blocks|key|1487463|text|我确认了head()的行为在版本0.12和0.13之间发生了变化。在我看来这是个bug。我创建了一个issue。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1487464|但是groupby操作实际上并不返回按组排序的DataFrame。.head()方法在这里有点误导--它只是一个方便的特性，允许您重新检查分组的对象(在本例中为df)。groupby的结果是一个单独类型的对象，一个GroupBy对象。您必须使用apply、transform或filter才能返回到DataFrame或系列。|1487465|如果您想要做的只是按列A中的值进行排序，则应该使用df.sort('A')。|1487466|entityMap|0|LINK|mutability|MUTABLE|url|https://github.com/pydata/pandas/issues/6721^0|4|6|1E|5|0|0|X|7|28|2|2C|7|2Z|7|3E|5|3K|9|3U|6|0|P|C|0^^$0|@$1|2|3|4|5|6|7|R|8|@$9|S|A|T|B|C]]|D|@$9|U|A|V|1|W]]|E|$]]|$1|F|3|G|5|6|7|X|8|@$9|Y|A|Z|B|C]|$9|10|A|11|B|C]|$9|12|A|13|B|C]|$9|14|A|15|B|C]|$9|16|A|17|B|C]|$9|18|A|19|B|C]|$9|1A|A|1B|B|C]]|D|@]|E|$]]|$1|H|3|I|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|J|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|K|$L|$5|M|N|O|E|$P|Q]]]]

I confirmed that the behavior of <code>head()</code> changes between version 0.12 and 0.13. That looks like a bug to me. I created an <a href="https://github.com/pydata/pandas/issues/6721" rel="noreferrer">issue</a>.

But a groupby operation doesn't actually return a DataFrame sorted by group. The <code>.head()</code> method is a little misleading here -- it's just a convenience feature to let you re-examine the object (in this case, <code>df</code>) that you grouped. The result of <code>groupby</code> is separate kind of object, a <code>GroupBy</code> object. You must <code>apply</code>, <code>transform</code>, or <code>filter</code> to get back to a DataFrame or Series.

If all you wanted to do was sort by the values in columns A, you should use <code>df.sort('A')</code>.

blocks|key|1738132|text|只需执行以下操作：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738133|grouped_df+=+df.groupby('A')

for+key,+item+in+grouped_df:
++++print(grouped_df.get_group(key),+"\n\n")|code-block|syntax|javascript|1738134|1738135|1738136|+弃用通知：ix是deprecated+in+0.20.0|blockquote|offset|length|style|BOLD|CODE|1738137|1738138|1738139|这也同样有效，|1738140|grouped_df+=+df.groupby('A')++++
gb+=+grouped_df.groups

for+key,+values+in+gb.iteritems():
++++print(df.ix[values],+"\n\n")|1738141|用于选择性键分组的：使用gb.keys()在key_list_from_gb中插入所需的键:例如，|1738142|gb+=+grouped_df.groups
gb.keys()

key_list_from_gb+=+[key1,+key2,+key3]

for+key,+values+in+gb.items():
++++if+key+in+key_list_from_gb:
++++++++print(df.ix[values],+"\n")|1738143|entityMap|0|LINK|mutability|MUTABLE|url|https://pandas.pydata.org/docs/whatsnew/v0.20.0.html#deprecate-ix^0|0|0|0|0|1|4|6|2|9|K|0|0|0|0|0|0|9|1|C|9|M|G|0|0^^$0|@$1|2|3|4|5|6|7|18|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|19|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|1A|8|@]|9|@]|A|$]]|$1|H|3|-4|5|6|7|1B|8|@]|9|@]|A|$]]|$1|I|3|J|5|K|7|1C|8|@$L|1D|M|1E|N|O]|$L|1F|M|1G|N|P]]|9|@$L|1H|M|1I|1|1J]]|A|$]]|$1|Q|3|-4|5|6|7|1K|8|@]|9|@]|A|$]]|$1|R|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|S|3|T|5|6|7|1M|8|@]|9|@]|A|$]]|$1|U|3|V|5|D|7|1N|8|@]|9|@]|A|$E|F]]|$1|W|3|X|5|6|7|1O|8|@$L|1P|M|1Q|N|O]|$L|1R|M|1S|N|P]|$L|1T|M|1U|N|P]]|9|@]|A|$]]|$1|Y|3|Z|5|D|7|1V|8|@]|9|@]|A|$E|F]]|$1|10|3|-4|5|6|7|1W|8|@]|9|@]|A|$]]]|11|$12|$5|13|14|15|A|$16|17]]]]

