另一个简单的程序。

def flatten_columns(df, sep='.'):
    def _remove_empty(column_name):
        return tuple(element for element in column_name if element)
    def _join(column_name):
        return sep.join(column_name)

    new_columns = [_join(_remove_empty(column)) for column in df.columns.values]
    df.columns = new_columns

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

处理多个级别和混合类型的通用解决方案:

df.columns = ['_'.join(tuple(map(str, t))) for t in df.columns.values]

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')

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

如果您想在级别之间的名称中使用分隔符，这个函数可以很好地工作。

def flattenHierarchicalCol(col,sep = '_'):
    if not type(col) is tuple:
        return col
    else:
        new_col = ''
        for leveli,level in enumerate(col):
            if not level == '':
                if not leveli == 0:
                    new_col += sep
                new_col += level
        return new_col

df.columns = df.columns.map(flattenHierarchicalCol)