blocks|key|2147493|text|这是我的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2147494|grpby+=+df.groupby(['date','Name'])
a+=+grpby.apply(lambda+x:+np.average(a+=+x['SharePrice'],weights=x['NoShares'])).to_frame(name='SharePrice')
b+=+grpby.agg({'NoShares':'sum','Volume':'mean','Relation':'max'})
print+b.join(a)

+++++++++++++Volume+Relation++NoShares++SharePrice
date+++Name+++++++++++++++++++++++++++++++++++++++
2/1/10+yyy+++5.0000++++++++o+++++++++1++++++2.0000
+++++++zzz+++1.5000++++++++d+++++++++3++++++3.6667
2/5/10+vvv+++1.0000++++++++o+++++++++5++++++5.0000
+++++++www+++1.0000++++++++d++++++++10++++++5.0000
+++++++xxx+++1.0000+++++++do+++++++++5++++++5.0000
2/6/10+zzz+++3.0000++++++++d+++++++++1++++++1.0000|code-block|syntax|javascript|2147495|之后只有reset_index()。|2147496|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]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

here's my solution:

<pre><code>grpby = df.groupby(['date','Name'])
a = grpby.apply(lambda x: np.average(a = x['SharePrice'],weights=x['NoShares'])).to_frame(name='SharePrice')
b = grpby.agg({'NoShares':'sum','Volume':'mean','Relation':'max'})
print b.join(a)

 Volume Relation NoShares SharePrice
date Name 
2/1/10 yyy 5.0000 o 1 2.0000
 zzz 1.5000 d 3 3.6667
2/5/10 vvv 1.0000 o 5 5.0000
 www 1.0000 d 10 5.0000
 xxx 1.0000 do 5 5.0000
2/6/10 zzz 3.0000 d 1 1.0000
</code></pre>

just reset_index() afterwards.

blocks|key|3438400|text|我在这里做了个假设。当你说组的日期和名称，并保持关系-我假设，滴答和关系也将是独特的，这些组。因此，为了简单起见，我将所有4项进行分组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3438401|df+=+pd.DataFrame([
++++++++++++++++['2/1/10',+'aaa',+'zzz',+1,+1,+1,+'d'],+
++++++++++++++++['2/1/10',+'aaa',+'yyy',+1,+2,+5,+'o'],
++++++++++++++++['2/1/10',+'aaa',+'zzz',+2,+5,+2,+'d'],++
++++++++++++++++['2/5/10',+'bbb',+'xxx',+5,+5,+1,+'do'],
++++++++++++++++['2/5/10',+'ccc',+'www',+5,+5,+1,+'d'],
++++++++++++++++['2/5/10',+'ccc',+'www',+5,+5,+1,+'d'],
++++++++++++++++['2/5/10',+'ddd',+'vvv',+5,+5,+1,+'o'],
++++++++++++++++['2/6/10',+'aaa',+'zzz',+1,+1,+3,+'d']],
+++++++++++++columns+=+['date',+'ticker',+'Name',+'NoShares',
++++++++++++++++++++++++'SharePrice',+'Volume',+'Relation'])

def+process_date(dg):
++++return+pd.DataFrame([[
++++++++++++++++++++++++dg['NoShares'].sum(),
++++++++++++++++++++++++(dg['NoShares']+*+dg['SharePrice']).sum()+/+dg['NoShares'].sum(),
++++++++++++++++++++++++dg['Volume'].mean(),
++++++++++++++++++++++++]],+columns=['NoShares',+'SharePrice',+'Volume'])

df.groupby(['date',+'ticker',+'Name',+'Relation']).apply(process_date).reset_index(4,+drop=True).reset_index(drop=False)|code-block|syntax|javascript|3438402|结果：|3438403|+++++date+ticker+Name+Relation++NoShares++SharePrice++Volume
0++2/1/10++++aaa++yyy++++++++o+++++++++1++++2.000000+++++5.0
1++2/1/10++++aaa++zzz++++++++d+++++++++3++++3.666667+++++1.5
2++2/5/10++++bbb++xxx+++++++do+++++++++5++++5.000000+++++1.0
3++2/5/10++++ccc++www++++++++d++++++++10++++5.000000+++++1.0
4++2/5/10++++ddd++vvv++++++++o+++++++++5++++5.000000+++++1.0
5++2/6/10++++aaa++zzz++++++++d+++++++++1++++1.000000+++++3.0|3438404|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|$]]

I made an assumption here. When you said group by date and Name and to keep relation - I am assuming that ticker and relation will also be unique to those groups. So for simplicity I am grouping by all 4.

