Prepend a level to a pandas MultiIndex

Question

I have a DataFrame with a MultiIndex created after some grouping:

import numpy as np
import pandas as pd
from numpy.random import randn

df = pd.DataFrame({'A' : ['a1', 'a1', 'a2', 'a3'], 
                   'B' : ['b1', 'b2', 'b3', 'b4'], 
                   'Vals' : randn(4)}
                 ).groupby(['A', 'B']).sum()

#            Vals
# A  B           
# a1 b1 -1.632460
#    b2  0.596027
# a2 b3 -0.619130
# a3 b4 -0.002009

How do I prepend a level to the MultiIndex so that I turn it into something like:

#                       Vals
# FirstLevel A  B           
# Foo        a1 b1 -1.632460
#               b2  0.596027
#            a2 b3 -0.619130
#            a3 b4 -0.002009

Answer 1

You can first add it as a normal column and then append it to the current index, so:

df['Firstlevel'] = 'Foo'
df.set_index('Firstlevel', append=True, inplace=True)

And change the order if needed with:

df.reorder_levels(['Firstlevel', 'A', 'B'])

Which results in:

                      Vals
Firstlevel A  B           
Foo        a1 b1  0.871563
              b2  0.494001
           a2 b3 -0.167811
           a3 b4 -1.353409

---

EDIT

Even shorter solution (saves the reordering of levels):

df['Firstlevel'] = 'Foo'
df.set_index(['Firstlevel', df.index], inplace=True)

Or alternatively (use axis=1 for columns):

df = pd.concat([df], keys=["Firstlevel"], axis=0)

Answer 2

Another answer using from_tuples(). This generalizes this previous answer.

key = "Foo"
name = "First"
# If df.index.nlevels > 1:
df.index = pd.MultiIndex.from_tuples(((key, *item) for item in df.index),
                                     names=[name]+df.index.names)
# If df.index.nlevels == 1:
# df.index = pd.MultiIndex.from_tuples(((key, item) for item in df.index),
#                                      names=[name]+df.index.names)

I like this approach because

• it only modifies the index (no unnecessary copy action of the body)
• it works for both axes (row and column indices)
• it still can be written as a one-liner

---

Wrapping the above in a function makes it easier to switch between row and column indexes, and between single-level and multi-level indexes:

def prepend_index_level(index, key, name=None):
    names = index.names
    if index.nlevels==1:
        # Sequence of tuples
        index = ((item,) for item in index)

    tuples_gen = ((key,)+item for item in index)
    return pd.MultiIndex.from_tuples(tuples_gen, names=[name]+names)

df.index = prepend_index_level(df.index, key="Foo", name="First")
df.columns = prepend_index_level(df.columns, key="Bar", name="Top")

# Top               Bar
#                  Vals
# First A  B
# Foo   a1 b1 -0.446066
#          b2 -0.248027
#       a2 b3  0.522357
#       a3 b4  0.404048

---

Finally, the above can be further generalized by inserting the key at any index level:

def insert_index_level(index, key, name=None, level=0):
    def insert_(pos, seq, value):
        seq = list(seq)
        seq.insert(pos, value)
        return tuple(seq)

    names = insert_(level, index.names, name)
    if index.nlevels==1:
        # Sequence of tuples.
        index = ((item,) for item in index)
    
    tuples_gen = (insert_(level, item, key) for item in index)
    return pd.MultiIndex.from_tuples(tuples_gen, names=names)

df.index = insert_index_level(df.index, key="Foo", name="Last", level=2)
df.columns = insert_index_level(df.columns, key="Bar", name="Top", level=0)

# Top              Bar
#                 Vals
# A  B  Last
# a1 b1 Foo  -0.595949
#    b2 Foo  -1.621233
# a2 b3 Foo  -0.748917
# a3 b4 Foo   2.147814

Answer 3

How about building it from scratch with pandas.MultiIndex.from_tuples?

df.index = p.MultiIndex.from_tuples(
    [(nl, A, B) for nl, (A, B) in
        zip(['Foo'] * len(df), df.index)],
    names=['FirstLevel', 'A', 'B'])

Similarly to cxrodger's solution, this is a flexible method and avoids modifying the underlying array for the dataframe.

