填補角度串列（數字）之間的空白

如果我有一個數字串列，我將用簡單的例子進行解釋，然后深入探討

t_original = [180,174,168,166,162,94,70,80,128,131,160,180]

如果我們繪制它，它會從 180 下降到 70，然后它會再次上升到 180

但如果我們突然將第四個值 (166) 更改為 450，那么串列將是

t = [180,174,168,700,162,94,70,80,128,131,160,180]

圖中沒有意義

我想將第四個值（700）視為錯誤值我想用相對值替換它，即使不是原始值而是相對于前兩個元素（168,174） 我想對整個串列做同樣的事情如果另一個再次出現錯誤值

我們可以稱之為[填補數字串列之間的空白]

所以我嘗試做同樣的想法，但更大的例子

我試過的方法

我將與輸出共享我的代碼，過濾意味著應用填充間隙功能

我的代碼

def preprocFN(*U):
prePlst=[] # after preprocessing list
#preprocessing Fc =| 2*LF1 prev by 1 - LF2 prev by 2 |
c0 = -2     #(previous) by 2
c1 =-1      #(previous)
c2 =0       #(current)
c3 = 1      #(next)
preP = U[0] # original list
if c2 == 0:
    prePlst.append(preP[0])
    prePlst.append(preP[1])
    c1 =2
    c2 =2
    c0 =2
oldlen = len(preP)
while oldlen > c2:
    Equ = abs(2*preP[c1] - preP[c0]) #fn of preprocessing #removed abs()
    formatted_float = "{:.2f}".format(Equ) #with .2 number only
    equu = float(formatted_float)          #from string float to float
    prePlst.insert(c2,equu)      # insert the preprocessed value to the List
    c1 =1
    c2 =1
    c0 =1

return prePlst

我的輸入：

我的代碼：

def outLiersFN(*U):
outliers=[] # after preprocessing list
#preprocessing Fc =| 2*LF1 prev by 1 - LF2 prev by 2 |
c0 = -2     #(previous) by 2 #from original
c1 =-1      #(previous)      #from original
c2 =0       #(current)       #from original
c3 = 1      #(next)          #from original
preP = U[0] # original list
if c2 == 0:
    outliers.append(preP[0])
    c1 =1
    c2 =1
    c0 =1
    c3 =1
oldlen = len(preP)
M_RangeOfMotion = 90
while oldlen > c2 :
    if c3 == oldlen:
        outliers.insert(c2, preP[c2]) #preP[c2] >> last element in old list
        break
    if (preP[c2] > M_RangeOfMotion and preP[c2] < (preP[c1]   preP[c3])/2) or (preP[c2] < M_RangeOfMotion and preP[c2] > (preP[c1]   preP[c3])/2): #Check Paper 3.3.1
        Equ = (preP[c1]   preP[c3])/2 #fn of preprocessing # From third index # ==== inserting current frame
        formatted_float = "{:.2f}".format(Equ) #with .2 number only
        equu = float(formatted_float)          #from string float to float
        outliers.insert(c2,equu)      # insert the preprocessed value to the List
        c1 =1
        c2 =1
        c0 =1
        c3 =1
    else :
        Equ = preP[c2] # fn of preprocessing #put same element (do nothing)
        formatted_float = "{:.2f}".format(Equ)  # with .2 number only
        equu = float(formatted_float)  # from string float to float
        outliers.insert(c2, equu)  # insert the preprocessed value to the List
        c1  = 1
        c2  = 1
        c0  = 1
        c3  = 1
return outliers

uj5u.com熱心網友回復：

我建議以下演算法：

• t[i]如果資料點偏離平均值t[i-2], t[i-1], t[i], t[i 1], t[i 2]超過這 5 個元素的標準偏差，則該資料點被視為例外值。
• 例外值被它們周圍的兩個元素的平均值替換。

import matplotlib.pyplot as plt
from statistics import mean, stdev

t = [180,174,168,700,162,94,70,80,128,131,160,180]

def smooth(t):
    new_t = []
    for i, x in enumerate(t):
        neighbourhood = t[max(i-2,0): i 3]
        m = mean(neighbourhood)
        s = stdev(neighbourhood, xbar=m)
        if abs(x - m) > s:
            x = ( t[i - 1   (i==0)*2]   t[i   1 - (i 1==len(t))*2] ) / 2
        new_t.append(x)
    return new_t

new_t = smooth(t)

plt.plot(t)
plt.plot(new_t)
plt.show()

標籤：Python 数学 数学优化 正常化 数据预处理

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