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Looking for extra enlightenment on *args and **kwargs, I came across the following short piece of code:

def concatenate(**kwargs):
    result = ""
    # Iterating over the Python kwargs dictionary
    for arg in kwargs.values():
        result += arg
    return result

print(concatenate(a="Real", b="Python", c="Is", d="Great", e="!"))

The page says it would produce:

RealPythonIsGreat!

And it indeed does. With Python 3.7.2. But that is what it spits when I run it with Python 2.7.16:

RealIsPython!Great

It shuffles the terms in a totally nonsensical, Yoda-like way. Why?

petezurich
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Fausto Arinos Barbuto
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    `kwargs` is a dictionary, and dictionaries didn't have a specific order until more recent Python versions. – mkrieger1 Dec 20 '19 at 13:34
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    And that is why we have `PEP 468` https://www.python.org/dev/peps/pep-0468/ – Raj Dec 20 '19 at 13:35
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    Python 2 isn't a thing anymore, don't even bother with it: https://pythonclock.org/ – cglacet Dec 20 '19 at 13:36
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    Does this answer your question? [Using an OrderedDict in \*\*kwargs](https://stackoverflow.com/questions/26748097/using-an-ordereddict-in-kwargs) – Raj Dec 20 '19 at 13:36
  • But I don't see why you would use keyword args if you don't care about their names, just use a list. – cglacet Dec 20 '19 at 13:37

1 Answers1

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A dictionary is a collection which is unordered. At least for older Python versions. Only Python 3.6 onwards keeps the insertion order.

Burkhard
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  • Yes, but it is also "orderly unordered", that is, I (obviously) always get the same result no matter how many times I execute the script. Why is that "unordered order" preferred to any other? If I remove the `.values()` I get "acbed", which is also nonsense. – Fausto Arinos Barbuto Dec 20 '19 at 15:21