我前段时间编写了一段代码，使用文本文件中的数据处理光谱并对其进行计算。我开始使用的代码只是在没有任何功能的情况下逐行完成，尽管很长，但它在2.11秒内完成运行（根据%%timeit）。下面是原始代码，标记为。
但是，我希望将代码放入函数中，以便将来更容易阅读和使用不同的模型。尽管我使用了与之前相同的所有步骤（但这次是在我的函数中），但速度要慢得多。此代码也在下面。现在，我必须等待大约15-20分钟才能获得相同的输出。为什么它这么慢，有什么方法可以让它明显更快但仍然使用函数？
原始代码：
import re
import matplotlib.pyplot as plt
import numpy as np
import scipy.interpolate
filename = 'bpass_spectra.txt'
extinctionfile = 'ExtinctionLawPoints.txt' # from R_V = 4.0
pointslist = []
datalist = []
speclist = []
# Constants
Msun = 1.98892e30 # solar mass [kg]
h = 4.1357e-15 # Planck's constant [eV s]
c = float(3e8) # speed of light [m/s]
# Read spectra file
f = open(filename, 'r')
rawspectra = f.readlines()
met = re.findall('Z\s=\s(\d*\.\d+)', rawspectra[0])
del rawspectra[0]
for i in range(len(rawspectra)):
newlist = rawspectra[i].split(' ')
datalist.append(newlist)
# Read extinction curve data file
rawpoints = open(extinctionfile, 'r').readlines()
for i in range(len(rawpoints)):
newlst = re.split('(?!\S)\s(?=\S)|(?!\S)\s+(?=\S)', rawpoints[i])
pointslist.append(newlst)
pointslist = pointslist[3:]
lambdalist = [float(item[0]) for item in pointslist]
k_abslist = [float(item[4]) for item in pointslist]
xvallist = [(c*h)/(lamb*1e-6) for lamb in lambdalist]
k_interp = scipy.interpolate.interp1d(xvallist, k_abslist)
# Create new lists
Elist = [float(item[0]) for item in datalist]
speclambdalist = [h*c*1e9/E for E in Elist]
z1list = [float(item[1]) for item in datalist]
speclist.extend(z1list)
met = met[0]
klist = [None]*len(speclist)
Loutlist = [None]*len(speclist)
Tlist = [None]*len(speclist)
# Define parameters
b = 2.0
R = 1.0
z = 1.0
Mgas = 1.0 # mass of gas, input
Mhalo = 2e41 # mass of dark matter halo, known
if float(met) > 0.0052:
DGRlist = [50.0*np.exp(-2.21)*float(met)]*len(speclist)
elif float(met) <= 0.0052:
DGRlist = [((50.0*float(met))**3.15)*np.exp(-0.96)]*len(speclist)
for i in range(len(speclist)):
if float(Elist[i]) <= 4.1357e-3: # frequencies <= 10^12 Hz
klist[i] = 0.1*(float(Elist[i])/(1000.0*h))**b # extinction law [cm^2/g]
elif float(Elist[i]) > 4.1357e-3: # frequencies > 10^12 Hz
klist[i] = k_interp(Elist[i]) # interpolated function's value at Elist[i]
Mdustlist = [Mgas*DGR for DGR in DGRlist] # dust mass
Rhalo = 0.784*(0.27**2.0)*(0.7**(-2.0/3.0))*float(10.0/(1.0+z))*((Mhalo/(1e8*Msun))**(1.0/3.0))
Rdust = 0.018*Rhalo # [kpc]
for i in range(len(speclist)):
Tlist[i] = 3*Mdustlist[i]*klist[i]/(4*np.pi*Rdust)
Linlist = [float(spectra)*R for spectra in speclist]
# Outgoing luminosity as function of wavelength
for i in range(len(Linlist)):
Loutlist[i] = Linlist[i]*np.exp(-Tlist[i])
# Test the calculation
print "LIN ELEMENTS 0 AND 1000:", Linlist[0], Linlist[1000]
print "LOUT ELEMENTS 0 AND 1000:", Loutlist[0], Loutlist[1000]
新功能＆＃34;功能化＆＃34;代码（慢得多）：
import re
import matplotlib.pyplot as plt
import numpy as np
import scipy.interpolate
# Required files and lists
filename = 'bpass_spectra.txt' # number of columns = 4
