blocks|key|506498|text|self.left+=+None出了什么问题|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|506499|entityMap^0|0|G|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|K|8|@]|D|@]|E|$]]]|G|$]]

What's wrong with <code>self.left = None</code>?

blocks|key|722629|text|del关键字就可以了。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|722630|>>>+a=1
>>>+a
1
>>>+del+a
>>>+a
Traceback+(most+recent+call+last):
++File+"<stdin>",+line+1,+in+<module>
NameError:+name+'a'+is+not+defined|code-block|syntax|javascript|722631|但在这种情况下，我投票给self.left+=+None|722632|entityMap^0|0|3|0|0|C|G|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|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|V|8|@]|D|@]|E|$]]]|N|$]]

The <code>del</code> keyword would do.

<pre><code>&gt;&gt;&gt; a=1
&gt;&gt;&gt; a
1
&gt;&gt;&gt; del a
&gt;&gt;&gt; a
Traceback (most recent call last):
 File "&lt;stdin&gt;", line 1, in &lt;module&gt;
NameError: name 'a' is not defined
</code></pre>

But in this case I vote for <code>self.left = None</code>

blocks|key|722700|text|var+=+None“清除该值”，将变量的值设置为"null“，类似于"None”的值，但是指向该变量的指针将保持不变。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|722701|del+var完全删除了变量的定义。|722702|如果你想在以后使用这个变量，比如为它设置一个新的值，也就是保留这个变量，None会更好。|722703|entityMap^0|0|A|0|0|7|0|10|4|0^^$0|@$1|2|3|4|5|6|7|L|8|@$9|M|A|N|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|U|8|@]|D|@]|E|$]]]|K|$]]

<code>var = None</code> "clears the value", setting the value of the variable to "null" like value of "None", however the pointer to the variable remains. 

<code>del var</code> removes the definition for the variable totally. 

In case you want to use the variable later, e.g. set a new value for it, i.e. retain the variable, <code>None</code> would be better.

blocks|key|506689|text|实际上，这并不会删除变量/属性。它所要做的就是将它的值设置为None，因此该变量仍然会占用内存中的空间。如果您想从内存中完全清除该变量的所有存在，您只需输入：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|506690|del+self.left|code-block|syntax|javascript|506691|entityMap^0|U|4|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

Actually, that does not delete the variable/property. All it will do is set its value to <code>None</code>, therefore the variable will still take up space in memory. If you want to completely wipe all existence of the variable from memory, you can just type:

<pre><code>del self.left
</code></pre>

blocks|key|506849|text|你想删除一个变量吗？|type|unstyled|depth|inlineStyleRanges|entityRanges|data|506850|好的，我想我有一个替代@bnaul's+answer的最好的主意|offset|length|506851|您可以使用del删除单个名称|style|CODE|506852|del+x|code-block|syntax|javascript|506853|或者，您可以从globals()对象中删除它们：|506854|for+name+in+dir():
++++if+not+name.startswith('_'):
++++++++del+globals()[name]|506855|这只是一个示例循环；它只删除不以下划线开头的名称，并假设您在解释器中只使用开头没有下划线的名称(并非没有理由)。如果你真的想要更彻底，你可以使用硬编码的名字列表来代替(白名单)。除了退出和重新启动解释器之外，没有内置的函数来为您执行清除。|506856|您导入的模块(如import+os)将保持导入状态，因为它们被sys.modules引用；后续导入将重用已经导入的模块对象。只是在当前的全局名称空间中不会有对它们的引用。|506857|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/a/8237659/4518341^0|0|B|F|0|0|5|3|0|0|7|9|0|0|0|8|9|V|B|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@$D|16|E|17|1|18]]|A|$]]|$1|F|3|G|5|6|7|19|8|@$D|1A|E|1B|H|I]]|9|@]|A|$]]|$1|J|3|K|5|L|7|1C|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1D|8|@$D|1E|E|1F|H|I]]|9|@]|A|$]]|$1|Q|3|R|5|L|7|1G|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|1H|8|@]|9|@]|A|$]]|$1|U|3|V|5|6|7|1I|8|@$D|1J|E|1K|H|I]|$D|1L|E|1M|H|I]]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1N|8|@]|9|@]|A|$]]]|X|$Y|$5|Z|10|11|A|$12|13]]]]