Simply do:
<pre><code>grouped_df = df.groupby('A')

for key, item in grouped_df:
 print(grouped_df.get_group(key), &quot;\n\n&quot;)
</code></pre>
<hr />
<blockquote>
Deprecation Notice: <code>ix</code> was <a href="https://pandas.pydata.org/docs/whatsnew/v0.20.0.html#deprecate-ix" rel="nofollow noreferrer">deprecated in 0.20.0</a>
</blockquote>
This also works,
<pre><code>grouped_df = df.groupby('A') 
gb = grouped_df.groups

for key, values in gb.iteritems():
 print(df.ix[values], &quot;\n\n&quot;)
</code></pre>
For selective key grouping: Insert the keys you want inside the <code>key_list_from_gb</code>, in following, using <code>gb.keys()</code>: For Example,
<pre><code>gb = grouped_df.groups
gb.keys()

key_list_from_gb = [key1, key2, key3]

for key, values in gb.items():
 if key in key_list_from_gb:
 print(df.ix[values], &quot;\n&quot;)
</code></pre>

blocks|key|1738164|text|此外，其他简单的替代方案可以是：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738165|gb+=+df.groupby("A")
gb.count()+#+or,
gb.get_group(your_key)|code-block|syntax|javascript|1738166|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Also, other simple alternative could be:

<pre><code>gb = df.groupby("A")
gb.count() # or,
gb.get_group(your_key)
</code></pre>

blocks|key|5265364|text|如果您只是在寻找一种显示它的方法，您可以使用describe()：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5265365|grp+=+df.groupby['colName']
grp.describe()|code-block|syntax|javascript|5265366|这会给你一张整齐的桌子。|5265367|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|$]]

If you're simply looking for a way to display it, you could use describe():

<pre><code>grp = df.groupby['colName']
grp.describe()
</code></pre>

This gives you a neat table.

blocks|key|3847845|text|另一个简单的选择：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3847846|for+name_of_the_group,+group+in+grouped_dataframe:
+++print+(name_of_the_group)
+++print+(group)|code-block|syntax|javascript|3847847|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Another simple alternative: 

<pre><code>for name_of_the_group, group in grouped_dataframe:
 print (name_of_the_group)
 print (group)
</code></pre>

blocks|key|3847865|text|感谢Surya的深刻见解。我会整理他的解决方案，简单地这样做：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3847866|for+key,+value+in+df.groupby('A'):
++++print(key,+value)|code-block|syntax|javascript|3847867|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Thanks to Surya for good insights. I'd clean up his solution and simply do:

<pre><code>for key, value in df.groupby('A'):
 print(key, value)
</code></pre>

blocks|key|3847885|text|我发现了一种巧妙的方法，仅供头脑风暴使用，请参阅代码：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3847886|df['a']+=+df['A']++#+create+a+shadow+column+for+MultiIndexing
df.sort_values('A',+inplace=True)
df.set_index(["A","a"],+inplace=True)
print(df)|code-block|syntax|javascript|3847887|输出：|3847888|+++++++++++++B
A+++++a
one+++one++++0
++++++one++++1
++++++one++++5
three+three++3
++++++three++4
two+++two++++2|3847889|pros非常容易打印，因为它返回一个dataframe，而不是Groupby对象。而且输出看起来很不错。缺点是它创建了一系列冗余数据。|3847890|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|S|8|@]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

I found a tricky way, just for brainstorm, see the code:

<pre><code>df['a'] = df['A'] # create a shadow column for MultiIndexing
df.sort_values('A', inplace=True)
df.set_index(["A","a"], inplace=True)
print(df)
</code></pre>

the output:

