
Middle = floor (middle);
if (Choice > Prob_array (middle))
low = middle;
else high = middle;
End
End
% finally remove low as the result of this screening
% is not explained here, and its thought is the same as in the picture above
This method is suitable for discrete numbers, for consecutive numbers I do not know how to deal with, personal conjecture may use probability density? Hope Big God answer! Method of Selection.
Do not worry about the number of calculations in particular: Do not worry about the number of calculations in particular:
Read a bit of Ndndsim source. The method of generating Zipf distribution inside is interesting. I hope I can give the main reference. Given cumulative distribution functions and ranges such as 1-50. Calculates the cumulative distribution function in an array of 1 to 50 function values. Then a random number is generated with a homogeneous distribution. Then compare the values of the 50 distribution functions in turn. Returns the subscript. Of course, the return subscript can vary with the requirements. 100/rand (1,100) rand (1,rand (1,rand (1,100)) so you can see.
