如果你想保留multiindex第二层的任何聚合信息，你可以试试这个:

In [1]: new_cols = [''.join(t) for t in df.columns]
Out[1]:
['USAF',
 'WBAN',
 'day',
 'month',
 's_CDsum',
 's_CLsum',
 's_CNTsum',
 's_PCsum',
 'tempfamax',
 'tempfamin',
 'year']

In [2]: df.columns = new_cols

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

也许有点晚了，但如果你不担心重复的列名:

df.columns = df.columns.tolist()

df.columns = ['_'.join(tup).rstrip('_') for tup in df.columns.values]

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)

Andy Hayden的答案当然是最简单的方法——如果你想避免重复的列标签，你需要稍微调整一下

In [34]: df
Out[34]: 
     USAF   WBAN  day  month  s_CD  s_CL  s_CNT  s_PC  tempf         year
                               sum   sum    sum   sum   amax   amin      
0  702730  26451    1      1    12     0     13     1  30.92  24.98  1993
1  702730  26451    2      1    13     0     13     0  32.00  24.98  1993
2  702730  26451    3      1     2    10     13     1  23.00   6.98  1993
3  702730  26451    4      1    12     0     13     1  10.04   3.92  1993
4  702730  26451    5      1    10     0     13     3  19.94  10.94  1993


In [35]: mi = df.columns

In [36]: mi
Out[36]: 
MultiIndex
[(USAF, ), (WBAN, ), (day, ), (month, ), (s_CD, sum), (s_CL, sum), (s_CNT, sum), (s_PC, sum), (tempf, amax), (tempf, amin), (year, )]


In [37]: mi.tolist()
Out[37]: 
[('USAF', ''),
 ('WBAN', ''),
 ('day', ''),
 ('month', ''),
 ('s_CD', 'sum'),
 ('s_CL', 'sum'),
 ('s_CNT', 'sum'),
 ('s_PC', 'sum'),
 ('tempf', 'amax'),
 ('tempf', 'amin'),
 ('year', '')]

In [38]: ind = pd.Index([e[0] + e[1] for e in mi.tolist()])

In [39]: ind
Out[39]: Index([USAF, WBAN, day, month, s_CDsum, s_CLsum, s_CNTsum, s_PCsum, tempfamax, tempfamin, year], dtype=object)

In [40]: df.columns = ind




In [46]: df
Out[46]: 
     USAF   WBAN  day  month  s_CDsum  s_CLsum  s_CNTsum  s_PCsum  tempfamax  tempfamin  \
0  702730  26451    1      1       12        0        13        1      30.92      24.98   
1  702730  26451    2      1       13        0        13        0      32.00      24.98   
2  702730  26451    3      1        2       10        13        1      23.00       6.98   
3  702730  26451    4      1       12        0        13        1      10.04       3.92   
4  702730  26451    5      1       10        0        13        3      19.94      10.94   




   year  
0  1993  
1  1993  
2  1993  
3  1993  
4  1993