How to iterate with condition based on previous rows/columns values?

I have the following dataframe df with datetime index.

                    open    high    low     close
DateTime                
2016-02-04 01:00:00 1.10886 1.10887 1.10702 1.10747
2016-02-04 02:00:00 1.10746 1.10893 1.10695 1.10810
2016-02-04 03:00:00 1.10809 1.11403 1.10788 1.11326
2016-02-04 04:00:00 1.11327 1.11673 1.11306 1.11628
2016-02-04 05:00:00 1.11628 1.11835 1.11523 1.11815
2016-02-04 06:00:00 1.11813 1.11848 1.11653 1.11723
2016-02-04 07:00:00 1.11712 1.12020 1.11705 1.11941

I need to filter out the indexes where the following condition is met. It’s written in mql4 code which is where I’m converting this from, so bear with me. Using numbers as indexes so the bar2Open == 'open' is the value in the second row prior to the current one and bar1Low == 'low' is the value of the previous row:

double buffer = 10;
double a = (bar2Open < bar2Close) ? bar1Low : bar1High;
double b = (bar2Open < bar2Close) ? bar3High : bar3Low;
   
if(abs(a-b) >= buffer)
{
  do something;
}

I’d like to do this with some form of itertuples() but I’m not sure how to access previous rows values in different columns. Output can also be in the form of another column that just gives 1 if the condition is met, 0 otherwise.

Answer

There is no need to iterate over the dataframe rows, that is usually a sign that you are doing something wrong when working with pandas. You can access values from previous rows using shift. The filtering can be done using something like:

import numpy as np
import pandas as pd

buffer = 10

cond = df['open'].shift(2) < df['close'].shift(2)
a = np.where(cond, df['low'].shift(1), df['high'].shift(1))
b = np.where(cond, df['high'].shift(3), df['low'].shift(3))

idx = np.abs(a - b) > 10
df = df.loc[idx]

Using the example dataframe in the question, this will result in an empty dataframe.

---

Assigning an 0 or 1 to a new column can be done using idx as follows:

df['filter'] = idx.astype(int)