Question

好的，所以我的辛普森规则被定义为：

def simpsonsRule(func, a, b, n, p0, r0):
    if n%2 == 1:
        return "Not applicable"
    else:
        h = (b - a) / float(n)
        s = func(a, p0, r0) + sum((4 if i%2 == 1 else 2) * func(a+i*h, p0, r0) for i in range(1,n)) + func(b, p0, r0)
        return s*h/3.0

然而，当我做类似的事情时：

def integrate_NFW(rx,ps,rs):
    rho = ps/((r/rs)*((1+(r/rs))**2))
    function_result = rho * 4.0 * np.pi * rx**2
    return function_result
def chisqfuncNFW(iter_vars):
    global v_model
    #Normalizes p0 (p0 is too large relative to rc)
    ps = iter_vars[0] * 3.85e+09
    rs = iter_vars[1]
    for index in range(0, am):
        integral_result = simpsonsRule(integrate_NFW, 0.0, r[index], 200, ps, rs)
        print(integral_result)

当你打印出integral_result时，它返回一个数字数组：

[  1.58771810e+13   3.68633515e+12   1.60346051e+12   8.81279407e+11
   5.37962555e+11   3.54826396e+11   2.49107306e+11   1.80747811e+11
   1.36318422e+11   1.05440828e+11   8.32851651e+10   6.66410643e+10
   5.41730944e+10   4.48302130e+10]

所以integral_result = simpsonsRule(integrate_NFW, 0.0, r[index], 200, ps, rs)返回一个数组而不是一个数字

我想补充说，对于我的另一个模型，它工作正常（它返回一个数字而不是数组）：

def integrate_Burk(rx,p0,r0):
    rho = (p0 * r0**3) / ( (rx + r0) * (rx**2 + r0**2) )
    function_result = rho * 4.0 * np.pi * rx**2
    return function_result
def chisqfuncBurk(iter_vars):
    global v_model
    #Normalizes p0 (p0 is too large relative to rc)
    p0 = iter_vars[0] * 3.85e+09
    r0 = iter_vars[1]
    v_model = []
    for index in range(0, am):
        integral_result = simpsonsRule(integrate_Burk, 0.0, r[index], 200, p0, r0)

此外，r是一个数字数组：

0.22
0.66
1.11
1.55
2.00
2.45
2.89
3.34
3.78
4.22
4.66
5.11
5.56
6.00

和am是r中的数字量（在这种情况下，我相信它是14）

如果我遗漏了任何内容或者您想要其他代码，请告诉我

编辑以下是一些复制错误的代码

from scipy.optimize import*
import numpy as np
am = 14
r = np.array([0.22,
0.66,
1.11,
1.55,
2.00,
2.45,
2.89,
3.34,
3.78,
4.22,
4.66,
5.11,
5.56,
6.00])
def simpsonsRule(func, a, b, n, p0, r0):
    if n%2 == 1:
        return "Not applicable"
    else:
        h = (b - a) / float(n)
        s = func(a, p0, r0) + sum((4 if i%2 == 1 else 2) * func(a+i*h, p0, r0) for i in range(1,n)) + func(b, p0, r0)
        return s*h/3.0

def integrate_NFW(rx,ps,rs):
    rho = ps/((r/rs)*((1+(r/rs))**2))
    function_result = rho * 4.0 * np.pi * rx**2
    return function_result
def chisqfuncNFW(iter_vars):
    global v_model
    #Normalizes p0 (p0 is too large relative to rc)
    ps = iter_vars[0] * 3.85e+09
    rs = iter_vars[1]
    for index in range(0, am):
        integral_result = simpsonsRule(integrate_NFW, 0.0, r[index], 200, ps, rs)
        print(integral_result)
initial_guess = np.array([1.0, 2.0])
resNFW = minimize(chisqfuncNFW, initial_guess,method = 'Nelder-Mead')

Answer 1

您可以访问函数中的全局数组r：

def integrate_NFW(rx,ps,rs):
    rho = ps/((r/rs)*((1+(r/rs))**2))

这会将rho变成数组。

Simpson的规则在Python中返回一个数组

1 个答案: