Problem: Say, you have a dictionary {0: 'Alice', 1: 'Bob'} and you want to create a list of dictionaries with copies of the original dictionary: [{0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}].
d = {0: 'Alice', 1: 'Bob'} dicts = [{**d} for _ in range(3)]
print(dicts)
# [{0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}]
You use list comprehension with a “throw-away” loop variable underscore _ to create a list of 3 elements. You can change the value 3 if you need more or fewer elements in your list.
The expression {**d} unpacks all (key, value) pairs from the original dictionary d into a new dictionary. For more information about the unpacking operator, see this Finxter blog tutorial.
The resulting list contains copies of the original dictionary. If you change one, the others won’t see that change:
dicts[0][2] = 'Frank'
print(dicts)
# [{0: 'Alice', 1: 'Bob', 2: 'Frank'}, {0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}]
Only the first dictionary in the list contains the new key value pair (2: 'Frank') which proves that the dictionaries don’t point to the same object in memory. This would be the case if you’d use the following method of copying a list with a single dictionary:
d2 = {0: 'Alice', 1: 'Bob'} dicts2 = [d2] * 3
print(dicts2)
# [{0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}, {0: 'Alice', 1: 'Bob'}]
The method looks right but all three dictionaries are essentially the same:
dicts2[0][2] = 'Frank'
print(dicts2)
# [{0: 'Alice', 1: 'Bob', 2: 'Frank'}, {0: 'Alice', 1: 'Bob', 2: 'Frank'}, {0: 'Alice', 1: 'Bob', 2: 'Frank'}]
If you change one, you change all.
You can see this effect yourself in the following memory visualizer tool:
Exercise: change the method to the correct one so that the change affects only the first dictionary!
Related articles: How to create a Python list?
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https://www.sickgaming.net/blog/2020/07/...in-python/