Access index of last element in data frame

The former answer is now superseded by .iloc:

>>> df = pd.DataFrame({"date": range(10, 64, 8)})
>>> df.index += 17
>>> df
    date
17    10
18    18
19    26
20    34
21    42
22    50
23    58
>>> df["date"].iloc[0]
10
>>> df["date"].iloc[-1]
58

The shortest way I can think of uses .iget():

>>> df = pd.DataFrame({"date": range(10, 64, 8)})
>>> df.index += 17
>>> df
    date
17    10
18    18
19    26
20    34
21    42
22    50
23    58
>>> df['date'].iget(0)
10
>>> df['date'].iget(-1)
58

Alternatively:

>>> df['date'][df.index[0]]
10
>>> df['date'][df.index[-1]]
58

There’s also .first_valid_index() and .last_valid_index(), but depending on whether or not you want to rule out NaNs they might not be what you want.

Remember that df.ix[0] doesn’t give you the first, but the one indexed by 0. For example, in the above case, df.ix[0] would produce

>>> df.ix[0]
Traceback (most recent call last):
  File "<ipython-input-489-494245247e87>", line 1, in <module>
    df.ix[0]
[...]
KeyError: 0