我正在将一些代码从python list原语迁移到pandas实现。对于某些时间序列，我想找到所有不连续的段及其持续时间。在熊猫中有这种干净的方法吗？
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
例如，第一次不连续事件是从5:40到7:40。
答案 0 :(得分：6)
只要您有系列或单列数据框，这就应该有效。
>>>pd.Series(df.isnull().index).diff()
可以通过以下方式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
答案 1 :(得分：2)
不像基于熊猫的解决方案那样优雅或简短，但考虑到性能，可以使用NumPy数组和函数。因此，为了解决这种情况并假设日期时间具有常规频率，这里采用基于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()
请注意这主要基于另一个NumPy solution to : Longest run/island of a number in Python.
运行时测试
我会计时@ilmarinen's pandas based solution和基于NumPy的方法在本文前面发布的足够大的数据框中填充了随机元素并随机放置了50％的NaN。
功能定义：
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