差异pandas.DateTimeIndex没有频率 [英] Difference pandas.DateTimeIndex without a frequency

问题描述

不规则的时间序列data存储在pandas.DataFrame中.设置了DatetimeIndex.我需要索引中连续条目之间的时间差.

An irregular time series data is stored in a pandas.DataFrame. A DatetimeIndex has been set. I need the time difference between consecutive entries in the index.

我以为会很简单

data.index.diff()

但是得到

AttributeError: 'DatetimeIndex' object has no attribute 'diff'

我尝试过

data.index - data.index.shift(1)

但是得到

ValueError: Cannot shift with no freq

在执行此操作之前，我不想首先推断或强制执行频率.时间序列中有很大的差距，这些差距会扩大到nan的大范围运行.关键是要首先找到这些差距.

I do not want to infer or enforce a frequency first before doing this operation. There are large gaps in the time series that would be expanded to large runs of nan. The point is to find these gaps first.

那么，执行此看似简单的操作的一种干净方法是什么?

So, what is a clean way to do this seemingly simple operation?

推荐答案

索引尚未实现diff函数.

但是，可以首先使用Series "rel =" nofollow noreferrer> Index.to_series ，如果您需要保留原始索引.如果需要默认索引，请使用不带索引参数的Series构造函数.

However, it is possible to convert the index to a Series first by using Index.to_series, if you need to preserve the original index. Use the Series constructor with no index parameter if the default index is needed.

代码示例:

rng = pd.to_datetime(['2015-01-10','2015-01-12','2015-01-13'])
data = pd.DataFrame({'a': range(3)}, index=rng)  
print(data)
             a
 2015-01-10  0
 2015-01-12  1
 2015-01-13  2

a = data.index.to_series().diff()
print(a)

2015-01-10      NaT
2015-01-12   2 days
2015-01-13   1 days
dtype: timedelta64[ns]

a = pd.Series(data.index).diff()
print(a)
 0      NaT
 1   2 days
 2   1 days
dtype: timedelta64[ns]

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