Question

我正在尝试使用与Python Numpy.random.Choice

相同的代码

关键部分是：probability

与a中的每个条目相关联的概率。如果没有给出，则样本假定在a。中的所有条目上均匀分布。

一些测试代码：

import numpy as np

n = 5

vocab_size = 3

p = np.array( [[ 0.65278451], [ 0.0868038725], [ 0.2604116175]])

print('Sum: ', repr(sum(p)))

for t in range(n):
    x = np.random.choice(range(vocab_size), p=p.ravel())
    print('x: %s x[x]: %s' % (x, p.ravel()[x]))

print(p.ravel())

这给出了输出：

Sum:  array([ 1.])
x: 0 x[x]: 0.65278451
x: 0 x[x]: 0.65278451
x: 0 x[x]: 0.65278451
x: 0 x[x]: 0.65278451
x: 0 x[x]: 0.65278451
[ 0.65278451  0.08680387  0.26041162]

有时。

这里有一个分布，它是一个部分随机的分布，但那里也有结构。

我想在C＃中实现这一点，说实话，我不确定这是否有效。

大约4年前，有一个很好的问题：Emulate Python's random.choice in .NET

现在这已经很老了，也没有真正深入研究统一概率分布，我想我会要求进行一些阐述？

现在时代已经改变，代码正在发生变化，我认为可能有更好的方法来实现.NET Random.Choice()方法。

public static int Choice(Vector sequence, int a = 0, int size = 0, bool replace = false)
{
    // F(x)
    var Fx = 1/(b - a)
    var p = (xmax - xmin) * Fx

    return random.Next(0, sequence.Length);
}

Vector只是一个双[]。

我如何随意从矢量中选择概率：

 p = np.array(
 [[ 0.01313731], [ 0.01315883], [ 0.01312814], [ 0.01316345], [ 0.01316839],
 [ 0.01314225], [ 0.01317578], [ 0.01312916], [ 0.01316344], [ 0.01317046],
 [ 0.01314973], [ 0.01314432], [ 0.01317042], [ 0.01314846], [ 0.01315124],
 [ 0.01316694], [ 0.0131816 ], [ 0.01315033], [ 0.0131645 ], [ 0.01314199],
 [ 0.01315199], [ 0.01314431], [ 0.01314458], [ 0.01314999], [ 0.01315409],
 [ 0.01316245], [ 0.01315008], [ 0.01314104], [ 0.01315215], [ 0.01317024],
 [ 0.01315993], [ 0.01318789], [ 0.0131677 ], [ 0.01316761], [ 0.01315658],
 [ 0.01315902], [ 0.01314266], [ 0.0131637 ], [ 0.01315702], [ 0.01315776],
 [ 0.01316194], [ 0.01316246], [ 0.01314769], [ 0.01315608], [ 0.01315487],
 [ 0.01316117], [ 0.01315083], [ 0.01315836], [ 0.0131665 ], [ 0.01314706],
 [ 0.01314923], [ 0.01317971], [ 0.01316373], [ 0.01314863], [ 0.01315498],
 [ 0.01315732], [ 0.01318195], [ 0.01315505], [ 0.01315979], [ 0.01315992],
 [ 0.01316072], [ 0.01314744], [ 0.0131638 ], [ 0.01315642], [ 0.01314933],
 [ 0.01316188], [ 0.01315458], [ 0.01315551], [ 0.01317907], [ 0.01316296],
 [ 0.01317765], [ 0.01316863], [ 0.01316804], [ 0.01314882], [ 0.01316548],
 [ 0.01315487]])

Python中的输出是：

Sum:  array([ 1.])
x: 21 x[x]: 0.01314431
x: 30 x[x]: 0.01315993
x: 54 x[x]: 0.01315498
x: 31 x[x]: 0.01318789
x: 27 x[x]: 0.01314104

有时。

修改喝完咖啡后睡觉，更有洞察力。文档说明：

从大小为3的np.arange（5）生成非均匀随机样本，无需替换：


np.random.choice（5,3，replace = False，p = [0.1,0,0.3,0.6,0]）     数组（[2,3,0]）

参数p为序列或选择引入了非均匀分布。

与a中的每个条目相关联的概率。如果没有给出，则样本假定在a中的所有条目上均匀分布。

所以我想，如果：

static int[] a = new int[] { 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20,
21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40,
41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60,
61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75};



