Why Does Changing a Copy Variable Affect the Original Variable in Python?

Python: Why Changing a Copy Variable Affects the Original Variable

In Python, you may encounter a peculiar behavior where modifications made to a copy variable appear to alter the original variable as well. This occurs because Python variables store references rather than actual values.

To understand this, consider the situation described:

org_list = ['y', 'c', 'gdp', 'cap']
copy_list = org_list       # Pass reference to org_list
copy_list.append('hum')

print(copy_list)           # ['y', 'c', 'gdp', 'cap', 'hum']
print(org_list)            # ['y', 'c', 'gdp', 'cap', 'hum']

When you assign copy_list to org_list, you are not creating a new list but rather establishing a reference to the same list object in memory. Hence, any changes to either copy_list or org_list directly affect both variables.

To create a truly independent copy, you need to pass a copy of the actual data, rather than a reference. This can be done by using the slice assignment operator:

copy_list = org_list[:]    # Create a deep copy by slicing

By slicing the original list, you create a new list object with its own copy of the data. Any modifications made to copy_list will not affect org_list, and vice versa.

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