pandas.groupby 的 group_keys 参数实际上是做什么的？
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What does the group_keys argument to pandas.groupby actually do?
提问by Paul
In pandas.DataFrame.groupby, there is an argument group_keys, which I gather is supposed to do something relating to how group keys are included in the dataframe subsets. According to the documentation:
在 中pandas.DataFrame.groupby，有一个参数group_keys，我收集它应该做一些与数据帧子集中如何包含组键相关的事情。根据文档：
group_keys: boolean, default True
When calling apply, add group keys to index to identify pieces
group_keys:布尔值，默认为 True
调用apply时，将组键添加到索引以识别碎片
However, I can't really find any examples where group_keysmakes an actual difference:
但是，我真的找不到任何group_keys有实际区别的例子：
import pandas as pd
df = pd.DataFrame([[0, 1, 3],
[3, 1, 1],
[3, 0, 0],
[2, 3, 3],
[2, 1, 0]], columns=list('xyz'))
gby = df.groupby('x')
gby_k = df.groupby('x', group_keys=False)
It doesn't make a difference in the output of apply:
它对以下输出没有影响apply：
ap = gby.apply(pd.DataFrame.sum)
# x y z
# x
# 0 0 1 3
# 2 4 4 3
# 3 6 1 1
ap_k = gby_k.apply(pd.DataFrame.sum)
# x y z
# x
# 0 0 1 3
# 2 4 4 3
# 3 6 1 1
And even if you print out the grouped subsets as you go, the results are still identical:
即使您随时打印出分组的子集，结果仍然相同：
def printer_func(x):
print(x)
return x
print('gby')
print('--------------')
gby.apply(printer_func)
print('--------------')
print('gby_k')
print('--------------')
gby_k.apply(printer_func)
print('--------------')
# gby
# --------------
# x y z
# 0 0 1 3
# x y z
# 0 0 1 3
# x y z
# 3 2 3 3
# 4 2 1 0
# x y z
# 1 3 1 1
# 2 3 0 0
# --------------
# gby_k
# --------------
# x y z
# 0 0 1 3
# x y z
# 0 0 1 3
# x y z
# 3 2 3 3
# 4 2 1 0
# x y z
# 1 3 1 1
# 2 3 0 0
# --------------
I considered the possibility that the default argument is actually True, but switching group_keysto explicitly Falsedoesn't make a difference either. What exactly is this argument for?
我考虑了默认参数实际上是 的可能性True，但切换group_keys到显式False也没有任何区别。这个论点究竟是为了什么？
(Run on pandasversion 0.18.1)
（在pandas版本上运行0.18.1）
Edit:I did find a way where group_keyschanges behavior, based on this answer:
编辑：group_keys根据这个答案，
我确实找到了一种改变行为的方法：
import pandas as pd
import numpy as np
row_idx = pd.MultiIndex.from_product(((0, 1), (2, 3, 4)))
d = pd.DataFrame([[4, 3], [1, 3], [1, 1], [2, 4], [0, 1], [4, 2]], index=row_idx)
df_n = d.groupby(level=0).apply(lambda x: x.nlargest(2, [0]))
# 0 1
# 0 0 2 4 3
# 3 1 3
# 1 1 4 4 2
# 2 2 4
df_k = d.groupby(level=0, group_keys=False).apply(lambda x: x.nlargest(2, [0]))
# 0 1
# 0 2 4 3
# 3 1 3
# 1 4 4 2
# 2 2 4
However, I'm still not clear on the intelligible principle behind what group_keysis supposed to do. This behavior does not seem intuitive based on @piRSquared's answer.
但是，我还没有背后的东西可以理解的原则明确的group_keys是应该做的。根据@piRSquared的回答，这种行为似乎并不直观。
回答by Nickil Maveli
group_keysparameter in groupbycomes handy during applyoperations that creates an additional index column corresponding to the grouped columns[group_keys=True] and eliminates in the case[group_keys=False] especially during the case when trying to perform operations on individual columns.
group_keys参数 ingroupby在apply创建对应于分组列 [ group_keys=True]的附加索引列的操作期间派上用场，并在 case[ group_keys=False] 中消除，尤其是在尝试对单个列执行操作的情况下。
One such instance:
一个这样的例子：
In [21]: gby = df.groupby('x',group_keys=True).apply(lambda row: row['x'])
In [22]: gby
Out[22]:
x
0 0 0
2 3 2
4 2
3 1 3
2 3
Name: x, dtype: int64
In [23]: gby_k = df.groupby('x', group_keys=False).apply(lambda row: row['x'])
In [24]: gby_k
Out[24]:
0 0
3 2
4 2
1 3
2 3
Name: x, dtype: int64
One of it's intended application could be to group by one of the levels of the hierarchy by converting it to a Multi-indexdataframe object.
它的预期应用程序之一可能是通过将其转换为Multi-index数据帧对象来按层次结构的一个级别进行分组。
In [27]: gby.groupby(level='x').sum()
Out[27]:
x
0 0
2 4
3 6
Name: x, dtype: int64
回答by piRSquared
If you are passing a function that preserves an index, pandas tries to keep that information. But if you pass a function that removes all semblance of index information, group_keys=Trueallows you to keep that information.
如果您传递一个保留索引的函数，pandas 会尝试保留该信息。但是，如果您传递一个删除所有索引信息外观的函数，则group_keys=True允许您保留该信息。
Use this instead
改用这个
f = lambda df: df.reset_index(drop=True)
Then the different groupby
那么不同的 groupby
gby.apply(lambda df: df.reset_index(drop=True))
gby_k.apply(lambda df: df.reset_index(drop=True))
