Pandas: slice one multiindex dataframe with multii

I have two multiindexed dataframes, one with two levels and one with three. The first two levels match in both dataframes. I would like to find all values from the first dataframe where the first two index levels match in the second dataframe. The second data frame does not have a third level.

The closest answer I have found is this: How to slice one MultiIndex DataFrame with the MultiIndex of another -- however the setup is slightly different and doesn't seem to translate to this case.

Consider the setup below

array_1 = [np.array(['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux']),
np.array(['one', 'two', 'one', 'two', 'one', 'two', 'one', 'two']),
np.array(['a', 'a','a', 'a','b','b','b','b' ])]

array_2 = [np.array(['bar', 'bar', 'baz', 'baz', 'foo', 'foo', 'qux', 'qux']),
      np.array(['one', 'two', 'three', 'one', 'two', 'two', 'one', 'two'])]

df_1 = pd.DataFrame(np.random.randn(8,4), index=array_1).sort_index()

print df_1
                  0         1         2         3
bar one a  1.092651 -0.325324  1.200960 -0.790002
    two a -0.415263  1.006325 -0.077898  0.642134
baz one a -0.343707  0.474817  0.396702 -0.379066
    two a  0.315192 -1.548431 -0.214253 -1.790330
foo one b  1.022050 -2.791862  0.172165  0.924701
    two b  0.622062 -0.193056 -0.145019  0.763185
qux one b -1.241954 -1.270390  0.147623 -0.301092
    two b  0.778022  1.450522  0.683487 -0.950528

df_2 = pd.DataFrame(np.random.randn(8,4), index=array_2).sort_index()

print df_2

                  0         1         2         3
bar one   -0.354889 -1.283470 -0.977933 -0.601868
    two   -0.849186 -2.455453  0.790439  1.134282
baz one   -0.143299  2.372440 -0.161744  0.919658
    three -1.008426 -0.116167 -0.268608  0.840669
foo two   -0.644028  0.447836 -0.576127 -0.891606
    two   -0.163497 -1.255801 -1.066442  0.624713
qux one   -1.545989 -0.422028 -0.489222 -0.357954
    two   -1.202655  0.736047 -1.084002  0.732150

Now I query the second, dataframe, returning a subset of the original indexes

df_2_selection = df_2[(df_2 > 1).any(axis=1)]
print df_2_selection

                0         1         2         3
bar two -0.849186 -2.455453  0.790439  1.134282
baz one -0.143299  2.372440 -0.161744  0.919658

I would like to find all the values in df_1 that match the indices found in df_2. The first two levels line up, but the third does not.

This problem is easy when the indices line up, and would be solved by something like df_1.loc[df_2_selection.index] #this works if indexes are the same

Also I can find thhe values which match one of the levels with something like df_1[df_1.index.isin(df_2_selection.index.get_level_values(0),level = 0)] but this does not solve the problem.

Chaining these statements together does not provide the desired functionality

df_1[(df_1.index.isin(df_2_selection.index.get_level_values(0),level = 0)) & (df_1.index.isin(df_2_selection.index.get_level_values(1),level = 1))]

I envision something along the lines of:

df_1_select = df_1[(df_1.index.isin(
    df_2_selection.index.get_level_values([0,1]),level = [0,1])) #Doesnt Work

print df_1_select

                  0         1         2         3
bar two a -0.415263  1.006325 -0.077898  0.642134
baz one a -0.343707  0.474817  0.396702 -0.379066

I have tried many other methods, all of which have not worked exactly how I wanted. Thank you for your consideration.

EDIT:

This df_1.loc[pd_idx[df_2_selection.index.get_level_values(0),df_2_selection.index.get_level_values(1),:],:] Also does not work

I want only the rows where both levels match. Not where either level match.

EDIT 2: This solution was posted by someone who has since deleted it

id=[x+([x for x in df_1.index.levels[-1]]) for x in df_2_selection.index.values]

pd.concat([df_1.loc[x] for x in id])

Which indeed does work! However on large dataframes it is prohibitively slow. Any help with new methods / speedup is greatly appreciated.

回答1:

You can use reset_index() and merge().

With df_2_selection as:

                0         1         2         3
foo two -0.530151  0.932007 -1.255259  2.441294
qux one  2.006270  1.087412 -0.840916 -1.225508

Merge with:

lvls = ["level_0","level_1"]

(df_1.reset_index()
 .merge(df_2_selection.reset_index()[lvls], on=lvls)
 .set_index(["level_0","level_1","level_2"])
 .rename_axis([None]*3)
)

Output:

                  0         1         2         3
foo two b -0.112696  0.287421 -0.380692 -0.035471
qux one b  0.658227  0.632667 -0.193224  1.073132

Note: The rename_axis() part just removes the level names, e.g. level_0. It's purely cosmetic, and not necessary to perform the actual matching procedure.

回答2:

Try this:

pd.concat([
    df_1.xs(key, drop_level=False)
    for key in df_2_selection.index.values])