static double[] p = new double[] { 0.01313731,  0.01315883,  0.01312814,  0.01316345,  0.01316839,  0.01314225,
0.01317578,  0.01312916,  0.01316344,  0.01317046,  0.01314973,  0.01314432,
0.01317042,  0.01314846,  0.01315124,  0.01316694,  0.0131816,   0.01315033,
0.0131645,   0.01314199,  0.01315199,  0.01314431,  0.01314458,  0.01314999,
0.01315409,  0.01316245,  0.01315008,  0.01314104,  0.01315215,  0.01317024,
0.01315993,  0.01318789,  0.0131677,   0.01316761,  0.01315658,  0.01315902,
0.01314266,  0.0131637,   0.01315702,  0.01315776,  0.01316194,  0.01316246,
0.01314769,  0.01315608,  0.01315487,  0.01316117,  0.01315083,  0.01315836,
0.0131665,   0.01314706,  0.01314923,  0.01317971,  0.01316373,  0.01314863,
0.01315498,  0.01315732,  0.01318195,  0.01315505,  0.01315979,  0.01315992,
0.01316072,  0.01314744,  0.0131638,   0.01315642,  0.01314933,  0.01316188,
0.01315458,  0.01315551,  0.01317907,  0.01316296,  0.01317765,  0.01316863,
0.01316804,  0.01314882,  0.01316548,  0.01315487 };

我如何有效地计算这种分布？

修改

虽然上述p参数可能没有明确的分布：

此p参数执行：

p = np.array(
[[  3.09571694e-03], [  6.62372261e-04], [  2.52917874e-04], [  6.93371978e-04],
[  2.22301291e-04], [  3.53796717e-02], [  2.36204398e-04], [  2.41100042e-04],
[  1.59093166e-02], [  5.17099025e-04], [  2.72037896e-04], [  1.29918769e-03],
[  2.68077696e-02], [  5.68696611e-04], [  5.32142704e-04], [  5.88432463e-05],
[  2.53700138e-02], [  2.51216588e-03], [  4.72895541e-04], [  4.20276848e-03],
[  5.65701874e-05], [  1.84972048e-03], [  8.46515331e-03], [  8.02505743e-02],
[  5.34274983e-04], [  5.18868535e-04], [  2.22580377e-04], [  2.50133462e-02],
[  3.70997917e-02], [  5.84941482e-05], [  6.49978323e-04], [  4.18675536e-01],
[  6.16371962e-02], [  3.82260752e-04], [  6.09901544e-04], [  2.54540201e-03],
[  2.46758824e-04], [  4.13621365e-04], [  5.23495532e-04], [  6.40675685e-03],
[  1.14165332e-03], [  1.89148994e-04], [  8.41715724e-04], [  8.65699032e-04],
[  6.71368283e-04], [  2.14908596e-03], [  5.80679210e-02], [  1.11176616e-02],
[  6.58134137e-05], [  2.38992622e-02], [  2.91388753e-04], [  1.93989753e-03],
[  1.82157325e-03], [  3.33691627e-03], [  5.69157244e-03], [  1.11033592e-04],
[  2.42448034e-04], [  8.42765356e-05], [  1.31656056e-02], [  1.68779684e-02],
[  2.72298244e-02], [  8.19056613e-04], [  1.14640462e-02], [  6.21846308e-05],
[  9.24618073e-04], [  3.63659515e-02], [  7.17286486e-05], [  6.24008652e-04],
[  2.59900890e-03], [  1.57848651e-04], [  5.71378707e-05], [  7.62828929e-04],
[  2.91648042e-04], [  1.67612579e-04], [  1.65455262e-04], [  1.01981563e-02]])

一些左倾斜的高斯分布。 PoyserMath的这段视频非常好：Stats: Finding Probability Using a Normal Distribution Table 解释为什么p必须总和为1.0

编辑：12.04.17 - 最后我找到了与此相关的python文件!!!

# Author: Hamzeh Alsalhi <ha258@cornell.edu>
#
# License: BSD 3 clause
from __future__ import division
import numpy as np
import scipy.sparse as sp
import operator
import array

from sklearn.utils import check_random_state
from sklearn.utils.fixes import astype
from ._random import sample_without_replacement

__all__ = ['sample_without_replacement', 'choice']


# This is a backport of np.random.choice from numpy 1.7
# The function can be removed when we bump the requirements to >=1.7
def choice(a, size=None, replace=True, p=None, random_state=None):
    """
    choice(a, size=None, replace=True, p=None)

    Generates a random sample from a given 1-D array

    .. versionadded:: 1.7.0

    Parameters
    -----------
    a : 1-D array-like or int
        If an ndarray, a random sample is generated from its elements.
        If an int, the random sample is generated as if a was np.arange(n)

    size : int or tuple of ints, optional
        Output shape. Default is None, in which case a single value is
        returned.

    replace : boolean, optional
        Whether the sample is with or without replacement.

    p : 1-D array-like, optional
        The probabilities associated with each entry in a.
        If not given the sample assumes a uniform distribution over all
        entries in a.

    random_state : int, RandomState instance or None, optional (default=None)
        If int, random_state is the seed used by the random number generator;
        If RandomState instance, random_state is the random number generator;
        If None, the random number generator is the RandomState instance used
        by `np.random`.


