我正在寻找一种模拟经典频率严重程度分布的方法,例如：
X = sum(i = 1..N,Y_i),其中N例如是泊松分布,Y是对数正态.

简单的天真的numpy脚本将是：

import numpy as np
SIM_NUM = 3

X = []
for _i in range(SIM_NUM):
    nr_claims = np.random.poisson(1)
    temp = []
    for _j in range(nr_claims):
         temp.append(np.random.lognormal(0, 1))
    X.append(sum(temp))

现在,我尝试将其矢量化以提高性能.以下是更好的一点：

N = np.random.poisson(1, SIM_NUM)
X = []
for n in N:
    X.append(sum(np.random.lognormal(0, 1, n)))

我的问题是我是否还可以向量化第二个循环？例如,通过模拟所有损失：

N = np.random.poisson(1, SIM_NUM)
# print(N) would lead to something like [1 3 0]
losses = np.random.lognormal(0,1, sum(N))
# print(N) would lead to something like 
#[ 0.56750244  0.84161871  0.41567216  1.04311742]

# X should now be [ 0.56750244, 0.84161871 + 0.41567216 + 1.04311742, 0] 

我的想法是“智能切片”或“ A = [[1,0,0,0]],[0,1,1,1],[0,0,0,0]]的矩阵乘法.但是我无法从这些想法中做出些聪明的事情.

我正在寻找最快的X计算.

解决方法:

我们可以使用np.bincount,这对于基于间隔/ ID的求和操作非常有效,特别是在处理一维数组时.实现看起来像这样-

# Generate all poisson distribution values in one go
pv = np.random.poisson(1,SIM_NUM)

# Use poisson values to get count of total for random lognormal needed.
# Generate all those random numbers again in vectorized way 
rand_arr = np.random.lognormal(0, 1, pv.sum())

# Finally create IDs using pv as extents for use with bincount to do
# ID based and thus effectively interval-based summing
out = np.bincount(np.arange(pv.size).repeat(pv),rand_arr,minlength=SIM_NUM)

运行时测试-

功能定义：

def original_app1(SIM_NUM):
    X = []
    for _i in range(SIM_NUM):
        nr_claims = np.random.poisson(1)
        temp = []
        for _j in range(nr_claims):
             temp.append(np.random.lognormal(0, 1))
        X.append(sum(temp))
    return X

def original_app2(SIM_NUM):
    N = np.random.poisson(1, SIM_NUM)
    X = []
    for n in N:
        X.append(sum(np.random.lognormal(0, 1, n)))
    return X

def vectorized_app1(SIM_NUM):
    pv = np.random.poisson(1,SIM_NUM)
    r = np.random.lognormal(0, 1,pv.sum())
    return np.bincount(np.arange(pv.size).repeat(pv),r,minlength=SIM_NUM)

大型数据集上的时间：

In [199]: SIM_NUM = 1000

In [200]: %timeit original_app1(SIM_NUM)
100 loops, best of 3: 2.6 ms per loop

In [201]: %timeit original_app2(SIM_NUM)
100 loops, best of 3: 6.65 ms per loop

In [202]: %timeit vectorized_app1(SIM_NUM)
1000 loops, best of 3: 252 µs per loop

In [203]: SIM_NUM = 10000

In [204]: %timeit original_app1(SIM_NUM)
10 loops, best of 3: 26.1 ms per loop

In [205]: %timeit original_app2(SIM_NUM)
10 loops, best of 3: 77.5 ms per loop

In [206]: %timeit vectorized_app1(SIM_NUM)
100 loops, best of 3: 2.46 ms per loop

因此,我们正在研究那里的10倍加速.