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Not using defaultdict()

When a dict is created using defaultdict(), the value for each key in the dict will default to the value provided as the first argument of defaultdict(). This is more concise and less error-prone than manually setting the value of each key.


The code below defines an empty dict and then manually initializes the keys of the dict. Although there is nothing wrong with this code, there is a more concise and less error-prone way to achieve the same idea, as explained in the solution below.

d = {}

if not "k" in d:
    d["k"] = 6

d["k"] += 1

print(d["k"])  # 7

Best practice

Use defaultdict() to initialize dict keys

The modified code below uses defaultdict to initialize the dict. Whenever a new key is created, the default value for that key is 6. This code is functionally equivalent to the previous code, but this one is more concise and less error-prone, because every key automatically initializes to 6 with no work on the part of the programmer.

from collections import defaultdict

d = defaultdict(lambda : 6)
d["k"] += 1

print(d["k"])  # 7