Do you want to delete a variable, don't you?
ok, I think I've got a best alternative idea to <a href="https://stackoverflow.com/a/8237659/4518341">@bnaul's answer</a>:
You can delete individual names with <code>del</code>:
<pre><code>del x
</code></pre>
or you can remove them from the <code>globals()</code> object:
<pre><code>for name in dir():
 if not name.startswith('_'):
 del globals()[name]
</code></pre>
This is just an example loop; it defensively only deletes names that do not start with an underscore, making a (not unreasoned) assumption that you only used names without an underscore at the start in your interpreter. You could use a hard-coded list of names to keep instead (whitelisting) if you really wanted to be thorough. There is no built-in function to do the clearing for you, other than just exit and restart the interpreter.
Modules you've imported (like <code>import os</code>) are going to remain imported because they are referenced by <code>sys.modules</code>; subsequent imports will reuse the already imported module object. You just won't have a reference to them in your current global namespace.

blocks|key|506888|text|删除其内容，方法是将其设置为None，然后使用del将其指针从内存中移除|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|506889|variable+=+None;+del+variable|code-block|syntax|javascript|506890|entityMap^0|E|4|N|3|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Delete its contents by setting it to <code>None</code> and then <code>del</code> to remove its pointer from memory
<pre><code>variable = None; del variable
</code></pre>

blocks|key|506749|text|我使用了上面提到的几个选项：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|506750|del+self.left|code-block|syntax|javascript|506751|或使用以下命令将值设置为None|506752|self.left+=+None|506753|在使用将值设置为None时，了解其中的差异并放置一些异常处理程序是很重要的。如果您使用模板打印条件语句的值，例如，|506754|print("The+value+of+the+variable+is+{}".format(self.left))|506755|您可能会看到变量的值打印"The+value+of+the+variable+is+None“。因此，您必须在那里放置一些异常处理程序：|506756|if+self.left:
++++#Then+only+print+stuff|506757|仅当self.left不为None时，上述命令才会打印值|offset|length|style|CODE|506758|entityMap^0|0|0|0|0|0|0|0|0|D|4|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|13|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|14|8|@]|9|@]|A|$]]|$1|M|3|N|5|D|7|15|8|@]|9|@]|A|$E|F]]|$1|O|3|P|5|6|7|16|8|@]|9|@]|A|$]]|$1|Q|3|R|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|18|8|@$U|19|V|1A|W|X]]|9|@]|A|$]]|$1|Y|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]]|Z|$]]

I used a few options mentioned above : 

<pre><code>del self.left
</code></pre>

or setting value to None using 

<pre><code>self.left = None
</code></pre>

It's important to know the differences and put a few exception handlers in place when you use set the value to None. If you're printing the value of the conditional statements using a template, say, 

<pre><code>print("The value of the variable is {}".format(self.left))
</code></pre>

you might see the value of the variable printing "The value of the variable is None". Thus, you'd have to put a few exception handlers there :

<pre><code>if self.left:
 #Then only print stuff

</code></pre>

The above command will only print values if self.left is not <code>None</code>

Is there a way to clear the value of a variable in python?
For example if I was implementing a binary tree:
<pre><code>class Node:
 self.left = somenode1
 self.right = somenode2
</code></pre>
If I wanted to remove some node from the tree, I would need to set <code>self.left</code> to empty.

Clear variable in python

解释器

有没有办法在python中清除变量的值？例如，如果我正在实现一个二叉树：class Node:    self.left = somenode1    self.right = somenode2如果我想从树中删除一些节点，我需要将self.left设置为空。

问清除python中的变量
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