<pre><code>df = pd.DataFrame([
 ['2/1/10', 'aaa', 'zzz', 1, 1, 1, 'd'], 
 ['2/1/10', 'aaa', 'yyy', 1, 2, 5, 'o'],
 ['2/1/10', 'aaa', 'zzz', 2, 5, 2, 'd'], 
 ['2/5/10', 'bbb', 'xxx', 5, 5, 1, 'do'],
 ['2/5/10', 'ccc', 'www', 5, 5, 1, 'd'],
 ['2/5/10', 'ccc', 'www', 5, 5, 1, 'd'],
 ['2/5/10', 'ddd', 'vvv', 5, 5, 1, 'o'],
 ['2/6/10', 'aaa', 'zzz', 1, 1, 3, 'd']],
 columns = ['date', 'ticker', 'Name', 'NoShares',
 'SharePrice', 'Volume', 'Relation'])

def process_date(dg):
 return pd.DataFrame([[
 dg['NoShares'].sum(),
 (dg['NoShares'] * dg['SharePrice']).sum() / dg['NoShares'].sum(),
 dg['Volume'].mean(),
 ]], columns=['NoShares', 'SharePrice', 'Volume'])

df.groupby(['date', 'ticker', 'Name', 'Relation']).apply(process_date).reset_index(4, drop=True).reset_index(drop=False)
</code></pre>

Results:

<pre><code> date ticker Name Relation NoShares SharePrice Volume
0 2/1/10 aaa yyy o 1 2.000000 5.0
1 2/1/10 aaa zzz d 3 3.666667 1.5
2 2/5/10 bbb xxx do 5 5.000000 1.0
3 2/5/10 ccc www d 10 5.000000 1.0
4 2/5/10 ddd vvv o 5 5.000000 1.0
5 2/6/10 aaa zzz d 1 1.000000 3.0
</code></pre>

blocks|key|3198553|text|Dickster和Leo的答案都很好，但请注意，默认情况下，.groupby有dropna=True设置。因此，如果您有一个数据集，并在多个列上执行groupby，其中一些列可能包含NaN's，那么Pandas将删除这些组。最后的DataFrame将有较少的行。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|3198554|Server上的相同SQL查询不会在group子句中的列中删除带空值的行。我不知道其他关系数据库管理系统是否如此，但请记住，在默认情况下，Pandas以不同的方式对待group+by。|3198555|entityMap^0|U|8|13|B|22|7|2J|5|37|9|0|2B|8|0^^$0|@$1|2|3|4|5|6|7|J|8|@$9|K|A|L|B|C]|$9|M|A|N|B|C]|$9|O|A|P|B|C]|$9|Q|A|R|B|C]|$9|S|A|T|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|X|8|@]|D|@]|E|$]]]|I|$]]

Both Dickster's and Leo's answers work well but just be aware that <code>.groupby</code> has <code>dropna=True</code> set by default. So if you have a dataset and perform <code>groupby</code> on multiple columns where some of those columns might contain <code>NaN's</code> Pandas will drop these groups. The final <code>DataFrame</code> will have less rows then.
The same SQL query on a SQL Server don't drop rows with NULL values in columns that are in a group by clause. I don't know if that's true for other RDBMS but just bear in mind that Pandas by default treat <code>group by</code> in a different way.

I've been searching extensively but can't get my head around this issue:

I have a dataframe in pandas that looks like this:

<pre><code>date ticker Name NoShares SharePrice Volume Relation
2/1/10 aaa zzz 1 1 1 d 
2/1/10 aaa yyy 1 2 5 o
2/1/10 aaa zzz 2 5 2 d 
2/5/10 bbb xxx 5 5 1 do
2/5/10 ccc www 5 5 1 d
2/5/10 ccc www 5 5 1 d
2/5/10 ddd vvv 5 5 1 o
2/6/10 aaa zzz 1 1 3 d
</code></pre>

Requirements

<ol>
<li>I want to group by date and Name and:</li>
<li>have the number of shares summed up</li>
<li>have a weighted mean column for the share price (the weights are the NoShares)</li>
<li>average the volume and have it as a column</li>
<li>Leave relation as it is</li>
</ol>

so my output would look like this:

<pre><code>date ticker Name NoShares SharePrice Volume Relation
2/1/10 aaa zzz 3 3.6 1 d 
2/1/10 aaa yyy 1 2 5 o
2/5/10 bbb xxx 5 5 1 do
2/5/10 ccc www 10 5 1 d
2/5/10 ddd vvv 5 5 1 o
2/6/10 aaa zzz 1 1 3 d
</code></pre>

I tried the documentation and other answers on stack overflow but don't seem to be able to get it right. Appreciate the help. Cheers.

reduce a panda dataframe by groups

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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我已经搜索了很多次了，但是我无法理解这个问题：我有一个熊猫的数据图，看起来是这样的：date    ticker Name NoShares SharePrice Volume Relation2/1/10  aaa    zzz  1        1          1      d 2/1/10  aaa   ...

问按组减少熊猫数据
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问按组减少熊猫数据EN