在 Python 中计算熵的最快方法

Question

在我的项目中，我需要多次计算 0-1 向量的熵。这是我的代码：

def entropy(labels):
    """ Computes entropy of 0-1 vector. """
    n_labels = len(labels)

    if n_labels <= 1:
        return 0

    counts = np.bincount(labels)
    probs = counts[np.nonzero(counts)] / n_labels
    n_classes = len(probs)

    if n_classes <= 1:
        return 0
    return - np.sum(probs * np.log(probs)) / np.log(n_classes)

有没有更快的方法？

Answer 1

聚会迟到了，但我偶然发现了这一点，所有答案似乎都依赖于 Kullback–Leibler 散度，它没有上限，因此不符合我的需求。

这里我有一个从 [0,1] 出发的熵函数的近似值（TODO! 可以改进）。

它计算单个列的偏差。

class Pandas_Dataframe_helper:
    #some other methods here...
    @staticmethod
    def column_biass(df_column):
        df_column_as_list           =   list(df_column)
        N                           =   len(df_column_as_list)
        values,counts               =   np.unique(df_column_as_list, return_counts=True)
        #generate synth list (TODO! what if not even number? Minimum Comun Multiple of(num_different_labels,[x for x in counts]))
        num_different_labels        =   len(values)
        num_items_per_label         =   N // num_different_labels
        synthetic_list              =   []
        for current_value in values:
            synthetic_list.extend([current_value] * num_items_per_label)
        #TODO! aproximacion
        if(len(synthetic_list) != len(df_column_as_list)):
            synthetic_list.extend([current_value] * (len(df_column_as_list) - len(synthetic_list)))
        #now, extrapolate differences between sorted-input-list and synsthetic_list
        df_column_as_list_sorted    =   sorted(df_column_as_list)
        counter_unmatches           =   0
        for i in range(0,N):
            if(df_column_as_list_sorted[i] != synthetic_list[i]):
                counter_unmatches   +=  1
        #upper_bound = g(N,num_different_labels)
        #((K-1)M)-1 K==num_different_labels , M==num theorically perfect distribution's items per label 
        upper_bound                 =   ((num_different_labels-1)*num_items_per_label)-1
        return counter_unmatches/upper_bound
    #---------------------------------------------------------------------------------------------------------------------

完整代码https://github.com/glezo1/pcommonlibs/blob/master/com/glezo/pandas_dataframe_helper/Pandas_Dataframe_Helper.py

Answer 2

这是迄今为止我发现的最快的 Python 实现：

import numpy as np

def entropy(labels):
    ps = np.bincount(labels) / len(labels)
    return -np.sum([p * np.log2(p) for p in ps if p > 0])

Answer 3

均匀分布的数据（高熵）：

s=range(0,256)

香农熵逐步计算：

import collections
import math

# calculate probability for each byte as number of occurrences / array length
probabilities = [n_x/len(s) for x,n_x in collections.Counter(s).items()]
# [0.00390625, 0.00390625, 0.00390625, ...]

# calculate per-character entropy fractions
e_x = [-p_x*math.log(p_x,2) for p_x in probabilities]
# [0.03125, 0.03125, 0.03125, ...]

# sum fractions to obtain Shannon entropy
entropy = sum(e_x)
>>> entropy 
8.0

单线（假设import collections）：

def H(s): return sum([-p_x*math.log(p_x,2) for p_x in [n_x/len(s) for x,n_x in collections.Counter(s).items()]])

适当的功能：

import collections
import math

def H(s):
    probabilities = [n_x/len(s) for x,n_x in collections.Counter(s).items()]
    e_x = [-p_x*math.log(p_x,2) for p_x in probabilities]    
    return sum(e_x)

测试用例 - 取自CyberChef entropy estimator的英文文本：

>>> H(range(0,256))
8.0
>>> H(range(0,64))
6.0
>>> H(range(0,128))
7.0
>>> H([0,1])
1.0
>>> H('Standard English text usually falls somewhere between 3.5 and 5')
4.228788210509104

Answer 4

BiEntropy 不会是计算熵的最快方法，但它是严格的，并且以明确定义的方式建立在 Shannon Entropy 之上。它已经在包括图像相关应用在内的各个领域进行了测试。 它是在 Github 上用 Python 实现的。

