Question

我有一个numpy数组（实际上是从GIS栅格地图导入的），其中包含物种出现的概率值如下例所示：

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阵列的离散版本。就像我有例如100个人位于该阵列的区域（1200个细胞）他们是怎么回事分散式？当然应根据概率分配它们，意思是较低的值表示较低的发生概率。然而，由于一切都是统计数据，个人仍有可能处于低概率细胞。多个人应该可以在牢房中占据......

就像将连续分布曲线再次转换为直方图一样。像许多不同的直方图可能会产生一定的分布曲线，它也应该是相反的。因此，应用我正在寻找的算法每次都会产生不同的离散值。

... python中有任何算法可以做到吗？由于我不熟悉离散化，也许有人可以提供帮助。

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，可以将random.choice替换为：

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

，并提供：

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“应用”的例子。考虑骰子有6张脸（6个数字）。每个数字/结果的概率为1/6。以数组的形式显示骰子可能如下所示：

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

因此，将骰子滚动100次(n=100)会导致以下模拟：

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

我认为这对于这样的应用来说可能是一种合适的方法。感谢ecatmur的帮助！

/约翰内斯

Answer 3

这与我本月早些时候的question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

重要的两条主线是上面代码中的最后两行

Python中概率数组的离散化

3 个答案: