blocks|key|193747|text|对于DataFrame+df|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|193748|import+numpy+as+np
index+=+df['b'].index[df['b'].apply(np.isnan)]|code-block|syntax|javascript|193749|将返回可用于索引回df的MultiIndex，例如：|193750|df['a'].ix[index[0]]
>>>+1.452354|193751|对于整数索引：|193752|df_index+=+df.index.values.tolist()
[df_index.index(i)+for+i+in+index]
>>>+[3,+6]|193753|entityMap^0|C|2|0|0|9|2|C|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@$9|Z|A|10|B|C]|$9|11|A|12|B|C]]|D|@]|E|$]]|$1|M|3|N|5|H|7|13|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|14|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

For DataFrame <code>df</code>:

<pre><code>import numpy as np
index = df['b'].index[df['b'].apply(np.isnan)]
</code></pre>

will give you back the <code>MultiIndex</code> that you can use to index back into <code>df</code>, e.g.:

<pre><code>df['a'].ix[index[0]]
&gt;&gt;&gt; 1.452354
</code></pre>

For the integer index:

<pre><code>df_index = df.index.values.tolist()
[df_index.index(i) for i in index]
&gt;&gt;&gt; [3, 6]
</code></pre>

blocks|key|193775|text|这里有一个更简单的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193776|inds+=+pd.isnull(df).any(1).nonzero()[0]|offset|length|style|CODE|193777|In+[9]:+df
Out[9]:+
++++++++++0+++++++++1
0++0.450319++0.062595
1+-0.673058++0.156073
2+-0.871179+-0.118575
3++0.594188+++++++NaN
4+-1.017903+-0.484744
5++0.860375++0.239265
6+-0.640070+++++++NaN
7+-0.535802++1.632932
8++0.876523+-0.153634
9+-0.686914++0.131185

In+[10]:+pd.isnull(df).any(1).nonzero()[0]
Out[10]:+array([3,+6])|code-block|syntax|javascript|193778|entityMap^0|0|0|14|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Here is a simpler solution:

<code>inds = pd.isnull(df).any(1).nonzero()[0]</code>

<pre><code>In [9]: df
Out[9]: 
 0 1
0 0.450319 0.062595
1 -0.673058 0.156073
2 -0.871179 -0.118575
3 0.594188 NaN
4 -1.017903 -0.484744
5 0.860375 0.239265
6 -0.640070 NaN
7 -0.535802 1.632932
8 0.876523 -0.153634
9 -0.686914 0.131185

In [10]: pd.isnull(df).any(1).nonzero()[0]
Out[10]: array([3, 6])
</code></pre>

blocks|key|412101|text|单行解决方案。但是，它只适用于一列。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|412102|df.loc[pandas.isna(df["b"]),+:].index|code-block|syntax|javascript|412103|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|$]]

One line solution. However it works for one column only.

<pre><code>df.loc[pandas.isna(df["b"]), :].index
</code></pre>

blocks|key|193834|text|以防万一，如果您想要找到所有列的'nan‘坐标(假设它们都是数字)，请执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193835|df+=+pd.DataFrame([[0,1,3,4,np.nan,2],[3,5,6,np.nan,3,3]])

df
+++0++1++2++++3++++4++5
0++0++1++3++4.0++NaN++2
1++3++5++6++NaN++3.0++3

np.where(np.asanyarray(np.isnan(df)))
(array([0,+1]),+array([4,+3]))|code-block|syntax|javascript|193836|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|$]]

And just in case, if you want to find the coordinates of 'nan' for all the columns instead (supposing they are all numericals), here you go:

<pre><code>df = pd.DataFrame([[0,1,3,4,np.nan,2],[3,5,6,np.nan,3,3]])

df
 0 1 2 3 4 5
0 0 1 3 4.0 NaN 2
1 3 5 6 NaN 3.0 3

np.where(np.asanyarray(np.isnan(df)))
(array([0, 1]), array([4, 3]))
</code></pre>

blocks|key|193927|text|不知道这是否太晚了，但您可以使用np.where来查找非值的索引：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193928|indices+=+list(np.where(df['b'].isna()[0]))|code-block|syntax|javascript|193929|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|$]]

Don't know if this is too late but you can use np.where to find the indices of non values as such:

<pre><code>indices = list(np.where(df['b'].isna()[0]))
</code></pre>

blocks|key|412138|text|在有datetime索引的情况下，您想要有以下值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|412139|df.loc[pd.isnull(df).any(1),+:].index.values|code-block|syntax|javascript|412140|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|$]]

in the case you have datetime index and you want to have the values:

<pre><code>df.loc[pd.isnull(df).any(1), :].index.values
</code></pre>

blocks|key|412168|text|以下是对几种方法的测试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|412169|%25timeit+np.where(np.isnan(df['b']))[0]
%25timeit+pd.isnull(df['b']).nonzero()[0]
%25timeit+np.where(df['b'].isna())[0]
%25timeit+df.loc[pd.isna(df['b']),+:].index|code-block|syntax|javascript|412170|以及它们对应的时间：|412171|333+µs+±+9.95+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+1000+loops+each)
280+µs+±+220+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+1000+loops+each)
313+µs+±+128+ns+per+loop+(mean+±+std.+dev.+of+7+runs,+1000+loops+each)
6.84+ms+±+1.59+µs+per+loop+(mean+±+std.+dev.+of+7+runs,+100+loops+each)|412172|看起来pd.isnull(df['DRGWeight']).nonzero()[0]在时间上赢得了胜利，但前三种方法中的任何一种都具有可比的性能。|offset|length|style|CODE|412173|entityMap^0|0|0|0|0|3|13|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|V|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Here are tests for a few methods:

