[英]Passing a dictionary to a function as keyword parameters
I'd like to call a function in python using a dictionary.我想使用字典在 python 中调用一个函数。
Here is some code:这是一些代码:
d = dict(param='test')
def f(param):
print(param)
f(d)
This prints {'param': 'test'}
but I'd like it to just print test
.这会打印{'param': 'test'}
但我希望它只打印test
。
I'd like it to work similarly for more parameters:我希望它对更多参数有类似的工作:
d = dict(p1=1, p2=2)
def f2(p1, p2):
print(p1, p2)
f2(d)
Is this possible?这可能吗?
Figured it out for myself in the end.最后自己想通了。 It is simple, I was just missing the ** operator to unpack the dictionary很简单,我只是缺少 ** 运算符来解包字典
So my example becomes:所以我的例子变成了:
d = dict(p1=1, p2=2)
def f2(p1,p2):
print p1, p2
f2(**d)
In[1]: def myfunc(a=1, b=2):
In[2]: print(a, b)
In[3]: mydict = {'a': 100, 'b': 200}
In[4]: myfunc(**mydict)
100 200
A few extra details that might be helpful to know (questions I had after reading this and went and tested):一些可能有助于了解的额外细节(我在阅读本文并进行测试后遇到的问题):
Examples:例子:
Number 1: The function can have parameters that are not included in the dictionary数字 1:函数可以有字典中没有的参数
In[5]: mydict = {'a': 100}
In[6]: myfunc(**mydict)
100 2
Number 2: You can not override a function parameter that is already in the dictionary数字 2:您不能覆盖字典中已经存在的函数参数
In[7]: mydict = {'a': 100, 'b': 200}
In[8]: myfunc(a=3, **mydict)
TypeError: myfunc() got multiple values for keyword argument 'a'
Number 3: The dictionary can not have values that aren't in the function.数字 3:字典不能包含不在函数中的值。
In[9]: mydict = {'a': 100, 'b': 200, 'c': 300}
In[10]: myfunc(**mydict)
TypeError: myfunc() got an unexpected keyword argument 'c'
How to use a dictionary with more keys than function arguments:如何使用键多于函数参数的字典:
A solution to #3, above, is to accept (and ignore) additional kwargs in your function (note, by convention _
is a variable name used for something being discarded, though technically it's just a valid variable name to Python):上面 #3 的解决方案是在你的函数中接受(并忽略)额外的 kwargs(注意,按照约定_
是用于被丢弃的东西的变量名,尽管从技术上讲它只是 Python 的有效变量名):
In[11]: def myfunc2(a=None, **_):
In[12]: print(a)
In[13]: mydict = {'a': 100, 'b': 200, 'c': 300}
In[14]: myfunc2(**mydict)
100
Another option is to filter the dictionary based on the keyword arguments available in the function:另一种选择是根据函数中可用的关键字参数过滤字典:
In[15]: import inspect
In[16]: mydict = {'a': 100, 'b': 200, 'c': 300}
In[17]: filtered_mydict = {k: v for k, v in mydict.items() if k in [p.name for p in inspect.signature(myfunc).parameters.values()]}
In[18]: myfunc(**filtered_mydict)
100 200
Example with both positional and keyword arguments:具有位置和关键字参数的示例:
Notice further than you can use positional arguments and lists or tuples in effectively the same way as kwargs, here's a more advanced example incorporating both positional and keyword args:请注意,您可以像使用 kwargs 一样有效地使用位置参数和列表或元组,这是一个结合位置参数和关键字参数的更高级示例:
In[19]: def myfunc3(a, *posargs, b=2, **kwargs):
In[20]: print(a, b)
In[21]: print(posargs)
In[22]: print(kwargs)
In[23]: mylist = [10, 20, 30]
In[24]: mydict = {'b': 200, 'c': 300}
In[25]: myfunc3(*mylist, **mydict)
10 200
(20, 30)
{'c': 300}
Here ya go - works just any other iterable:在这里你去 - 适用于任何其他可迭代:
d = {'param' : 'test'}
def f(dictionary):
for key in dictionary:
print key
f(d)
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