How to access the last element in a Pandas series
Question:
Let us consider the following data frame:
import pandas as pd
d = {'col1': [1, 2, 3], 'col2': [3, 4, 5]}
df=pd.DataFrame(data=d)
If I want to access the first element in pandas series df['col1'], I can simply go df['col1'][0].
But how can I access the last element in this series?
I have tried df['col1'][-1] which returns the following error:
KeyError: -1L
I know that I could go for something like df['col1'][len(df)-1] but why is reverse indexing impossible here?
Answers:
For select last value need Series.iloc or Series.iat, because df['col1'] return Series:
print (df['col1'].iloc[-1])
3
print (df['col1'].iat[-1])
3
Or convert Series to numpy array and select last:
print (df['col1'].values[-1])
3
Or use DataFrame.iloc or DataFrame.iat – but is necessary position of column by Index.get_loc:
print (df.iloc[-1, df.columns.get_loc('col1')])
3
print (df.iat[-1, df.columns.get_loc('col1')])
3
Or is possible use last value of index (necessary not duplicated) and select by DataFrame.loc:
print (df.loc[df.index[-1], 'col1'])
3
You can also use tail:
print(df['col1'].tail(1).item())
Output:
3
To take last through values is the most quick way:
%timeit df['code'].values[-1]. )
5.58 µs ± 985 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df.loc[df.index[-1], 'code']
12 µs ± 2.71 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df['code'].iat[-1]
5.71 µs ± 896 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df['code'].tail(1).item()
36 µs ± 3.23 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit df.iloc[-1, df.columns.get_loc('code')]
33.7 µs ± 5.23 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit df['code'].iloc[-1]
8.08 µs ± 496 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Alternatively you can use take:
In [6]: df['col1'].take([-1]).item()
Out[6]: 3
You can also use:
df.iloc[-1, 0]
Let us consider the following data frame:
import pandas as pd
d = {'col1': [1, 2, 3], 'col2': [3, 4, 5]}
df=pd.DataFrame(data=d)
If I want to access the first element in pandas series df['col1'], I can simply go df['col1'][0].
But how can I access the last element in this series?
I have tried df['col1'][-1] which returns the following error:
KeyError: -1L
I know that I could go for something like df['col1'][len(df)-1] but why is reverse indexing impossible here?
For select last value need Series.iloc or Series.iat, because df['col1'] return Series:
print (df['col1'].iloc[-1])
3
print (df['col1'].iat[-1])
3
Or convert Series to numpy array and select last:
print (df['col1'].values[-1])
3
Or use DataFrame.iloc or DataFrame.iat – but is necessary position of column by Index.get_loc:
print (df.iloc[-1, df.columns.get_loc('col1')])
3
print (df.iat[-1, df.columns.get_loc('col1')])
3
Or is possible use last value of index (necessary not duplicated) and select by DataFrame.loc:
print (df.loc[df.index[-1], 'col1'])
3
You can also use tail:
print(df['col1'].tail(1).item())
Output:
3
To take last through values is the most quick way:
%timeit df['code'].values[-1]. )
5.58 µs ± 985 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df.loc[df.index[-1], 'code']
12 µs ± 2.71 µs per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df['code'].iat[-1]
5.71 µs ± 896 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
%timeit df['code'].tail(1).item()
36 µs ± 3.23 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit df.iloc[-1, df.columns.get_loc('code')]
33.7 µs ± 5.23 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit df['code'].iloc[-1]
8.08 µs ± 496 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
Alternatively you can use take:
In [6]: df['col1'].take([-1]).item()
Out[6]: 3
You can also use:
df.iloc[-1, 0]