blocks|key|805728|text|你不需要迭代数组。只要：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|805729|>>>+x+=+['ala@ala.com',+'bala@bala.com']
>>>+x
['ala@ala.com',+'bala@bala.com']
>>>+x.remove('ala@ala.com')
>>>+x
['bala@bala.com']|code-block|syntax|javascript|805730|这将删除与该字符串匹配的第一个匹配项。|805731|编辑:编辑之后，您仍然不需要遍历。只需这样做：|805732|index+=+initial_list.index(item1)
del+initial_list[index]
del+other_list[index]|805733|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

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

<pre><code>&gt;&gt;&gt; x = ['ala@ala.com', 'bala@bala.com']
&gt;&gt;&gt; x
['ala@ala.com', 'bala@bala.com']
&gt;&gt;&gt; x.remove('ala@ala.com')
&gt;&gt;&gt; x
['bala@bala.com']
</code></pre>

This will remove the first occurence that matches the string.

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

<pre><code>index = initial_list.index(item1)
del initial_list[index]
del other_list[index]
</code></pre>

blocks|key|590044|text|最明智的方法是使用zip()和列表理解/生成器表达式：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|590045|filtered+=+(
++++(email,+other)+
++++++++for+email,+other+in+zip(emails,+other_list)+
++++++++++++if+email+==+'something@something.com')

new_emails,+new_other_list+=+zip(*filtered)|code-block|syntax|javascript|590046|此外，如果您没有使用array.array()或numpy.array()，那么很可能您使用的是[]或list()，它们给出的是列表，而不是数组。不是一回事。|590047|entityMap^0|9|5|0|0|A|D|O|D|1C|2|1F|6|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@$9|T|A|U|B|C]|$9|V|A|W|B|C]|$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|11|8|@]|D|@]|E|$]]]|N|$]]

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

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

new_emails, new_other_list = zip(*filtered)
</code></pre>

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

blocks|key|590135|text|您的for循环不正确，如果您需要在for循环中使用索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590136|for+index,+item+in+enumerate(emails):
++++#+whatever+(but+you+can't+remove+element+while+iterating)|code-block|syntax|javascript|590137|在您的情况下，Bogdan解决方案是可以的，但您的数据结构选择不是很好。必须维护这两个列表，其中一个列表中的数据与另一个列表中的数据在同一索引中相关，这是很笨拙的。|590138|元组列表(电子邮件，其他数据)可能更好，或者以电子邮件为关键字的字典。|590139|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

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

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

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.

blocks|key|805950|text|使用filter()和lambda将提供一种简洁的方法来删除不需要的值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|805951|newEmails+=+list(filter(lambda+x+:+x+!=+'something@something.com',+emails))|code-block|syntax|javascript|805952|这不会修改电子邮件。它创建一个新的列表newEmails，其中只包含匿名函数返回True的元素。|805953|entityMap^0|2|8|B|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

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

<pre><code>newEmails = list(filter(lambda x : x != 'something@something.com', emails))
</code></pre>

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

blocks|key|590249|text|对于这个问题，有一个替代的解决方案，它也处理重复的匹配。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590250|我们从两个等长的列表开始：emails，otherarray。目标是为每个索引i+where+emails[i]+==+'something@something.com'从两个列表中删除项。|offset|length|style|CODE|590251|这可以通过使用列表理解，然后通过zip拆分来实现|590252|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']|code-block|syntax|javascript|590253|entityMap^0|0|D|6|K|A|13|1|1B|12|0|G|3|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@$D|S|E|T|F|G]|$D|U|E|V|F|G]|$D|W|E|X|F|G]|$D|Y|E|Z|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|10|8|@$D|11|E|12|F|G]]|9|@]|A|$]]|$1|J|3|K|5|L|7|13|8|@]|9|@]|A|$M|N]]|$1|O|3|-4|5|6|7|14|8|@]|9|@]|A|$]]]|P|$]]

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

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

This can be achieved using a list comprehension and then splitting via <code>zip</code>:

<pre><code>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']
</code></pre>

blocks|key|590299|text|如果您想删除数组的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|590300|使用array_name.pop(index_no.)|590301|例如：-|590302|>>>+arr+=+[1,2,3,4]
>>>+arr.pop(2)
>>>arr
[1,2,4]|code-block|syntax|javascript|590303|如果要从数组中删除特定的字符串/元素，则|590304|>>>+arr1+=+['python3.6'+,+'python2'+,'python3']
>>>+arr1.remove('python2')
>>>+arr1
['python3.6','python3']|590305|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|S|8|@]|9|@]|A|$]]|$1|F|3|G|5|H|7|T|8|@]|9|@]|A|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|H|7|V|8|@]|9|@]|A|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

<h1>If you want to delete the index of array:</h1>
Use array_name.pop(index_no.)
ex:-
<pre><code>&gt;&gt;&gt; arr = [1,2,3,4]
&gt;&gt;&gt; arr.pop(2)
&gt;&gt;&gt;arr
[1,2,4]
</code></pre>
<h1>If you want to delete a particular string/element from the array then</h1>
<pre><code>&gt;&gt;&gt; arr1 = ['python3.6' , 'python2' ,'python3']
&gt;&gt;&gt; arr1.remove('python2')
&gt;&gt;&gt; arr1
['python3.6','python3']
</code></pre>

I want to write something that removes a specific element from an array. I know that I have to <code>for</code> 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:

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

How to remove specific element from an array using python

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我想写一些从数组中移除特定元素的代码。我知道我必须对数组进行for循环才能找到与内容匹配的元素。假设我有一个电子邮件数组，我想去掉与某个电子邮件字符串匹配的元素。我实际上喜欢使用for循环结构，因为我也需要对其他数组使用相同的索引。下面是我的代码：for index, item in emails:    if ema...

问如何使用python从数组中删除特定元素
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