Python 3.6.9 - 列表 - 在 for 循环中就地修改

Question

Background :

I have a list of names and email addresses stored in a Python list.

user_data_list = [['Adam', 'adam@abc.com'],
                  ['Alice', 'alice@stu.edu'],
                  ['Eve', 'eve@abc.com'],
                  ['Bob', 'bob@abc.com']]

I have two lists containing the old and the updated email addresses, old_domain_email_list and new_domain_email_list , respectively. The contents of the lists are:

old_domain_email_list = ['adam@abc.com', 'eve@abc.com', 'bob@abc.com']
new_domain_email_list = ['adam@xyz.com', 'eve@xyz.com', 'bob@xyz.com']

I intend to update the email addresses to now contain xyz.com instead of abc.com . The below lines of code iterates through the user_data_list and updates the matched entries in the list.

for user in user_data_list:
    for old_domain_email, new_domain_email in zip(old_domain_email_list, new_domain_email_list):
        if user[1] == old_domain_email:
            user[1] = new_domain_email

The changed name and email address pairs are then written to a CSV file.

with open(report_file_location, 'w+') as output_file:
        writer = csv.writer(output_file)
        writer.writerows(user_data_list)
        output_file.close()

Query :

Are the user details in the user_data_list updated by the line user[1] = ' ' + new_domain_email happening in-place ? The code doesn't explicitly tries to modify the user_data_list , but still, the writer.writerows(user_data_list) method correctly creates the new file with the updated domain names.

Python version used: 3.6.9

I hope I am able to make the question clear.

Answer 1

Yes, the modification is happening in place. user[i] =... invokes user 's __setitem__ method which does not change the object to which the user reference points.

Answer 2

Yes, it is. You can simplify the email address replacement using list comprehension.

[[i[0], i[1].split('@')[0]+'@xyz.com'] for i in user_data_list]

result:

[['Adam', 'adam@xyz.com'], ['Alice', 'alice@xyz.com'], ['Eve', 'eve@xyz.com'], ['Bob', 'bob@xyz.com']]