Python 计算等效 A 声级

import math

with open("./data.txt") as noise_data:
    lines = noise_data.readlines()
    number_string = '\t'.join(lines)

number = number_string .split('\t')
number = [x.strip() for x in number if x.strip()!='']
number.pop(0)
number.pop(200)
# number.pop(400)

def L_number(order):
    for i in number[:order]:
        l = 0
        l = l + (1/order)*pow(order, (0.1)*float(i))
    L_10 = 10*math.log(10,l)
    
    for i in number[:order+40]:
        l = 0
        l = l + (1/(order + 40))*pow(order, (0.1)*float(i))
    L_50 = 10*math.log(10,l)

    for i in number[:order+80]:
        l = 0
        l = l + (1/(order + 80))*pow(order, (0.1)*float(i))
    L_90 = 10*math.log(10,l)

    L_eq = L_50 + ((L_10 - L_90)**2)/60