blocks|key|2135978|text|您可以使用按位“或”%7C“来完成该操作。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2135979|df_filtered+=+df[~((df['Borough']+==+'Not+assigned')+%7C+
+++++++++++++++++++(df['Neighbourhood']+==+'Not+assigned'))]|code-block|syntax|javascript|2135980|然后，对样本数据集的结果是：|2135981|++Postcode+++Borough+Neighbourhood
0+++++++M1++++Ottawa++++++++Ottawa
2+++++++M3++Montreal++++++Montreal
3+++++++M4+++Toronto+++++Barrhaven|2135982|entityMap^0|A|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@$9|R|A|S|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|T|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|U|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|V|8|@]|D|@]|E|$I|J]]|$1|O|3|-4|5|6|7|W|8|@]|D|@]|E|$]]]|P|$]]

You can do it using bitwise "or" <code>|</code>:

<pre><code>df_filtered = df[~((df['Borough'] == 'Not assigned') | 
 (df['Neighbourhood'] == 'Not assigned'))]
</code></pre>

The result then with your sample dataset is:

<pre><code> Postcode Borough Neighbourhood
0 M1 Ottawa Ottawa
2 M3 Montreal Montreal
3 M4 Toronto Barrhaven
</code></pre>

blocks|key|3150565|text|尝试：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3150566|df+=+df[~(df['Borough'].eq('Not+assigned')+%7C+df['Borough'].eq('Not+assigned'))]|code-block|syntax|javascript|3150567|++Postcode+++Borough+Neighbourhood
0+++++++M1++++Ottawa++++++++Ottawa
2+++++++M3++Montreal++++++Montreal
3+++++++M4+++Toronto+++++Barrhaven|3150568|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|D|7|M|8|@]|9|@]|A|$E|F]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

Try:

<pre><code>df = df[~(df['Borough'].eq('Not assigned') | df['Borough'].eq('Not assigned'))]
</code></pre>

<pre><code> Postcode Borough Neighbourhood
0 M1 Ottawa Ottawa
2 M3 Montreal Montreal
3 M4 Toronto Barrhaven
</code></pre>

blocks|key|852035|text|我们可以在axis+=+1中使用DataFrame.ne+%2B+来执行boolean+indexing|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|CODE|entityRanges|data|852036|df_filtered+=+df[df[['Borough','Neighbourhood']].ne('Not+assigned').all(axis=1)]
print(df_filtered)|code-block|syntax|javascript|852037|输出|852038|++Postcode+++Borough+Neighbourhood
0+++++++M1++++Ottawa++++++++Ottawa
2+++++++M3++Montreal++++++Montreal
3+++++++M4+++Toronto+++++Barrhaven|852039|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.ne.html|1|https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#boolean-indexing^0|5|8|5|8|G|C|Y|G|G|C|0|Y|G|1|0|0|0|2|0|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|D]|$9|14|A|15|B|D]|$9|16|A|17|B|D]]|E|@$9|18|A|19|1|1A]|$9|1B|A|1C|1|1D]]|F|$]]|$1|G|3|H|5|I|7|1E|8|@]|E|@]|F|$J|K]]|$1|L|3|M|5|6|7|1F|8|@$9|1G|A|1H|B|C]]|E|@]|F|$]]|$1|N|3|O|5|I|7|1I|8|@]|E|@]|F|$J|K]]|$1|P|3|-4|5|6|7|1J|8|@]|E|@]|F|$]]]|Q|$R|$5|S|T|U|F|$V|W]]|X|$5|S|T|U|F|$V|Y]]]]

We can use <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.ne.html" rel="nofollow noreferrer"><code>DataFrame.ne</code></a> + <a href="https://pandas.pydata.org/pandas-docs/stable/reference/api/pandas.DataFrame.all.html" rel="nofollow noreferrer"><code>DataFrame.all</code></a> with <code>axis = 1</code> to performance a <a href="https://pandas.pydata.org/pandas-docs/stable/user_guide/indexing.html#boolean-indexing" rel="nofollow noreferrer"><code>boolean indexing</code></a>:

<pre><code>df_filtered = df[df[['Borough','Neighbourhood']].ne('Not assigned').all(axis=1)]
print(df_filtered)
</code></pre>

Output

<pre><code> Postcode Borough Neighbourhood
0 M1 Ottawa Ottawa
2 M3 Montreal Montreal
3 M4 Toronto Barrhaven
</code></pre>

I have a dataset consisting of 3 columns (Postcode, Borough and Neighbourhood) set up as follows: 

<pre><code> df = pd.DataFrame({'Postcode' : ['M1', 'M2', 'M3', 'M4', 'M5'], 
 'Borough' : ['Ottawa', 'Not assigned', 'Montreal', 'Toronto', 'Kent'],
 'Neighbourhood' : ['Ottawa', 'Toronto', 'Montreal', 'Barrhaven', 'Not assigned']})
</code></pre>

which looks like this: 

<img src="https://i.stack.imgur.com/3IVrM.png" alt="Dataframe">

The values in the columns Borough and Neighbourhood can either be "Not assigned" or a valid text - the value "Not assigned" could be in both cells, or one or the other.

What I want to do is remove the rows in the entire dataset where it has "Not assigned" in either column. 

I am very new to Python ... I thought I would try to create an additional column based on the value of one of the cells, giving True or False, so I tried this... 

<pre><code> df['Outcome'] = ["True" if x =='Not assigned' else "False" for x in df['Borough']] 
</code></pre>

... which added an extra column successfully

<img src="https://i.stack.imgur.com/dHmjl.png" alt="Dataframe2">

Then I guess I would try and remove those TRUE rows with the drop() function and repeat the process on the Neighbourhood column. But this seems a messy way to do it and I will end up with 20 lines of code where I am sure it can be done much more efficiently. 

Can someone please advise me the simplest way to get rid of those rows?

How to delete rows in a dataframe based on 2 variables in either column

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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我有一个由3列组成的数据集(邮政编码、Borough和Neighbourhood)，设置如下：    df = pd.DataFrame({'Postcode' : ['M1', 'M2', 'M3', 'M4', 'M5'],             'Borough' : ['Ottawa', 'Not assig...

问如何根据两列中的两个变量删除数据帧中的行
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问如何根据两列中的两个变量删除数据帧中的行EN