Answer 5

此方法通过允许分箱扩展了其他解决方案。例如，bin=None（默认）不会对 x 进行分箱，而是为x 的每个元素计算经验概率，而bin=256 在计算经验概率之前将x 分块到 256 个分箱中。

import numpy as np

def entropy(x, bins=None):
    N   = x.shape[0]
    if bins is None:
        counts = np.bincount(x)
    else:
        counts = np.histogram(x, bins=bins)[0] # 0th idx is counts
    p   = counts[np.nonzero(counts)]/N # avoids log(0)
    H   = -np.dot( p, np.log2(p) )
    return H 

Answer 6

def entropy(base, prob_a, prob_b ):
  import math
  base=2
  x=prob_a
  y=prob_b
  expression =-((x*math.log(x,base)+(y*math.log(y,base))))    
  return [expression]

Answer 7

一个不依赖于 numpy 的答案：

import math
from collections import Counter

def eta(data, unit='natural'):
    base = {
        'shannon' : 2.,
        'natural' : math.exp(1),
        'hartley' : 10.
    }

    if len(data) <= 1:
        return 0

    counts = Counter()

    for d in data:
        counts[d] += 1

    ent = 0

    probs = [float(c) / len(data) for c in counts.values()]
    for p in probs:
        if p > 0.:
            ent -= p * math.log(p, base[unit])

    return ent

这将接受您可以扔给它的任何数据类型：

>>> eta(['mary', 'had', 'a', 'little', 'lamb'])
1.6094379124341005

>>> eta([c for c in "mary had a little lamb"])
2.311097886212714

@Jarad 提供的答案也建议了时间。为此：

repeat_number = 1000000
e = timeit.repeat(
    stmt='''eta(labels)''', 
    setup='''labels=[1,3,5,2,3,5,3,2,1,3,4,5];from __main__ import eta''', 
    repeat=3, 
    number=repeat_number)

Timeit 结果：（我相信这比最好的 numpy 方法快约 4 倍）

print('Method: {}, Avg.: {:.6f}'.format("eta", np.array(e).mean()))

Method: eta, Avg.: 10.461799

Answer 8

这是我的方法：

labels = [0, 0, 1, 1]

from collections import Counter
from scipy import stats

stats.entropy(list(Counter(labels).values()), base=2)

Answer 9

from collections import Counter
from scipy import stats

labels = [0.9, 0.09, 0.1]
stats.entropy(list(Counter(labels).keys()), base=2)

Answer 10

将数据作为pd.Series 和scipy.stats，计算给定数量的熵非常简单：

import pandas as pd
import scipy.stats

def ent(data):
    """Calculates entropy of the passed `pd.Series`
    """
    p_data = data.value_counts()           # counts occurrence of each value
    entropy = scipy.stats.entropy(p_data)  # get entropy from counts
    return entropy

注意：scipy.stats 将规范化提供的数据，所以这不需要明确地完成，即传递一个计数数组可以正常工作。

Answer 11

@Sanjeet Gupta 的回答很好，但可以精简。这个问题专门询问“最快”的方式，但我只看到一个答案的时间，所以我将发布使用 scipy 和 numpy 与原始发帖人的 entropy2 答案的比较，稍作改动。

四种不同的方法：scipy/numpy、numpy/math、pandas/numpy、numpy

import numpy as np
from scipy.stats import entropy
from math import log, e
import pandas as pd

import timeit

def entropy1(labels, base=None):
  value,counts = np.unique(labels, return_counts=True)
  return entropy(counts, base=base)

def entropy2(labels, base=None):
  """ Computes entropy of label distribution. """

  n_labels = len(labels)

  if n_labels <= 1:
    return 0

  value,counts = np.unique(labels, return_counts=True)
  probs = counts / n_labels
  n_classes = np.count_nonzero(probs)

  if n_classes <= 1:
    return 0

  ent = 0.

