blocks|key|61018|text|请参阅此增强issue|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|61019|简而言之，答案是肯定的，groupby将保留传入的排序。你可以像这样用你的例子来证明这一点：|61020|In+[20]:+df.sort_index(ascending=False).groupby('A').agg([np.mean,+lambda+x:+x.iloc[1]+])
Out[20]:+
+++++++++++B+++++++++++++C+++++++++
++++++++mean+<lambda>+mean+<lambda>
A++++++++++++++++++++++++++++++++++
group1++11.0+++++++10++101++++++100
group2++17.5+++++++10++175++++++100
group3++11.0+++++++10++101++++++100|code-block|syntax|javascript|61021|然而，这不适用于重采样，因为它需要单调索引(它将使用非单调索引，但将首先对其进行排序)。|61022|他们是groupby的sort=标志，但这与组本身的排序有关，而不是组内的观察结果。|style|CODE|61023|仅供参考：df.groupby('A').nth(1)是获取组的第二个值的安全方法(因为如果一个组有<2个元素，上面的方法将失败)|61024|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/pydata/pandas/issues/8588^0|6|5|0|0|0|0|0|B|5|0|5|M|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@$A|11|B|12|1|13]]|C|$]]|$1|D|3|E|5|6|7|14|8|@]|9|@]|C|$]]|$1|F|3|G|5|H|7|15|8|@]|9|@]|C|$I|J]]|$1|K|3|L|5|6|7|16|8|@]|9|@]|C|$]]|$1|M|3|N|5|6|7|17|8|@$A|18|B|19|O|P]]|9|@]|C|$]]|$1|Q|3|R|5|6|7|1A|8|@$A|1B|B|1C|O|P]]|9|@]|C|$]]|$1|S|3|-4|5|6|7|1D|8|@]|9|@]|C|$]]]|T|$U|$5|V|W|X|C|$Y|Z]]]]

See this enhancement <a href="https://github.com/pydata/pandas/issues/8588" rel="noreferrer">issue</a>

The short answer is yes, the groupby will preserve the orderings as passed in. You can prove this by using your example like this:

<pre><code>In [20]: df.sort_index(ascending=False).groupby('A').agg([np.mean, lambda x: x.iloc[1] ])
Out[20]: 
 B C 
 mean &lt;lambda&gt; mean &lt;lambda&gt;
A 
group1 11.0 10 101 100
group2 17.5 10 175 100
group3 11.0 10 101 100
</code></pre>

This is NOT true for resample however as it requires a monotonic index (it WILL work with a non-monotonic index, but will sort it first).

Their is a <code>sort=</code> flag to groupby, but this relates to the sorting of the groups themselves and not the observations within a group.

FYI: <code>df.groupby('A').nth(1)</code> is a safe way to get the 2nd value of a group (as your method above will fail if a group has &lt; 2 elements)

blocks|key|61052|text|更简单：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|61053|++import+pandas+as+pd
++pd.pivot_table(df,index='A',aggfunc=(np.mean))|code-block|syntax|javascript|61054|输出：|61055|++++++++++++B++++C
+++++A++++++++++++++++
+++group1++11.0++101
+++group2++17.5++175
+++group3++11.0++101|61056|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Even easier:

<pre><code> import pandas as pd
 pd.pivot_table(df,index='A',aggfunc=(np.mean))
</code></pre>

output:

<pre><code> B C
 A 
 group1 11.0 101
 group2 17.5 175
 group3 11.0 101
</code></pre>

blocks|key|4567367|text|Panda的0.19.1文档说"groupby保留了每个组中行的顺序“，所以这是有保证的行为。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4567368|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html|offset|length|4567369|entityMap|0|LINK|mutability|MUTABLE|url^0|0|0|2B|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|N|8|@]|9|@$D|O|E|P|1|Q]]|A|$]]|$1|F|3|-4|5|6|7|R|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|C]]]]

Panda's 0.19.1 doc says "groupby preserves the order of rows within each group", so this is guaranteed behavior.

