What does single(not double) asterisk * means when unpacking dictionary in Python?











up vote
2
down vote

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Can anyone explain the difference when unpacking the dictionary using single or double asterisk? You can mention their difference when used in function parameters, only if it is relevant here, which I don't think so.



However, there may be some relevance, because they share the same asterisk syntax.



def foo(a,b)
return a+b

tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp) #typeError: keyword should be string. Why it bothers to check the type of keyword?


Besides, why the key of dictionary is not allowed to be non-string when passed as function arguments in THIS situation? Are there any exceptions? Why they design Python in this way, is it because the compiler can't deduce the types in here or something?



Thanks!










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  • 1




    Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
    – Alexander Reynolds
    Nov 5 at 3:55












  • Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
    – ShadowRanger
    2 days ago















up vote
2
down vote

favorite












Can anyone explain the difference when unpacking the dictionary using single or double asterisk? You can mention their difference when used in function parameters, only if it is relevant here, which I don't think so.



However, there may be some relevance, because they share the same asterisk syntax.



def foo(a,b)
return a+b

tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp) #typeError: keyword should be string. Why it bothers to check the type of keyword?


Besides, why the key of dictionary is not allowed to be non-string when passed as function arguments in THIS situation? Are there any exceptions? Why they design Python in this way, is it because the compiler can't deduce the types in here or something?



Thanks!










share|improve this question


















  • 1




    Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
    – Alexander Reynolds
    Nov 5 at 3:55












  • Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
    – ShadowRanger
    2 days ago













up vote
2
down vote

favorite









up vote
2
down vote

favorite











Can anyone explain the difference when unpacking the dictionary using single or double asterisk? You can mention their difference when used in function parameters, only if it is relevant here, which I don't think so.



However, there may be some relevance, because they share the same asterisk syntax.



def foo(a,b)
return a+b

tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp) #typeError: keyword should be string. Why it bothers to check the type of keyword?


Besides, why the key of dictionary is not allowed to be non-string when passed as function arguments in THIS situation? Are there any exceptions? Why they design Python in this way, is it because the compiler can't deduce the types in here or something?



Thanks!










share|improve this question













Can anyone explain the difference when unpacking the dictionary using single or double asterisk? You can mention their difference when used in function parameters, only if it is relevant here, which I don't think so.



However, there may be some relevance, because they share the same asterisk syntax.



def foo(a,b)
return a+b

tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp) #typeError: keyword should be string. Why it bothers to check the type of keyword?


Besides, why the key of dictionary is not allowed to be non-string when passed as function arguments in THIS situation? Are there any exceptions? Why they design Python in this way, is it because the compiler can't deduce the types in here or something?



Thanks!







python dictionary iterable-unpacking






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asked Nov 5 at 3:45









Han XIAO

380111




380111








  • 1




    Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
    – Alexander Reynolds
    Nov 5 at 3:55












  • Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
    – ShadowRanger
    2 days ago














  • 1




    Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
    – Alexander Reynolds
    Nov 5 at 3:55












  • Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
    – ShadowRanger
    2 days ago








1




1




Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
– Alexander Reynolds
Nov 5 at 3:55






Single * unpacking grabs the keys, so it's adding 1 and 3. With double ** it's trying to call foo(1=2, 3=4) which doesn't make any sense. Keyword arguments must be a valid identifier.
– Alexander Reynolds
Nov 5 at 3:55














Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
– ShadowRanger
2 days ago




Responding to "Why it bothers to check the type of keyword?", named arguments to functions can only be strings, so trying to use non-string names guarantees the names can't match. CPython takes advantage of this by using special purpose lookup functions for dicts guaranteed to be composed exclusively of strings (which a lot of implementation internals happen to be), so the rejecting non-strings ensures strings can go through the fastest code paths (speeding up all Python code).
– ShadowRanger
2 days ago












4 Answers
4






active

oldest

votes

















up vote
3
down vote



accepted










When dictionaries are iterated as lists the iteration takes the keys of it, for example



for key in tmp:
print(key)


is the same as



for key in tmp.keys():
print(key)


in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().



Double asterisk is used for when you define a function with keyword parameters such as



def foo(a, b):


or



def foo(**kwargs):


here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.






share|improve this answer



















  • 1




    It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
    – Brad Solomon
    Nov 5 at 4:07






  • 2




    Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
    – Alexander Reynolds
    Nov 5 at 4:09






  • 1




    "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
    – Han XIAO
    Nov 5 at 4:09




















up vote
2
down vote













def foo(a,b)
return a+b

tmp = {1:2,3:4}
foo(*tmp) #you get 4
foo(**tmp)


In this case:
foo(*tmp) mean foo(1, 3)
foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks @ Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:



def foo(1=2, 3=4):
<your code>


or



def foo('1'=2, '3'=4):
<your code>


See python_basic_syntax for more details.






share|improve this answer



















  • 1




    You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
    – Alexander Reynolds
    Nov 5 at 4:05










  • Thanks for your comment, I've just made an edit
    – enamoria
    Nov 5 at 4:20


















up vote
1
down vote













It is a Extended Iterable Unpacking.



