我有一個 n_years by n_repeats 計數(shù)數(shù)據(jù)數(shù)組。對于每個元素 ( e )，我想從損失嚴(yán)重性數(shù)組中抽取e次并取抽取的總和。以下是迄今為止我能做的最好的。它幾乎不比forpython 中的兩個嵌套循環(huán)快。在我的實(shí)際用例中，我的數(shù)組是 100,000 x 1,000。有誰知道如何使用純 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)每個循環(huán) 291 μs ± 10.6 μs（7 次運(yùn)行的平均值 ± 標(biāo)準(zhǔn)偏差，每次 1000 次循環(huán)）編輯：帶有嵌套循環(huán)的更簡單的代碼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)
1 回答
慕斯王
TA貢獻(xiàn)1864條經(jīng)驗(yàn) 獲得超2個贊
你可以像這樣更快地做到這一點(diǎn)：
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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