Python3如果不是条件简化

这两个条件检查是否相同？我想不出如何检查它们是否相同

l1 = []
l2 = []

if not l1 and not l2:
    print ('y')

if not (l1 and l2):
    print ('y')

感谢所有回复的人，我已经做了一些基本的计时，看看哪个更快

import time
l1 = []
l2 = []

st = time.time()
for i in range(100000000):
    if not l1 and not l2:
        pass
end = time.time()
print ('if not l1 and not l2: '+str(end-st))

st = time.time()
for i in range(100000000):
    if not (l1 or l2):
        pass
end = time.time()
print ('if not (l1 or l2): '+str(end-st))

印刷品：

if not l1 and not l2: 8.533874750137329
if not (l1 or l2): 7.91820216178894

---

不，它们不一样。看

反例是：

l1 = [0]
l2 = []

---

它们不一样。您需要修改第二个条件，如下所示，使其等效：

l1 = []
l2 = []

if not l1 and not l2:
    print ('y')

if not (l1 or l2):
    print ('y')

if not (l1 or l2):
    print ('y')

---

如果要保持相同，请使用或操作：

l1 = []
l2 = []

if not l1 and not l2:
    print ('y')

等价物：

l1 = []
l2 = []

if not l1 and not l2:
    print ('y')

if not (l1 or l2):
    print ('y')

if not (l1 or l2):
    print ('y')