Python 实用宝典

Question 1

I want to write something that removes a specific element from an array. I know that I have to for loop through the array to find the element that matches the content.

Let’s say that I have an array of emails and I want to get rid of the element that matches some email string.

I’d actually like to use the for loop structure because I need to use the same index for other arrays as well.

Here is the code that I have:

for index, item in emails:
    if emails[index] == 'something@something.com':
         emails.pop(index)
         otherarray.pop(index)

Question 2

You don’t need to iterate the array. Just:

>>> x = ['ala@ala.com', 'bala@bala.com']
>>> x
['ala@ala.com', 'bala@bala.com']
>>> x.remove('ala@ala.com')
>>> x
['bala@bala.com']

This will remove the first occurence that matches the string.

EDIT: After your edit, you still don’t need to iterate over. Just do:

index = initial_list.index(item1)
del initial_list[index]
del other_list[index]

Question 3

Using filter() and lambda would provide a neat and terse method of removing unwanted values:

newEmails = list(filter(lambda x : x != 'something@something.com', emails))

This does not modify emails. It creates the new list newEmails containing only elements for which the anonymous function returned True.

Question 4

Your for loop is not right, if you need the index in the for loop use:

for index, item in enumerate(emails):
    # whatever (but you can't remove element while iterating)

In your case, Bogdan solution is ok, but your data structure choice is not so good. Having to maintain these two lists with data from one related to data from the other at same index is clumsy.

A list of tupple (email, otherdata) may be better, or a dict with email as key.

Question 5

The sane way to do this is to use zip() and a List Comprehension / Generator Expression:

filtered = (
    (email, other) 
        for email, other in zip(emails, other_list) 
            if email == 'something@something.com')

new_emails, new_other_list = zip(*filtered)

Also, if your’e not using array.array() or numpy.array(), then most likely you are using [] or list(), which give you Lists, not Arrays. Not the same thing.

Question 6

There is an alternative solution to this problem which also deals with duplicate matches.

We start with 2 lists of equal length: emails, otherarray. The objective is to remove items from both lists for each index i where emails[i] == 'something@something.com'.

This can be achieved using a list comprehension and then splitting via zip:

emails = ['abc@def.com', 'something@something.com', 'ghi@jkl.com']
otherarray = ['some', 'other', 'details']

from operator import itemgetter

res = [(i, j) for i, j in zip(emails, otherarray) if i!= 'something@something.com']
emails, otherarray = map(list, map(itemgetter(0, 1), zip(*res)))

print(emails)      # ['abc@def.com', 'ghi@jkl.com']
print(otherarray)  # ['some', 'details']

Python 实用宝典

如何使用python从数组中删除特定元素

问题：如何使用python从数组中删除特定元素

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