遇到不支持的op_LOAD_CLOSURE使用如何处理？

Question

这是我的MWE：

from numba import njit
import numpy as np

@njit   
def solve(n):
    count = np.zeros(n + 1, dtype=int)
    res = np.array([0], dtype=int)

    def search(sz=0, max_val=1, single=0, previous=None):
        nonlocal res
        if sz == 4 * n:
            res[0] += 1
            return
        if single and count[0] < 2 * n:
            count[0] += 1
            search(sz + 1, max_val, single)
            count[0] -= 1
        for i in range(1, max_val + 1):
            if i != previous and count[i] < 2:
                count[i] += 1
                search(sz + 1, max_val + (i == max_val and max_val < n), single + (count[i] == 1) - (count[i] == 2), i)
                count[i] -= 1
 
    search()
    return res[0]

for i in range(1, 6):
    print(solve(i))

这给出：

NotImplementedError: Failed in nopython mode pipeline (step: analyzing bytecode)
Unsupported use of op_LOAD_CLOSURE encountered

让它与 numba 一起工作的正确方法是什么？如果删除 @njit 行，代码会正确运行，但运行速度会很慢。

Answer 1

Numba 不支持递归内部函数：

Numba 现在支持内部函数只要它们是非递归的并且仅在本地调用，但不作为参数传递或作为结果返回。

编写没有递归内部函数的代码。

Answer 2

我将

@njit

放入内部函数：

import numpy as np
from numba import njit


def solve(n):
    @njit
    def search(count, res, sz=0, max_val=1, single=0, previous=None):
        if sz == 4 * n:
            res[0] += 1
            return
        if single and count[0] < 2 * n:
            count[0] += 1
            search(count, res, sz + 1, max_val, single)
            count[0] -= 1
        for i in range(1, max_val + 1):
            if i != previous and count[i] < 2:
                count[i] += 1
                search(
                    count,
                    res,
                    sz + 1,
                    max_val + (i == max_val and max_val < n),
                    single + (count[i] == 1) - (count[i] == 2),
                    i,
                )
                count[i] -= 1

    count = np.zeros(n + 1, dtype=int)
    res = np.array([0], dtype=int)
    search(count, res)
    return res[0]


for i in range(1, 6):
    print(solve(i))