我不理解下面的例子，假设我有这些函数:
# python likes
def save(filename, data, **kwargs):
fo = openX(filename, "w", **kwargs) # <- #1
fo.write(data)
fo.close()
# python doesnt like
def save2(filename, data, **kwargs):
fo = openX(filename, "w", kwargs) # <- #2
fo.write(data)
fo.close()
def openX(filename, mode, **kwargs):
#doing something fancy and returning a file object
为什么第一条是正确的，第二条是错误的?**kwargs基本上是一个字典，所以如果我想把参数传递给openX，我认为正确的方法是不使用**，只给出字典。但是Python显然不喜欢第二个，它告诉我我给了3个而不是2个参数。
这背后的原因是什么呢?
扩展@gecco的回答，下面是一个例子，将向您展示两者的区别:
def foo(**kwargs):
for entry in kwargs.items():
print("Key: {}, value: {}".format(entry[0], entry[1]))
# call using normal keys:
foo(a=1, b=2, c=3)
# call using an unpacked dictionary:
foo(**{"a": 1, "b":2, "c":3})
# call using a dictionary fails because the function will think you are
# giving it a positional argument
foo({"a": 1, "b": 2, "c": 3})
# this yields the same error as any other positional argument
foo(3)
foo("string")
在这里，您可以看到解包字典是如何工作的，以及为什么发送实际的字典会失败
下面的代码使用kwargs并将其转移到另一个函数:
def myprint( kwargs ):
# default values
a = kwargs.get('a', None)
b = kwargs.get('b', None)
# print both
print('a={}, b={}'.format(a,b))
def mytest( **kwargs ):
myprint( kwargs )
mytest()
mytest(b=2)
收益率:
a=None, b=None
a=None, b=2
