Python: Random Number Generator With Mean And Standard Deviation
Solution 1:
You are looking for stats.truncnorm:
import scipy.statsas stats
a, b = 500, 600
mu, sigma = 550, 30
dist = stats.truncnorm((a - mu) / sigma, (b - mu) / sigma, loc=mu, scale=sigma)
values = dist.rvs(1000)
Solution 2:
There are other choices for your problem too. Wikipedia has a list of continuous distributions with bounded intervals, depending on the distribution you may be able to get your required characteristics with the right parameters. For example, if you want something like "a bounded Gaussian bell" (not truncated) you can pick the (scaled) beta distribution:
import numpy as np
import scipy.stats
import matplotlib.pyplot as plt
defmy_distribution(min_val, max_val, mean, std):
scale = max_val - min_val
location = min_val
# Mean and standard deviation of the unscaled beta distribution
unscaled_mean = (mean - min_val) / scale
unscaled_var = (std / scale) ** 2# Computation of alpha and beta can be derived from mean and variance formulas
t = unscaled_mean / (1 - unscaled_mean)
beta = ((t / unscaled_var) - (t * t) - (2 * t) - 1) / ((t * t * t) + (3 * t * t) + (3 * t) + 1)
alpha = beta * t
# Not all parameters may produce a valid distributionif alpha <= 0or beta <= 0:
raise ValueError('Cannot create distribution for the given parameters.')
# Make scaled beta distribution with computed parametersreturn scipy.stats.beta(alpha, beta, scale=scale, loc=location)
np.random.seed(100)
min_val = 1.5
max_val = 35
mean = 9.87
std = 3.1
my_dist = my_distribution(min_val, max_val, mean, std)
# Plot distribution PDF
x = np.linspace(min_val, max_val, 100)
plt.plot(x, my_dist.pdf(x))
# Statsprint('mean:', my_dist.mean(), 'std:', my_dist.std())
# Get a large sample to check bounds
sample = my_dist.rvs(size=100000)
print('min:', sample.min(), 'max:', sample.max())
Output:
mean: 9.87 std: 3.100000000000001
min: 1.9290674232087306 max: 25.03903889816994
Probability density function plot:
Note that not every possible combination of bounds, mean and standard deviation will produce a valid distribution in this case, though, and depending on the resulting values of alpha
and beta
the probability density function may look like an "inverted bell" instead (even though mean and standard deviation would still be correct).
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