python - 重新分配链表元素的 'next' 链接时 Numba 降低错误

Question

下面的代码是我用numba实现的链表(相关例子可以看here、here和here)。remove 函数删除位于 index 位置的元素。(请注意，这里假设 index 始终是有效位置)要删除这样的元素 (index)，我们会执行类似 element[index-1].next -> element[index+1] 的赋值。但是 numba 似乎不喜欢我这样做:

from numba import deferred_type,optional
from numba import int64
from numba import jitclass,njit

list_type = deferred_type()

spec = [
    ('data',int64),
    ('next',optional(list_type))
]
@jitclass(spec)
class List(object):
    def __init__(self,data,next):
        self.data = data
        self.next = next

    def prepend(self, data):
        return List(data, self)

list_type.define(List.class_type.instance_type)

def length(stack):
    i = 0
    while stack is not None:
        stack = stack.next
        i+=1
    return i

@njit
def remove(stack,index):
    prev = None
    if index == 0:
        stack = stack.next
    else:
        cur = stack
        i = 0
        while cur is not None:
            if index == i:
                break
            i = i+1
            prev = cur
            cur = cur.next
        prev.next = cur.next
    return stack

def runme():
    from numpy.random import randint
    a = randint(0,100,10)

    list_ = None

    for n in a:
        if list_ is None:
            list_ = List(n,None)
        else:
            list_ = list_.prepend(n)
    print(length(list_))

    indexes = list(range(len(a)))
    for i in indexes[::-1]:
        list_ = remove(list_,i)
    print(length(list_))

if __name__ == '__main__':
    runme()

它在 prev.next = cur.next 行中断并出现以下错误:

numba.errors.LoweringError: Failed at nopython (nopython mode backend)
No definition for lowering ?instance.jitclass.List#14bebc8<data:int64,next:?DeferredType#140432587762264>.next = ?DeferredType#140432587762264
File "numba_test.py", line 43
[1] During: lowering "(prev).next = $70.2" at numba_test.py (43)

看起来 prev.next 无法重新分配，但我不太确定这里发生了什么。

我正在使用 numba 0.35.0 和 python 3.6.2。

有什么想法吗？

谢谢。

---

现状

该错误是 Numba 中的错误。更多信息，请前往相应问题:https://github.com/numba/numba/issues/2606

Answer 1

看起来 nopython 模式确实支持此功能。只需使用“普通”jit 装饰器。如果需要，这会回退到纯 Python:

from numba import jit

@jit
def remove(stack,index):
    prev = None
    if index == 0:
        stack = stack.next
    else:
        cur = stack
        i = 0
        while cur is not None:
            if index == i:
                break
            i = i+1
            prev = cur
            cur = cur.next
        prev.next = cur.next
    return stack

因为 njit 等同于 jit(nopython=True)。

测试输出:

10
0