tplquad:TypeError:'float'对象不可调用

时间:2015-04-12 18:20:09

标签: python scipy integrate

我正在尝试运行此代码,但我在标题中收到错误。我查找了关于tplquad的文档和示例,但我无法理解我的问题。非常感谢你提前!

这里是我的代码:

from numpy import *
from pylab import *

import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from math import *
from scipy.integrate import quad,dblquad,tplquad
from scipy.integrate import nquad

fig_width = 6.
fig_height = fig_width*0.75
fig_size =  [fig_width,fig_height]
params = {'backend': 'TkAgg',
          'axes.labelsize': 30,
          'text.fontsize': 20,
          'title.fontsize': 20,
          'legend.fontsize': 20,
          'xtick.labelsize': 20,
          'ytick.labelsize': 20,
          'text.usetex': False,
          'font.family': 'sans-serif',
          'figure.figsize': fig_size}
rcParams.update(params)

pi=3.14
pt_T=3.
#T=0.47
thetaP= -pi
precision=5
y=0


M_T=linspace(1.,7.,precision)
integral1d=[0]*precision


#chi now is def with a plus instead of the minus in the article
def chi(thetap1,p1,thetaP,T,M_T):
    return abs((2*p1*T*sqrt(pt_T**2+(M_T**2+pt_T**2)*sinh(y)**2)*sin(thetaP)*sin(thetap1))**2 - (2*p1*T*(sqrt(M_T**2+pt_T**2)*cosh(y)- sqrt(pt_T**2+(M_T**2+pt_T**2)*sinh(y)**2)*cos(thetaP) *cos(thetap1) )-(T**2)* M_T**2)**2)+1

def p1max(thetaP, thetap1,T,M_T):
    return (M_T**2)*T/(2*(sqrt(M_T**2+pt_T**2)*cosh(y)- sqrt(pt_T**2+(M_T**2+pt_T**2))*sinh(y)**2*cos(thetaP-thetap1)))-0.1

def p1min(thetaP, thetap1,T,M_T):
    #NOT SURE ABOUT THE T AT DENOMINATOR
    return (M_T**2)*T/(2*(sqrt(M_T**2+pt_T**2)*cosh(y)- sqrt(pt_T**2+(M_T**2+pt_T**2))*sinh(y)**2*cos(thetaP+thetap1))) +0.1

def integral(thetaP,T,M_T): 
    area =dblquad(lambda p1, thetap1: 5*(1/(18*pi**5))*sin(thetap1)*(p1/(sqrt(chi(thetap1,p1,thetaP,T,M_T))))*(1/(exp(p1/T) + 1))*(1/(exp((sqrt(M_T**2 + pt_T**2)*cosh(y) - p1/T) +1))) , -pi+0.1, -0.1, lambda p1: p1min(thetaP, p1,T,M_T), lambda p1: p1max(thetaP,p1,T,M_T))  #CHANGE   1., lambda p1:10.)
    return area[0]

def integrand(M_T, p1,thetap1,T):
    return pt_T*T*2*pi*5*(1/(18*pi**5))*sin(thetap1)*(p1/(sqrt(chi(thetap1,p1,thetaP,T,M_T))))*(1/(exp(p1/T) + 1))*(1/(exp((sqrt(M_T**2 + pt_T**2)*cosh(y) - p1/T) +1)))


def formula151(M_T):
    area =tplquad(lambda  p1, thetap1,T:  pt_T*T*2*pi*5*(1/(18*pi**5))*sin(thetap1)*(p1/(sqrt(chi(thetap1,p1,thetaP,T,M_T))))*(1/(exp(p1/T) + 1))*(1/(exp((sqrt(M_T**2 + pt_T**2)*cosh(y) - p1/T) +1))) ,0.333, 20./3,lambda thetap1: -pi+0.1, -0.1, lambda thetap1, p1: p1min(thetaP, p1,T,M_T),lambda thetap1,p1: p1max(thetaP,p1,T,M_T) ) 
    return area[0]


#solving the integral
for ind in range(0, precision):
    integral1d[ind]=formula151( M_T[ind])
    print integral1d[ind]



plot(M_T,integral1d)
xlabel('M/T')
ylabel('prod rate')
title('thetaP =-3.12')
plt.yscale('log')
#plt.xscale('log')
show()

错误来自第57行,其中使用了tplquad,完整的追溯是

---------------------------------------------------------------------------
TypeError                                 Traceback (most recent call last)
/usr/lib/python2.7/dist-packages/IPython/utils/py3compat.pyc in execfile(fname, *where)
    202             else:
    203                 filename = fname
--> 204             __builtin__.execfile(filename, *where)

/home/chiara/Scrivania/formula15a.py in <module>()
     61 #solving the integral
     62 for ind in range(0, precision):
---> 63         integral1d[ind]=formula151( M_T[ind])
     64         print integral1d[ind]
     65 

