I have beginner two questions
- What does
*zor*fooor**foomean regarding function in Python. - This works -
a = lambda *z :zBut this does not -a = lambda **z: z. Because it is supposed to take 0 arguments. What does this actually mean?
- stackoverflow.com/questions/5218993/…Jaroslav Jandek– Jaroslav Jandek2011-03-08 12:56:10 +00:00Commented Mar 8, 2011 at 12:56
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4 Answers
*z and **z in Python refer to args and kwargs. args are positional arguments and kwargs are keyword arguments. lambda **z doesn't work in your example because z isn't a keyword argument: it's merely positional. Compare these different results:
>>> a = lambda z: z >>> b = lambda *z: z >>> c = lambda **z: z >>> a([1,2,3]) [1, 2, 3] >>> b([1,2,3]) ([1, 2, 3],) >>> c([1,2,3]) # list arg passed, **kwargs expected Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: <lambda>() takes exactly 0 arguments (1 given) >>> c(z=[1,2,3]) # explicit kwarg {'z': [1, 2, 3]} >>> c(x=[1,2,3]) # explicit kwarg {'x': [1, 2, 3]} >>> c({'x':[1,2,3]}) # dict called as a single arg Traceback (most recent call last): File "<stdin>", line 1, in <module> TypeError: <lambda>() takes exactly 0 arguments (1 given) >>> c(**{'x':[1,2,3]}) # dict called as **kwargs {'x': [1, 2, 3]} >>> b(*[1,2,3]) # list called as *args (1, 2, 3) 
Comments
Check out the link its a good blog post on How to use *args and **kwargs in Python http://www.saltycrane.com/blog/2008/01/how-to-use-args-and-kwargs-in-python/
Comments
When you see something like def foo(*args) or def foo(**kwargs) they're telling you they're expecting something more than a single argument. They're expecting several arguments or several named arguments. Often, however, you see this:
def foo(item1, item2, item3, *args, **kwargs) Which tells you that they're expecting at least item's 1, 2, and 3 but "args" and "kwargs" are optional parameters.
To answer your question regarding a = lambda **z: z try passing in a "named argument list" like so:
> a = lambda **z:z > a(b=1, c=2) >> {b : 1, c : 2} The output will be a dictionary, just as you inadvertently defined. Named argument lists are, in essence, dictionaries.
Comments
*args can be used to make a function take an arbitrary number of arguments. **kwargs can be used to make a function take arbitrary keyword arguments. This means that:
>>> foo = lambda *args: args >>> bar = lambda **kwargs: kwargs >>> foo(1,2,3) (1,2,3) >>> bar(asdf=1) {'asdf':1} >>> bar(1,2,3) TypeError: <lambda>() takes exactly 0 arguments (3 given) The last error happens because bar can only take keyword arguments.
2 Comments
Mark Dickinson
Does
foo(1, 2, 3) really produce a list for you? It gives a tuple here.
Björn Pollex
Sorry, hasty there. It is actually a tuple. Thanks.