Dictionaries and default values

PythonDictionaryCoding Style

Python Problem Overview


Assuming connectionDetails is a Python dictionary, what's the best, most elegant, most "pythonic" way of refactoring code like this?

if "host" in connectionDetails:
    host = connectionDetails["host"]
else:
    host = someDefaultValue

Python Solutions


Solution 1 - Python

Like this:

host = connectionDetails.get('host', someDefaultValue)

Solution 2 - Python

You can also use the defaultdict like so:

from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"

You can pass any ordinary function instead of lambda:

from collections import defaultdict
def a():
  return 4

b = defaultdict(a, key="some_value")
b['absent'] => 4
b['key'] => "some_value"

Solution 3 - Python

While .get() is a nice idiom, it's slower than if/else (and slower than try/except if presence of the key in the dictionary can be expected most of the time):

>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.07691968797894333
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.4583777282275605
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(1, 10)")
0.17784020746671558
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="a=d.get(2, 10)")
0.17952161730158878
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.10071221458065338
>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}", 
... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.06966537335119938

Solution 4 - Python

For multiple different defaults try this:

connectionDetails = { "host": "www.example.com" }
defaults = { "host": "127.0.0.1", "port": 8080 }

completeDetails = {}
completeDetails.update(defaults)
completeDetails.update(connectionDetails)
completeDetails["host"]  # ==> "www.example.com"
completeDetails["port"]  # ==> 8080

Solution 5 - Python

This is not exactly the question asked for but there is a method in python dictionaries: dict.setdefault

    host = connectionDetails.setdefault('host',someDefaultValue)

However this method sets the value of connectionDetails['host'] to someDefaultValue if key host is not already defined, unlike what the question asked.

Solution 6 - Python

(this is a late answer)

An alternative is to subclass the dict class and implement the __missing__() method, like this:

class ConnectionDetails(dict):
    def __missing__(self, key):
        if key == 'host':
            return "localhost"
        raise KeyError(key)

Examples:

>>> connection_details = ConnectionDetails(port=80)

>>> connection_details['host']
'localhost'

>>> connection_details['port']
80

>>> connection_details['password']
Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 6, in __missing__
KeyError: 'password'

Solution 7 - Python

Testing @Tim Pietzcker's suspicion about the situation in PyPy (5.2.0-alpha0) for Python 3.3.5, I find that indeed both .get() and the if/else way perform similar. Actually it seems that in the if/else case there is even only a single lookup if the condition and the assignment involve the same key (compare with the last case where there is two lookups).

>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[1]\nexcept KeyError:\n a=10")
0.011889292989508249
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="try:\n a=d[2]\nexcept KeyError:\n a=10")
0.07310474599944428
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(1, 10)")
0.010391917996457778
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="a=d.get(2, 10)")
0.009348208011942916
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 1 in d:\n a=d[1]\nelse:\n a=10")
0.011475925013655797
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=10")
0.009605801998986863
>>>> timeit.timeit(setup="d={1:2, 3:4, 5:6, 7:8, 9:0}",
.... stmt="if 2 in d:\n a=d[2]\nelse:\n a=d[1]")
0.017342638995614834

Solution 8 - Python

You can use a lamba function for this as a one-liner. Make a new object connectionDetails2 which is accessed like a function...

connectionDetails2 = lambda k: connectionDetails[k] if k in connectionDetails.keys() else "DEFAULT"

Now use

connectionDetails2(k)

instead of

connectionDetails[k]

which returns the dictionary value if k is in the keys, otherwise it returns "DEFAULT"

Attributions

All content for this solution is sourced from the original question on Stackoverflow.

The content on this page is licensed under the Attribution-ShareAlike 4.0 International (CC BY-SA 4.0) license.

Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionmnowotkaView Question on Stackoverflow
Solution 1 - PythonMattHView Answer on Stackoverflow
Solution 2 - PythontamerlahaView Answer on Stackoverflow
Solution 3 - PythonTim PietzckerView Answer on Stackoverflow
Solution 4 - PythonJerome BaumView Answer on Stackoverflow
Solution 5 - PythonSriramView Answer on Stackoverflow
Solution 6 - PythonLaurent LAPORTEView Answer on Stackoverflow
Solution 7 - PythonTillView Answer on Stackoverflow
Solution 8 - PythonCasualScienceView Answer on Stackoverflow