/home/chiara/Scrivania/formula15a.py in formula151(M_T)
     55 
     56 def formula151(M_T):
---> 57         area =tplquad(lambda  p1, thetap1,T:  pt_T*T*2*pi*5*(1/(18*pi**5))*sin(thetap1)*(p1/(sqrt(chi(thetap1,p1,thetaP,T,M_T))))*(1/(exp(p1/T) + 1))*(1/(exp((sqrt(M_T**2 + pt_T**2)*cosh(y) - p1/T) +1))) ,0.333, 20./3,lambda thetap1: -pi+0.1, -0.1, lambda thetap1, p1: p1min(thetaP, p1,T,M_T),lambda thetap1,p1: p1max(thetaP,p1,T,M_T) )
     58         return area[0]
     59 

/usr/lib/python2.7/dist-packages/scipy/integrate/quadpack.pyc in tplquad(func, a, b, gfun, hfun, qfun, rfun, args, epsabs, epsrel)
    498 
    499     """
--> 500     return dblquad(_infunc2,a,b,gfun,hfun,(func,qfun,rfun,args),epsabs=epsabs,epsrel=epsrel)
    501 
    502 

/usr/lib/python2.7/dist-packages/scipy/integrate/quadpack.pyc in dblquad(func, a, b, gfun, hfun, args, epsabs, epsrel)
    433 
    434     """
--> 435     return quad(_infunc,a,b,(func,gfun,hfun,args),epsabs=epsabs,epsrel=epsrel)
    436 
    437 

/usr/lib/python2.7/dist-packages/scipy/integrate/quadpack.pyc in quad(func, a, b, args, full_output, epsabs, epsrel, limit, points, weight, wvar, wopts, maxp1, limlst)
    252         args = (args,)
    253     if (weight is None):
--> 254         retval = _quad(func,a,b,args,full_output,epsabs,epsrel,limit,points)
    255     else:
    256         retval = _quad_weight(func,a,b,args,full_output,epsabs,epsrel,limlst,limit,maxp1,weight,wvar,wopts)

/usr/lib/python2.7/dist-packages/scipy/integrate/quadpack.pyc in _quad(func, a, b, args, full_output, epsabs, epsrel, limit, points)
    317     if points is None:
    318         if infbounds == 0:
--> 319             return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
    320         else:
    321             return _quadpack._qagie(func,bound,infbounds,args,full_output,epsabs,epsrel,limit)

/usr/lib/python2.7/dist-packages/scipy/integrate/quadpack.pyc in _infunc(x, func, gfun, hfun, more_args)
    379 def _infunc(x,func,gfun,hfun,more_args):
    380     a = gfun(x)
--> 381     b = hfun(x)
    382     myargs = (x,) + more_args
    383     return quad(func,a,b,args=myargs)[0]

TypeError: 'float' object is not callable

注意:功能&#34;积分&#34;和&#34; integrand&#34;是定义但最终没有使用......我只是把它们留在那里

1 个答案:

答案 0 :(得分:1)

scipy.integrate docs开始,tplquad的签名(计算三重积分的数值近似值)为:

  

scipy.integrate.tplquad(func,a,b,gfun,hfun,qfun,rfun,args =(),epsabs = 1.49e-08,epsrel = 1.49e-08)

其中func是要集成的三个变量的函数,ab是外积分的gfunhfun的浮点限制是一个变量的函数给出中间积分的极限,qfunrfun是两个变量的函数,给出了最里面积分的极限。

我很难弄清楚发生了什么,直到我重新格式化你的代码以使其更具可读性。这是您对tplquad的调用,重新格式化以使线条长度更短:

area = tplquad(
    lambda p1, thetap1, T: (
        pt_T*T*2*pi*5*(1/(18*pi**5))*sin(thetap1)*
        (p1/(sqrt(chi(thetap1,p1,thetaP,T,M_T))))*
        (1/(exp(p1/T) + 1))*
        (1/(exp((sqrt(M_T**2 + pt_T**2)*cosh(y) - p1/T) + 1)))
    ),
    0.333,  # a
    20./3,  # b
    lambda thetap1: -pi + 0.1,  # gfun
    -0.1,                       # hfun
    lambda thetap1, p1: p1min(thetaP, p1, T, M_T),  # qfun
    lambda thetap1, p1: p1max(thetaP, p1, T, M_T),  # rfun
)

(正如@ user2357112建议的那样,它还有助于将这些lambda表达式从调用中拉出来并将它们定义为单独的函数。特别是,如果为被积函数定义一个单独的函数,则可以使用它。能够逐个执行计算,并且不需要将一个巨大的表达式放在一行上。)

重新格式化后,您可以更轻松地查看问题所在:在致电tplquad时,您正在为-0.1传递常量hfun 。这不会起作用:在数学中,人们可以(ab)使用常数值来表示一个常数函数,但是编程语言(以及一些数学家,那些)往往会更挑剔:你&#39;这里需要一个实际的功能。将-0.1替换为lambda thetap1: -0.1

顺便说一下,我对你的变量订单也有点怀疑。 doc页面说明,虽然被积函数的输入顺序应该是(z, y, x)gfunhfun应该只是xqfun的函数, rfun应该是(x, y)的功能(按此顺序)。这似乎与您所拥有的不相符。