>>> def add(a=0, b=0):
... return a + b
...
>>> d = {'a': 2, 'b': 3}
>>> add(**d)#corresponding to add(a=2,b=3)
5


For single *,



def add(a=0, b=0):
... return a + b
...
>>> d = {'a': 2, 'b': 3}
>>> add(*d)#corresponding to add(a='a',b='b')
ab


Learn more here.






share|improve this answer























  • Extended iterable unpacking is different from *args, **kwargs unpacking.
    – Alexander Reynolds
    Nov 5 at 4:01


















up vote
0
down vote













I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:



#suppose you have this function
def foo(a,**b):
print(a)
for x in b:
print(x,"...",b[x])
#suppose you call this function in the following form
foo(whatever,m=1,n=2)
#the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
#so you can also do foo(whatever,**{"m":1,"n":2})
#the reason for this syntax is you actually do
**b is m=1,n=2 #something like pattern matching mechanism
so b is {"m":1,"n":2}, note "m" and "n" are now in string form
#the function is actually this:
def foo(a,**b): # b = {"m":1,"n":2}
print(a)
for x in b: #for x in b.keys(), thanks to @vlizana answer
print(x,"...",b[x])


All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.






share|improve this answer























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    4 Answers
    4






    active

    oldest

    votes








    4 Answers
    4






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    When dictionaries are iterated as lists the iteration takes the keys of it, for example



    for key in tmp:
    print(key)


    is the same as



    for key in tmp.keys():
    print(key)


    in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().



    Double asterisk is used for when you define a function with keyword parameters such as



    def foo(a, b):


    or



    def foo(**kwargs):


    here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.






    share|improve this answer



















    • 1




      It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
      – Brad Solomon
      Nov 5 at 4:07






    • 2




      Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
      – Alexander Reynolds
      Nov 5 at 4:09






    • 1




      "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
      – Han XIAO
      Nov 5 at 4:09

















    up vote
    3
    down vote



    accepted










    When dictionaries are iterated as lists the iteration takes the keys of it, for example



    for key in tmp:
    print(key)


    is the same as



    for key in tmp.keys():
    print(key)


    in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().



    Double asterisk is used for when you define a function with keyword parameters such as



    def foo(a, b):


    or



    def foo(**kwargs):


    here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.






    share|improve this answer



















    • 1




      It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
      – Brad Solomon
      Nov 5 at 4:07






    • 2




      Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
      – Alexander Reynolds
      Nov 5 at 4:09






    • 1




      "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
      – Han XIAO
      Nov 5 at 4:09















    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    When dictionaries are iterated as lists the iteration takes the keys of it, for example



    for key in tmp:
    print(key)


    is the same as



    for key in tmp.keys():
    print(key)


    in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().



    Double asterisk is used for when you define a function with keyword parameters such as



    def foo(a, b):


    or



    def foo(**kwargs):


    here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.






    share|improve this answer














    When dictionaries are iterated as lists the iteration takes the keys of it, for example



    for key in tmp:
    print(key)


    is the same as



    for key in tmp.keys():
    print(key)


    in this case, unpacking as *tmp is equivalent to *tmp.keys(), ignoring the values. If you want to use the values you can use *tmp.values().



    Double asterisk is used for when you define a function with keyword parameters such as



    def foo(a, b):


    or



    def foo(**kwargs):


    here you can store the parameters in a dictionary and pass it as **tmp. In the first case keys must be strings with the names of the parameter defined in the function firm. And in the second case you can work with kwargs as a dictionary inside the function.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 5 at 4:11

























    answered Nov 5 at 4:05









    vlizana

    239111




    239111








    • 1




      It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
      – Brad Solomon
      Nov 5 at 4:07






    • 2




      Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
      – Alexander Reynolds
      Nov 5 at 4:09






    • 1




      "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
      – Han XIAO
      Nov 5 at 4:09
















    • 1




      It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
      – Brad Solomon
      Nov 5 at 4:07






    • 2




      Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
      – Alexander Reynolds
      Nov 5 at 4:09






    • 1




      "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
      – Han XIAO
      Nov 5 at 4:09










    1




    1




    It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
    – Brad Solomon
    Nov 5 at 4:07




    It would probably be helpful to link to the docs on expressions where both of these are formally explained. ("If the syntax *expression appears in the function call, expression must evaluate to an iterable.")
    – Brad Solomon
    Nov 5 at 4:07




    2




    2




    Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
    – Alexander Reynolds
    Nov 5 at 4:09




    Not a bad response but I want to poke to improve your answer (esp. since this topic is actually annoyingly complex). You can require keyword arguments without setting defaults, called keyword-only args. Also collecting keyword arguments with **kwargs is a little bit different than they would be with kw-only args or kwargs with default values---in particular, you can send in strings that are not valid python identifiers.
    – Alexander Reynolds
    Nov 5 at 4:09




    1




    1




    "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
    – Han XIAO
    Nov 5 at 4:09






    "Double asterisk is used for when you define a function with keyword parameters (with a default value)" is somewhat not precise. If I do def add(a,b): return a+b and tmp = {'a':1,'b':2} and call 'add(**tmp)', I can still get the value of three. So the function parameters does not need default values?
    – Han XIAO
    Nov 5 at 4:09














    up vote
    2
    down vote













    def foo(a,b)
    return a+b

    tmp = {1:2,3:4}
    foo(*tmp) #you get 4
    foo(**tmp)


    In this case:
    foo(*tmp) mean foo(1, 3)
    foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks @ Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:



    def foo(1=2, 3=4):
    <your code>


    or



    def foo('1'=2, '3'=4):
    <your code>


    See python_basic_syntax for more details.






    share|improve this answer



















    • 1




      You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
      – Alexander Reynolds
      Nov 5 at 4:05










    • Thanks for your comment, I've just made an edit
      – enamoria
      Nov 5 at 4:20















    up vote
    2
    down vote













    def foo(a,b)
    return a+b

    tmp = {1:2,3:4}
    foo(*tmp) #you get 4
    foo(**tmp)


    In this case:
    foo(*tmp) mean foo(1, 3)
    foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks @ Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:



    def foo(1=2, 3=4):
    <your code>


    or



    def foo('1'=2, '3'=4):
    <your code>


    See python_basic_syntax for more details.






    share|improve this answer



















    • 1




      You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
      – Alexander Reynolds
      Nov 5 at 4:05










    • Thanks for your comment, I've just made an edit
      – enamoria
      Nov 5 at 4:20













    up vote
    2
    down vote










    up vote
    2
    down vote









    def foo(a,b)
    return a+b

    tmp = {1:2,3:4}
    foo(*tmp) #you get 4
    foo(**tmp)


    In this case:
    foo(*tmp) mean foo(1, 3)
    foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks @ Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:



    def foo(1=2, 3=4):
    <your code>


    or



    def foo('1'=2, '3'=4):
    <your code>


    See python_basic_syntax for more details.






    share|improve this answer














    def foo(a,b)
    return a+b

    tmp = {1:2,3:4}
    foo(*tmp) #you get 4
    foo(**tmp)


    In this case:
    foo(*tmp) mean foo(1, 3)
    foo(**tmp) mean foo(1=2, 3=4), which will raise an error since 1 can't be an argument. Arg must be strings and (thanks @ Alexander Reynolds for pointing this out) must start with underscore or alphabetical character. An argument must be a valid Python identifier. This mean you can't even do something like this:



    def foo(1=2, 3=4):
    <your code>


    or



    def foo('1'=2, '3'=4):
    <your code>


    See python_basic_syntax for more details.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 5 at 4:19

























    answered Nov 5 at 4:00









    enamoria

    424210




    424210








    • 1




      You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
      – Alexander Reynolds
      Nov 5 at 4:05










    • Thanks for your comment, I've just made an edit
      – enamoria
      Nov 5 at 4:20














    • 1




      You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
      – Alexander Reynolds
      Nov 5 at 4:05










    • Thanks for your comment, I've just made an edit
      – enamoria
      Nov 5 at 4:20








    1




    1




    You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
    – Alexander Reynolds
    Nov 5 at 4:05




    You can improve your answer by linking to documentation; as it stands right now it is not quite correct (e.g., '2' is a string but is not a valid python identifier).
    – Alexander Reynolds
    Nov 5 at 4:05












    Thanks for your comment, I've just made an edit
    – enamoria
    Nov 5 at 4:20




    Thanks for your comment, I've just made an edit
    – enamoria
    Nov 5 at 4:20










    up vote
    1
    down vote













    It is a Extended Iterable Unpacking.



    >>> def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(**d)#corresponding to add(a=2,b=3)
    5


    For single *,



    def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(*d)#corresponding to add(a='a',b='b')
    ab


    Learn more here.






    share|improve this answer























    • Extended iterable unpacking is different from *args, **kwargs unpacking.
      – Alexander Reynolds
      Nov 5 at 4:01















    up vote
    1
    down vote













    It is a Extended Iterable Unpacking.



    >>> def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(**d)#corresponding to add(a=2,b=3)
    5


    For single *,



    def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(*d)#corresponding to add(a='a',b='b')
    ab


    Learn more here.






    share|improve this answer























    • Extended iterable unpacking is different from *args, **kwargs unpacking.
      – Alexander Reynolds
      Nov 5 at 4:01













    up vote
    1
    down vote










    up vote
    1
    down vote









    It is a Extended Iterable Unpacking.



