python - 按 timedelta 修剪 TimeSeries

Question

我正在尝试从 pandas TimeSeries 中删除所有“旧”值，例如所有超过 1 天的值(相对于最新值)。

我天真地尝试过这样的事情:

from datetime import timedelta
def trim(series):
    return series[series.index.max() - series.index < timedelta(days=1)]

报错:

TypeError: ufunc 'subtract' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule 'safe'

很明显，这个表达式有问题:series.index.max() - series.index

然后我发现这个有效:

def trim(series):
    return series[series.index > series.index.max() - timedelta(days=1)]

有人可以解释为什么后者有效而前者引发错误吗？

编辑:我使用的是 pandas 版本 0.12.0

Answer 1

这是 0.13 中的示例(to_timedelta 在 0.12 中不可用，因此你必须做 np.timedelta64(4,'D'))

In [12]: rng = pd.date_range('1/1/2011', periods=10, freq='D')

In [13]: ts = pd.Series(randn(len(rng)), index=rng)

In [14]: ts
Out[14]: 
2011-01-01   -0.348362
2011-01-02    1.782487
2011-01-03    1.146537
2011-01-04   -0.176308
2011-01-05   -0.185240
2011-01-06    1.767135
2011-01-07    0.615911
2011-01-08    2.459799
2011-01-09    0.718081
2011-01-10   -0.520741
Freq: D, dtype: float64

In [15]: x = ts.index.to_series().max()-ts.index.to_series()

In [16]: x
Out[16]: 
2011-01-01   9 days
2011-01-02   8 days
2011-01-03   7 days
2011-01-04   6 days
2011-01-05   5 days
2011-01-06   4 days
2011-01-07   3 days
2011-01-08   2 days
2011-01-09   1 days
2011-01-10   0 days
Freq: D, dtype: timedelta64[ns]

In [17]: x[x>pd.to_timedelta('4 days')]
Out[17]: 
2011-01-01   9 days
2011-01-02   8 days
2011-01-03   7 days
2011-01-04   6 days
2011-01-05   5 days
Freq: D, dtype: timedelta64[ns]