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name

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If I define a class method with a keyword argument thus:

class foo(object):
  def foodo(thing=None, thong='not underwear'):
    print thing if thing else "nothing" 
    print 'a thong is',thong

calling the method generates a TypeError:

myfoo = foo()
myfoo.foodo(thing="something")

...
TypeError: foodo() got multiple values for keyword argument 'thing'

What's going on?

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    • You will never get a satisfactory answer as to why explicit self is better than implicit this.

The problem is that the first argument passed to class methods in python is always a copy of the class instance on which the method is called, typically labelled self. If the class is declared thus:

class foo(object):
  def foodo(self, thing=None, thong='not underwear'):
    print thing if thing else "nothing" 
    print 'a thong is',thong

it behaves as expected.

Explanation:

Without self as the first parameter, when myfoo.foodo(thing="something") is executed, the foodo method is called with arguments (myfoo, thing="something"). The instance myfoo is then assigned to thing (since thing is the first declared parameter), but python also attempts to assign "something" to thing, hence the Exception.

To demonstrate, try running this with the original code:

myfoo.foodo("something")
print
print myfoo

You'll output like:

<__main__.foo object at 0x321c290>
a thong is something

<__main__.foo object at 0x321c290>

You can see that 'thing' has been assigned a reference to the instance 'myfoo' of the class 'foo'. This section of the docs explains how function arguments work a bit more.

  • 163
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    • of note: you can get the same type of error if your function def includes self as the first parameter, and then you accidentally call the function also with self as the first parameter.

Thanks for the instructive posts. I'd just like to keep a note that if you're getting "TypeError: foodo() got multiple values for keyword argument 'thing'", it may also be that you're mistakenly passing the 'self' as a parameter when calling the function (probably because you copied the line from the class declaration - it's a common error when one's in a hurry).

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      • 1
    • That's what happened to me, thanks for adding to this answer. This might be the more common mistake to make, which is why you're getting my upvote.

This might be obvious, but it might help someone who has never seen it before. This also happens for regular functions if you mistakenly assign a parameter by position and explicitly by name.

>>> def foodo(thing=None, thong='not underwear'):
...     print thing if thing else "nothing"
...     print 'a thong is',thong
...
>>> foodo('something', thing='everything')
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: foodo() got multiple values for keyword argument 'thing'
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just add 'staticmethod' decorator to function and problem is fixed

class foo(object):
    @staticmethod
    def foodo(thing=None, thong='not underwear'):
        print thing if thing else "nothing" 
        print 'a thong is',thong
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      • 2
    • staticmethod stops the method receiving self as a first argument. So now, if you call myfoo.foodo(thing="something"), thing="something" will be assigned to the first argument, rather than the implicit self argument.
      • 1
    • it also means that you can't access class variables within the function, which are usually accessed through self
      • 1
    • I just got this error and this solution solved my problem. But can you elaborate on how or why this decorator solved the problem?

I want to add one more answer :

It happens when you try to pass positional parameter with wrong position order along with keyword argument in calling function.

there is difference between parameter and argument you can read in detail about here Arguments and Parameter in python

def hello(a,b=1, *args):
   print(a, b, *args)


hello(1, 2, 3, 4,a=12)

since we have three parameters :

a is positional parameter

b=1 is keyword and default parameter

*args is variable length parameter

so we first assign a as positional parameter , means we have to provide value to positional argument in its position order, here order matter. but we are passing argument 1 at the place of a in calling function and then we are also providing value to a , treating as keyword argument. now a have two values :

one is positional value: a=1

second is keyworded value which is a=12

Solution

We have to change hello(1, 2, 3, 4,a=12) to hello(1, 2, 3, 4,12) so now a will get only one positional value which is 1 and b will get value 2 and rest of values will get *args (variable length parameter)

additional information

if we want that *args should get 2,3,4 and a should get 1 and b should get 12

then we can do like this
def hello(a,*args,b=1): pass hello(1, 2, 3, 4,b=12)

Something more :

def hello(a,*c,b=1,**kwargs):
    print(b)
    print(c)
    print(a)
    print(kwargs)

hello(1,2,1,2,8,9,c=12)

output :

1

(2, 1, 2, 8, 9)

1

{'c': 12}
  • 4
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This error can also happen if you pass a key word argument for which one of the keys is similar (has same string name) to a positional argument.

>>> class Foo():
...     def bar(self, bar, **kwargs):
...             print(bar)
... 
>>> kwgs = {"bar":"Barred", "jokes":"Another key word argument"}
>>> myfoo = Foo()
>>> myfoo.bar("fire", **kwgs)
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
TypeError: bar() got multiple values for argument 'bar'
>>> 

"fire" has been accepted into the 'bar' argument. And yet there is another 'bar' argument present in kwargs.

You would have to remove the keyword argument from the kwargs before passing it to the method.

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Also this can happen in Django if you are using jquery ajax to url that reverses to a function that doesn't contain 'request' parameter

$.ajax({
  url: '{{ url_to_myfunc }}',
});


def myfunc(foo, bar):
    ...
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