<a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html" rel="noreferrer">http://pandas.pydata.org/pandas-docs/stable/generated/pandas.DataFrame.groupby.html</a>

blocks|key|4567401|text|为了保持秩序，您需要传递.groupby(...,+sort=False)。在您的示例中，分组列已经排序，因此没有区别，但通常必须使用sort=False标志：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4567402|+df.groupby('A',+sort=False).agg([np.mean,+lambda+x:+x.iloc[1]+])|code-block|syntax|javascript|4567403|entityMap^0|C|P|1V|A|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|$]]

In order to preserve order, you'll need to pass <code>.groupby(..., sort=False)</code>. In your case the grouping column is already sorted, so it does not make difference, but generally one must use the <code>sort=False</code> flag:

<pre><code> df.groupby('A', sort=False).agg([np.mean, lambda x: x.iloc[1] ])
</code></pre>

blocks|key|61170|text|不幸的是，这个问题的答案是否定的。在过去的几天里，我创建了一个非均匀分块的算法，发现它不可能保持顺序，因为groupby引入了子帧，其中每个帧的关键是groupby输入。所以你最终会得到：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|61171|allSubFrames+=+df.groupby("myColumnToOrderBy")
for+orderKey,+individualSubFrame+in+allSubFrames:
+++++do+something...|code-block|syntax|javascript|61172|因为它使用字典，所以你失去了排序。|61173|如果您随后执行排序，如上所述，我刚刚测试了一个海量数据集，您最终将得到O(n+log+n)计算。|61174|但是，我发现，例如，如果您有按顺序排序的时间序列数据，而您希望保留顺序，则最好将排序列更改为列表，然后创建一个计数器来记录每个时间序列中的第一项。这将导致O(n)计算。|61175|因此，基本上，如果您使用相对较小的数据集，上述建议的答案是合理的，但如果使用大数据集，则需要考虑避免groupby和sort。取而代之的是：list(df['myColumnToOrderBy'])和迭代器。|offset|length|style|CODE|61176|entityMap^0|0|0|0|0|0|1Y|T|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]]|$1|G|3|H|5|6|7|W|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|Y|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|Z|8|@$O|10|P|11|Q|R]]|9|@]|A|$]]|$1|S|3|-4|5|6|7|12|8|@]|9|@]|A|$]]]|T|$]]

Unfortunately the answer to this question is NO. In the past few days I've created an algorithm for non-uniform chunking and found that is cannot possibly retain order because a groupby introduces subframes where the key to each frame is the groupby input. So you end up with:
<pre><code>allSubFrames = df.groupby(&quot;myColumnToOrderBy&quot;)
for orderKey, individualSubFrame in allSubFrames:
 do something...
</code></pre>
Because its using dictionaries you lose the ordering.
If you perform a sort afterwards, as mentioned above, which I've just tested for a massive dataset, you end up with an O(n log n) computation.
However, I found that if you have for instance ordered time series data in order, where you want to preserve the order, it is better to change the ordering column into a list and then create a counter that records the first item in each time series. This results in a O(n) calculation.
So, essentially if you are using a relatively small dataset the proposed answers above are reasonable, but if using a big data set you need to consider avoiding groupby and sort. Instead use: <code>list(df['myColumnToOrderBy'])</code> and iterator over it.

I've frequented used pandas' <code>agg()</code> function to run summary statistics on every column of a data.frame. For example, here's how you would produce the mean and standard deviation:

<pre><code>df = pd.DataFrame({'A': ['group1', 'group1', 'group2', 'group2', 'group3', 'group3'],
 'B': [10, 12, 10, 25, 10, 12],
 'C': [100, 102, 100, 250, 100, 102]})

&gt;&gt;&gt; df
[output]
 A B C
0 group1 10 100
1 group1 12 102
2 group2 10 100
3 group2 25 250
4 group3 10 100
5 group3 12 102
</code></pre>

In both of those cases, the order that individual rows are sent to the agg function does not matter. But consider the following example, which:

<pre><code>df.groupby('A').agg([np.mean, lambda x: x.iloc[1] ])

[output]

 mean &lt;lambda&gt; mean &lt;lambda&gt;
A 
group1 11.0 12 101 102
group2 17.5 25 175 250
group3 11.0 12 101 102
</code></pre>

In this case the lambda functions as intended, outputting the second row in each group. However, I have not been able to find anything in the pandas documentation that implies that this is guaranteed to be true in all cases. I want use <code>agg()</code> along with a weighted average function, so I want to be sure that the rows that come into the function will be in the same order as they appear in the original data frame.

Does anyone know, ideally via somewhere in the docs or pandas source code, if this is guaranteed to be the case?

Python Pandas: Is Order Preserved When Using groupby() and agg()?

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我经常使用pandas的agg()函数对data.frame的每一列运行汇总统计信息。例如，下面是生成平均值和标准差的方法：df = pd.DataFrame({'A': ['group1', 'group1', 'group2', 'group2', 'group3', 'group3'],             ...

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