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数组中的 Python Numba 值

转载 作者:行者123 更新时间:2023-12-04 01:13:14 27 4
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我正在尝试检查一个数字是否在 int8 的 NumPy 数组中。我试过这个,但它不起作用。

from numba import njit
import numpy as np

@njit
def c(b):
return 9 in b

a = np.array((9, 10, 11), 'int8')
print(c(a))

我得到的错误是

Invalid use of Function(<built-in function contains>) with argument(s) of type(s): (array(int8, 1d, C), Literal[int](9))
* parameterized
In definition 0:
All templates rejected with literals.
In definition 1:
All templates rejected without literals.
In definition 2:
All templates rejected with literals.
In definition 3:
All templates rejected without literals.
In definition 4:
All templates rejected with literals.
In definition 5:
All templates rejected without literals.
This error is usually caused by passing an argument of a type that is unsupported by the named function.
[1] During: typing of intrinsic-call at .\emptyList.py (6)

如何在保持性能的同时解决这个问题?将检查数组是否有两个值,1 和 -1,长度为 32 项。它们没有排序。

最佳答案

检查两个值是否在数组中

为了仅检查两个值是否出现在数组中,我建议使用简单的暴力算法。

代码

import numba as nb
import numpy as np

@nb.njit(fastmath=True)
def isin(b):
for i in range(b.shape[0]):
res=False
if (b[i]==-1):
res=True
if (b[i]==1):
res=True
return res

#Parallelized call to isin if the data is an array of shape (n,m)
@nb.njit(fastmath=True,parallel=True)
def isin_arr(b):
res=np.empty(b.shape[0],dtype=nb.boolean)
for i in nb.prange(b.shape[0]):
res[i]=isin(b[i,:])

return res

性能

#Create some data (320MB)
A=(np.random.randn(10000000,32)-0.5)*5
A=A.astype(np.int8)
res=isin_arr(A) 11ms per call

因此,通过这种方法,我获得了大约 29GB/s 的吞吐量,这与内存带宽相差不远。您还可以尝试减少 Testdatasize,使其适合 L3 缓存以避免内存带宽限制。使用 3.2 MB 测试数据时,我获得了 100 GB/s 的吞吐量(远远超出了我的内存带宽),这清楚地表明此实现的内存带宽有限。

关于数组中的 Python Numba 值,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54930852/

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