blocks|key|3701951|text|这种类型的错误通常表示Pandas已经创建了dataframe子集的轻量级副本，然后您可以尝试在适当的位置对其进行修改。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3701952|例如，假设您有一个dataframe，它的列是整数1..10，并创建了一组偶数行：|3701953|df+=+pandas.DataFrame({+'int':+range(10),
++++++++++++++++++++++++'label':+[+'x'%2Bstr(i)+for+i+in+range(10)+]+})
sub+=+df[(df.int+%25+2)+==+0]|code-block|syntax|javascript|3701954|dataframe+sub现在包含：|offset|length|style|CODE|3701955|0++++x0
2++++x2
4++++x4
6++++x6
8++++x8
Name:+label,+dtype:+object|3701956|但是，sub实际上只是原始数据帧df的内存占用情况的视图。如果我们现在尝试修改sub|3701957|sub.label+=+[+'y'+%2B+str(j)+for+j+in+range(5)+]|3701958|我们得到了您已经看到的错误消息：|3701959|A+value+is+trying+to+be+set+on+a+copy+of+a+slice+from+a+DataFrame.
Try+using+.loc[row_indexer,col_indexer]+=+value+instead|3701960|如果要避免这种情况，可能需要使用.copy()方法：|3701961|newsub+=+df[(df.int+%25+2)+==+0].copy()
newsub.label+=+[+'y'+%2B+str(j)+for+j+in+range(5)+]|3701962|entityMap^0|0|0|0|A|3|0|0|3|3|G|2|13|3|0|0|0|0|G|7|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|15|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|17|8|@$K|18|L|19|M|N]]|9|@]|A|$]]|$1|O|3|P|5|F|7|1A|8|@]|9|@]|A|$G|H]]|$1|Q|3|R|5|6|7|1B|8|@$K|1C|L|1D|M|N]|$K|1E|L|1F|M|N]|$K|1G|L|1H|M|N]]|9|@]|A|$]]|$1|S|3|T|5|F|7|1I|8|@]|9|@]|A|$G|H]]|$1|U|3|V|5|6|7|1J|8|@]|9|@]|A|$]]|$1|W|3|X|5|F|7|1K|8|@]|9|@]|A|$G|H]]|$1|Y|3|Z|5|6|7|1L|8|@$K|1M|L|1N|M|N]]|9|@]|A|$]]|$1|10|3|11|5|F|7|1O|8|@]|9|@]|A|$G|H]]|$1|12|3|-4|5|6|7|1P|8|@]|9|@]|A|$]]]|13|$]]

This type of error is usually a sign that Pandas has created a lightweight copy of a subset of a dataframe, which you then try to modify in place.

For example, suppose you had a dataframe which had a column of the integers 1..10, and created a selection of rows that were even:

<pre><code>df = pandas.DataFrame({ 'int': range(10),
 'label': [ 'x'+str(i) for i in range(10) ] })
sub = df[(df.int % 2) == 0]
</code></pre>

The dataframe <code>sub</code> now contains:

<pre><code>0 x0
2 x2
4 x4
6 x6
8 x8
Name: label, dtype: object
</code></pre>

However, <code>sub</code> is really only a view onto the memory footprint of the original dataframe, <code>df</code>. If we now try to modify <code>sub</code>:

<pre><code>sub.label = [ 'y' + str(j) for j in range(5) ]
</code></pre>

We get the error message you've been seeing:

<pre><code>A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
</code></pre>

If you want to avoid this, you may need to use the <code>.copy()</code> method:

<pre><code>newsub = df[(df.int % 2) == 0].copy()
newsub.label = [ 'y' + str(j) for j in range(5) ]
</code></pre>

blocks|key|2897754|text|问题出在过滤上：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2897755|separ+=+tweets[tweets['TradeWar']+==+True]+|code-block|syntax|javascript|2897756|什么与以下内容相同：|2897757|separ+=+tweets[tweets['TradeWar']]|2897758|并且需要copy|offset|length|style|CODE|2897759|separ+=+tweets[tweets['TradeWar']].copy()|2897760|如果您稍后修改separ中的值，您会发现修改不会传播回原始数据(tweets)，并且Pandas会发出警告。|2897761|同样，代码也应该简化：|2897762|separ['RT']+=+separ['text'].str.contains(r"%5E(?=.*\bRT\b\s@\b).*$",+case=False)+|2897763|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.copy.html^0|0|0|0|0|4|4|4|4|0|0|0|7|5|W|6|0|0|0^^$0|@$1|2|3|4|5|6|7|16|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|17|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|18|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|1A|8|@$M|1B|N|1C|O|P]]|9|@$M|1D|N|1E|1|1F]]|A|$]]|$1|Q|3|R|5|D|7|1G|8|@]|9|@]|A|$E|F]]|$1|S|3|T|5|6|7|1H|8|@$M|1I|N|1J|O|P]|$M|1K|N|1L|O|P]]|9|@]|A|$]]|$1|U|3|V|5|6|7|1M|8|@]|9|@]|A|$]]|$1|W|3|X|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|Y|3|-4|5|6|7|1O|8|@]|9|@]|A|$]]]|Z|$10|$5|11|12|13|A|$14|15]]]]

Problem is in filtering:

<pre><code>separ = tweets[tweets['TradeWar'] == True] 
</code></pre>

What is same as:

<pre><code>separ = tweets[tweets['TradeWar']]
</code></pre>

And need <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.copy.html" rel="nofollow noreferrer"><code>copy</code></a>:

<pre><code>separ = tweets[tweets['TradeWar']].copy()
</code></pre>

If you modify values in <code>separ</code> later you will find that the modifications do not propagate back to the original data (<code>tweets</code>), and that Pandas does warning.

Also code shoud be simplify too:

<pre><code>separ['RT'] = separ['text'].str.contains(r"^(?=.*\bRT\b\s@\b).*$", case=False) 
</code></pre>

I am using ipython3 notebook and running the following command:

<pre><code>separ['RT'] = separ['text'].str.contains(r"^(?=.*\bRT\b\s@\b).*$", 
case=False) == True
</code></pre>

The warning I get is as follows:

<pre><code>-c:3: SettingWithCopyWarning: 
A value is trying to be set on a copy of a slice from a DataFrame.
Try using .loc[row_indexer,col_indexer] = value instead
</code></pre>

I have no idea what is means and how to handle this.

How to deal with SettingWithCopyWarning:

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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我正在使用ipython3 notebook并运行以下命令：separ['RT'] = separ['text'].str.contains(r"^(?=.*\bRT\b\s@\b).*$", case=False) == True我得到的警告如下：-c:3: SettingWithCopyWarning: A val...

问如何处理SettingWithCopyWarning：
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问如何处理SettingWithCopyWarning：EN