I have a DF in the following format:

 col1 col2
ID Date
 1 1993-12-31 4 6
 1994-12-31 8 5
 1995-12-31 4 7
 1996-12-31 3 3
 2 2000-12-31 7 8
 2001-12-31 5 9
 2002-12-31 8 4

And I want to reset the 'Date' index giving the following:

 col1 col2
ID Date
 1 0 4 6
 1 8 5
 2 4 7
 3 3 3
 2 0 7 8
 1 5 9
 2 8 4

I thought simply df.reset_index(level='Date', inplace=True, drop=True) would do it, but it does not.

Using pd.MultiIndex.from_arrays and groupby + cumcount.

df.index = pd.MultiIndex.from_arrays(
 [df.index.get_level_values(0), df.groupby(level=0).cumcount()],
 names=['ID', 'Date'])

df
 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 0 7 8
 1 5 9
 2 8 4

This won't generalise to N levels, but there should be a df.index.set_levels equivalent I'm forgetting...

You can groupby ID, then reset the index on each group using apply:

new_df = (df.groupby(df.index.get_level_values('ID'))
 .apply(lambda x: x.reset_index()).drop(['ID','Date'],1))

new_df.index = new_df.index.rename(['ID','Date'])

>>> new_df
 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 0 7 8
 1 5 9
 2 8 4

Using set_index and cumcount:

tmp = df.reset_index('Date', drop=True)
tmp.set_index(df.groupby(level=0).cumcount().rename('Date'), append=True)

 col1 col2
ID Date
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 0 7 8
 1 5 9
 2 8 4

New Answer

Not as cool as the old answer but I'd rather be accurate than cool.

from collections import defaultdict
from itertools import count
d = defaultdict(count)

lbl = 
for a, *_ in df.index.values:
 lbl.append(next(d[a]))

lvl = pd.RangeIndex(max(lbl) + 1)

df.set_index(df.index.set_labels(lbl, 1).set_levels(lvl, 1))

 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 0 7 8
 1 5 9
 2 8 4

OLD ANSWER

Do Not Use

I misread the question. I didn't see that the new index needed to reset for every group.

Hopefully useful to someone.

You can use pandas.MultiIndex.set_levels

n = 1
lvl = df.index.levels[n]
new_lvl = pd.RangeIndex(len(lvl))
new_idx = df.index.set_levels(new_lvl, n)
df.set_index(new_idx)

 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 4 7 8
 5 5 9
 6 8 4

One-line

Yay! o/

df.set_index(df.index.set_levels(pd.RangeIndex(len(df.index.levels[1])), 1))

 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 4 7 8
 5 5 9
 6 8 4

In place

df.index.set_levels(pd.RangeIndex(len(df.index.levels[1])), 1, inplace=True)
df

 col1 col2
ID Date 
1 0 4 6
 1 8 5
 2 4 7
 3 3 3
2 4 7 8
 5 5 9
 6 8 4