<pre><code> B
A a
one one 0
 one 1
 one 5
three three 3
 three 4
two two 2
</code></pre>

The pros is so easy to print, as it returns a dataframe, instead of Groupby Object. And the output looks nice.
While the con is that it create a series of redundant data.

blocks|key|3847922|text|不能直接通过print语句查看groupBy数据，但是可以通过迭代group+using+for循环来查看group+by数据|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3847923|group+=+df.groupby('A')+#group+variable+contains+groupby+data
for+A,A_df+in+group:+#+A+is+your+column+and+A_df+is+group+of+one+kind+at+a+time
++print(A)
++print(A_df)|code-block|syntax|javascript|3847924|在尝试将其作为groupby结果后，您将获得输出|3847925|我希望它能帮上忙|3847926|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]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|P|8|@]|9|@]|A|$]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

you cannot see the groupBy data directly by print statement but you can see by iterating over the group using for loop 
try this code to see the group by data

<pre><code>group = df.groupby('A') #group variable contains groupby data
for A,A_df in group: # A is your column and A_df is group of one kind at a time
 print(A)
 print(A_df)
</code></pre>

you will get an output after trying this as a groupby result

I hope it helps

blocks|key|5265526|text|GroupBy对象上的调用列表()|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5265527|print(list(df.groupby('A')))|code-block|syntax|javascript|5265528|为您提供：|5265529|[('one',++++++A++B
0++one++0
1++one++1
5++one++5),+('three',++++++++A++B
3++three++3
4++three++4),+('two',++++++A++B
2++two++2)]|5265530|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|$]]

Call list() on the GroupBy object

<pre><code>print(list(df.groupby('A')))
</code></pre>

gives you:

<pre><code>[('one', A B
0 one 0
1 one 1
5 one 5), ('three', A B
3 three 3
4 three 4), ('two', A B
2 two 2)]
</code></pre>

blocks|key|3847995|text|在python+3中|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3847996|k+=+None
for+name_of_the_group,+group+in+dict(df_group):
++++if(k+!=+name_of_the_group):
++++++++print+('\n',+name_of_the_group)
++++++++print('..........','\n')
++++print+(group)
++++k+=+name_of_the_group|code-block|syntax|javascript|3847997|以更具交互性的方式|3847998|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|$]]

In python 3

<pre><code>k = None
for name_of_the_group, group in dict(df_group):
 if(k != name_of_the_group):
 print ('\n', name_of_the_group)
 print('..........','\n')
 print (group)
 k = name_of_the_group
</code></pre>

In more interactive way

blocks|key|5265595|text|除了前面的答案之外：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5265596|以你的例子为例，|5265597|df+=+pd.DataFrame({'A':+['one',+'one',+'two',+'three',+'three',+'one'],+'B':+range(6)})|code-block|syntax|javascript|5265598|然后是简单的单行代码|5265599|df.groupby('A').apply(print)|5265600|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Q|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

In addition to previous answers:

Taking your example,

<pre><code>df = pd.DataFrame({'A': ['one', 'one', 'two', 'three', 'three', 'one'], 'B': range(6)})
</code></pre>

Then simple 1 line code

<pre><code>df.groupby('A').apply(print)
</code></pre>

blocks|key|5265602|text|要打印分组的df的所有(或任意多个)行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5265603|import+pandas+as+pd
pd.set_option('display.max_rows',+500)

grouped_df+=+df.group(['var1',+'var2'])
print(grouped_df)|code-block|syntax|javascript|5265604|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

to print all (or arbitrarily many) lines of the grouped df: 

