blocks|key|367416|text|我发现最简单的方法是重置索引，删除NaNs，然后再重新设置索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|367417|In+[26]:+dfA.reset_index()
Out[26]:+
++index+special_name
0+++NaN++++++++Apple
1+++OMG+++++++Banana

In+[30]:+df+=+dfA.reset_index().dropna().set_index('index')

In+[31]:+df
Out[31]:+
++++++special_name
index+++++++++++++
OMG+++++++++Banana|code-block|syntax|javascript|367418|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I've found that the easiest way is to reset the index, drop the NaNs, and then reset the index again.

<pre><code>In [26]: dfA.reset_index()
Out[26]: 
 index special_name
0 NaN Apple
1 OMG Banana

In [30]: df = dfA.reset_index().dropna().set_index('index')

In [31]: df
Out[31]: 
 special_name
index 
OMG Banana
</code></pre>

blocks|key|184190|text|编辑:以下内容可能只适用于MultiIndex__s，并且在任何情况下都会被新的df.index.isnull()函数淘汰(参见其他答案)。为了历史的利益，我会留下这个答案。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|184191|对于现在开始这样做的人来说，不需要重新编制索引就可以直接做到这一点，因为索引中的can将用标签-1表示。所以：|184192|df+=+dfA[dfA.index.labels!=-1]|code-block|syntax|javascript|184193|更好的是，在Pandas>0.16.1中，可以在不复制的情况下使用drop()在内部执行此操作：|184194|dfA.drop(labels=[-1],+level='index',+inplace=True)|184195|注:索引级别名为“index”有点误导:它通常是一些特定用途的内容，如“日期”或“实验性_run”。|184196|entityMap^0|D|A|14|H|0|1B|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|14|8|@]|D|@]|E|$K|L]]|$1|Q|3|R|5|6|7|15|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

Edit: the following probably only applies to <code>MultiIndex</code>s, and is in any case obsoleted by the new <code>df.index.isnull()</code> function (see other answers). I'll leave this answer just for historical interest.

For people who come to this now, one can do this directly without reindexing by relying on the fact that NaNs in the index will be represented with the label <code>-1</code>. So:

<pre><code>df = dfA[dfA.index.labels!=-1]
</code></pre>

Even better, in Pandas>0.16.1, one can use drop() to do this inplace without copying:

<pre><code>dfA.drop(labels=[-1], level='index', inplace=True)
</code></pre>

NB: It's a bit misleading that the index level is called 'index': it would usually be something more use-specific like 'date' or 'experimental_run'..

blocks|key|367490|text|对于熊猫版本>=+0.20.0，您可以：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|367491|df+=+df[df.index.notnull()]|code-block|syntax|javascript|367492|旧版本：|367493|df+=+df[pandas.notnull(df.index)]|367494|要把它分解：|367495|notnull生成一个布尔掩码，例如[False,+False,+True]，其中True表示对应位置为null的值(numpy.nan或None)。然后使用df[boolean_mask]选择其索引与掩码中的真值相对应的行。|offset|length|style|CODE|367496|entityMap^0|0|0|0|0|0|0|7|I|K|1N|9|1X|4|27|G|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|X|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$O|10|P|11|Q|R]|$O|12|P|13|Q|R]|$O|14|P|15|Q|R]|$O|16|P|17|Q|R]|$O|18|P|19|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]]|T|$]]

With pandas version &gt;= 0.20.0 you can:
<pre class="lang-py prettyprint-override"><code>df = df[df.index.notnull()]
</code></pre>
With older versions:
<pre><code>df = df[pandas.notnull(df.index)]
</code></pre>
To break it down:
<code>notnull</code> generates a boolean mask, e.g. <code>[False, False, True]</code>, where True denotes the value at the corresponding position is null (<code>numpy.nan</code> or <code>None</code>). We then select the rows whose index corresponds to a true value in the mask by using <code>df[boolean_mask]</code>.

blocks|key|332896|text|测试这是有效的：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|332897|df.reset_index(inplace=True)|offset|length|style|CODE|332898|df.drop(df[df['index'].isnull()].index,+inplace=True)|332899|332900|我是如何检查上面的|332901|使用df=pd.DataFrame(data=['Banana',+'Apple'],+index=[np.nan,+'OMG'],columns=['Special_name'])复制原始问题中的表|332902|然后输入上述两行代码--我试着用人类语言解释如下：|332903|第一行将索引重置为整数，NaN现在位于一个列中，列名以索引的原始名称命名(在上面的示例中，“index”，因为没有指定名称)--熊猫使用reset_index()命令自动做到这一点。|unordered-list-item|332904|最里面括号中的第2行：df[df['index'].isnull()]对列“index”使用isnull()命令显示'NaN‘值的行进行筛选。.index用于将指向所有'+index+'=NaN行的明确索引对象传递给表达式最外层的df.drop(。|332905|332906|332907|注意:测试上述命令以处理列中的多个NaN值|332908|使用Python+3.5.1，Pandas+0.17.1通过Anaconda包32位|332909|entityMap^0|0|0|S|0|0|1H|0|0|0|2|2H|0|0|0|B|O|1Z|6|37|8|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|13|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|14|8|@$D|15|E|16|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|17|8|@$D|18|E|19|F|G]]|9|@]|A|$]]|$1|J|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|1B|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|1C|8|@$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1F|8|@]|9|@]|A|$]]|$1|Q|3|R|5|S|7|1G|8|@]|9|@]|A|$]]|$1|T|3|U|5|S|7|1H|8|@$D|1I|E|1J|F|G]|$D|1K|E|1L|F|G]|$D|1M|E|1N|F|G]]|9|@]|A|$]]|$1|V|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]|$1|W|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]|$1|X|3|Y|5|6|7|1Q|8|@]|9|@]|A|$]]|$1|Z|3|10|5|6|7|1R|8|@]|9|@]|A|$]]|$1|11|3|-4|5|6|7|1S|8|@]|9|@]|A|$]]]|12|$]]

