Dataframe Slice does not remove Index Values

I understand your concern, but I believe you have to see what is happening in pandas low-level application.

First, we must declare that indexes are supposed to be immutable. You can check more of its documentation here -> http://pandas.pydata.org/pandas-docs/stable/indexing.html#setting-metadata

When you create a dataframe object, let’s name it df and you want to access its rows, basically all you do is passing a boolean series that Pandas will match with its corresponding index.

Follow this example:

index = pd.MultiIndex.from_product([['A','B'],[5,6]])
df = pd.DataFrame(data=np.random.randint(1,100,(4)), index=index, columns=["P"])

      P
A 5   5
  6  51
B 5  93
  6  76

Now, let’s say we want to select the rows with P > 90. How would you do that? df[df["P"] > 90], right? But look at what df[“P”] > 90 actually returns.

A  5     True
   6     True
B  5     True
   6    False
Name: P, dtype: bool

As you can see, it returns a boolean series matching the original index. Why? Because Pandas needs to map which index values have an equivalent true value, so it can select the proper outcome. So basically, during your slice opperations, you will always carry this index, because it is a mapping element for the object.

However, hope is not gone. Depending on your application, if you believe it is actually taking a huge portion of your memory, you can spend a little time doing the following:

def df_sliced_index(df):
    new_index = []
    rows = []
    for ind, row in df.iterrows():
        new_index.append(ind)
        rows.append(row)
    return pd.DataFrame(data=rows, index=pd.MultiIndex.from_tuples(new_index))

df_sliced_index(df[df['P'] > 90]).index

Which yields what I believe, is the desired output:

MultiIndex(levels=[[u'B'], [5]], labels=[[0], [0]])

But if data is too large to worry you about the size of index, I wonder how much it may cost you in terms of time.