    Returns
    --------
    samples : 1-D ndarray, shape (size,)
    The generated random samples

    Raises
    -------
    ValueError
    If a is an int and less than zero, if a or p are not 1-dimensional,
    if a is an array-like of size 0, if p is not a vector of
    probabilities, if a and p have different lengths, or if
    replace=False and the sample size is greater than the population
    size

    See Also
    ---------
    randint, shuffle, permutation

    Examples
    ---------
    Generate a uniform random sample from np.arange(5) of size 3:

    >>> np.random.choice(5, 3)  # doctest: +SKIP
    array([0, 3, 4])
    >>> #This is equivalent to np.random.randint(0,5,3)

    Generate a non-uniform random sample from np.arange(5) of size 3:

    >>> np.random.choice(5, 3, p=[0.1, 0, 0.3, 0.6, 0])  # doctest: +SKIP
    array([3, 3, 0])

    Generate a uniform random sample from np.arange(5) of size 3 without
    replacement:

    >>> np.random.choice(5, 3, replace=False)  # doctest: +SKIP
    array([3,1,0])
    >>> #This is equivalent to np.random.shuffle(np.arange(5))[:3]

    Generate a non-uniform random sample from np.arange(5) of size
    3 without replacement:

    >>> np.random.choice(5, 3, replace=False, p=[0.1, 0, 0.3, 0.6, 0])
    ... # doctest: +SKIP
    array([2, 3, 0])

    Any of the above can be repeated with an arbitrary array-like
    instead of just integers. For instance:

    >>> aa_milne_arr = ['pooh', 'rabbit', 'piglet', 'Christopher']
    >>> np.random.choice(aa_milne_arr, 5, p=[0.5, 0.1, 0.1, 0.3])
    ... # doctest: +SKIP
    array(['pooh', 'pooh', 'pooh', 'Christopher', 'piglet'],
    dtype='|S11')

    """
    random_state = check_random_state(random_state)

    # Format and Verify input
    a = np.array(a, copy=False)
    if a.ndim == 0:
        try:
            # __index__ must return an integer by python rules.
            pop_size = operator.index(a.item())
        except TypeError:
            raise ValueError("a must be 1-dimensional or an integer")
        if pop_size <= 0:
            raise ValueError("a must be greater than 0")
    elif a.ndim != 1:
        raise ValueError("a must be 1-dimensional")
    else:
        pop_size = a.shape[0]
        if pop_size is 0:
            raise ValueError("a must be non-empty")

    if p is not None:
        p = np.array(p, dtype=np.double, ndmin=1, copy=False)
        if p.ndim != 1:
            raise ValueError("p must be 1-dimensional")
        if p.size != pop_size:
            raise ValueError("a and p must have same size")
        if np.any(p < 0):
            raise ValueError("probabilities are not non-negative")
        if not np.allclose(p.sum(), 1):
            raise ValueError("probabilities do not sum to 1")

    shape = size
    if shape is not None:
        size = np.prod(shape, dtype=np.intp)
    else:
        size = 1

    # Actual sampling
    if replace:
        if p is not None:
            cdf = p.cumsum()
            cdf /= cdf[-1]
            uniform_samples = random_state.random_sample(shape)
            idx = cdf.searchsorted(uniform_samples, side='right')
            # searchsorted returns a scalar
            idx = np.array(idx, copy=False)
        else:
            idx = random_state.randint(0, pop_size, size=shape)
    else:
        if size > pop_size:
            raise ValueError("Cannot take a larger sample than "
                             "population when 'replace=False'")

        if p is not None:
            if np.sum(p > 0) < size:
                raise ValueError("Fewer non-zero entries in p than size")
            n_uniq = 0
            p = p.copy()
            found = np.zeros(shape, dtype=np.int)
            flat_found = found.ravel()
            while n_uniq < size:
                x = random_state.rand(size - n_uniq)
                if n_uniq > 0:
                    p[flat_found[0:n_uniq]] = 0
                cdf = np.cumsum(p)
                cdf /= cdf[-1]
                new = cdf.searchsorted(x, side='right')
                _, unique_indices = np.unique(new, return_index=True)
                unique_indices.sort()
                new = new.take(unique_indices)
                flat_found[n_uniq:n_uniq + new.size] = new
                n_uniq += new.size
            idx = found
        else:
            idx = random_state.permutation(pop_size)[:size]
            if shape is not None:
                idx.shape = shape

    if shape is None and isinstance(idx, np.ndarray):
        # In most cases a scalar will have been made an array
        idx = idx.item(0)

