pandas rolling appy on a dataframe

Question

I have Yahoo Stock data, that I would like to manipulate, like so,

import pandas as pd import pandas.io.data as web data = web.DataReader('SPY','yahoo') data.head() Out[13]: Open High Low Close Volume Adj Close Date 2010-01-04 112.37 113.39 111.51 113.33 118944600 103.44 2010-01-05 113.26 113.68 112.85 113.63 111579900 103.71 2010-01-06 113.52 113.99 113.43 113.71 116074400 103.79 2010-01-07 113.50 114.33 113.18 114.19 131091100 104.23 2010-01-08 113.89 114.62 113.66 114.57 126402800 104.57

For any given date, I would like to look forward 2 days and find the lowest quote for it. So, for 2010-1-4, the correct answer would be 112.85.

Now, I could iterate over all the dates with a for loop and get what I want. But I would like to figure out if I could do this in a vectorized manner. Maybe by using a rolling_apply lambda function. This is what I have done so far...

def foo(x): today = x[0] forward = x[1:] return (forward.min()) pd.rolling_apply(data,2,foo)

This does not work since the rolling_apply works on a Series and does not have access to the other columns on the data frame.

Is this some neat way to this?

Doesn't pd.rolling_apply(data['Low'],2,min) do what you want? — EdChum
– EdChum, Commented Aug 29, 2014 at 19:41

EdChum · Accepted Answer · 2014-08-29 20:22:49Z

Rather than calling rolling_apply on the whole dataframe, just call it on the column of interest and call min:

pd.rolling_apply(data['Low'],2,min)

Interestingly the global min function outperforms the numpy min, perhaps not that surprising given that all we are doing is finding the lowest value of a 2 element array:

In [26]: %timeit pd.rolling_apply(data['Low'],2,np.min) %timeit pd.rolling_apply(data['Low'],2,min) 10 loops, best of 3: 15.4 ms per loop 1000 loops, best of 3: 1.44 ms per loop

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