Python中概率数组的离散化

Question

我有一个numpy数组（实际上是从GIS光栅图导入的），它包含物种出现的概率值，如以下示例：

a = random.randint(1.0,20.0,1200).reshape(40,30)
b = (a*1.0)/sum(a)

现在我想再次获得该数组的离散版本。就像我有例如，100个位于该阵列区域（1200个细胞）的个体分布式？当然，它们应该根据它们的概率来分布，意味着较低的值表示发生的概率较低。然而，由于一切都是统计数据，因此个人被定位的可能性仍然很低单间牢房应该有可能多个个体可以占据细胞。。。

这就像将连续分布曲线再次转换为直方图。就像许多不同的直方图可能导致某个分布曲线一样，它也应该是相反的。因此，应用我正在寻找的算法将每次产生不同的离散值。

python中有什么算法可以做到这一点吗？由于我对离散化不太熟悉，也许有人可以帮忙。

Answer 1

将random.choice与bincount:一起使用

np.bincount(np.random.choice(b.size, 100, p=b.flat),
            minlength=b.size).reshape(b.shape)

如果您没有NumPy 1.7，您可以将random.choice替换为：

np.searchsorted(np.cumsum(b), np.random.random(100))

给予：

np.bincount(np.searchsorted(np.cumsum(b), np.random.random(100)),
            minlength=b.size).reshape(b.shape)

Answer 2

到目前为止，我认为ecatmur的答案似乎非常合理和简单。

我只想添加一个更"实用"的例子。考虑骰子具有6个面（6个数字）。每个数字/结果的概率为1/6。以阵列的形式显示骰子可能看起来像：

b = np.array([[1,1,1],[1,1,1]])/6.0

因此掷骰子100次(n=100)得到以下模拟：

np.bincount(np.searchsorted(np.cumsum(b), np.random.random(n)),minlength=b.size).reshape(b.shape)

我认为这可能是一个适当的方法，为这样的申请。因此，感谢您ecatmur的帮助！

/Johannes

Answer 3

这与我本月早些时候提出的问题类似。

import random
def RandFloats(Size):
    Scalar = 1.0
    VectorSize = Size
    RandomVector = [random.random() for i in range(VectorSize)]
    RandomVectorSum = sum(RandomVector)
    RandomVector = [Scalar*i/RandomVectorSum for i in RandomVector]
    return RandomVector
from numpy.random import multinomial
import math
def RandIntVec(ListSize, ListSumValue, Distribution='Normal'):
    """
    Inputs:
    ListSize = the size of the list to return
    ListSumValue = The sum of list values
    Distribution = can be 'uniform' for uniform distribution, 'normal' for a normal distribution ~ N(0,1) with +/- 5 sigma  (default), or a list of size 'ListSize' or 'ListSize - 1' for an empirical (arbitrary) distribution. Probabilities of each of the p different outcomes. These should sum to 1 (however, the last element is always assumed to account for the remaining probability, as long as sum(pvals[:-1]) <= 1).  
    Output:
    A list of random integers of length 'ListSize' whose sum is 'ListSumValue'.
    """
    if type(Distribution) == list:
        DistributionSize = len(Distribution)
        if ListSize == DistributionSize or (ListSize-1) == DistributionSize:
            Values = multinomial(ListSumValue,Distribution,size=1)
            OutputValue = Values[0]
    elif Distribution.lower() == 'uniform': #I do not recommend this!!!! I see that it is not as random (at least on my computer) as I had hoped
        UniformDistro = [1/ListSize for i in range(ListSize)]
        Values = multinomial(ListSumValue,UniformDistro,size=1)
        OutputValue = Values[0]
    elif Distribution.lower() == 'normal':
        """
            Normal Distribution Construction....It's very flexible and hideous
            Assume a +-3 sigma range.  Warning, this may or may not be a suitable range for your implementation!
            If one wishes to explore a different range, then changes the LowSigma and HighSigma values
            """
            LowSigma    = -3#-3 sigma
            HighSigma   = 3#+3 sigma
            StepSize    = 1/(float(ListSize) - 1)
            ZValues     = [(LowSigma * (1-i*StepSize) +(i*StepSize)*HighSigma) for i in range(int(ListSize))]
            #Construction parameters for N(Mean,Variance) - Default is N(0,1)
            Mean        = 0
            Var         = 1
            #NormalDistro= [self.NormalDistributionFunction(Mean, Var, x) for x in ZValues]
            NormalDistro= list()
            for i in range(len(ZValues)):
                if i==0:
                    ERFCVAL = 0.5 * math.erfc(-ZValues[i]/math.sqrt(2))
                    NormalDistro.append(ERFCVAL)
                elif i ==  len(ZValues) - 1:
                    ERFCVAL = NormalDistro[0]
                    NormalDistro.append(ERFCVAL)
                else:
                    ERFCVAL1 = 0.5 * math.erfc(-ZValues[i]/math.sqrt(2))
                    ERFCVAL2 = 0.5 * math.erfc(-ZValues[i-1]/math.sqrt(2))
                    ERFCVAL = ERFCVAL1 - ERFCVAL2
                    NormalDistro.append(ERFCVAL)  
            #print "Normal Distribution sum = %f"%sum(NormalDistro)
            Values = multinomial(ListSumValue,NormalDistro,size=1)
            OutputValue = Values[0]
        else:
            raise ValueError ('Cannot create desired vector')
        return OutputValue
    else:
        raise ValueError ('Cannot create desired vector')
    return OutputValue
ProbabilityDistibution = RandFloats(1200)#This is your probability distribution for your 1200 cell array
SizeDistribution = RandIntVec(1200,100,Distribution=ProbabilityDistribution)#for a 1200 cell array, whose sum is 100 with given probability distribution 

两条重要的主线是上代码的最后两行