Question

我是python的新手，我被分配创建自己的算法来解决线性回归而不使用任何导入。问题是，当我尝试我的程序计算错误时，它给出了奇怪的值（我将它与微软excel的计算进行比较）。这是我的计划：

x=[1.,1.,2.,2.,2.,2.,2.,2.,2.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,3.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,4.,5.,5.,5.,5.,5.,5.,5.,5.,6.]
y=[67.,62.,109.,83.,91.,88.,123.,100.,109.,137.,131.,122.,122.,118.,115.,131.,143.,142.,122.,140.,150.,140.,150,150.,140.,150.,130.,130.,138.,135.,146.,146.,145.,145.,144.,140.,150.,152.,157.,155.,153.,154.,158.,162.,161.,162.,165.,171.,162.,169.,167.,150.,170.,140.,140.,150.,150.,150.,160.,150.,150.,150.,150.,140.,160.,170.,160.,160.,170.,171.,188.,170.,150.,150.,160.,160.,180.,170.]
sumx = 0
sumxdoubled = 0
sumxsquare = 0
sumxy = 0
meanx = 0
sumy = 0
sumerror = 0
n= 78

for i in range(78):
   sumxy = sumxy + (x[i] * y[i])
print("Total (x.y) : ",sumxy)

for i in range(78):
   sumx = sumx + x[i]
print("Total x : ",sumx)

for i in range(78):
   sumxsquare = sumxsquare + (x[i] ** 2)
print("Total (x^2) : ",sumxsquare)

sumxdoubled = sumx ** 2
print("(Total x)^2 : ",sumxdoubled)

meanx = sumx / n
print("Average x : ",meanx)

for i in range(78):
   sumy = sumy + y[i]
print("Total y : ",sumy)

meany = sumy / n
print("Average y : ",meany)

a1 = ((n*sumxy) - (sumx * sumy)) / ((n*sumxsquare) - sumxdoubled)
print("a1 = ",a1)

a0 = meany - a1 * meanx
print("a0 = ",a0)
for i in range (78):
   sumerror = sumerror + (y[i] - a0 - (a1 * x[i]))
print("Total error = ",sumerror)

输出是：

Total (x.y) :  42117.0
Total x :  283.0
Total (x^2) :  1093.0
(Total x)^2 :  80089.0
Average x :  3.628205128205128
Total y :  11201.0
Average y :  143.60256410256412
a1 =  22.312294288480153
a0 =  62.64898354307843
Total error =  -7.673861546209082e-13

使用microsoft excel尝试相同数据时的错误值为-14.25

为什么python提供的值甚至不接近excel值-14.25？我无法猜出程序有什么问题，因为我确定我使用正确的算法来计算错误。

Answer 1

你的问题不在于python，而在于你的数学问题。计算错误时，首先必须添加括号以确保正在进行正确的计算：

sumerror = sumerror + (y[i] - a0 - (a1 * x[i])) # <-- missing brackets
sumerror = sumerror + (y[i] - (a0 - (a1 * x[i])))

但是你甚至没有完成，你需要将这个结果除以n然后取平方根。

>>> sumerror = (sumerror / n)**0.5
>>> print("Total error = ",sumerror)
Total error = 12.724274483009689

由于这是编程论坛中的一个问题，我会指出，当你在那里时，你可以使用一些内置函数来让自己更轻松。

for i in range(78):
   sumxy = sumxy + (x[i] * y[i])

很糟糕，您已经硬编码了每次使用新列表时需要更新的列表长度。有一个内置函数len（），它将为您提供此功能。在这种情况下，甚至不需要，您可以使用sum（）和稍微更高级的zip来将列表连接在一起。

# zip(x, y) returns an iterator like [(x0, y0), (x1, y1), ..., (xn, yn)]
>>> sumxy = sum(x*y for x, y in zip(x, y))
>>> print("Total (x.y) : ",sumxy)
Total (x.y) :  42117.0

Python，尝试计算线性回归误差时的奇怪值

1 个答案: