在将代码从VBA转换为Python时遇到麻烦

Question

我有问题将此代码从VBA转换为python，这是一个需要转换为python而不是VBA的函数

Function NC(SPL, pond) As Single
Dim A As Single
Dim B As Single
Dim I As Integer
Dim SPL1(8) As Single
B = 0

If pond = "A" Then
SPL1(1) = SPL(1) + 26.2228
SPL1(2) = SPL(2) + 16.1897
SPL1(3) = SPL(3) + 8.6748
SPL1(4) = SPL(4) + 3.2478
SPL1(5) = SPL(5)
SPL1(6) = SPL(6) - 1.2017
SPL1(7) = SPL(7) - 0.9636
SPL1(8) = SPL(8) + 1.1469
Else
SPL1(1) = SPL(1)
SPL1(2) = SPL(2)
SPL1(3) = SPL(3)
SPL1(4) = SPL(4)
SPL1(5) = SPL(5)
SPL1(6) = SPL(6)
SPL1(7) = SPL(7)
SPL1(8) = SPL(8)
End If

For I = 1 To 8
If I = 1 Then
A = 1.5215 * SPL1(1) - 57.029
ElseIf I = 2 Then
A = 1.2855 * SPL1(2) - 31.628
ElseIf I = 3 Then
A = 1.1853 * SPL1(3) - 18.938
ElseIf I = 4 Then
A = 1.0888 * SPL1(4) - 8.5807
ElseIf I = 5 Then
A = 1.019 * SPL1(5) - 2.0793
ElseIf I = 6 Then
A = 0.9922 * SPL1(6) + 1.2421
ElseIf I = 7 Then
A = 0.9738 * SPL1(7) + 3.2226
ElseIf I = 8 Then
A = 0.9738 * SPL1(8) + 4.1964
End If
If A > B Then
B = A
End If
Next I

NC = Int(Round(B + 0.5))

End Function

这是我目前在python中得到的但是它在python中给了我一个errorcode错误错误说这是python的索引问题我想解决这个错误，使函数工作但不确定如何解决

def NC(SPL,pond):
SPL1 =[(0.0 for x in range(0,8))]
B = 0.0
if pond == 'A':
SPL1[0] = SPL[0] + 26.228
SPL1[1] = SPL[1] + 16.1897
SPL1[2] = SPL[2] + 8.6748
SPL1[3] = SPL[3] + 3.2478
SPL1[4] = SPL[4]
SPL1[5] = SPL[5] - 1.2017
SPL1[6] = SPL[6] - 0.9636
SPL1[7] = SPL[7] + 1.1469
else:
SPL1[0] = SPL[0]
SPL1[1] = SPL[1] 
SPL1[2] = SPL[2] 
SPL1[3] = SPL[3] 
SPL1[4] = SPL[4]
SPL1[5] = SPL[5] 
SPL1[6] = SPL[6] 
SPL1[7] = SPL[7] 
for i in range(0, 8):
if i == 0:
A = 1.5215 * SPL1[0] -57.029
if i == 1:
A = 1.2855 * SPL1[1] -31.628
if i == 2:
A = 1.1853 * SPL1[2] -18.938   
if i == 3:
A = 1.0888 * SPL1[3] -8.5807     
if i == 4:
A = 1.019 * SPL1[4] -2.0793
if i == 5:
A = 0.9922 * SPL1[5] +1.2421
if i == 6:
A = 0.9738 * SPL1[6] +3.2226
if i == 7:
A = 0.9738 * SPL1[7] +4.1964    
if A > B:
B = A
return int(B+0.5)

Answer 1

Python是动态类型化和0基数索引的。按照惯例，也使用小写的变量名。所以你的代码应该是这样的:

def nc(spl, pond):

spl1 =[0.0 for _ in range(8))]

b = 0.0

if pond == 'A':
spl1[0] = spl[0] + 26.228
#and so on
else:
spl1[0] = spl[0]
# and so on
for i in range(0, 8):
if i == 0:
a = 1.5215 * spl1[0] -57.029
# and so on
if a > b:
b = a

return int(b+0.5)