ATR 是给定时间段内真实范围的平均值。真实范围是(高-低)，这意味着我用以下方法计算了这个:
df['High'].subtract(df['Low']).rolling(distance).mean()
但是，如果需要短周期(或上例中的“距离”)，则 ATR 可能会非常跳跃，即某些数字之间会出现较大的零星间隙。
真正的 ATR 方程可以识别这一点并通过执行以下操作来平滑它:
Current ATR = [(Prior ATR x 13) + Current TR] / 14
但是我不确定如何以与上面相同的方式执行此操作，即列宽操作。
示例数据，包括来 self 的原始方法的 TR 和 ATR(10):
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
请您参考如下方法:
对于其他正在研究如何执行此操作的人，这是我的答案。
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
