Packing and unpacking dictionary

Question

Obviously I am missing something very easy here but I cannot get the answer.

The question is why the code:

def func1(arg1, *arg2):
    print arg1
    print arg2
arg1=1
arg2 = [1,2,3]
func1(arg1, *arg2)

gives 1 (1, 2, 3)

while

def func2(arg1, **arg2):
    print arg1
    print arg2
arg1=1
arg2 = {'arg2_1':1,'arg2_2':2,'arg2_3':3}
func2(arg1, **arg2)

gives 1 {} instead of 1 {'arg2_1':1,'arg2_2':2,'arg2_3':3}.

How can I pack and unpack the dictionary without having to write all of it's elements neither on the function definition nor on the function call? (In the real case the dictionary has a lot of elements and is defined by comprehension.)

You are already doing it, as far as I can tell. Or I don't understand the question. — wim, Apr 03 '17 at 15:04
I can't reproduce this behaviour (printing of a blank dictionary). What version of Python are you using? — asongtoruin, Apr 03 '17 at 15:06
I run it on http://www.codeskulptor.org/ Now I did it on my pc runs as expected.. — akotronis, Apr 03 '17 at 16:23
As they say in the docs: "It implements a subset of Python 2" — Felix, Apr 03 '17 at 16:29

score 0 · Accepted Answer · answered Apr 03 '17 at 15:14

0

Actually it does. Using python 2.7.12:

>>> def func2(arg1, **arg2):
...     print arg1
...     print arg2
...
>>> arg1=1
>>> arg2 = {'arg2_1':1,'arg2_2':2,'arg2_3':3}
>>> func2(arg1, **arg2)
1
{'arg2_1': 1, 'arg2_3': 3, 'arg2_2': 2}

answered Apr 03 '17 at 15:14

Felix

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Packing and unpacking dictionary

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