Since a variable in Python is merely a label or reference to an object in the memory, a simple assignment operator will not create copy of object.
Example 1
In this example, we have a dictionary "d1" and we assign it to another variable "d2". If "d1" is updated, the changes also reflect in "d2".
d1 = {"a":11, "b":22, "c":33} d2 = d1 print ("id:", id(d1), "dict: ",d1) print ("id:", id(d2), "dict: ",d2) d1["b"] = 100 print ("id:", id(d1), "dict: ",d1) print ("id:", id(d2), "dict: ",d2)
Output
id: 2215278891200 dict: {'a': 11, 'b': 22, 'c': 33}
id: 2215278891200 dict: {'a': 11, 'b': 22, 'c': 33}
id: 2215278891200 dict: {'a': 11, 'b': 100, 'c': 33}
id: 2215278891200 dict: {'a': 11, 'b': 100, 'c': 33}
To avoid this, and make a shallow copy of a dictionary, use the copy() method instead of assignment.
Example 2
d1 = {"a":11, "b":22, "c":33} d2 = d1.copy() print ("id:", id(d1), "dict: ",d1) print ("id:", id(d2), "dict: ",d2) d1["b"] = 100 print ("id:", id(d1), "dict: ",d1) print ("id:", id(d2), "dict: ",d2)
Output
When "d1" is updated, "d2" will not change now because "d2" is the copy of dictionary object, not merely a reference.
id: 1586671734976 dict: {'a': 11, 'b': 22, 'c': 33}
id: 1586673973632 dict: {'a': 11, 'b': 22, 'c': 33}
id: 1586671734976 dict: {'a': 11, 'b': 100, 'c': 33}
id: 1586673973632 dict: {'a': 11, 'b': 22, 'c': 33}
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