Is there a difference between using a dict literal and a dict constructor?

Question:

Using PyCharm, I noticed it offers to convert a dict literal:

d = {
    'one': '1',
    'two': '2',
}

into a dict constructor:

d = dict(one='1', two='2')

Do these different approaches differ in some significant way?

(While writing this question I noticed that using dict() it seems impossible to specify a numeric key .. d = {1: 'one', 2: 'two'} is possible, but, obviously, dict(1='one' ...) is not. Anything else?)

Asked By: maligree

||

Answers:

I think you have pointed out the most obvious difference. Apart from that,

the first doesn’t need to lookup dict which should make it a tiny bit faster

the second looks up dict in locals() and then globals() and the finds the builtin, so you can switch the behaviour by defining a local called dict for example although I can’t think of anywhere this would be a good idea apart from maybe when debugging

Answered By: John La Rooy

These two approaches produce identical dictionaries, except, as you’ve noted, where the lexical rules of Python interfere.

Dictionary literals are a little more obviously dictionaries, and you can create any kind of key, but you need to quote the key names. On the other hand, you can use variables for keys if you need to for some reason:

a = "hello"
d = {
    a: 'hi'
    }

The dict() constructor gives you more flexibility because of the variety of forms of input it takes. For example, you can provide it with an iterator of pairs, and it will treat them as key/value pairs.

I have no idea why PyCharm would offer to convert one form to the other.

Answered By: Ned Batchelder

They look pretty much the same on Python 3.2.

As gnibbler pointed out, the first doesn’t need to lookup dict, which should make it a tiny bit faster.

>>> def literal():
...   d = {'one': 1, 'two': 2}
...
>>> def constructor():
...   d = dict(one='1', two='2')
...
>>> import dis
>>> dis.dis(literal)
  2           0 BUILD_MAP                2
              3 LOAD_CONST               1 (1)
              6 LOAD_CONST               2 ('one')
              9 STORE_MAP
             10 LOAD_CONST               3 (2)
             13 LOAD_CONST               4 ('two')
             16 STORE_MAP
             17 STORE_FAST               0 (d)
             20 LOAD_CONST               0 (None)
             23 RETURN_VALUE
>>> dis.dis(constructor)
  2           0 LOAD_GLOBAL              0 (dict)
              3 LOAD_CONST               1 ('one')
              6 LOAD_CONST               2 ('1')
              9 LOAD_CONST               3 ('two')
             12 LOAD_CONST               4 ('2')
             15 CALL_FUNCTION          512
             18 STORE_FAST               0 (d)
             21 LOAD_CONST               0 (None)
             24 RETURN_VALUE
Answered By: Paolo Moretti

From python 2.7 tutorial:

A pair of braces creates an empty
dictionary: {}. Placing a
comma-separated list of key:value
pairs within the braces adds initial
key:value pairs to the dictionary;
this is also the way dictionaries are
written on output.

tel = {'jack': 4098, 'sape': 4139}
data = {k:v for k,v in zip(xrange(10), xrange(10,20))}

While:

The dict() constructor builds
dictionaries directly from lists of
key-value pairs stored as tuples. When
the pairs form a pattern, list
comprehensions can compactly specify
the key-value list.

tel = dict([('sape', 4139), ('guido', 4127), ('jack', 4098)]) {'sape': 4139, 'jack': 4098, 'guido': 4127}
data = dict((k,v) for k,v in zip(xrange(10), xrange(10,20)))

When the keys are simple strings, it
is sometimes easier to specify pairs
using keyword arguments:

dict(sape=4139, guido=4127, jack=4098)
>>>  {'sape': 4139, 'jack':4098, 'guido': 4127}

So both {} and dict() produce dictionary but provide a bit different ways of dictionary data initialization.

Answered By: Artsiom Rudzenka

Literal is much faster, since it uses optimized BUILD_MAP and STORE_MAP opcodes rather than generic CALL_FUNCTION:

> python2.7 -m timeit "d = dict(a=1, b=2, c=3, d=4, e=5)"
1000000 loops, best of 3: 0.958 usec per loop

> python2.7 -m timeit "d = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5}"
1000000 loops, best of 3: 0.479 usec per loop

> python3.2 -m timeit "d = dict(a=1, b=2, c=3, d=4, e=5)"
1000000 loops, best of 3: 0.975 usec per loop

> python3.2 -m timeit "d = {'a':1, 'b':2, 'c':3, 'd':4, 'e':5}"
1000000 loops, best of 3: 0.409 usec per loop
Answered By: Daniel Kluev

I find the dict literal d = {'one': '1'} to be much more readable, your defining data, rather than assigning things values and sending them to the dict() constructor.

On the other hand i have seen people mistype the dict literal as d = {'one', '1'} which in modern python 2.7+ will create a set.

Despite this i still prefer to all-ways use the set literal because i think its more readable, personal preference i suppose.

Answered By: lee penkman

One big difference with python 3.4 + pycharm is that the dict() constructor
produces a “syntax error” message if the number of keys exceeds 256.

I prefer using the dict literal now.

Answered By: Michel Boiron

the dict() literal is nice when you are copy pasting values from something else (none python)
For example a list of environment variables.
if you had a bash file, say

FOO='bar'
CABBAGE='good'

you can easily paste then into a dict() literal and add comments. It also makes it easier to do the opposite, copy into something else. Whereas the {'FOO': 'bar'} syntax is pretty unique to python and json. So if you use json a lot, you might want to use {} literals with double quotes.

Answered By: Nick Humrich

Also consider the fact that tokens that match for operators can’t be used in the constructor syntax, i.e. dasherized keys.

>>> dict(foo-bar=1)
File "<stdin>", line 1
SyntaxError: keyword can't be an expression

>>> {'foo-bar': 1}
{'foo-bar': 1}
Answered By: Brian Whitton

There is no dict literal to create dict-inherited classes, custom dict classes with additional methods. In such case custom dict class constructor should be used, for example:

class NestedDict(dict):

    # ... skipped

state_type_map = NestedDict(**{
    'owns': 'Another',
    'uses': 'Another',
})
Answered By: Dmitriy Sintsov

Super late to the party here, but if you have a kwargs function:

def foo(a=None, b=None):
    ...

And your splatting a dict like so:

d_1 = { 'a': 1, 'b': 2 }
d_2 = dict(a=1, b=2)

# This works
foo(**d_1)

# And this as well
foo(**d_2)

But d_2 may be better suited for refactoring argument names that may change in your foo signature. Since in d_1 they are strings.

Answered By: Jonathan Nazario
Categories: questions Tags: , ,
Answers are sorted by their score. The answer accepted by the question owner as the best is marked with
at the top-right corner.