<pre class="lang-py prettyprint-override"><code>%timeit np.where(np.isnan(df['b']))[0]
%timeit pd.isnull(df['b']).nonzero()[0]
%timeit np.where(df['b'].isna())[0]
%timeit df.loc[pd.isna(df['b']), :].index
</code></pre>

And their corresponding timings:

<pre><code>333 µs ± 9.95 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
280 µs ± 220 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
313 µs ± 128 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
6.84 ms ± 1.59 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
</code></pre>

It would appear that <code>pd.isnull(df['DRGWeight']).nonzero()[0]</code> wins the day in terms of timing, but that any of the top three methods have comparable performance.

blocks|key|194055|text|另一个简单的解决方案是list(np.where(df['b'].isnull())[0])|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|194056|entityMap^0|B|Z|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|$]]

Another simple solution is <code>list(np.where(df['b'].isnull())[0])</code>

blocks|key|193876|text|下面是另一个更简单的方法：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193877|df+=+pd.DataFrame([[0,1,3,4,np.nan,2],[3,5,6,np.nan,3,3]])

inds+=+np.asarray(df.isnull()).nonzero()

(array([0,+1],+dtype=int64),+array([4,+3],+dtype=int64))|code-block|syntax|javascript|193878|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|$]]

Here is another simpler take:

<pre><code>df = pd.DataFrame([[0,1,3,4,np.nan,2],[3,5,6,np.nan,3,3]])

inds = np.asarray(df.isnull()).nonzero()

(array([0, 1], dtype=int64), array([4, 3], dtype=int64))
</code></pre>

blocks|key|193956|text|我正在查找具有NaN值的行的所有索引。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|193957|我的工作解决方案：|193958|def+get_nan_indexes(data_frame):
++++indexes+=+[]
++++print(data_frame)
++++for+column+in+data_frame:
++++++++index+=+data_frame[column].index[data_frame[column].apply(np.isnan)]
++++++++if+len(index):
++++++++++++indexes.append(index[0])
++++df_index+=+data_frame.index.values.tolist()
++++return+[df_index.index(i)+for+i+in+set(indexes)]|code-block|syntax|javascript|193959|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|L|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|M|8|@]|9|@]|A|$G|H]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

I was looking for all indexes of rows with NaN values. 
My working solution:

<pre><code>def get_nan_indexes(data_frame):
 indexes = []
 print(data_frame)
 for column in data_frame:
 index = data_frame[column].index[data_frame[column].apply(np.isnan)]
 if len(index):
 indexes.append(index[0])
 df_index = data_frame.index.values.tolist()
 return [df_index.index(i) for i in set(indexes)]
</code></pre>

blocks|key|194118|text|这将为您提供每一列中nan的索引值：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|194119|df.loc[pd.isna(df).any(1),+:].index|code-block|syntax|javascript|194120|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|$]]

This will give you the index values for nan in every column:
<pre><code>df.loc[pd.isna(df).any(1), :].index
</code></pre>

blocks|key|194126|text|++++index_nan+=+[]
++++++++for+index,+bool_v+in+df["b"].iteritems().isna():
+++++++++++if+bool_v+==+True:
+++++++++++++++index_nan.append(index)
++++print(index_nan)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|194127|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code> index_nan = []
 for index, bool_v in df[&quot;b&quot;].iteritems().isna():
 if bool_v == True:
 index_nan.append(index)
 print(index_nan)
</code></pre>

I have a pandas DataFrame like this:

<pre><code> a b
2011-01-01 00:00:00 1.883381 -0.416629
2011-01-01 01:00:00 0.149948 -1.782170
2011-01-01 02:00:00 -0.407604 0.314168
2011-01-01 03:00:00 1.452354 NaN
2011-01-01 04:00:00 -1.224869 -0.947457
2011-01-01 05:00:00 0.498326 0.070416
2011-01-01 06:00:00 0.401665 NaN
2011-01-01 07:00:00 -0.019766 0.533641
2011-01-01 08:00:00 -1.101303 -1.408561
2011-01-01 09:00:00 1.671795 -0.764629
</code></pre>

Is there an efficient way to find the "integer" index of rows with NaNs? In this case the desired output should be <code>[3, 6]</code>.

Find integer index of rows with NaN in pandas dataframe

我有一个像这样的熊猫DataFrame：                    a         b2011-01-01 00:00:00 1.883381  -0.4166292011-01-01 01:00:00 0.149948  -1.7821702011-01-01 02:00:00 -0.407604 0.3141682011-01-01 03:00:00 1.452354  NaN

问在pandas数据帧中查找具有NaN的行的整数索引
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回答 12

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问在pandas数据帧中查找具有NaN的行的整数索引EN

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问在pandas数据帧中查找具有NaN的行的整数索引
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