<pre><code>import pandas as pd
pd.set_option('display.max_rows', 500)

grouped_df = df.group(['var1', 'var2'])
print(grouped_df)
</code></pre>

blocks|key|5265625|text|在Jupyter+Notebook中，如果您执行以下操作，它将打印对象的一个很好的分组版本。apply方法有助于创建多索引数据帧。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5265626|by+=+'A'++#+groupby+'by'+argument
df.groupby(by).apply(lambda+a:+a[:])|code-block|syntax|javascript|5265627|输出：|5265628|+++++++++++++A++B
A++++++++++++++++
one+++0++++one++0
++++++1++++one++1
++++++5++++one++5
three+3++three++3
++++++4++three++4
two+++2++++two++2|5265629|如果希望by列不出现在输出中，只需删除这些列，如下所示。|5265630|df.groupby(by).apply(lambda+a:+a.drop(by,+axis=1)[:])|5265631|5265632|+++++++++B
A+++++++++
one+++0++0
++++++1++1
++++++5++5
three+3++3
++++++4++4
two+++2++2|5265633|在这里，我不确定为什么在最后.iloc[:]不能工作而不能用[:]。因此，如果将来(或目前)由于更新而出现一些问题，.iloc[:len(a)]也可以工作。|5265634|entityMap^0|1A|5|0|0|0|0|4|2|0|0|0|0|E|8|U|3|1M|E|0^^$0|@$1|2|3|4|5|6|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|12|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|13|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|14|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|15|8|@$9|16|A|17|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|H|7|18|8|@]|D|@]|E|$I|J]]|$1|S|3|L|5|6|7|19|8|@]|D|@]|E|$]]|$1|T|3|U|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|V|3|W|5|6|7|1B|8|@$9|1C|A|1D|B|C]|$9|1E|A|1F|B|C]|$9|1G|A|1H|B|C]]|D|@]|E|$]]|$1|X|3|-4|5|6|7|1I|8|@]|D|@]|E|$]]]|Y|$]]

In Jupyter Notebook, if you do the following, it prints a nice grouped version of the object. The <code>apply</code> method helps in creation of a multiindex dataframe.
<pre><code>by = 'A' # groupby 'by' argument
df.groupby(by).apply(lambda a: a[:])
</code></pre>
Output:
<pre><code> A B
A 
one 0 one 0
 1 one 1
 5 one 5
three 3 three 3
 4 three 4
two 2 two 2
</code></pre>
If you want the <code>by</code> column(s) to not appear in the output, just drop the column(s), like so.
<pre><code>df.groupby(by).apply(lambda a: a.drop(by, axis=1)[:])
</code></pre>
Output:
<pre><code> B
A 
one 0 0
 1 1
 5 5
three 3 3
 4 4
two 2 2
</code></pre>
Here, I am not sure as to why <code>.iloc[:]</code> does not work instead of <code>[:]</code> at the end. So, if there are some issues in future due to updates (or at present), <code>.iloc[:len(a)]</code> also works.

blocks|key|1492006|text|df.groupby('key+you+want+to+group+by').apply(print)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1492007|正如其他成员所提到的，这是可视化groupby对象的最简单和最简单的解决方案。|unstyled|1492008|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>df.groupby('key you want to group by').apply(print)
</code></pre>
As mentioned by an other member, this is the easiest and simplest solution to visualize a groupby object.

blocks|key|5265701|text|这是一个更好的通用答案。此功能将打印所有组名称和值，或选择一个或多个组进行显示。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|5265702|def+print_pd_groupby(X,+grp=None):
++++'''Display+contents+of+a+Panda+groupby+object
++++:param+X:+Pandas+groupby+object
++++:param+grp:+a+list+with+one+or+more+group+names
++++'''
++++if+grp+is+None:
++++++++for+k,i+in+X:
++++++++++++print("group:",+k)
++++++++++++print(i)
++++else:
++++++++for+j+in+grp:
++++++++++++print("group:",+j)
++++++++++++print(X.get_group(j))|code-block|syntax|javascript|5265703|在您的示例中，以下是会话输出|5265704|In+[116]:+df+=+pd.DataFrame({'A':+['one',+'one',+'two',+'three',+'three',+'one'],+'B':+range(6)})

In+[117]:+dfg+=+df.groupby('A')

In+[118]:+print_pd_groupby(dfg)
group:+one
+++++A++B
0++one++0
1++one++1
5++one++5
group:+three
+++++++A++B
3++three++3
4++three++4
group:+two
+++++A++B
2++two++2

In+[119]:+print_pd_groupby(dfg,+grp+=+["one",+"two"])
group:+one
+++++A++B
0++one++0
1++one++1
5++one++5
group:+two
+++++A++B
2++two++2|5265705|这是一个更好的答案，因为函数是可重用的内容，将其放入包或函数集合中，再也不会重写这种“脚本”方法。|5265706|天哪，像这样的东西应该是Pandas+groupby中的内置方法。|5265707|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

This is a better general purpose answer. This function will print all group names and values, or optionally selects one or more groups for display.
<pre><code>def print_pd_groupby(X, grp=None):
 '''Display contents of a Panda groupby object
 :param X: Pandas groupby object
 :param grp: a list with one or more group names
 '''
 if grp is None:
 for k,i in X:
 print(&quot;group:&quot;, k)
 print(i)
 else:
 for j in grp:
 print(&quot;group:&quot;, j)
 print(X.get_group(j))
</code></pre>
In your example case, here's session output
<pre><code>In [116]: df = pd.DataFrame({'A': ['one', 'one', 'two', 'three', 'three', 'one'], 'B': range(6)})

In [117]: dfg = df.groupby('A')

In [118]: print_pd_groupby(dfg)
group: one
 A B
0 one 0
1 one 1
5 one 5
group: three
 A B
3 three 3
4 three 4
group: two
 A B
2 two 2

In [119]: print_pd_groupby(dfg, grp = [&quot;one&quot;, &quot;two&quot;])
group: one
 A B
0 one 0
1 one 1
5 one 5
group: two
 A B
2 two 2
</code></pre>
This is a better answer because a function is re-usable content, put it in your package or function collection and never re-write that &quot;scriptish&quot; approach again.
IMHO, something like this should be a built in method in Pandas groupby.

blocks|key|1738554|text|df.groupby('A').apply(display)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1738555|提供整洁的格式|unstyled|1738556|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>df.groupby('A').apply(display)
</code></pre>
gives neat formatting

blocks|key|1738569|text|您只需要将DataFrameGroupBy对象转换为list，您就可以简单地打印它。ls_grouped_df+=列表(df.groupby(‘A’))打印(Ls_grouped_df)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1738570|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

you just need to convert the DataFrameGroupBy object to list and you can simply print it..
ls_grouped_df = list(df.groupby('A'))
print(ls_grouped_df)

blocks|key|3848123|text|使用get_group()方法可以得到如下所示的结果|type|unstyled|depth|inlineStyleRanges|entityRanges|data|3848124|new_group+=+df.groupby(['A'])
get_group('')|code-block|syntax|javascript|3848125|将要获取的组的名称放入该方法中|3848126|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|$]]

use the get_group() method
you can have something like this
<pre><code>new_group = df.groupby(['A'])
get_group('')
</code></pre>
put the name of the group you want to get inside the method

I want to print the result of grouping with Pandas.

I have a dataframe:

<pre><code>import pandas as pd
df = pd.DataFrame({'A': ['one', 'one', 'two', 'three', 'three', 'one'], 'B': range(6)})
print(df)

 A B
0 one 0
1 one 1
2 two 2
3 three 3
4 three 4
5 one 5
</code></pre>

When printing after grouping by 'A' I have the following:

<pre><code>print(df.groupby('A'))

&lt;pandas.core.groupby.DataFrameGroupBy object at 0x05416E90&gt;
</code></pre>

How can I print the dataframe grouped?

If I do:

<pre><code>print(df.groupby('A').head())
</code></pre>

I obtain the dataframe as if it was not grouped:

<pre><code> A B
A 
one 0 one 0
 1 one 1
two 2 two 2
three 3 three 3
 4 three 4
one 5 one 5
</code></pre>

I was expecting something like:

<pre><code> A B
A 
one 0 one 0
 1 one 1
 5 one 5
two 2 two 2
three 3 three 3
 4 three 4
</code></pre>

How to print a groupby object

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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我想打印Pandas分组的结果。我有一个数据框架：import pandas as pddf = pd.DataFrame({'A': ['one', 'one', 'two', 'three', 'three', 'one'], 'B': range(6)})print(df)       A  B0    one ...

问如何打印groupby对象
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问如何打印groupby对象EN