Python 求最大折扣连续性的长度

我正在将一些代码从pythonlist原语迁移到pandas实现。对于某些时间序列，我想找出所有不连续的片段及其持续时间。在熊猫身上有干净的方法吗

我的数据框如下所示：

In [23]: df
Out[23]:
2016-07-01 05:35:00    60.466667
2016-07-01 05:40:00          NaN
2016-07-01 05:45:00          NaN
2016-07-01 05:50:00          NaN
2016-07-01 05:55:00          NaN
2016-07-01 06:00:00          NaN
2016-07-01 06:05:00          NaN
2016-07-01 06:10:00          NaN
2016-07-01 06:15:00          NaN
2016-07-01 06:20:00          NaN
2016-07-01 06:25:00          NaN
2016-07-01 06:30:00          NaN
2016-07-01 06:35:00          NaN
2016-07-01 06:40:00          NaN
2016-07-01 06:45:00          NaN
2016-07-01 06:50:00          NaN
2016-07-01 06:55:00          NaN
2016-07-01 07:00:00          NaN
2016-07-01 07:05:00          NaN
2016-07-01 07:10:00          NaN
2016-07-01 07:15:00          NaN
2016-07-01 07:20:00          NaN
2016-07-01 07:25:00          NaN
2016-07-01 07:30:00          NaN
2016-07-01 07:35:00          NaN
2016-07-01 07:40:00          NaN
2016-07-01 07:45:00    63.500000
2016-07-01 07:50:00    67.293333
2016-07-01 07:55:00    67.633333
2016-07-01 08:00:00    68.306667
                         ...
2016-07-01 11:20:00          NaN
2016-07-01 11:25:00          NaN
2016-07-01 11:30:00    62.000000
2016-07-01 11:35:00    69.513333
2016-07-01 11:40:00    64.931298
2016-07-01 11:45:00    51.980000
2016-07-01 11:50:00    55.253333
2016-07-01 11:55:00    51.273333
2016-07-01 12:00:00    52.080000
2016-07-01 12:05:00    54.580000
2016-07-01 12:10:00    55.306667
2016-07-01 12:15:00    55.200000
2016-07-01 12:20:00    57.140000
2016-07-01 12:25:00    57.020000
2016-07-01 12:30:00    57.526667
2016-07-01 12:35:00    57.880000
2016-07-01 12:40:00    67.286667
2016-07-01 12:45:00    58.153333
2016-07-01 12:50:00    57.460000
2016-07-01 12:55:00    54.413333
2016-07-01 13:00:00    55.526667
2016-07-01 13:05:00    56.120000
2016-07-01 13:10:00    55.620000
2016-07-01 13:15:00    56.420000
2016-07-01 13:20:00    51.893333
2016-07-01 13:25:00    74.451613
2016-07-01 13:30:00    54.898551
2016-07-01 13:35:00          NaN
2016-07-01 13:40:00    63.355140
2016-07-01 13:45:00    61.000000
Freq: 5T, dtype: float64

---

其中，例如，第一个不连续事件是从5:40到7:40。

只要您有一个系列或单列数据帧，这应该可以工作

>>>pd.Series(df.isnull().index).diff()

可通过以下方式进行改进以获得有用的输出：

MIN_GAP_TIMEDELTA = Timedelta(minutes=30)
discontinuities = pd.Series(df.isnull().index).diff()
discontinuities.sort(ascending=False)
discontinuities[discontinuities > MIN_GAP_TIMEDELTA].size

---

不像基于pandas的解决方案那样优雅或简洁，但考虑到性能，可以考虑使用NumPy阵列和函数。为了解决这种情况，假设日期时间有一个固定的频率，这里有一个基于NumPy的方法来获得不连续长度，最大长度和阈值计数-

# Get indices of start and stop indices of discontinuities signified by NaNs
idx = np.where(np.diff(np.hstack(([False],np.isnan(df[0]),[False]))))[0]

# Do differentiation on those indices which would give us the length of 
# intervals of discontinuities. These could be used in various ways.
discontinuity_lens = np.diff(idx.reshape(-1,2),axis=1)

# Max discontinuity length
discontinuity_maxlen = discontinuity_lens.max()

# Count of discontinuities that are greater than a threshold of 30 mins as
# listed with threshold parameter : MIN_GAP_TIMEDELTA = Timedelta(minutes=30)
# (in terms of steps that would be 6 because freq of input dataframe is 5 mins)
thresholded_count = (discontinuity_lens>=6).sum()

请注意，这主要是基于另一个原因

运行时测试

我将在一个足够大的数据帧上发布基于NumPy的计时方法，该数据帧中填充了随机元素，并随机放置了50%的NAN

函数定义：

def thresholdedcount_pandas(df):
    MIN_GAP_TIMEDELTA = pd.Timedelta(minutes=30)
    discontinuities = df.dropna().reset_index()['index'].diff()
    return (discontinuities > MIN_GAP_TIMEDELTA).sum()

def thresholdedcount_numpy(df):
    idx = np.where(np.diff(np.hstack(([False],np.isnan(df[0]),[False]))))[0]
    nan_interval_lens = np.diff(idx.reshape(-1,2),axis=1)
    return (nan_interval_lens>=6).sum()

时间：

In [325]: # Random dataframe with 5 min interval data and filled with 50% NaNs
     ...: rng = pd.date_range('1/1/2011', periods=10000, freq='5Min')
     ...: df = pd.DataFrame(np.random.randn(len(rng)), index=rng)
     ...: df[0][np.random.randint(0,df.shape[0],(int(df.shape[0]/2)))] = np.nan
     ...: 

In [326]: np.allclose(thresholdedcount_pandas(df),thresholdedcount_numpy(df))
Out[326]: True

In [327]: %timeit thresholdedcount_pandas(df)
100 loops, best of 3: 3 ms per loop

In [328]: %timeit thresholdedcount_numpy(df)
1000 loops, best of 3: 318 µs per loop

---

这看起来像一个系列，而不是一个数据帧。我们将研究这个解决方案。现在我更倾向于使用一个更简单的解决方案，因为性能对于这些人来说并不是一个大问题，因为它主要是在后台工作上。