You could try iloc
method of dataframe:
In [26]: df
Out[26]:
a b c d e
0 -1.079547 -0.722903 0.457495 -0.687271 -0.787058
1 1.326133 1.359255 -0.964076 -1.280502 1.460792
2 0.479599 -1.465210 -0.058247 -0.984733 -0.348068
3 -0.608238 -1.238068 -0.126889 0.572662 -1.489641
4 -1.533707 -0.218298 -0.877619 0.679370 0.485987
5 -0.864651 -0.180165 -0.528939 0.270885 1.313946
6 0.747612 -1.206509 0.616815 -1.758354 -0.158203
7 -2.309582 -0.739730 -0.004303 0.125640 -0.973230
8 1.735822 -0.750698 1.225104 0.431583 -1.483274
9 -0.374557 -1.132354 0.875028 0.032615 -1.131971
In [27]: df['e'].iloc[-1]
Out[27]: -1.1319705662711321
Or if you want just scalar you could use iat
which is faster. From docs:
If you only want to access a scalar value, the fastest way is to use the at
and iat
methods, which are implemented on all of the data structures
In [28]: df.e.iat[-1]
Out[28]: -1.1319705662711321
Benchmarking:
In [31]: %timeit df.e.iat[-1]
100000 loops, best of 3: 18 μs per loop
In [32]: %timeit df.e.iloc[-1]
10000 loops, best of 3: 24 μs per loop
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