这个帖子上的所有答案都有点过时了。在pandas 0.24.0版本中，.to_flat_index()可以满足您的需要。

来自panda自己的文档:

MultiIndex.to_flat_index () 将MultiIndex转换为包含关卡值的元组索引。

文档中的一个简单例子:

import pandas as pd
print(pd.__version__) # '0.23.4'
index = pd.MultiIndex.from_product(
        [['foo', 'bar'], ['baz', 'qux']],
        names=['a', 'b'])

print(index)
# MultiIndex(levels=[['bar', 'foo'], ['baz', 'qux']],
#           codes=[[1, 1, 0, 0], [0, 1, 0, 1]],
#           names=['a', 'b'])

应用to_flat_index ():

index.to_flat_index()
# Index([('foo', 'baz'), ('foo', 'qux'), ('bar', 'baz'), ('bar', 'qux')], dtype='object')

用它代替现有的熊猫柱

一个你如何在dat上使用它的例子，这是一个带MultiIndex列的DataFrame:

dat = df.loc[:,['name','workshop_period','class_size']].groupby(['name','workshop_period']).describe()
print(dat.columns)
# MultiIndex(levels=[['class_size'], ['count', 'mean', 'std', 'min', '25%', '50%', '75%', 'max']],
#            codes=[[0, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 3, 4, 5, 6, 7]])

dat.columns = dat.columns.to_flat_index()
print(dat.columns)
# Index([('class_size', 'count'),  ('class_size', 'mean'),
#     ('class_size', 'std'),   ('class_size', 'min'),
#     ('class_size', '25%'),   ('class_size', '50%'),
#     ('class_size', '75%'),   ('class_size', 'max')],
#  dtype='object')

就地扁化和重命名

可能值得注意的是，如何将它与一个简单的列表理解(感谢@Skippy和@mmann1123)结合起来连接元素，这样你得到的列名就是简单的字符串，例如用下划线分隔:

dat.columns = ["_".join(a) for a in dat.columns.to_flat_index()]

我认为最简单的方法是将列设置为顶层:

df.columns = df.columns.get_level_values(0)

注意:如果to级别有名称，您也可以通过this访问它，而不是0。

.

如果你想合并/加入你的MultiIndex到一个索引(假设你的列中只有字符串条目)，你可以:

df.columns = [' '.join(col).strip() for col in df.columns.values]

注意:当没有第二个索引时，我们必须去掉空白。

In [11]: [' '.join(col).strip() for col in df.columns.values]
Out[11]: 
['USAF',
 'WBAN',
 'day',
 'month',
 's_CD sum',
 's_CL sum',
 's_CNT sum',
 's_PC sum',
 'tempf amax',
 'tempf amin',
 'year']

在读完所有的答案后，我想到了这个:

def __my_flatten_cols(self, how="_".join, reset_index=True):
    how = (lambda iter: list(iter)[-1]) if how == "last" else how
    self.columns = [how(filter(None, map(str, levels))) for levels in self.columns.values] \
                    if isinstance(self.columns, pd.MultiIndex) else self.columns
    return self.reset_index() if reset_index else self
pd.DataFrame.my_flatten_cols = __my_flatten_cols

用法:

给定一个数据帧:

df = pd.DataFrame({"grouper": ["x","x","y","y"], "val1": [0,2,4,6], 2: [1,3,5,7]}, columns=["grouper", "val1", 2])

  grouper  val1  2
0       x     0  1
1       x     2  3
2       y     4  5
3       y     6  7

