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对数组中的每个元素 n 快速执行 n 次函数

对数组中的每个元素 n 快速执行 n 次函数

www说 2023-04-18 14:27:16
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。以下是迄今为止我能做的最好的。它几乎不比forpython 中的两个嵌套循环快。在我的实际用例中,我的数组是 100,000 x 1,000。有谁知道如何使用纯 numpy 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)每个循环 291 µs ± 10.6 µs(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的代码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 lossescalculate_insured_losses(freq, sev)
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1 回答

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慕斯王

TA贡献1864条经验 获得超2个赞

你可以像这样更快地做到这一点:


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)


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反对 回复 2023-04-18
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