问题描述
ATR 是给定时间段内真实范围的平均值.真实范围是(高-低)意味着我已经用以下方法计算了这个:
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()
但是，如果需要短时间段(或上例中的距离")，ATR 可能会非常不稳定，即某些数字之间会出现较大的零星间隙.
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.
真正的 ATR 方程认识到这一点，并通过执行以下操作使其平滑:
The real ATR equation recognises this and smooths it out by doing the following:
Current ATR = [(Prior ATR x 13) + Current TR] / 14
但是，我不确定如何以与上述相同的方式执行此操作，即列范围操作.
However I am unsure how to do this in the same manner as I did above, i.e. a column wide operation.
包括来自我原始方法的 TR 和 ATR(10) 的样本数据:
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
推荐答案
这不是 TR 的正确计算，请参阅 - ATR，但我会这样做:
That's not the right calculation for TR see - ATR, but here is how I would do it:
其中 alpha = 2/(span+1)
Where alpha = 2 / (span+1)
df['ATR'] = df['TR'].ewm(span = 10).mean()
否则，您应该能够像这样轻松地进行自己的平滑处理:
Otherwise you should be able to easily do you're own smoothing like this:
df['ATR'] = (df['ATR'].shift(1)*13 + df['TR'])/14
这篇关于使用 Python 计算 OHLC 数据的平均真实范围 (ATR)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！