random

您所在的位置:网站首页 randomrandint取值范围 random

random

2024-01-02 11:03| 来源: 网络整理| 查看: 265

random — Generate pseudo-random numbers¶

Source code: Lib/random.py

This module implements pseudo-random number generators for various distributions.

For integers, there is uniform selection from a range. For sequences, there is uniform selection of a random element, a function to generate a random permutation of a list in-place, and a function for random sampling without replacement.

On the real line, there are functions to compute uniform, normal (Gaussian), lognormal, negative exponential, gamma, and beta distributions. For generating distributions of angles, the von Mises distribution is available.

Almost all module functions depend on the basic function random(), which generates a random float uniformly in the half-open range 0.0 > deck = 'ace two three four'.split() >>> shuffle(deck) # Shuffle a list >>> deck ['four', 'two', 'ace', 'three'] >>> sample([10, 20, 30, 40, 50], k=4) # Four samples without replacement [40, 10, 50, 30]

Simulations:

>>> # Six roulette wheel spins (weighted sampling with replacement) >>> choices(['red', 'black', 'green'], [18, 18, 2], k=6) ['red', 'green', 'black', 'black', 'red', 'black'] >>> # Deal 20 cards without replacement from a deck >>> # of 52 playing cards, and determine the proportion of cards >>> # with a ten-value: ten, jack, queen, or king. >>> deal = sample(['tens', 'low cards'], counts=[16, 36], k=20) >>> deal.count('tens') / 20 0.15 >>> # Estimate the probability of getting 5 or more heads from 7 spins >>> # of a biased coin that settles on heads 60% of the time. >>> sum(binomialvariate(n=7, p=0.6) >= 5 for i in range(10_000)) / 10_000 0.4169 >>> # Probability of the median of 5 samples being in middle two quartiles >>> def trial(): ... return 2_500 > sum(trial() for i in range(10_000)) / 10_000 0.7958

Example of statistical bootstrapping using resampling with replacement to estimate a confidence interval for the mean of a sample:

# https://www.thoughtco.com/example-of-bootstrapping-3126155 from statistics import fmean as mean from random import choices data = [41, 50, 29, 37, 81, 30, 73, 63, 20, 35, 68, 22, 60, 31, 95] means = sorted(mean(choices(data, k=len(data))) for i in range(100)) print(f'The sample mean of {mean(data):.1f} has a 90% confidence ' f'interval from {means[5]:.1f} to {means[94]:.1f}')

Example of a resampling permutation test to determine the statistical significance or p-value of an observed difference between the effects of a drug versus a placebo:

# Example from "Statistics is Easy" by Dennis Shasha and Manda Wilson from statistics import fmean as mean from random import shuffle drug = [54, 73, 53, 70, 73, 68, 52, 65, 65] placebo = [54, 51, 58, 44, 55, 52, 42, 47, 58, 46] observed_diff = mean(drug) - mean(placebo) n = 10_000 count = 0 combined = drug + placebo for i in range(n): shuffle(combined) new_diff = mean(combined[:len(drug)]) - mean(combined[len(drug):]) count += (new_diff >= observed_diff) print(f'{n} label reshufflings produced only {count} instances with a difference') print(f'at least as extreme as the observed difference of {observed_diff:.1f}.') print(f'The one-sided p-value of {count / n:.4f} leads us to reject the null') print(f'hypothesis that there is no difference between the drug and the placebo.')

Simulation of arrival times and service deliveries for a multiserver queue:

from heapq import heapify, heapreplace from random import expovariate, gauss from statistics import mean, quantiles average_arrival_interval = 5.6 average_service_time = 15.0 stdev_service_time = 3.5 num_servers = 3 waits = [] arrival_time = 0.0 servers = [0.0] * num_servers # time when each server becomes available heapify(servers) for i in range(1_000_000): arrival_time += expovariate(1.0 / average_arrival_interval) next_server_available = servers[0] wait = max(0.0, next_server_available - arrival_time) waits.append(wait) service_duration = max(0.0, gauss(average_service_time, stdev_service_time)) service_completed = arrival_time + wait + service_duration heapreplace(servers, service_completed) print(f'Mean wait: {mean(waits):.1f} Max wait: {max(waits):.1f}') print('Quartiles:', [round(q, 1) for q in quantiles(waits)])

See also

Statistics for Hackers a video tutorial by Jake Vanderplas on statistical analysis using just a few fundamental concepts including simulation, sampling, shuffling, and cross-validation.

Economics Simulation a simulation of a marketplace by Peter Norvig that shows effective use of many of the tools and distributions provided by this module (gauss, uniform, sample, betavariate, choice, triangular, and randrange).

A Concrete Introduction to Probability (using Python) a tutorial by Peter Norvig covering the basics of probability theory, how to write simulations, and how to perform data analysis using Python.

Recipes¶

These recipes show how to efficiently make random selections from the combinatoric iterators in the itertools module:

def random_product(*args, repeat=1): "Random selection from itertools.product(*args, **kwds)" pools = [tuple(pool) for pool in args] * repeat return tuple(map(random.choice, pools)) def random_permutation(iterable, r=None): "Random selection from itertools.permutations(iterable, r)" pool = tuple(iterable) r = len(pool) if r is None else r return tuple(random.sample(pool, r)) def random_combination(iterable, r): "Random selection from itertools.combinations(iterable, r)" pool = tuple(iterable) n = len(pool) indices = sorted(random.sample(range(n), r)) return tuple(pool[i] for i in indices) def random_combination_with_replacement(iterable, r): "Choose r elements with replacement. Order the result to match the iterable." # Result will be in set(itertools.combinations_with_replacement(iterable, r)). pool = tuple(iterable) n = len(pool) indices = sorted(random.choices(range(n), k=r)) return tuple(pool[i] for i in indices)

The default random() returns multiples of 2⁻⁵³ in the range 0.0 ≤ x < 1.0. All such numbers are evenly spaced and are exactly representable as Python floats. However, many other representable floats in that interval are not possible selections. For example, 0.05954861408025609 isn’t an integer multiple of 2⁻⁵³.

The following recipe takes a different approach. All floats in the interval are possible selections. The mantissa comes from a uniform distribution of integers in the range 2⁵² ≤ mantissa < 2⁵³. The exponent comes from a geometric distribution where exponents smaller than -53 occur half as often as the next larger exponent.

from random import Random from math import ldexp class FullRandom(Random): def random(self): mantissa = 0x10_0000_0000_0000 | self.getrandbits(52) exponent = -53 x = 0 while not x: x = self.getrandbits(32) exponent += x.bit_length() - 32 return ldexp(mantissa, exponent)

All real valued distributions in the class will use the new method:

>>> fr = FullRandom() >>> fr.random() 0.05954861408025609 >>> fr.expovariate(0.25) 8.87925541791544

The recipe is conceptually equivalent to an algorithm that chooses from all the multiples of 2⁻¹⁰⁷⁴ in the range 0.0 ≤ x < 1.0. All such numbers are evenly spaced, but most have to be rounded down to the nearest representable Python float. (The value 2⁻¹⁰⁷⁴ is the smallest positive unnormalized float and is equal to math.ulp(0.0).)

See also

Generating Pseudo-random Floating-Point Values a paper by Allen B. Downey describing ways to generate more fine-grained floats than normally generated by random().



【本文地址】


今日新闻


推荐新闻


CopyRight 2018-2019 办公设备维修网 版权所有 豫ICP备15022753号-3