我想做的事情是:
foo = {
'foo': 1,
'zip': 2,
'zam': 3,
'bar': 4
}
if ("foo", "bar") in foo:
#do stuff
我如何检查是否foo和酒吧都在dict foo?
我想做的事情是:
foo = {
'foo': 1,
'zip': 2,
'zam': 3,
'bar': 4
}
if ("foo", "bar") in foo:
#do stuff
我如何检查是否foo和酒吧都在dict foo?
当前回答
my_dict = {
'name': 'Askavy',
'country': 'India',
'age': 30
}
if set(('name', 'country','age')).issubset(my_dict.keys()):
print("All keys are present in the dictionary")
else:
print("All keys are not present in the dictionary")
其他回答
你可以这样做:
>>> if all(k in foo for k in ("foo","bar")):
... print "They're there!"
...
They're there!
我认为这是最聪明和最精辟的。
{'key1','key2'} <= my_dict.keys()
使用集:
if set(("foo", "bar")).issubset(foo):
#do stuff
另外:
if set(("foo", "bar")) <= set(foo):
#do stuff
在确定是否只有一些键匹配的情况下,这是有效的:
any_keys_i_seek = ["key1", "key2", "key3"]
if set(my_dict).intersection(any_keys_i_seek):
# code_here
pass
还有另一个选项,如果只有一些键匹配:
any_keys_i_seek = ["key1", "key2", "key3"]
if any_keys_i_seek & my_dict.keys():
# code_here
pass
你也可以使用. is子集()
>>> {"key1", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
True
>>> {"key4", "key2"}.issubset({"key1":1, "key2":2, "key3": 3})
False
>>>