如何获得大熊猫系列中最接近零的n？

Question

如何获得与n最接近的0值，类似于如何使用nsmallest()获得n的最小值。例如。与

series = pd.Series([-1.0,-0.75,-0.5,-0.25,0.25,0.5,0.75,1.0])
series

0   -1.00
1   -0.75
2   -0.50
3   -0.25
4    0.25
5    0.50
6    0.75
7    1.00
dtype: float64

例如n=4我想得到以下内容。

0   -0.25
1   0.25
2   -0.50
3   0.50
dtype: float64

Answer 1

使用loc，abs和nsmallest：

series.loc[series.abs().nsmallest(4).index]

3   -0.25
4    0.25
2   -0.50
5    0.50
dtype: float64

Answer 2

将Series.abs与Series.argsort一起用于排名，过滤n并按Series.iloc选择，如果性能很重要：

n = 4
series = series.iloc[series.abs().argsort()[:n]]
print (series)
3   -0.25
4    0.25
2   -0.50
5    0.50
dtype: float64

最后一个需要默认索引的地方：

n = 4
series = series.iloc[series.abs().argsort()[:n]].reset_index(drop=True)
print (series)
0   -0.25
1    0.25
2   -0.50
3    0.50
dtype: float64

性能：

series = pd.Series([-1.0,-0.75,-0.5,-0.25,0.25,0.5,0.75,1.0] * 10000)

n = 4000
series = series.iloc[series.abs().argsort()[:n]]
print (series)

In [114]: %timeit series.iloc[series.abs().argsort()[:n]]
794 µs ± 19.4 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

In [115]: %timeit series.loc[series.abs().nsmallest(n).index]
2.09 ms ± 34.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)