I have two sets of temperature date, which have readings at regular (but different) time intervals. I'm trying to get the correlation between these two sets of data.
I've been playing with Pandas to try to do this. I've created two timeseries, and am using TimeSeriesA.corr(TimeSeriesB)
. However, if the times in the 2 timeSeries do not match up exactly (they're generally off by seconds), I get Null as an answer. I could get a decent answer if I could:
a) Interpolate/fill missing times in each TimeSeries (I know this is possible in Pandas, I just don't know how to do it)
b) strip the seconds out of python datetime objects (Set seconds to 00, without changing minutes). I'd lose a degree of accuracy, but not a huge amount
c) Use something else in Pandas to get the correlation between two timeSeries
d) Use something in python to get the correlation between two lists of floats, each float having a corresponding datetime object, taking into account the time.
Anyone have any suggestions?
The serial correlation or autocorrelation of lag , , of a second order stationary time series is given by the autocovariance of the series normalised by the product of the spread. That is, ρ k = C k σ 2 . Note that ρ 0 = C 0 σ 2 = E [ ( x t − μ ) 2 ] σ 2 = σ 2 σ 2 = 1 .
By using corr() function we can get the correlation between two columns in the dataframe.
You have a number of options using pandas, but you have to make a decision about how it makes sense to align the data given that they don't occur at the same instants.
Use the values "as of" the times in one of the time series, here's an example:
In [15]: ts
Out[15]:
2000-01-03 00:00:00 -0.722808451504
2000-01-04 00:00:00 0.0125041039477
2000-01-05 00:00:00 0.777515530539
2000-01-06 00:00:00 -0.35714026263
2000-01-07 00:00:00 -1.55213541118
2000-01-10 00:00:00 -0.508166334892
2000-01-11 00:00:00 0.58016097981
2000-01-12 00:00:00 1.50766289013
2000-01-13 00:00:00 -1.11114968643
2000-01-14 00:00:00 0.259320239297
In [16]: ts2
Out[16]:
2000-01-03 00:00:30 1.05595278907
2000-01-04 00:00:30 -0.568961755792
2000-01-05 00:00:30 0.660511172645
2000-01-06 00:00:30 -0.0327384421979
2000-01-07 00:00:30 0.158094407533
2000-01-10 00:00:30 -0.321679671377
2000-01-11 00:00:30 0.977286027619
2000-01-12 00:00:30 -0.603541295894
2000-01-13 00:00:30 1.15993249209
2000-01-14 00:00:30 -0.229379534767
you can see these are off by 30 seconds. The reindex
function enables you to align data while filling forward values (getting the "as of" value):
In [17]: ts.reindex(ts2.index, method='pad')
Out[17]:
2000-01-03 00:00:30 -0.722808451504
2000-01-04 00:00:30 0.0125041039477
2000-01-05 00:00:30 0.777515530539
2000-01-06 00:00:30 -0.35714026263
2000-01-07 00:00:30 -1.55213541118
2000-01-10 00:00:30 -0.508166334892
2000-01-11 00:00:30 0.58016097981
2000-01-12 00:00:30 1.50766289013
2000-01-13 00:00:30 -1.11114968643
2000-01-14 00:00:30 0.259320239297
In [18]: ts2.corr(ts.reindex(ts2.index, method='pad'))
Out[18]: -0.31004148593302283
note that 'pad' is also aliased by 'ffill' (but only in the very latest version of pandas on GitHub as of this time!).
Strip seconds out of all your datetimes. The best way to do this is to use rename
In [25]: ts2.rename(lambda date: date.replace(second=0))
Out[25]:
2000-01-03 00:00:00 1.05595278907
2000-01-04 00:00:00 -0.568961755792
2000-01-05 00:00:00 0.660511172645
2000-01-06 00:00:00 -0.0327384421979
2000-01-07 00:00:00 0.158094407533
2000-01-10 00:00:00 -0.321679671377
2000-01-11 00:00:00 0.977286027619
2000-01-12 00:00:00 -0.603541295894
2000-01-13 00:00:00 1.15993249209
2000-01-14 00:00:00 -0.229379534767
Note that if rename causes there to be duplicate dates an Exception
will be thrown.
For something a little more advanced, suppose you wanted to correlate the mean value for each minute (where you have multiple observations per second):
In [31]: ts_mean = ts.groupby(lambda date: date.replace(second=0)).mean()
In [32]: ts2_mean = ts2.groupby(lambda date: date.replace(second=0)).mean()
In [33]: ts_mean.corr(ts2_mean)
Out[33]: -0.31004148593302283
These last code snippets may not work if you don't have the latest code from https://github.com/wesm/pandas. If .mean()
doesn't work on a GroupBy
object per above try .agg(np.mean)
Hope this helps!
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