Fix Python – Why doesn’t a python dict.update() return the object?


Asked By – Paul Tarjan

I ‘m trying to do :

award_dict = {
    "url" : "",
    "imageurl" : "",
    "count" : 1,

def award(name, count, points, desc_string, my_size, parent) :
    if my_size > count :
        a = {
            "name" : name,
            "description" : desc_string % count,
            "points" : points,
            "parent_award" : parent,
        return self.add_award(a, siteAlias, alias).award

But if felt really cumbersome in the function, and I would have rather done :

        return self.add_award({
            "name" : name,
            "description" : desc_string % count,
            "points" : points,
            "parent_award" : parent,
        }.update(award_dict), siteAlias, alias).award

Why doesn’t update return the object so you can chain?

JQuery does this to do chaining. Why isn’t it acceptable in python?

Now we will see solution for issue: Why doesn’t a python dict.update() return the object?


Python’s mostly implementing a pragmatically tinged flavor of command-query separation: mutators return None (with pragmatically induced exceptions such as pop😉 so they can’t possibly be confused with accessors (and in the same vein, assignment is not an expression, the statement-expression separation is there, and so forth).

That doesn’t mean there aren’t a lot of ways to merge things up when you really want, e.g., dict(a, **award_dict) makes a new dict much like the one you appear to wish .update returned — so why not use THAT if you really feel it’s important?

Edit: btw, no need, in your specific case, to create a along the way, either:

dict(name=name, description=desc % count, points=points, parent_award=parent,

creates a single dict with exactly the same semantics as your a.update(award_dict) (including, in case of conflicts, the fact that entries in award_dict override those you’re giving explicitly; to get the other semantics, i.e., to have explicit entries “winning” such conflicts, pass award_dict as the sole positional arg, before the keyword ones, and bereft of the ** form — dict(award_dict, name=name etc etc).

This question is answered By – Alex Martelli

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