Question

我需要检查数组realtime中是否有时间样本不是增量的（时间倒退）

realtime
Out[2]: 
array([datetime.datetime(2017, 11, 3, 20, 25, 10, 724000),
       datetime.datetime(2017, 11, 3, 20, 25, 10, 744000),
       datetime.datetime(2017, 11, 3, 20, 25, 10, 764000), ...,
       datetime.datetime(2017, 11, 4, 2, 13, 44, 704000),
       datetime.datetime(2017, 11, 4, 2, 13, 44, 724000),
       datetime.datetime(2017, 11, 4, 2, 13, 44, 744000)], dtype=object)

实时是1045702L！

我试着做了

d = pd.DataFrame(np.zeros((len(realtime), 1)))
for i in range(len(realtime)):
    if any(realtime[i] <= x for x in realtime[:i]): # smaller/equal than any prior
        d.iloc[i] = True

但它需要永远...... 有没有更快的方法来检查数组中的元素是否是增量的，如果没有标记它们？

Answer 1

您可以使用array timedelta numpy.diff进行comapre 0创建：

b = np.diff(realtime) > datetime.timedelta(0)
print (b)

[ True  True  True  True  True]

在pandas中，您可以转换为pd.Series对象并使用diff：

b = pd.Series(realtime).diff()
#replace first NaN value to 1
b.iat[0] = 1

print (b > pd.Timedelta(0))
0    True
1    True
2    True
3    True
4    True
5    True
dtype: bool

realtime会自动投放到np.datetime64，diff会从Timedelta生成realtime = np.array([datetime.datetime(2017, 11, 3, 20, 25, 10, 724000), datetime.datetime(2017, 11, 3, 20, 25, 10, 744000), datetime.datetime(2017, 11, 3, 20, 25, 10, 764000), datetime.datetime(2017, 11, 4, 2, 13, 44, 704000), datetime.datetime(2017, 11, 4, 2, 13, 44, 724000), datetime.datetime(2017, 11, 4, 2, 13, 44, 744000)], dtype=object) realtime = np.random.choice(realtime, size=1045702) In [256]: %timeit[x.total_seconds() > 0 for x in np.diff(realtime)] 1 loop, best of 3: 382 ms per loop In [257]: %timeit np.diff(realtime) > datetime.timedelta(0) 10 loops, best of 3: 88.2 ms per loop In [258]: %timeit (pd.Series(realtime).diff() > pd.Timedelta(0)) 10 loops, best of 3: 147 ms per loop In [259]: %%timeit ...: b = pd.Series(realtime).diff() ...: b.iat[0] = 1 ...: ...: b > pd.Timedelta(0) ...: 10 loops, best of 3: 149 ms per loop个对象。

<强>计时：

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检查数组的元素是否是单调的

1 个答案: