Calculating Average True Range (ATR) on OHLC data with Python

Question

The ATR is the average of the True Range for a given period. True Range is (High-Low) meaning I have computed this with the following:

df['High'].subtract(df['Low']).rolling(distance).mean()

However if a short period (or 'distance' in the example above) is required the ATR can be very jumpy, i.e. with large sporadic gaps appearing between some numbers.

The real ATR equation recognises this and smooths it out by doing the following:

Current ATR = [(Prior ATR x 13) + Current TR] / 14

However I am unsure how to do this in the same manner as I did above, i.e. a column wide operation.

Sample data including the TR and ATR(10) from my original method:

Date        Time            Open    High    Low     Close   TR      ATR
30/09/16    14:45:00+00:00  1.1216  1.1221  1.1208  1.1209  0.0013  0.0013
30/09/16    15:00:00+00:00  1.1209  1.1211  1.1203  1.1205  0.0008  0.0013
30/09/16    15:15:00+00:00  1.1205  1.1216  1.1204  1.1216  0.0012  0.0013
30/09/16    15:30:00+00:00  1.1217  1.1222  1.1213  1.1216  0.0008  0.0013
30/09/16    15:45:00+00:00  1.1216  1.1240  1.1216  1.1240  0.0025  0.0015
30/09/16    16:00:00+00:00  1.1239  1.1246  1.1228  1.1242  0.0019  0.0015
30/09/16    16:15:00+00:00  1.1242  1.1251  1.1235  1.1240  0.0016  0.0016
30/09/16    16:30:00+00:00  1.1240  1.1240  1.1234  1.1236  0.0007  0.0014
30/09/16    16:45:00+00:00  1.1237  1.1245  1.1235  1.1238  0.0009  0.0012
30/09/16    17:00:00+00:00  1.1238  1.1239  1.1231  1.1233  0.0008  0.0012
30/09/16    17:15:00+00:00  1.1233  1.1245  1.1232  1.1240  0.0013  0.0012
30/09/16    17:30:00+00:00  1.1240  1.1242  1.1228  1.1230  0.0013  0.0013
30/09/16    17:45:00+00:00  1.1230  1.1230  1.1221  1.1227  0.0009  0.0013
30/09/16    18:00:00+00:00  1.1227  1.1232  1.1227  1.1232  0.0005  0.0012
30/09/16    18:15:00+00:00  1.1232  1.1232  1.1227  1.1227  0.0005  0.0010
30/09/16    18:30:00+00:00  1.1227  1.1231  1.1225  1.1231  0.0006  0.0009
30/09/16    18:45:00+00:00  1.1231  1.1237  1.1230  1.1232  0.0007  0.0008
30/09/16    19:00:00+00:00  1.1232  1.1233  1.1229  1.1231  0.0004  0.0008
30/09/16    19:15:00+00:00  1.1231  1.1234  1.1230  1.1230  0.0004  0.0007
30/09/16    19:30:00+00:00  1.1231  1.1234  1.1230  1.1234  0.0004  0.0007
30/09/16    19:45:00+00:00  1.1233  1.1240  1.1230  1.1239  0.0010  0.0007
30/09/16    20:00:00+00:00  1.1239  1.1242  1.1237  1.1238  0.0005  0.0006
30/09/16    20:15:00+00:00  1.1238  1.1240  1.1235  1.1237  0.0005  0.0006
30/09/16    20:30:00+00:00  1.1237  1.1238  1.1235  1.1235  0.0003  0.0005
30/09/16    20:45:00+00:00  1.1235  1.1236  1.1233  1.1233  0.0003  0.0005
30/09/16    21:00:00+00:00  1.1233  1.1238  1.1233  1.1237  0.0006  0.0005
30/09/16    21:15:00+00:00  1.1237  1.1244  1.1237  1.1242  0.0008  0.0005
30/09/16    21:30:00+00:00  1.1242  1.1243  1.1239  1.1239  0.0004  0.0005
30/09/16    21:45:00+00:00  1.1239  1.1244  1.1236  1.1241  0.0008  0.0006

Answer 1

For anyone else looking on how to do this, here is my answer.

def wwma(values, n):
    """
     J. Welles Wilder's EMA 
    """
    return values.ewm(alpha=1/n, adjust=False).mean()

def atr(df, n=14):
    data = df.copy()
    high = data[HIGH]
    low = data[LOW]
    close = data[CLOSE]
    data['tr0'] = abs(high - low)
    data['tr1'] = abs(high - close.shift())
    data['tr2'] = abs(low - close.shift())
    tr = data[['tr0', 'tr1', 'tr2']].max(axis=1)
    atr = wwma(tr, n)
    return atr