    >>> def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(**d)#corresponding to add(a=2,b=3)
    5


    For single *,



    def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(*d)#corresponding to add(a='a',b='b')
    ab


    Learn more here.






    share|improve this answer














    It is a Extended Iterable Unpacking.



    >>> def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(**d)#corresponding to add(a=2,b=3)
    5


    For single *,



    def add(a=0, b=0):
    ... return a + b
    ...
    >>> d = {'a': 2, 'b': 3}
    >>> add(*d)#corresponding to add(a='a',b='b')
    ab


    Learn more here.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 5 at 4:02

























    answered Nov 5 at 3:56









    Ashutosh Chapagain

    664213




    664213












    • Extended iterable unpacking is different from *args, **kwargs unpacking.
      – Alexander Reynolds
      Nov 5 at 4:01


















    • Extended iterable unpacking is different from *args, **kwargs unpacking.
      – Alexander Reynolds
      Nov 5 at 4:01
















    Extended iterable unpacking is different from *args, **kwargs unpacking.
    – Alexander Reynolds
    Nov 5 at 4:01




    Extended iterable unpacking is different from *args, **kwargs unpacking.
    – Alexander Reynolds
    Nov 5 at 4:01










    up vote
    0
    down vote













    I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:



    #suppose you have this function
    def foo(a,**b):
    print(a)
    for x in b:
    print(x,"...",b[x])
    #suppose you call this function in the following form
    foo(whatever,m=1,n=2)
    #the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
    #so you can also do foo(whatever,**{"m":1,"n":2})
    #the reason for this syntax is you actually do
    **b is m=1,n=2 #something like pattern matching mechanism
    so b is {"m":1,"n":2}, note "m" and "n" are now in string form
    #the function is actually this:
    def foo(a,**b): # b = {"m":1,"n":2}
    print(a)
    for x in b: #for x in b.keys(), thanks to @vlizana answer
    print(x,"...",b[x])


    All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.






    share|improve this answer



























      up vote
      0
      down vote













      I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:



      #suppose you have this function
      def foo(a,**b):
      print(a)
      for x in b:
      print(x,"...",b[x])
      #suppose you call this function in the following form
      foo(whatever,m=1,n=2)
      #the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
      #so you can also do foo(whatever,**{"m":1,"n":2})
      #the reason for this syntax is you actually do
      **b is m=1,n=2 #something like pattern matching mechanism
      so b is {"m":1,"n":2}, note "m" and "n" are now in string form
      #the function is actually this:
      def foo(a,**b): # b = {"m":1,"n":2}
      print(a)
      for x in b: #for x in b.keys(), thanks to @vlizana answer
      print(x,"...",b[x])


      All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.






      share|improve this answer

























        up vote
        0
        down vote










        up vote
        0
        down vote









        I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:



        #suppose you have this function
        def foo(a,**b):
        print(a)
        for x in b:
        print(x,"...",b[x])
        #suppose you call this function in the following form
        foo(whatever,m=1,n=2)
        #the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
        #so you can also do foo(whatever,**{"m":1,"n":2})
        #the reason for this syntax is you actually do
        **b is m=1,n=2 #something like pattern matching mechanism
        so b is {"m":1,"n":2}, note "m" and "n" are now in string form
        #the function is actually this:
        def foo(a,**b): # b = {"m":1,"n":2}
        print(a)
        for x in b: #for x in b.keys(), thanks to @vlizana answer
        print(x,"...",b[x])


        All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.






        share|improve this answer














        I think the ** double asterisk in function parameter and unpacking dictionary means intuitively in this way:



        #suppose you have this function
        def foo(a,**b):
        print(a)
        for x in b:
        print(x,"...",b[x])
        #suppose you call this function in the following form
        foo(whatever,m=1,n=2)
        #the m=1 syntax actually means assign parameter by name, like foo(a = whatever, m = 1, n = 2)
        #so you can also do foo(whatever,**{"m":1,"n":2})
        #the reason for this syntax is you actually do
        **b is m=1,n=2 #something like pattern matching mechanism
        so b is {"m":1,"n":2}, note "m" and "n" are now in string form
        #the function is actually this:
        def foo(a,**b): # b = {"m":1,"n":2}
        print(a)
        for x in b: #for x in b.keys(), thanks to @vlizana answer
        print(x,"...",b[x])


        All the syntax make sense now. And it is the same for single asterisk. It is only worth noting that if you use single asterisk to unpack dictionary, you are actually trying to unpack it in a list way, and only key of dictionary are unpacked.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited Nov 5 at 4:54

























        answered Nov 5 at 4:40









        Han XIAO

        380111




        380111






























             

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