extinctionfile = 'ExtinctionLawPoints.txt' # R_V = 4.0
datalist = []
if filename == 'bpass_spectra.txt':
filetype = 4
else:
filetype = 1
if extinctionfile == 'ExtinctionLawPoints.txt':
R_V = 4.0
else:
R_V = 1.0 #to be determined
# Constants
M_sun = 1.98892e30 # solar mass [kg]
h = 4.1357e-15 # Planck's constant [eV s]
c = float(3e8) # speed of light [m/s]
# Inputs
beta = 2.0
R = 1.0
z = 1.0
M_gas = 1.0
M_halo = 2e41
# Read spectra file
f = open(filename, 'r')
rawlines = f.readlines()
met = re.findall('Z\s=\s(\d*\.\d+)', rawlines[0])
del rawlines[0]
for i in range(len(rawlines)):
newlist = rawlines[i].split(' ')
datalist.append(newlist)
# Read extinction curve data file
rawpoints = open(extinctionfile, 'r').readlines()
def interpolate(R_V, rawpoints, Elist, j):
pointslist = []
if R_V == 4.0:
for i in range(len(rawpoints)):
newlst = re.split('(?!\S)\s(?=\S)|(?!\S)\s+(?=\S)', rawpoints[i])
pointslist.append(newlst)
pointslist = pointslist[3:]
lambdalist = [float(item[0]) for item in pointslist]
k_abslist = [float(item[4]) for item in pointslist]
xvallist = [(c*h)/(lamb*1e-6) for lamb in lambdalist]
k_interp = scipy.interpolate.interp1d(xvallist, k_abslist)
return k_interp(Elist[j])
# Dust extinction function
def dust(interpolate, filetype, datalist, beta, R, z, M_gas, M_halo, met):
speclist = []
if filetype == 4:
metallicity = float(met[0])
Elist = [float(item[0]) for item in datalist]
speclambdalist = [h*c*1e9/E for E in Elist]
met1list = [float(item[1]) for item in datalist]
speclist.extend(met1list)
klist, Tlist = [None]*len(speclist), [None]*len(speclist)
if metallicity > 0.0052:
DGRlist = [50.0*np.exp(-2.21)*metallicity]*len(speclist) # dust to gas ratio
elif metallicity <= 0.0052:
DGRlist = [((50.0*metallicity)**3.15)*np.exp(-0.96)]*len(speclist)
for i in range(len(speclist)):
if Elist[i] <= 4.1357e-3: # frequencies <= 10^12 Hz
klist[i] = 0.1*(float(Elist[i])/(1000.0*h))**beta # extinction law [cm^2/g]
elif Elist[i] > 4.1357e-3: # frequencies > 10^12 Hz
klist[i] = interpolate(R_V, rawpoints, Elist, i) # interpolated function's value at Elist[i]
Mdustlist = [M_gas*DGR for DGR in DGRlist] # dust mass
R_halo = 0.784*(0.27**2.0)*(0.7**(-2.0/3.0))*float(10/(1+z))*((M_halo/(1e8*M_sun))**(1.0/3.0))
R_dust = 0.018*R_halo # [kpc]
# Optical depth calculation
Tlist = [3*Mdustlist[i]*klist[i]/(4*np.pi*R_dust) for i in range(len(speclist))]
# Ingoing and outgoing luminosities as functions of wavelength
Linlist = [float(spectra)*R for spectra in speclist]
Loutlist = [Linlist[i]*np.exp(-Tlist[i]) for i in range(len(speclist))]
return speclambdalist, Linlist, Loutlist
print dust(interpolate, filetype, datalist, beta, R, z, M_gas, M_halo, met)
即使我只有函数返回Loutlist而不是3个列表的元组，它仍然非常慢。有关为什么会这样的想法？此外，我想要返回元组，然后绘制speclambdalist与Linlist的关系，并在同一个地块上绘制speclambdalist与Loutlist的关系。但我的印象是，每当我致电dust(interpolate, filetype, datalist, beta, R, z, M_gas, M_halo, met)[i] i = 0, 1, or 2 {我将多次这样做时），它都必须再次运行该功能时间。有没有办法绕过这些额外的运行来进一步提高速度？谢谢！