为了获取所在位置而不是所在索引,对于index推荐使用两种方式index.get_loc(ele)
和index.to_list().index(ele)
,第一种语句更短,第二种更快
In [1]: import pandas as pd
In [2]: import numpy as np
In [3]: data = [406400, 203200, 101600, 76100, 50800, 25400, 19050, 12700,
...: 9500, 6700, 4750, 3350, 2360, 1700, 1180, 850,
...: 600, 425, 300, 212, 150, 106, 75, 53,
...: 38]
In [4]: myseries = pd.Series(data, index=range(1,26))
In [5]: myseries[21]
Out[5]: 150
In [7]: %timeit myseries[myseries == 150].index[0]
416 µs ± 5.05 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [8]: %timeit myseries[myseries == 150].first_valid_index()
585 µs ± 32.5 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [9]: %timeit myseries.where(myseries == 150).first_valid_index()
652 µs ± 23.3 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
In [10]: %timeit myseries.index[np.where(myseries == 150)[0][0]]
195 µs ± 1.18 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [11]: %timeit pd.Series(myseries.index, index=myseries)[150]
178 µs ± 9.35 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [12]: %timeit myseries.index[pd.Index(myseries).get_loc(150)]
77.4 µs ± 1.41 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [13]: %timeit myseries.index[list(myseries).index(150)]
12.7 µs ± 42.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
In [14]: %timeit myseries.index[myseries.tolist().index(150)]
9.46 µs ± 19.2 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)
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