Answer 4

A nice way to do this in one line using pandas.concat():

import pandas as pd

pd.concat([df], keys=['Foo'], names=['Firstlevel'])

An even shorter way:

pd.concat({'Foo': df}, names=['Firstlevel'])

This can be generalized to many data frames, see the docs.

Answer 5

I made a little function out of cxrodgers answer, which IMHO is the best solution since it works purely on an index, independent of any data frame or series.

There is one fix I added: the to_frame() method will invent new names for index levels that don't have one. As such the new index will have names that don't exist in the old index. I added some code to revert this name-change.

Below is the code, I've used it myself for a while and it seems to work fine. If you find any issues or edge cases, I'd be much obliged to adjust my answer.

import pandas as pd

def _handle_insert_loc(loc: int, n: int) -> int:
    """
    Computes the insert index from the right if loc is negative for a given size of n.
    """
    return n + loc + 1 if loc < 0 else loc


def add_index_level(old_index: pd.Index, value: Any, name: str = None, loc: int = 0) -> pd.MultiIndex:
    """
    Expand a (multi)index by adding a level to it.

    :param old_index: The index to expand
    :param name: The name of the new index level
    :param value: Scalar or list-like, the values of the new index level
    :param loc: Where to insert the level in the index, 0 is at the front, negative values count back from the rear end
    :return: A new multi-index with the new level added
    """
    loc = _handle_insert_loc(loc, len(old_index.names))
    old_index_df = old_index.to_frame()
    old_index_df.insert(loc, name, value)
    new_index_names = list(old_index.names)  # sometimes new index level names are invented when converting to a df,
    new_index_names.insert(loc, name)        # here the original names are reconstructed
    new_index = pd.MultiIndex.from_frame(old_index_df, names=new_index_names)
    return new_index

It passed the following unittest code:

import unittest

import numpy as np
import pandas as pd

class TestPandaStuff(unittest.TestCase):

    def test_add_index_level(self):
        df = pd.DataFrame(data=np.random.normal(size=(6, 3)))
        i1 = add_index_level(df.index, "foo")

        # it does not invent new index names where there are missing
        self.assertEqual([None, None], i1.names)

        # the new level values are added
        self.assertTrue(np.all(i1.get_level_values(0) == "foo"))
        self.assertTrue(np.all(i1.get_level_values(1) == df.index))

        # it does not invent new index names where there are missing
        i2 = add_index_level(i1, ["x", "y"]*3, name="xy", loc=2)
        i3 = add_index_level(i2, ["a", "b", "c"]*2, name="abc", loc=-1)
        self.assertEqual([None, None, "xy", "abc"], i3.names)

        # the new level values are added
        self.assertTrue(np.all(i3.get_level_values(0) == "foo"))
        self.assertTrue(np.all(i3.get_level_values(1) == df.index))
        self.assertTrue(np.all(i3.get_level_values(2) == ["x", "y"]*3))
        self.assertTrue(np.all(i3.get_level_values(3) == ["a", "b", "c"]*2))

        # df.index = i3
        # print()
        # print(df)

Answer 6

I think this is a more general solution:

# Convert index to dataframe
old_idx = df.index.to_frame()

# Insert new level at specified location
old_idx.insert(0, 'new_level_name', new_level_values)

# Convert back to MultiIndex
df.index = pandas.MultiIndex.from_frame(old_idx)

Some advantages over the other answers:

• The new level can be added at any location, not just the top.
• It is purely a manipulation on the index and doesn't require manipulating the data, like the concatenation trick.
• It doesn't require adding a column as an intermediate step, which can break multi-level column indexes.