指數平滑平均值

Question

我正在瀏覽本網站以瞭解有關指數平滑平均值的更多信息，但不確定關於代碼的1部分。

import pandas, numpy as np 
ewma = pandas.stats.moments.ewma 

# make a hat function, and add noise 
x = np.linspace(0,1,100) 
x = np.hstack((x,x[::-1])) 
x += np.random.normal(loc=0, scale=0.1, size=200) 
plot(x, alpha=0.4, label='Raw') 

# take EWMA in both directions with a smaller span term 
fwd = ewma(x, span=15) # take EWMA in fwd direction 
bwd = ewma(x[::-1], span=15) # take EWMA in bwd direction 
c = np.vstack((fwd, bwd[::-1])) # lump fwd and bwd together 
c = np.mean(c, axis=0) # average 

# regular EWMA, with bias against trend 
plot(ewma(x, span=20), 'b', label='EWMA, span=20') 

# "corrected" (?) EWMA 
plot(c, 'r', label='Reversed-Recombined') 

我不明白的是本節

# take EWMA in both directions with a smaller span term 
fwd = ewma(x, span=15) # take EWMA in fwd direction 
bwd = ewma(x[::-1], span=15) # take EWMA in bwd direction 
c = np.vstack((fwd, bwd[::-1])) # lump fwd and bwd together 
c = np.mean(c, axis=0) # average 

可能有人請您解釋一下這到底是怎麼回事呢？

該網站的完整的源代碼是：http://connor-johnson.com/2014/02/01/smoothing-with-exponentially-weighted-moving-averages/

Answer 1

我想這裏的主要問題是什麼bwd[::-1]手段？查看添加的其他評論。

# take EWMA in both directions with a smaller span term 

fwd = ewma(x, span=15) # take EWMA in fwd direction 
## This part should not be a problem, right? 

bwd = ewma(x[::-1], span=15) # take EWMA in bwd direction 
## x[::-1] means to go thr x, from end to beginning(!), with a step of -1 
## (hence it is going from the back to the front) 

c = np.vstack((fwd, bwd[::-1])) # lump fwd and bwd together 
c = np.mean(c, axis=0) # average 
## Then we reverse ewma into a beginning-to-end order 
## and take the average of fwd and bwd 
## IMO, better written just as: 

#c = 0.5*(fwd + bwd[::-1]) 

這個想法是，在前向EWMA中，當前值受早期值影響，但越來越少。另一方面，後向EWMA受後面的值影響。最後，通過取前向和後向EWMA的平均值，可以創建一些受周圍值影響的東西（如果我們稱之爲），但越來越少，因爲當您離開當前位置時。