如何比较两个numba jitclass对象是否相同?
我有以下代码
from numba import jitclass
import numba
node_type = numba.deferred_type()
DoubleLinkedNode_spec = [
('value', numba.optional(numba.typeof(1.0))),
('prev', numba.optional(node_type)),
('next', numba.optional(node_type))
]
@jitclass(DoubleLinkedNode_spec)
class DoubleLinkedNode(object):
def __init__(self, value, prev, next):
self.value = value
self.prev = prev
self.next = next
node_type.define(DoubleLinkedNode.class_type.instance_type)
n1 = DoubleLinkedNode(1.0, None, None)
n2 = DoubleLinkedNode(2.0, n1, None)
n1.next = n2
print(f'{n2}\n{n2.prev.next}')
#outputs:
# <numba.jitclass.boxing.DoubleLinkedNode object at 0x7fbf26923850>
# <numba.jitclass.boxing.DoubleLinkedNode object at 0x7fbf256b3cf0>
print(f'Next is None. n1: {n1.next is None} n2: {n2.next is None}')
#outputs:
# Next is None. n1: False n2: True
这是一个双链接列表的标准节点。
列表中的 is
操作符不工作,因为它们不在内存的同一个地址。
is None
似乎是可行的。但是我可以相信它吗?Dunder方法在jitclasses中不能正常工作,所以我实现了类似的东西。__eq__
.
from numba.experimental import jitclass
import numba
node_type = numba.deferred_type()
DoubleLinkedNode_spec = [
('value', numba.optional(numba.typeof(1.0))),
('prev', numba.optional(node_type)),
('next', numba.optional(node_type))
]
@jitclass(DoubleLinkedNode_spec)
class DoubleLinkedNode(object):
def __init__(self, value, prev, next):
self.value = value
self.prev = prev
self.next = next
def equal(self, other):
return self.prevs_eq(other) and self.nexts_eq(other)
def prevs_eq(self, other):
while True:
if self.prev is None:
return self.value == other.value
if not self.value == other.value:
return False
self = self.prev
other = other.prev
def nexts_eq(self, other):
while True:
if self.next is None:
return self.value == other.value
if not self.value == other.value:
return False
self = self.next
other = other.next
node_type.define(DoubleLinkedNode.class_type.instance_type)
n1 = DoubleLinkedNode(1.0, None, None)
n2 = DoubleLinkedNode(2.0, n1, None)
n1.next = n2
print(f'{n2.equal(n2.prev.next)}')
# True
你必须打电话给... equal
方法,这是不优雅的,但如果你把它改名为 __eq__
它根本就不会工作。这对我目前测试的情况来说是可行的。