Rename MultiIndex columns in Pandas

df = pd.DataFrame([[1,2,3], [10,20,30], [100,200,300]])
df.columns = pd.MultiIndex.from_tuples((("a", "b"), ("a", "c"), ("d", "f")))
df

returns

     a         d
b    c    f
0    1    2    3
1   10   20   30
2  100  200  300

and

df.columns.levels[1]

returns

Index([u'b', u'c', u'f'], dtype='object')

I want to rename "f" to "e". According to pandas.MultiIndex.rename I run:

df.columns.rename(["b1", "c1", "f1"], level=1)

But it raises

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
<ipython-input-110-b171a2b5706c> in <module>()
----> 1 df.columns.rename(["b1", "c1", "f1"], level=1)


C:\Users\USERNAME\AppData\Local\Continuum\Miniconda2\lib\site-packages\pandas\indexes\base.pyc in set_names(self, names, level, inplace)
994         if level is not None and not is_list_like(level) and is_list_like(
995                 names):
--> 996             raise TypeError("Names must be a string")
997
998         if not is_list_like(names) and level is None and self.nlevels > 1:


TypeError: Names must be a string

I use Python 2.7.12 |Continuum Analytics, Inc.| (default, Jun 29 2016, 11:07:13) [MSC v.1500 64 bit (AMD64)]' and pandas 0.19.1

111690 次浏览
小开
最佳答案

Use set_levels:

In [22]:
df.columns.set_levels(['b1','c1','f1'],level=1,inplace=True)
df


Out[22]:
a         d
b1   c1   f1
0    1    2    3
1   10   20   30
2  100  200  300

rename sets the name for the index, it doesn't rename the column names:

In [26]:
df.columns = df.columns.rename("b1", level=1)
df


Out[26]:
a         d
b1    b    c    f
0     1    2    3
1    10   20   30
2   100  200  300

This is why you get the error

小开

In pandas 0.21.0+ use parameter level=1:

d = dict(zip(df.columns.levels[1], ["b1", "c1", "f1"]))
print (d)
{'c': 'c1', 'b': 'b1', 'f': 'f1'}


df = df.rename(columns=d, level=1)
print (df)
a         d
b1   c1   f1
0    1    2    3
1   10   20   30
2  100  200  300
小开

There is also index.set_names (code)

df.index.set_names(["b1", "c1", "f1"], inplace=True)
小开

Another thing you can't do is df.rename(columns={('d', 'f'): ('e', 'g')}), even though it seems correct. In other words: .rename() does not do what one expects, <...>

-- Lukas at comment

The "hacky" way is something like this (as far as for pandas 1.0.5)

def rename_columns(df, columns, inplace=False):
"""Rename dataframe columns.


Parameters
----------
df : pandas.DataFrame
Dataframe.
columns : dict-like
Alternative to specifying axis. If `df.columns` is
:obj: `pandas.MultiIndex`-object and has a few levels, pass equal-size tuples.


Returns
-------
pandas.DataFrame or None
Returns dataframe with modifed columns or ``None`` (depends on `inplace` parameter value).
    

Examples
--------
>>> columns = pd.Index([1, 2, 3])
>>> df = pd.DataFrame([[1, 2, 3], [10, 20, 30]], columns=columns)
...     1   2   3
... 0   1   2   3
... 1  10  20  30
>>> rename_columns(df, columns={1 : 10})
...    10   2   3
... 0   1   2   3
... 1  10  20  30
    

MultiIndex
    

>>> columns = pd.MultiIndex.from_tuples([("A0", "B0", "C0"), ("A1", "B1", "C1"), ("A2", "B2", "")])
>>> df = pd.DataFrame([[1, 2, 3], [10, 20, 30]], columns=columns)
>>> df
...    A0  A1  A2
...    B0  B1  B2
...    C0  C1
... 0   1   2   3
... 1  10  20  30
>>> rename_columns(df, columns={("A2", "B2", "") : ("A3", "B3", "")})
...    A0  A1  A3
...    B0  B1  B3
...    C0  C1
... 0   1   2   3
... 1  10  20  30
"""
columns_new = []
for col in df.columns.values:
if col in columns:
columns_new.append(columns[col])
else:
columns_new.append(col)
columns_new = pd.Index(columns_new, tupleize_cols=True)


if inplace:
df.columns = columns_new
else:
df_new = df.copy()
df_new.columns = columns_new
return df_new

So just

>>> df = pd.DataFrame([[1,2,3], [10,20,30], [100,200,300]])
>>> df.columns = pd.MultiIndex.from_tuples((("a", "b"), ("a", "c"), ("d", "f")))
>>> rename_columns(df, columns={('d', 'f'): ('e', 'g')})
...      a         e
...      b    c    g
... 0    1    2    3
... 1   10   20   30
... 2  100  200  300

What does the pandas-team think about this? Why is this behavior not default?

小开

You can use pandas.DataFrame.rename() directly

Say you have the following dataframe

print(df)


a         d
b    c    f
0    1    2    3
1   10   20   30
2  100  200  300

df = df.rename(columns={'f': 'f1', 'd': 'd1'})

print(df)


a        d1
b    c   f1
0    1    2    3
1   10   20   30
2  100  200  300

You see, column name mapper doesn't relate with level.

Say you have the following dataframe

     a         d
b    f    f
0    1    2    3
1   10   20   30
2  100  200  300

If you want to rename the f under a, you can do

df.columns = df.columns.values
df.columns = pd.MultiIndex.from_tuples(df.rename(columns={('a', 'f'): ('a', 'af')}))

print(df)


a         d
b   af    f
0    1    2    3
1   10   20   30
2  100  200  300
小开

Using dicts to rename tuples

Since multi-index stores values as tuples, and python dicts accept tuples as keys and values, we can replace them using a dict.

mapping_dict = {("d","f"):("d","e")}


# Dictionary allows using tuples as keys and values
def rename_tuple(tuple_, dict_):
"""Replaces tuple if present in tuple dict"""
if tuple_ in dict_.keys():
return dict_[tuple_]
return tuple_


# Rename chosen elements from list of tuples from df.columns
altered_index_list = [rename_tuple(tuple_,mapping_dict) for tuple_ in df.columns.to_list()]


# Update columns with new renamed columns
df.columns = pd.Index(altered_index_list)

Which returns the intended df

     a         d
b    c    e
0    1    2    3
1   10   20   30
2  100  200  300

Aggregating in a function

This could then be aggregated in a function to simplify things

def rename_multi_index(index,mapper):
"""Renames pandas multi_index"""
return pd.Index([rename_tuple(tuple_,mapper) for tuple_ in index])


# And now simply do
df.columns = rename_multi_index(df.columns,mapping_dict)
小开

Another way to do that is with pandas.Series.map and a lambda function as follows

df.columns = df.columns.map(lambda x: (x[0], "e") if x[1] == "f" else x)


[Out]:
a         d
b    c    e
0    1    2    3
1   10   20   30
2  100  200  300

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