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The former answer is now superseded by .iloc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 NaN
s 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
Combining @comte's answer and dmdip's answer in Get index of a row of a pandas dataframe as an integer
df.tail(1).index.item()
gives you the value of the index.
Note that indices are not always well defined not matter they are multi-indexed or single indexed. Modifying dataframes using indices might result in unexpected behavior. We will have an example with a multi-indexed case but note this is also true in a single-indexed case.
Say we have
df = pd.DataFrame({'x':[1,1,3,3], 'y':[3,3,5,5]}, index=[11,11,12,12]).stack()
11 x 1
y 3
x 1
y 3
12 x 3
y 5 # the index is (12, 'y')
x 3
y 5 # the index is also (12, 'y')
df.tail(1).index.item() # gives (12, 'y')
Trying to access the last element with the index df[12, "y"]
yields
(12, y) 5
(12, y) 5
dtype: int64
If you attempt to modify the dataframe based on the index (12, y)
, you will modify two rows rather than one. Thus, even though we learned to access the value of last row's index, it might not be a good idea if you want to change the values of last row based on its index as there could be many that share the same index. You should use df.iloc[-1]
to access last row in this case though.
Reference
https://pandas.pydata.org/pandas-docs/stable/generated/pandas.Index.item.html
df.tail(1).index
seems the most readable
It may be too late now, I use index
method to retrieve last index of a DataFrame, then use [-1]
to get the last values:
For example,
df = pd.DataFrame(np.zeros((4, 1)), columns=['A'])
print(f'df:\n{df}\n')
print(f'Index = {df.index}\n')
print(f'Last index = {df.index[-1]}')
The output is
df:
A
0 0.0
1 0.0
2 0.0
3 0.0
Index = RangeIndex(start=0, stop=4, step=1)
Last index = 3