get first and last values in a groupby

This could be on of the easy solution.

df.groupby(level = 0, as_index= False).nth([0,-1])

      X   Y
a a   0   1
  d   6   7
b e   8   9
  g  12  13
c h  14  15
  i  16  17
d j  18  19

Hope this helps. (Y)

Option 1

def first_last(df):
    return df.ix[[0, -1]]

df.groupby(level=0, group_keys=False).apply(first_last)

---

Option 2 - only works if index is unique

idx = df.index.to_series().groupby(level=0).agg(['first', 'last']).stack()
df.loc[idx]

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Option 3 - per notes below, this only makes sense when there are no NAs

I also abused the agg function. The code below works, but is far uglier.

df.reset_index(1).groupby(level=0).agg(['first', 'last']).stack() \
    .set_index('level_1', append=True).reset_index(1, drop=True) \
    .rename_axis([None, None])

---

Note

per @unutbu: agg(['first', 'last']) take the firs non-na values.

I interpreted this as, it must then be necessary to run this column by column. Further, forcing index level=1 to align may not even make sense.

Let's include another test

df = pd.DataFrame(np.arange(20).reshape(10, -1),
                  [list('aaaabbbccd'),
                   list('abcdefghij')],
                  list('XY'))

df.loc[tuple('aa'), 'X'] = np.nan

---

def first_last(df):
    return df.ix[[0, -1]]

df.groupby(level=0, group_keys=False).apply(first_last)

df.reset_index(1).groupby(level=0).agg(['first', 'last']).stack() \
    .set_index('level_1', append=True).reset_index(1, drop=True) \
    .rename_axis([None, None])

Sure enough! This second solution is taking the first valid value in column X. It is now nonsensical to have forced that value to align with the index a.

Please try this:

For last value: df.groupby('Column_name').nth(-1),

For first value: df.groupby('Column_name').nth(0)