  # Compute entropy
  base = e if base is None else base
  for i in probs:
    ent -= i * log(i, base)

  return ent

def entropy3(labels, base=None):
  vc = pd.Series(labels).value_counts(normalize=True, sort=False)
  base = e if base is None else base
  return -(vc * np.log(vc)/np.log(base)).sum()

def entropy4(labels, base=None):
  value,counts = np.unique(labels, return_counts=True)
  norm_counts = counts / counts.sum()
  base = e if base is None else base
  return -(norm_counts * np.log(norm_counts)/np.log(base)).sum()

Timeit 操作：

repeat_number = 1000000

a = timeit.repeat(stmt='''entropy1(labels)''',
                  setup='''labels=[1,3,5,2,3,5,3,2,1,3,4,5];from __main__ import entropy1''',
                  repeat=3, number=repeat_number)

b = timeit.repeat(stmt='''entropy2(labels)''',
                  setup='''labels=[1,3,5,2,3,5,3,2,1,3,4,5];from __main__ import entropy2''',
                  repeat=3, number=repeat_number)

c = timeit.repeat(stmt='''entropy3(labels)''',
                  setup='''labels=[1,3,5,2,3,5,3,2,1,3,4,5];from __main__ import entropy3''',
                  repeat=3, number=repeat_number)

d = timeit.repeat(stmt='''entropy4(labels)''',
                  setup='''labels=[1,3,5,2,3,5,3,2,1,3,4,5];from __main__ import entropy4''',
                  repeat=3, number=repeat_number)

Timeit 结果：

# for loop to print out results of timeit
for approach,timeit_results in zip(['scipy/numpy', 'numpy/math', 'pandas/numpy', 'numpy'], [a,b,c,d]):
  print('Method: {}, Avg.: {:.6f}'.format(approach, np.array(timeit_results).mean()))

Method: scipy/numpy, Avg.: 63.315312
Method: numpy/math, Avg.: 49.256894
Method: pandas/numpy, Avg.: 884.644023
Method: numpy, Avg.: 60.026938

获胜者：numpy/math（熵2）

还值得注意的是，上面的entropy2 函数可以处理数字和文本数据。例如：entropy2(list('abcdefabacdebcab'))。原始发帖人的答案来自 2013 年，并且有一个特定的用例用于合并整数，但它不适用于文本。

Answer 12

我最喜欢的熵函数如下：

def entropy(labels):
    prob_dict = {x:labels.count(x)/len(labels) for x in labels}
    probs = np.array(list(prob_dict.values()))

    return - probs.dot(np.log2(probs))

我仍在寻找一种更好的方法来避免 dict -> values -> list -> np.array 转换。如果我找到它会再次评论。

Answer 13

上面的答案很好，但是如果您需要一个可以沿不同轴操作的版本，这里有一个可行的实现。

def entropy(A, axis=None):
    """Computes the Shannon entropy of the elements of A. Assumes A is 
    an array-like of nonnegative ints whose max value is approximately 
    the number of unique values present.

    >>> a = [0, 1]
    >>> entropy(a)
    1.0
    >>> A = np.c_[a, a]
    >>> entropy(A)
    1.0
    >>> A                   # doctest: +NORMALIZE_WHITESPACE
    array([[0, 0], [1, 1]])
    >>> entropy(A, axis=0)  # doctest: +NORMALIZE_WHITESPACE
    array([ 1., 1.])
    >>> entropy(A, axis=1)  # doctest: +NORMALIZE_WHITESPACE
    array([[ 0.], [ 0.]])
    >>> entropy([0, 0, 0])
    0.0
    >>> entropy([])
    0.0
    >>> entropy([5])
    0.0
    """
    if A is None or len(A) < 2:
        return 0.

    A = np.asarray(A)

    if axis is None:
        A = A.flatten()
        counts = np.bincount(A) # needs small, non-negative ints
        counts = counts[counts > 0]
        if len(counts) == 1:
            return 0. # avoid returning -0.0 to prevent weird doctests
        probs = counts / float(A.size)
        return -np.sum(probs * np.log2(probs))
    elif axis == 0:
        entropies = map(lambda col: entropy(col), A.T)
        return np.array(entropies)
    elif axis == 1:
        entropies = map(lambda row: entropy(row), A)
        return np.array(entropies).reshape((-1, 1))
    else:
        raise ValueError("unsupported axis: {}".format(axis))

Answer 14

根据 unutbu 的建议，我创建了一个纯 python 实现。

def entropy2(labels):
 """ Computes entropy of label distribution. """
    n_labels = len(labels)

    if n_labels <= 1:
        return 0

    counts = np.bincount(labels)
    probs = counts / n_labels
    n_classes = np.count_nonzero(probs)

    if n_classes <= 1:
        return 0

    ent = 0.

    # Compute standard entropy.
    for i in probs:
        ent -= i * log(i, base=n_classes)

    return ent

我缺少的一点是标签是一个大数组，但概率是 3 或 4 个元素长。使用纯 python，我的应用程序现在速度是原来的两倍。