Why doesn't a python dict.update() return the object?

Question

I have this code:

award_dict = {
    "url": "http://facebook.com",
    "imageurl": "http://farm4.static.flickr.com/3431/3939267074_feb9eb19b1_o.png",
    "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,
        }
        a.update(award_dict)
        return self.add_award(a, siteAlias, alias).award

But the code felt rather cumbersome. I would have preferred to be able to write:

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

Why doesn't the update method return the original dictionary, so as to allow chaining, like how it works in JQuery? Why isn't it acceptable in python?

---

_{See How do I merge two dictionaries in a single expression in Python? for workarounds.}

Answer 1

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,
     **award_dict)

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).