对数组中的每个元素 n 快速执行 n 次函数

Question

我有一个 n_years by n_repeats 计数数据数组。

对于每个元素 (e)，我想从损失严重性数组中抽取 e 次并取抽取的总和.

以下是目前为止我能做的最好的。它比 python 中的两个嵌套 for 循环快不了多少。在我的实际用例中，我的数组是 100,000 x 1,000。

有没有人知道如何使用纯 numpy 来完成此操作？

frequency = np.array(
    [
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 1],
        [1, 2, 1],
        [1, 2, 1],
        [2, 4, 2],
        [2, 4, 2],
        [3, 5, 2],
    ]
)
sev = np.array([1,1,2,2,1,2,3,4,5,1,1,2])

def calculate_insured_losses(frequency, severity_array):

    def yearly_loss(element, severity_array=severity_array):  
        return 0 if element == 0 else np.random.choice(severity_array, size=element, replace=True).sum()

    return np.vectorize(yearly_loss)(frequency.flatten()).reshape(frequency.shape)

calculate_insured_losses(freq, sev)

291 µs ± 10.6 µs 每个循环（7 次运行的平均值 ± 标准偏差，每次 1000 次循环）

编辑：带有嵌套循环的更简单代码

def calculate_insured_losses(frequency, severity):
    
    def yearly_loss(element, severity_array=severity):
        if element == 0:
            return 0
        else:
            return np.random.choice(severity_array, size=element, replace=True).sum()
    
    n_years, n_repeats = frequency.shape
    
    losses = np.empty(shape=frequency.shape)
    
    for year in range(n_years):
        for repeat in range(n_repeats):
            losses[year, repeat] = yearly_loss(frequency[year, repeat])

    return losses

calculate_insured_losses(freq, sev)

Answer 1

你可以这样做更快：

import numpy as np

def calculate_insured_losses(frequency, severity_array):
    # Flattened frequencies table
    r = frequency.ravel()
    # Accumulate
    rcum = np.cumsum(r)
    # Take all ramdom samples at once
    c = np.random.choice(severity_array, rcum[-1], replace=True)
    # Sum segments
    res = np.add.reduceat(c, rcum - r)
    # Make zero elements
    res *= r.astype(bool)
    # Return reshaped result
    return res.reshape(frequency.shape)

# For comparison
def calculate_insured_losses_loop(frequency, severity_array):
    def yearly_loss(element, severity_array=severity_array):  
        return 0 if element == 0 else np.random.choice(severity_array, size=element, replace=True).sum()
    return np.vectorize(yearly_loss)(frequency.flatten()).reshape(frequency.shape)

# Test
frequency = np.array(
    [
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 0],
        [0, 0, 1],
        [1, 2, 1],
        [1, 2, 1],
        [2, 4, 2],
        [2, 4, 2],
        [3, 5, 2],
    ]
)
sev = np.array([1, 1, 2, 2, 1, 2, 3, 4, 5, 1, 1, 2])
# Check results from functions match
np.random.seed(0)
res = calculate_insured_losses(frequency, sev)
np.random.seed(0)
res_loop = calculate_insured_losses_loop(frequency, sev)
print(np.all(res == res_loop))
# True

# Benchmark
%timeit calculate_insured_losses(frequency, sev)
# 32.4 µs ± 220 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit calculate_insured_losses_loop(frequency, sev)
# 383 µs ± 11.6 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

对数组中的每个元素 n 快速执行 n 次函数

Quickly performing a function n times for each element n in an array

python

numpy

montecarlo