Tested this to be working : 

<code>df.reset_index(inplace=True)</code>

<code>df.drop(df[df['index'].isnull()].index, inplace=True)</code>

<hr>

<h2>How I checked the above</h2>

Replicated the table in the original question using 
<code>df=pd.DataFrame(data=['Banana', 'Apple'], index=[np.nan, 'OMG'],columns=['Special_name'])</code> 

then input the above two code lines- which I try to explain in human language below:

<ul>
<li>1st line resets the index to integers, and the NaN is now in a column named after the original name of the index ('index' in the example above as there was no name specified) - pandas does this automatically with the reset_index() command.</li>
<li>2nd line from innermost brackets: <code>df[df['index'].isnull()]</code> filters rows for which column named 'index' shows 'NaN' values using isnull() command. <code>.index</code> is used to pass an unambiguous index object pointing to all 'index'=NaN rows to the <code>df.drop(</code> in the outermost part of the expression. </li>
</ul>

nb: tested the above command to work on multiple NaN values in a column

Using Python 3.5.1 , Pandas 0.17.1 via Anaconda package 32bits

blocks|key|367574|text|没有一个答案对我是百分之百有效的。以下是起作用的原因：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|367575|In+[26]:+print+df
Out[26]:++++++++++++
++++++++++site_id++++++special_name
0+++++++++OMG++++++++++Apple
1+++++++++NaN++++++++++Banana
2+++++++++RLY++++++++++Orange


In+[27]:+df.dropna(inplace=True)
Out[27]:++++++++++++
++++++++++site_id++++++special_name
0+++++++++OMG++++++++++Apple
2+++++++++RLY++++++++++Orange

In+[28]:+df.reset_index(inplace=True)
Out[28]:++++++++++++
++++++++++index+++++site_id++++++special_name
0+++++++++0+++++++++OMG++++++++++Apple
1+++++++++2+++++++++RLY++++++++++Orange

In+[29]:+df.drop('index',+axis='columns',+inplace=True)
Out[29]:+++++++++++++
++++++++++site_id++++++special_name
0+++++++++OMG++++++++++Apple
1+++++++++RLY++++++++++Orange|code-block|syntax|javascript|367576|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

None of the answers worked 100% for me. Here's what worked:

<pre><code>In [26]: print df
Out[26]: 
 site_id special_name
0 OMG Apple
1 NaN Banana
2 RLY Orange


In [27]: df.dropna(inplace=True)
Out[27]: 
 site_id special_name
0 OMG Apple
2 RLY Orange

In [28]: df.reset_index(inplace=True)
Out[28]: 
 index site_id special_name
0 0 OMG Apple
1 2 RLY Orange

In [29]: df.drop('index', axis='columns', inplace=True)
Out[29]: 
 site_id special_name
0 OMG Apple
1 RLY Orange
</code></pre>

blocks|key|332921|text|或者，您可以使用query|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|332922|In+[4]:+df.query('index+==+index')
Out[4]:+
++++++++special_name
site_id+++++++++++++
OMG++++++++++++Apple|code-block|syntax|javascript|332923|这在与其本身相比返回False时作为NaN工作。|332924|In+[5]:+np.nan+==+np.nan
Out[5]:+False|332925|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.query.html^0|8|5|8|5|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@$9|Z|A|10|1|11]]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|15|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

Alternatively you can use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.query.html" rel="nofollow noreferrer"><code>query</code></a>:
<pre><code>In [4]: df.query('index == index')
Out[4]: 
 special_name
site_id 
OMG Apple
</code></pre>
This works as NaN when compared to itself returns False:
<pre><code>In [5]: np.nan == np.nan
Out[5]: False
</code></pre>

blocks|key|367598|text|另一个版本：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|367599|df[df.index.notna()]|code-block|syntax|javascript|367600|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another version:
<pre><code>df[df.index.notna()]
</code></pre>

I'm using python 2.7.3 and Pandas version 0.12.0.

I want to drop the row with the NaN index so that I only have valid site_id values.

<pre><code>print df.head()
 special_name
site_id
NaN Banana
OMG Apple

df.drop(df.index[0])

TypeError: 'NoneType' object is not iterable
</code></pre>

If I try dropping a range, like this:

<pre><code>df.drop(df.index[0:1])
</code></pre>

I get this error:

<pre><code>AttributeError: 'DataFrame' object has no attribute 'special_name'
</code></pre>

Trying to drop NaN indexed row in dataframe

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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我使用python2.7.3和Pandas版本0.12.0。我希望删除带有NaN索引的行，以便只有有效的site_id值。print df.head()            special_namesite_idNaN          BananaOMG          Appledf.drop(df.index...

问尝试删除NaN索引行中的数据
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问尝试删除NaN索引行中的数据EN