CODEWITHSUNDEEP
pythonpandas
usual me
usual me

Reputation: 8788

How to find holes in a Pandas time index

I am interesting in knowing if a time index contains holes. Say I have the following Series

ss = pd.Series( nr.randn(3), index=[ '2014-01-01', '2014-01-02', '2014-01-03' ] )
ss.index = pd.to_datetime( ss.index )
ss

Output 

2014-01-01    0.976455
2014-01-02   -0.610322
2014-01-03   -0.631592
dtype: float64

I thought I could do what I would do with lists (l[1:]-l[:-1])

ss.index[1:] - ss.index[:-1]

But here is the output, which I don't understand

<class 'pandas.tseries.index.DatetimeIndex'>
[2014-01-03]
Length: 1, Freq: None, Timezone: None

I ended up doing this (which is kinda ugly)

tmp = pd.Series( ss.index[1:] ) - pd.Series( ss.index[:-1] )
(tmp[0] == tmp ).all()

So I have 2 questions:

Upvotes: 2

Views: 1923

Answers (3)

Jeff
Jeff

Reputation: 129018

This is a slightly different method. A frequency will be returned if it can (e.g. its daily if the values are daily spaced with no holes). None otherwise.

In [14]: pd.infer_freq(Series(np.random.randn(3),index=['20140101','20140102','20140103']).index)
Out[14]: 'D'

In [15]: pd.infer_freq(Series(np.random.randn(3),index=['20140101','20140102','20140104']).index)

In [31]: pd.infer_freq(Series(np.random.randn(3),index=['20140101','20140201','20140301']).index)
Out[31]: 'MS'

Upvotes: 1

Guillaume Jacquenot
Guillaume Jacquenot

Reputation: 11717

You can try

tDelta = ss.index.date[1:]-ss.index.date[:-1]
secondBetweenEachEntries = [t.total_seconds() for t in tDelta]

That gives

import pandas as pd
import numpy.random as nr
ss = pd.Series( nr.randn(3), index=[ '2014-01-01', '2014-01-02', '2014-01-03' ] )
ss.index = pd.to_datetime( ss.index )
tDelta = ss.index.date[1:]-ss.index.date[:-1]

Upvotes: 1

John Zwinck
John Zwinck

Reputation: 249394

You can do it with numpy.diff():

np.diff(np.array(ss.index))

There is probably some slightly slicker way to do this, but the above works. It gives you:

array([86400000000000, 86400000000000], dtype='timedelta64[ns]')

Upvotes: 1

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