Single aggregation method: resulting variables named the same as source: df.groupby(by="grouper").agg("min").my_flatten_cols() Same as df.groupby(by="grouper", as_index=False) or .agg(...).reset_index() ----- before ----- val1 2 grouper ------ after ----- grouper val1 2 0 x 0 1 1 y 4 5 Single source variable, multiple aggregations: resulting variables named after statistics: df.groupby(by="grouper").agg({"val1": [min,max]}).my_flatten_cols("last") Same as a = df.groupby(..).agg(..); a.columns = a.columns.droplevel(0); a.reset_index(). ----- before ----- val1 min max grouper ------ after ----- grouper min max 0 x 0 2 1 y 4 6 Multiple variables, multiple aggregations: resulting variables named (varname)_(statname): df.groupby(by="grouper").agg({"val1": min, 2:[sum, "size"]}).my_flatten_cols() # you can combine the names in other ways too, e.g. use a different delimiter: #df.groupby(by="grouper").agg({"val1": min, 2:[sum, "size"]}).my_flatten_cols(" ".join) Runs a.columns = ["_".join(filter(None, map(str, levels))) for levels in a.columns.values] under the hood (since this form of agg() results in MultiIndex on columns). If you don't have the my_flatten_cols helper, it might be easier to type in the solution suggested by @Seigi: a.columns = ["_".join(t).rstrip("_") for t in a.columns.values], which works similarly in this case (but fails if you have numeric labels on columns) To handle the numeric labels on columns, you could use the solution suggested by @jxstanford and @Nolan Conaway (a.columns = ["_".join(tuple(map(str, t))).rstrip("_") for t in a.columns.values]), but I don't understand why the tuple() call is needed, and I believe rstrip() is only required if some columns have a descriptor like ("colname", "") (which can happen if you reset_index() before trying to fix up .columns) ----- before ----- val1 2 min sum size grouper ------ after ----- grouper val1_min 2_sum 2_size 0 x 0 4 2 1 y 4 12 2 You want to name the resulting variables manually: (this is deprecated since pandas 0.20.0 with no adequate alternative as of 0.23) df.groupby(by="grouper").agg({"val1": {"sum_of_val1": "sum", "count_of_val1": "count"}, 2: {"sum_of_2": "sum", "count_of_2": "count"}}).my_flatten_cols("last") Other suggestions include: setting the columns manually: res.columns = ['A_sum', 'B_sum', 'count'] or .join()ing multiple groupby statements. ----- before ----- val1 2 count_of_val1 sum_of_val1 count_of_2 sum_of_2 grouper ------ after ----- grouper count_of_val1 sum_of_val1 count_of_2 sum_of_2 0 x 2 2 2 4 1 y 2 10 2 12

由helper函数处理的情况

level names can be non-string, e.g. Index pandas DataFrame by column numbers, when column names are integers, so we have to convert with map(str, ..) they can also be empty, so we have to filter(None, ..) for single-level columns (i.e. anything except MultiIndex), columns.values returns the names (str, not tuples) depending on how you used .agg() you may need to keep the bottom-most label for a column or concatenate multiple labels (since I'm new to pandas?) more often than not, I want reset_index() to be able to work with the group-by columns in the regular way, so it does that by default

对我来说，最简单、最直观的解决方案是使用get_level_values组合列名。这可以防止在同一列上进行多个聚合时出现重复的列名:

level_one = df.columns.get_level_values(0).astype(str)
level_two = df.columns.get_level_values(1).astype(str)
df.columns = level_one + level_two

如果您想在列之间使用分隔符，可以这样做。这将返回与Seiji Armstrong对只包含两个索引级别值的列的下划线的接受答案的评论相同的东西:

level_one = df.columns.get_level_values(0).astype(str)
level_two = df.columns.get_level_values(1).astype(str)
column_separator = ['_' if x != '' else '' for x in level_two]
df.columns = level_one + column_separator + level_two

我知道这和上面安迪·海登的答案是一样的，但我认为这种方式更直观，更容易记住(所以我不需要一直引用这个线程)，特别是对于熊猫新手用户。

在您可能有3个列级别的情况下，此方法也更具有可扩展性。

level_one = df.columns.get_level_values(0).astype(str)
level_two = df.columns.get_level_values(1).astype(str)
level_three = df.columns.get_level_values(2).astype(str)
df.columns = level_one + level_two + level_three

在@jxstanford和@tvt173之后，我写了一个快速函数，无论字符串/int列名如何，它都应该做到这一点:

def flatten_cols(df):
    df.columns = [
        '_'.join(tuple(map(str, t))).rstrip('_') 
        for t in df.columns.values
        ]
    return df

最python化的方法是使用map函数。

df.columns = df.columns.map(' '.join).str.strip()

输出打印(df.columns):

Index(['USAF', 'WBAN', 'day', 'month', 's_CD sum', 's_CL sum', 's_CNT sum',
       's_PC sum', 'tempf amax', 'tempf amin', 'year'],
      dtype='object')

使用Python 3.6+和f string更新:

df.columns = [f'{f} {s}' if s != '' else f'{f}' 
              for f, s in df.columns]

print(df.columns)

输出:

Index(['USAF', 'WBAN', 'day', 'month', 's_CD sum', 's_CL sum', 's_CNT sum',
       's_PC sum', 'tempf amax', 'tempf amin', 'year'],
      dtype='object')