希望这也能帮助其他人理解:

dict1 = {'a': 'dict1', 'b': 'dict1', 'c': 'dict1'}
dict2 = {'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}

dict3 = dict1.copy()
dict3 = dict3 | dict2
dict4 = dict1.copy()
dict4 |= dict2
print(f'dict1:\n {dict1}')
print(f'dict2:\n {dict2}')
print(f'dict1 after dict1 = dict1 | dict2 (dict2 index c replaces dict1 index c, items in dict1 are discarded if present in dict2):\n {dict3}')
print(f'dict1 after dict1 |= dict2 (same behaviour as dict1 = dict1 | dict2):\n {dict4}')

dict5 = dict1.copy()
dict5 = dict2 | dict5
dict6 = dict2.copy()
dict6 |= dict1
print(f'dict1 after dict1 = dict2 | dict1 (dict2 index c is missing, dict1 index c was retained, items in dict2 are discarded if present in dict1):\n {dict5}')
print(f'dict2 after dict2 |= dict1 (same behaviour as dict2 = dict2 | dict1):\n {dict6}')


dict1:
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict1'}
dict2:
 {'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 = dict1 | dict2 (dict2 index c replaces dict1 index c, items in dict1 are discarded if present in dict2):
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 |= dict2 (same behaviour as dict1 = dict1 | dict2):
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 = dict2 | dict1 (dict2 index c is missing, dict1 index c was retained, items in dict2 are discarded if present in dict1):
 {'c': 'dict1', 'd': 'dict2', 'e': 'dict2', 'a': 'dict1', 'b': 'dict1'}
dict2 after dict2 |= dict1 (same behaviour as dict2 = dict2 | dict1):
 {'c': 'dict1', 'd': 'dict2', 'e': 'dict2', 'a': 'dict1', 'b': 'dict1'}

当与set一起使用时，它执行联合操作。

在Python中，|=(ior)类似于联合运算。 例如，如果x=5和x|=5，那么两个值都将首先转换为二进制值，然后执行并集操作，我们得到答案5。

给出一个用例(在花时间分析其他答案之后):

def process(item):
   return bool(item) # imagine some sort of complex processing taking place above

def any_success(data): # return True if at least one is successful
    at_least_one = False
    for item in data:
       at_least_one |= process(item)
    return at_least_one

>>> any_success([False, False, False])
False
>>> any_success([True, False, False])
True
>>> any_success([False, True, False])
True

基本上没有短路:如果你需要处理每个项目并记录至少一次成功，可能会有用。

请参见此回答中的注意事项

希望这也能帮助其他人理解:

dict1 = {'a': 'dict1', 'b': 'dict1', 'c': 'dict1'}
dict2 = {'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}

dict3 = dict1.copy()
dict3 = dict3 | dict2
dict4 = dict1.copy()
dict4 |= dict2
print(f'dict1:\n {dict1}')
print(f'dict2:\n {dict2}')
print(f'dict1 after dict1 = dict1 | dict2 (dict2 index c replaces dict1 index c, items in dict1 are discarded if present in dict2):\n {dict3}')
print(f'dict1 after dict1 |= dict2 (same behaviour as dict1 = dict1 | dict2):\n {dict4}')

dict5 = dict1.copy()
dict5 = dict2 | dict5
dict6 = dict2.copy()
dict6 |= dict1
print(f'dict1 after dict1 = dict2 | dict1 (dict2 index c is missing, dict1 index c was retained, items in dict2 are discarded if present in dict1):\n {dict5}')
print(f'dict2 after dict2 |= dict1 (same behaviour as dict2 = dict2 | dict1):\n {dict6}')


dict1:
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict1'}
dict2:
 {'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 = dict1 | dict2 (dict2 index c replaces dict1 index c, items in dict1 are discarded if present in dict2):
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 |= dict2 (same behaviour as dict1 = dict1 | dict2):
 {'a': 'dict1', 'b': 'dict1', 'c': 'dict2', 'd': 'dict2', 'e': 'dict2'}
dict1 after dict1 = dict2 | dict1 (dict2 index c is missing, dict1 index c was retained, items in dict2 are discarded if present in dict1):
 {'c': 'dict1', 'd': 'dict2', 'e': 'dict2', 'a': 'dict1', 'b': 'dict1'}
dict2 after dict2 |= dict1 (same behaviour as dict2 = dict2 | dict1):
 {'c': 'dict1', 'd': 'dict2', 'e': 'dict2', 'a': 'dict1', 'b': 'dict1'}

|是位或。所以x |= y等价于x = x | y。

对于集合，|有一个相关的含义:集合并集。与你在数学中使用OR求两个集合的交集相同，你可以在python中使用|

*注意:这两个表达式不是100%等效的。在x |= y后，id(x) == id(y)。x = x | y后，id(x) != id(y)