Python 使用numba.jit编译条件数组需要很长时间

如果我试图用numba的jit编译器编译一个包含一系列条件的函数，它需要很长时间。这个程序看起来基本上像

from numba import jit
import numpy as np

@jit(nopython=True)
def foo(a, b):
    valid = [
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0),
        (a - 1 >= 0) and (b - 1 >= 0)
    ]

foo(1, 1)

我排除了不会显著改变编译时间的所有内容。如果我使用超过20个元素，问题就会出现

| elements | time |
-------------------
|    21    | 2.7s |
|    22    | 5.1s |
|    23    |  10s |
|   ...    |  ... |
-------------------

尽管如此，该功能运行良好。有人知道，为什么用numba编译这样的函数需要这么长时间吗？以类似的方式使用整数或浮点数组合创建数组不会产生任何问题

您可能希望在numba上报告这一点，因为它的伸缩性很差，感觉编译器中出现了一些错误

<>你也可以考虑如果你真的需要大量这样的数组语句，如果问题可以更清楚地重构。例如，

valid

是否可以根据需要调用函数，而不是布尔数组

综上所述，当前版本的numba中的一个解决方案是展开条件

例如：

# "codegen"
for i in range(23):
    print(f'    valid[{i}] = (a - 1 >= 0) and (b - 1 >= 0)')

@jit(nopython=True)
def foo(a, b):
    valid = np.empty(23, dtype=np.bool_)
    valid[0] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[1] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[2] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[3] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[4] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[5] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[6] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[7] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[8] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[9] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[10] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[11] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[12] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[13] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[14] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[15] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[16] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[17] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[18] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[19] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[20] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[21] = (a - 1 >= 0) and (b - 1 >= 0)
    valid[22] = (a - 1 >= 0) and (b - 1 >= 0)

%time foo(1,1)
Wall time: 274 ms

---

你能提供一个可运行的完整示例吗？@JoshAdel我更改了示例，它现在可运行且更简单。在我原来的函数中，所有条件都不同，我必须使用

有效的

索引多个其他数组。使用数组而不是if语句似乎是更快、更可读的选择。我提出了一个问题，如果有什么变化，我会更新我的问题。