    # Use samples as indices for a if a is array-like
    if a.ndim == 0:
        return idx

    if shape is not None and idx.ndim == 0:
        # If size == () then the user requested a 0-d array as opposed to
        # a scalar object when size is None. However a[idx] is always a
        # scalar and not an array. So this makes sure the result is an
        # array, taking into account that np.array(item) may not work
        # for object arrays.
        res = np.empty((), dtype=a.dtype)
        res[()] = a[idx]
        return res

    return a[idx]


def random_choice_csc(n_samples, classes, class_probability=None,
                      random_state=None):
    """Generate a sparse random matrix given column class distributions

    Parameters
    ----------
    n_samples : int,
        Number of samples to draw in each column.

    classes : list of size n_outputs of arrays of size (n_classes,)
        List of classes for each column.

    class_probability : list of size n_outputs of arrays of size (n_classes,)
        Optional (default=None). Class distribution of each column. If None the
        uniform distribution is assumed.

    random_state : int, RandomState instance or None, optional (default=None)
        If int, random_state is the seed used by the random number generator;
        If RandomState instance, random_state is the random number generator;
        If None, the random number generator is the RandomState instance used
        by `np.random`.

    Returns
    -------
    random_matrix : sparse csc matrix of size (n_samples, n_outputs)

    """
    data = array.array('i')
    indices = array.array('i')
    indptr = array.array('i', [0])

    for j in range(len(classes)):
        classes[j] = np.asarray(classes[j])
        if classes[j].dtype.kind != 'i':
            raise ValueError("class dtype %s is not supported" %
                             classes[j].dtype)
        classes[j] = astype(classes[j], np.int64, copy=False)

        # use uniform distribution if no class_probability is given
        if class_probability is None:
            class_prob_j = np.empty(shape=classes[j].shape[0])
            class_prob_j.fill(1 / classes[j].shape[0])
        else:
            class_prob_j = np.asarray(class_probability[j])

        if np.sum(class_prob_j) != 1.0:
            raise ValueError("Probability array at index {0} does not sum to "
                             "one".format(j))

        if class_prob_j.shape[0] != classes[j].shape[0]:
            raise ValueError("classes[{0}] (length {1}) and "
                             "class_probability[{0}] (length {2}) have "
                             "different length.".format(j,
                                                        classes[j].shape[0],
                                                        class_prob_j.shape[0]))

        # If 0 is not present in the classes insert it with a probability 0.0
        if 0 not in classes[j]:
            classes[j] = np.insert(classes[j], 0, 0)
            class_prob_j = np.insert(class_prob_j, 0, 0.0)

        # If there are nonzero classes choose randomly using class_probability
        rng = check_random_state(random_state)
        if classes[j].shape[0] > 1:
            p_nonzero = 1 - class_prob_j[classes[j] == 0]
            nnz = int(n_samples * p_nonzero)
            ind_sample = sample_without_replacement(n_population=n_samples,
                                                    n_samples=nnz,
                                                    random_state=random_state)
            indices.extend(ind_sample)

            # Normalize probabilites for the nonzero elements
            classes_j_nonzero = classes[j] != 0
            class_probability_nz = class_prob_j[classes_j_nonzero]
            class_probability_nz_norm = (class_probability_nz /
                                         np.sum(class_probability_nz))
            classes_ind = np.searchsorted(class_probability_nz_norm.cumsum(),
                                          rng.rand(nnz))
            data.extend(classes[j][classes_j_nonzero][classes_ind])
        indptr.append(len(indices))

    return sp.csc_matrix((data, indices, indptr),
                         (n_samples, len(classes)),
                         dtype=int)

