blocks|key|813641|text|您可以使用lambda中的x+(具体来说，使用它是.index来获取您想要的值)。例如：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|813642|import+pandas+as+pd
import+numpy+as+np


def+weighted_avg(group_df,+whole_df,+values,+weights):
++++v+=+whole_df.loc[group_df.index,+values]
++++w+=+whole_df.loc[group_df.index,+weights]
++++return+(v+*+w).sum()+/+w.sum()


dfr+=+pd.DataFrame(np.random.randint(1,+50,+size=(4,+4)),+columns=list("ABCD"))
dfr["group"]+=+[1,+1,+0,+1]

print(dfr)
dfr+=+(
++++dfr.groupby("group")
++++.agg(
++++++++{"A":+"mean",+"B":+"sum",+"C":+lambda+x:+weighted_avg(x,+dfr,+"D",+"C")}
++++)
++++.reset_index()
)
print(dfr)|code-block|syntax|javascript|813643|指纹：|813644|++++A+++B+++C+++D++group
0++32+++2++34++29++++++1
1++33++32++15++49++++++1
2+++4++43++41++10++++++0
3++39++33+++7++31++++++1

+++group++++++++++A+++B++++++++++C
0++++++0+++4.000000++43++10.000000
1++++++1++34.666667++67++34.607143|813645|编辑:正如@enke在注释中所述，您可以使用已过滤的数据enke调用您的weighted_avg函数：|813646|weighted_avg(dfr.loc[x.index],+'D',+'C')|813647|entityMap^0|D|1|P|6|0|0|0|0|10|C|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]|$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|12|8|@$9|13|A|14|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|15|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|16|8|@]|D|@]|E|$]]]|T|$]]

You can use <code>x</code> you have in lambda (specifically, use it's <code>.index</code> to get values you want). For example:
<pre class="lang-py prettyprint-override"><code>import pandas as pd
import numpy as np


def weighted_avg(group_df, whole_df, values, weights):
 v = whole_df.loc[group_df.index, values]
 w = whole_df.loc[group_df.index, weights]
 return (v * w).sum() / w.sum()


dfr = pd.DataFrame(np.random.randint(1, 50, size=(4, 4)), columns=list(&quot;ABCD&quot;))
dfr[&quot;group&quot;] = [1, 1, 0, 1]

print(dfr)
dfr = (
 dfr.groupby(&quot;group&quot;)
 .agg(
 {&quot;A&quot;: &quot;mean&quot;, &quot;B&quot;: &quot;sum&quot;, &quot;C&quot;: lambda x: weighted_avg(x, dfr, &quot;D&quot;, &quot;C&quot;)}
 )
 .reset_index()
)
print(dfr)
</code></pre>
Prints:
<pre class="lang-none prettyprint-override"><code> A B C D group
0 32 2 34 29 1
1 33 32 15 49 1
2 4 43 41 10 0
3 39 33 7 31 1

 group A B C
0 0 4.000000 43 10.000000
1 1 34.666667 67 34.607143
</code></pre>
<hr />
EDIT: As @enke stated in comments, you can call your <code>weighted_avg</code> function with already filtered dataframe:
<pre class="lang-py prettyprint-override"><code>weighted_avg(dfr.loc[x.index], 'D', 'C')
</code></pre>

blocks|key|813664|text|你写过lambda+x:+weighted_avg(dfr,+'D',+'C')的地方|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|813665|这将计算dfr上的加权平均值，即整个表。|813666|如果您将其更改为lambda+group:+weighted_avg(group,+"D",+"C")|813667|那我觉得也许能行。|813668|(我已将lambda变量的名称更改为group，因为x描述性不强)|813669|entityMap^0|3|11|0|4|3|0|8|17|0|0|I|5|Q|1|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|J|3|K|5|6|7|Y|8|@]|D|@]|E|$]]|$1|L|3|M|5|6|7|Z|8|@$9|10|A|11|B|C]|$9|12|A|13|B|C]]|D|@]|E|$]]|$1|N|3|-4|5|6|7|14|8|@]|D|@]|E|$]]]|O|$]]

Where you have written
<code>lambda x: weighted_avg(dfr, 'D', 'C')</code>
this will calculate the weighted average over <code>dfr</code>, i.e. the whole table.
If you change it to
<code>lambda group: weighted_avg(group, &quot;D&quot;, &quot;C&quot;)</code>
then I think it may work.
(I've changed the name of the lambda variable to <code>group</code> since <code>x</code> is not very descriptive)

blocks|key|1626500|text|对于这种情况，我通常为计算的中间阶段添加列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1626501|df['product']+=+df['value']+*+df['weight']
weighted_avg+=+sum(df['product'])+/+sum(df['weight'])|code-block|syntax|javascript|1626502|然后，您可以正常地进行分组和子集选择：|1626503|df0+=+df[df['group']==0]
df1+=+df[df['group']==1]|1626504|并分别计算各组的weighted_avg。|offset|length|style|CODE|1626505|entityMap^0|0|0|0|0|8|C|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|$]]

For this sort of thing, I usually add columns for the intermediate stages of the calculation:
<pre><code>df['product'] = df['value'] * df['weight']
weighted_avg = sum(df['product']) / sum(df['weight'])
</code></pre>
You can then do grouping and subset-selction as normal:
<pre><code>df0 = df[df['group']==0]
df1 = df[df['group']==1]
</code></pre>
and calculate <code>weighted_avg</code> separately for each group

I want the ability to use custom functions in pandas groupby agg(). I Know there is the option of using apply but doing several aggregations is what I want. Below is my test code that I tried to get working for the weighted average.
Python Code
<pre><code>import pandas as pd
import numpy as np

def weighted_avg(df, values, weights):
 '''To calculate a weighted average in Pandas. Demo see https://www.statology.org/pandas-weighted-average/
 Example: df.groupby('Group Names').apply(w_avg, 'Results', 'AFY')'''
 v = df[values]
 w = df[weights]
 return (v * w).sum() / w.sum()

# below creates a dataframe.
dfr = pd.DataFrame(np.random.randint(1,50,size=(4,4)), columns=list('ABCD'))
dfr['group'] = [1, 1, 0, 1]

print(dfr)
dfr = dfr.groupby('group').agg({'A':'mean', 'B':'sum',
 'C': lambda x: weighted_avg(dfr, 'D', 'C')}).reset_index()
print(dfr)
</code></pre>
Results - Output
<pre><code> A B C D group
0 5 2 17 38 1
1 35 30 22 32 1
2 15 18 16 11 0
3 46 6 20 34 1
 group A B C
0 0 15.000000 18 29.413333
1 1 28.666667 38 29.413333
</code></pre>
The problem: The weighted average is returning the value for the whole table and not the 'group' column. How can I get the weighted average by group working?
I did try placing the groupby inside the function like <a href="https://www.geeksforgeeks.org/how-to-calculate-weighted-average-in-pandas/" rel="nofollow noreferrer">shown here</a> but no success.
Thank you for taking a look.

python pandas weighted average with the use of groupby agg()

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。

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问使用groupby agg()对python大熊猫加权平均
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