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

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

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.

>>> 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}

import itertools
dict_merge = lambda *args: dict(itertools.chain(*[d.iteritems() for d in args]))

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.

i.e.

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

vs

>>> dict({1:2}, **{'3':4})
{1: 2, '3': 4}

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

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}

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())

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()
    result.update(dict2)
    return result


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

Merge by concatenating list of items:

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