Answer 1

如果我理解正确 - 你想根据双精度数组给出的分布概率从Y元素列表中随机选择X元素，其中每个元素表示返回相同索引的元素的概率。我能想到的最直接的方式就是这个（见评论）：

using System;
using System.Collections.Generic;
using System.Linq;
using System.Threading;

static readonly ThreadLocal<Random> _random = new ThreadLocal<Random>(() => new Random());
static IEnumerable<T> Choice<T>(IList<T> sequence, int size, double[] distribution) {
    double sum = 0;
    // first change shape of your distribution probablity array
    // we need it to be cumulative, that is:
    // if you have [0.1, 0.2, 0.3, 0.4] 
    // we need     [0.1, 0.3, 0.6, 1  ] instead
    var cumulative = distribution.Select(c => {
        var result = c + sum;
        sum += c;
        return result;
    }).ToList();
    for (int i = 0; i < size; i++) {
        // now generate random double. It will always be in range from 0 to 1
        var r = _random.Value.NextDouble();
        // now find first index in our cumulative array that is greater or equal generated random value
        var idx = cumulative.BinarySearch(r);
        // if exact match is not found, List.BinarySearch will return index of the first items greater than passed value, but in specific form (negative)
        // we need to apply ~ to this negative value to get real index
        if (idx < 0)
            idx = ~idx; 
        if (idx > cumulative.Count - 1)
           idx = cumulative.Count - 1; // very rare case when probabilities do not sum to 1 becuase of double precision issues (so sum is 0.999943 and so on)
        // return item at given index
        yield return sequence[idx];
    }
}

我很难用简单的语言解释这一点，但我认为从代码中应该是相对明显的。也许最简单的解释一下。假设我们有分布[0.1,0.4,0.4,0.1]。累积版本（当我们将所有先前项目的总和添加到当前项目时）将如下所示：[0.1,0.5,0.9,1]。现在我们生成0到1范围内的随机数。它的分布是均匀的，所以任何值都是相同的。它在0-0.1范围内的可能性是多少？ 0.1。在0.1-0.5范围内？ 0.4。因此，您可以看到，在给定范围内均匀分布的0-1数的概率与我们在概率分布数组中的概率完全相同。

像这样使用：

var result = Choice(Enumerable.Range(0, 5).ToArray(), 3, new double[] {0.01, 0.01, 0.48, 0.48, 0.02}).ToArray();

将导致：

[3,3,3] // 
[2,3,2] // most often result with contain 2 and 3, because they both have 0.48 probablity and the rest elements have just 0.01
[1,3,2] // very rare other elements will appear

如果您需要没有重复的版本 - 也可以对此代码稍作修改。

如果您需要一个项目 - 使用size = 1调用上方函数或为方便起见创建重载。如果要传递单个整数而不是序列，则相同：

static T Choice<T>(IList<T> sequence, double[] distribution) {
    return Choice(sequence, 1, distribution).First();
}

static int Choice(int upTo, double[] distribution) {
    return Choice(Enumerable.Range(0, upTo).ToArray(), distribution);
}

Answer 2

您可以尝试NumSharp，它是numpy到C＃的端口

double[] distribution = new double[] { 0.1, 0.2, 0.3, 0.4 };

//Simple choice
for (int i = 0; i < 80; i++)
{
    int choice = NumSharp.np.random.choice(distribution.Count(), probabilities:distribution)[0].GetInt32();
    Console.Write(choice.ToString());
}
Console.WriteLine();

12222312301223333323323223022211111233123323333133211233333233013312213033030033

NumSharp包含附加参数（替换...），但到目前为止，仍然存在性能问题：

EvalTime("NumSharp Single thread",() =>
{
    for (int i = 0; i < 100000; i++)
        NumSharp.np.random.choice(distribution.Count(), probabilities: distribution)[0].GetInt32();
});

EvalTime("NumSharp Multithread", () =>
{
    Parallel.For(0, 100000, i => 
        NumSharp.np.random.choice(distribution.Count(), probabilities: distribution)[0].GetInt32());
});

EvalTime("Evk Single thread", () =>
{
    for (int i = 0; i < 100000; i++)
        Choice(upTo: distribution.Count(), distribution: distribution);
});

EvalTime("Evk Multithread", () =>
{
    Parallel.For(0, 100000, i =>
        Choice(upTo: distribution.Count(), distribution: distribution));
});

public static void EvalTime(string Name,Action a)
{
    Stopwatch sw = new Stopwatch();
    sw.Start();
    a();
    sw.Stop();
    Console.WriteLine("{0} : {1}ms", Name,sw.ElapsedMilliseconds);
}

以下是比较我的PC（24个逻辑内核）上的NumSharp和Evk的实现的输出

NumSharp Single thread : 10701ms
NumSharp Multithread : 19350ms
Evk Single thread : 117ms
Evk Multithread : 61ms

C＃中的Python numpy.random.choice具有非均匀概率分布

2 个答案: