blocks|key|4134048|text|使用value_counts并返回第一个索引值：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4134049|df["responseCount"]+=+(df.groupby("item")["response"]
++++++++++++++++++++++++.transform(lambda+x:+x.value_counts().index[0]))

print+(df)
+++item+response+responseCount
0+++++1++++++++A+++++++++++++A
1+++++1++++++++A+++++++++++++A
2+++++1++++++++B+++++++++++++A
3+++++2++++++++C+++++++++++++C
4+++++2++++++++C+++++++++++++C|code-block|syntax|javascript|4134050|或collections.Counter.most_common|4134051|from+collections+import+Counter

df["responseCount"]+=+(df.groupby("item")["response"]
+++++++++++++++++++++++++.transform(lambda+x:+Counter(x).most_common(1)[0][0]))

print+(df)
+++item+response+responseCount
0+++++1++++++++A+++++++++++++A
1+++++1++++++++A+++++++++++++A
2+++++1++++++++B+++++++++++++A
3+++++2++++++++C+++++++++++++C
4+++++2++++++++C+++++++++++++C|4134052|编辑：|4134053|问题是一个或多个NaN的唯一组，解决方案是使用if-else筛选|4134054|print+(df)
+++item+response
0+++++1++++++++A
1+++++1++++++++A
2+++++2++++++NaN
3+++++2++++++NaN
4+++++3++++++NaN

def+f(x):
++++s+=+x.value_counts()
++++print+(s)

++++A++++2
++++Name:+1,+dtype:+int64
++++Series([],+Name:+2,+dtype:+int64)
++++Series([],+Name:+3,+dtype:+int64)

++++#return+np.nan+if+s.empty+else+s.index[0]
++++return+np.nan+if+len(s)+==+0+else+s.index[0]

df["responseCount"]+=+df.groupby("item")["response"].transform(f)
print+(df)
+++item+response+responseCount
0+++++1++++++++A+++++++++++++A
1+++++1++++++++A+++++++++++++A
2+++++2++++++NaN+++++++++++NaN
3+++++2++++++NaN+++++++++++NaN
4+++++3++++++NaN+++++++++++NaN|4134055|entityMap|0|LINK|mutability|MUTABLE|url|http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.value_counts.html|1|https://docs.python.org/3.6/library/collections.html#collections.Counter.most_common^0|2|C|2|C|0|0|0|1|V|1|V|1|0|0|0|8|3|N|7|0|0^^$0|@$1|2|3|4|5|6|7|14|8|@$9|15|A|16|B|C]]|D|@$9|17|A|18|1|19]]|E|$]]|$1|F|3|G|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1B|8|@$9|1C|A|1D|B|C]]|D|@$9|1E|A|1F|1|1G]]|E|$]]|$1|M|3|N|5|H|7|1H|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|1I|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1J|8|@$9|1K|A|1L|B|C]|$9|1M|A|1N|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1O|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1P|8|@]|D|@]|E|$]]]|V|$W|$5|X|Y|Z|E|$10|11]]|12|$5|X|Y|Z|E|$10|13]]]]

Use <a href="http://pandas.pydata.org/pandas-docs/stable/generated/pandas.Series.value_counts.html" rel="nofollow noreferrer"><code>value_counts</code></a> and return first index value:

<pre><code>df["responseCount"] = (df.groupby("item")["response"]
 .transform(lambda x: x.value_counts().index[0]))

print (df)
 item response responseCount
0 1 A A
1 1 A A
2 1 B A
3 2 C C
4 2 C C
</code></pre>

Or <a href="https://docs.python.org/3.6/library/collections.html#collections.Counter.most_common" rel="nofollow noreferrer"><code>collections.Counter.most_common</code></a>:

<pre><code>from collections import Counter

df["responseCount"] = (df.groupby("item")["response"]
 .transform(lambda x: Counter(x).most_common(1)[0][0]))

print (df)
 item response responseCount
0 1 A A
1 1 A A
2 1 B A
3 2 C C
4 2 C C
</code></pre>

EDIT:

Problem is with one or multiple <code>NaN</code>s only groups, solution is filter with <code>if-else</code>:

<pre><code>print (df)
 item response
0 1 A
1 1 A
2 2 NaN
3 2 NaN
4 3 NaN

def f(x):
 s = x.value_counts()
 print (s)

 A 2
 Name: 1, dtype: int64
 Series([], Name: 2, dtype: int64)
 Series([], Name: 3, dtype: int64)

 #return np.nan if s.empty else s.index[0]
 return np.nan if len(s) == 0 else s.index[0]

df["responseCount"] = df.groupby("item")["response"].transform(f)
print (df)
 item response responseCount
0 1 A A
1 1 A A
2 2 NaN NaN
3 2 NaN NaN
4 3 NaN NaN
</code></pre>

blocks|key|4133667|text|有一个pd.Series.mode|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4133668|df.groupby('item').response.transform(pd.Series.mode)
Out[28]:+
0++++A
1++++A
2++++A
3++++C
4++++C
Name:+response,+dtype:+object|code-block|syntax|javascript|4133669|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/pandas-docs/version/0.22/generated/pandas.Series.mode.html^0|3|E|3|E|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

There is <a href="https://pandas.pydata.org/pandas-docs/version/0.22/generated/pandas.Series.mode.html" rel="noreferrer"><code>pd.Series.mode</code></a>:

<pre><code>df.groupby('item').response.transform(pd.Series.mode)
Out[28]: 
0 A
1 A
2 A
3 C
4 C
Name: response, dtype: object
</code></pre>

blocks|key|4133693|text|您可以使用标准库中的statistics.mode：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4133694|from+statistics+import+mode

df['mode']+=+df.groupby('item')['response'].transform(mode)

print(df)

+++item+response+mode
0+++++1++++++++A++++A
1+++++1++++++++A++++A
2+++++1++++++++B++++A
3+++++2++++++++C++++C
4+++++2++++++++C++++C|code-block|syntax|javascript|4133695|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/statistics.html#statistics.mode^0|A|F|A|F|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@$9|V|A|W|1|X]]|E|$]]|$1|F|3|G|5|H|7|Y|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|L|$M|$5|N|O|P|E|$Q|R]]]]

You can use <a href="https://docs.python.org/3/library/statistics.html#statistics.mode" rel="nofollow noreferrer"><code>statistics.mode</code></a> from standard library:

<pre><code>from statistics import mode

df['mode'] = df.groupby('item')['response'].transform(mode)

print(df)

 item response mode
0 1 A A
1 1 A A
2 1 B A
3 2 C C
4 2 C C
</code></pre>

I have the following DataFrame:

<pre><code>item response
1 A 
1 A 
1 B 
2 A 
2 A 
</code></pre>

I want to add a column with the most given response for an item. which should result in:

<pre><code>item response mostGivenResponse
1 A A
1 A A 
1 B A 
2 C C
2 C C
</code></pre>

I tried something like this: 

<pre><code>df["responseCount"] = df.groupby(["ItemCode", "Response"])["Response"].transform("count")

df["mostGivenResponse"] = df.groupby(['ItemCode'])['responseCount'].transform(max)
</code></pre>

But mostGivenResponse is now the count of the response in stead of the response itself.

Pandas: get string value with most occurrence in group

我有以下DataFrame：item    response1       A       1       A       1       B       2       A       2       A   我想添加一个列，其中包含针对某项的最多给定响应。这应该会导致：item    response  mostG...

问Pandas:获取组中出现次数最多的字符串值
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