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


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,
        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.

Asked By: Paul Tarjan



Its not that it isn’t acceptable, but rather that dicts weren’t implemented that way.

If you look at Django’s ORM, it makes extensive use of chaining. Its not discouraged, you could even inherit from dict and only override update to do update and return self, if you really want it.

class myDict(dict):
    def update(self, *args):
        dict.update(self, *args)
        return self
Answered By: Esteban Küber

Python’s API, by convention, distinguishes between procedures and functions. Functions compute new values out of their parameters (including any target object); procedures modify objects and don’t return anything (i.e. they return None). So procedures have side effects, functions don’t. update is a procedure, hence it doesn’t return a value.

The motivation for doing it that way is that otherwise, you may get undesirable side effects. Consider

bar = foo.reverse()

If reverse (which reverses the list in-place) would also return the list, users may think that reverse returns a new list which gets assigned to bar, and never notice that foo also gets modified. By making reverse return None, they immediately recognize that bar is not the result of the reversal, and will look more close what the effect of reverse is.

Answered By: Martin v. Löwis

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).

Answered By: Alex Martelli
>>> dict_merge = lambda a,b: a.update(b) or a
>>> dict_merge({'a':1, 'b':3},{'c':5})
{'a': 1, 'c': 5, 'b': 3}

Note that as well as returning the merged dict, it modifies the first parameter in-place. So dict_merge(a,b) will modify a.

Or, of course, you can do it all inline:

>>> (lambda a,b: a.update(b) or a)({'a':1, 'b':3},{'c':5})
{'a': 1, 'c': 5, 'b': 3}
Answered By: Crispin Wellington
import itertools
dict_merge = lambda *args: dict(itertools.chain(*[d.iteritems() for d in args]))
Answered By: Matt

not enough reputation for comment left on top answer

@beardc this doesn’t seem to be CPython thing. PyPy gives me “TypeError: keywords must be strings”

The solution with **kwargs only works because the dictionary to be merged only has keys of type string.


>>> dict({1:2}, **{3:4})
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: keyword arguments must be strings


>>> dict({1:2}, **{'3':4})
{1: 2, '3': 4}
Answered By: Stephan Scheller

as close to your proposed solution as I could get

from collections import ChainMap

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

Just been trying this myself in Python 3.4 (so wasn’t able to use the fancy {**dict_1, **dict_2} syntax).

I wanted to be able to have non-string keys in dictionaries as well as provide an arbitrary amount of dictionaries.

Also, I wanted to make a new dictionary so I opted to not use collections.ChainMap (kinda the reason I didn’t want to use dict.update initially.

Here’s what I ended up writing:

def merge_dicts(*dicts):
    all_keys  = set(k for d in dicts for k in d.keys())
    chain_map = ChainMap(*reversed(dicts))
    return {k: chain_map[k] for k in all_keys}

merge_maps({'1': 1}, {'2': 2, '3': 3}, {'1': 4, '3': 5})
# {'1': 4, '3': 5, '2': 2}
Answered By: freebie

This is easy as:

(lambda d: d.update(dict2) or d)(d1)

Or, if it is important not to modify the dictionary:

(lambda d: d.update(dict2) or d)(d1.copy())
Answered By: Kostya Goloveshko

For those coming late to the party, I had put some timing together (Py 3.7), showing that .update() based methods look a bit (~5%) faster when inputs are preserved and noticeably (~30%) faster when just updating in-place.

As usual, all the benchmarks should be taken with a grain of salt.

def join2(dict1, dict2, inplace=False):
    result = dict1 if inplace else dict1.copy()
    return result

def join(*items):
    iter_items = iter(items)
    result = next(iter_items).copy()
    for item in iter_items:
    return result

def update_or(dict1, dict2):
    return dict1.update(dict2) or dict1

d1 = {i: str(i) for i in range(1000000)}
d2 = {str(i): i for i in range(1000000)}

%timeit join2(d1, d2)
# 258 ms ± 1.47 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit join(d1, d2)
# 262 ms ± 2.97 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit dict(d1, **d2)
# 267 ms ± 2.74 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit {**d1, **d2}
# 267 ms ± 1.84 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

The timings for the in-place operations are a bit trickier, so it would need to be modified along an extra copy operation (the first timing is just for reference):

%timeit dd = d1.copy()
# 44.9 ms ± 495 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit dd = d1.copy(); join2(dd, d2)
# 296 ms ± 2.05 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit dd = d1.copy(); join2(dd, d2, True)
# 234 ms ± 1.02 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)

%timeit dd = d1.copy(); update_or(dd, d2)
# 235 ms ± 1.6 ms per loop (mean ± std. dev. of 7 runs, 1 loop each)
Answered By: norok2

Merge by concatenating list of items:

d1 = {1: "one"}
d2 = {2: "two"}
dict(list(d1.items()) + list(d2.items()))
# {1: 'one', 2: 'two'}
Answered By: matt91t