Question

我使用嵌套的scipy.integrate.quad调用来集成2维被积函数。被积函数由numpy函数组成 - 因此传递一个输入数组要高得多，而不是循环输入并为每个输入调用一次 - 由于numpy的数组，它快了~2个数量级。 / p>

然而....如果我想将我的被积函数只在一个维度上集成 - 但是在其他维度上输入数组的东西会掉下来 - 看起来像'scipy'的quadpack包不能做任何事情这是numpy处理阵列输入。有没有其他人看过这个 - 或者找到了修复它的方法 - 或者我误解了它。我从quad得到的错误是：

Traceback (most recent call last):
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 159, in <module>
    fnIntegrate_x(0, 1, NCALLS_SET, True)
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 35, in fnIntegrate_x
    I = Integrate_x(yarray)
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 23, in Integrate_x
    return quad(Integrand, 0, np.pi/2, args=(y))[0]
  File "C:\Python27\lib\site-packages\scipy\integrate\quadpack.py", line 247, in quad
    retval = _quad(func,a,b,args,full_output,epsabs,epsrel,limit,points)
  File "C:\Python27\lib\site-packages\scipy\integrate\quadpack.py", line 312, in _quad
    return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
quadpack.error: Supplied function does not return a valid float.

我已经把我正在尝试做的卡通版本放在下面 - 我实际上做的是一个更复杂的被积函数，但这是一个陀螺。

肉位于顶部 - 底部正在进行基准测试以显示我的观点。

import numpy as np
import time

from scipy.integrate import quad


def Integrand(x, y):
    '''
        Integrand
    '''
    return np.sin(x)*np.sin( y )

def Integrate_x(y):
    '''
        Integrate over x given (y)
    '''
    return quad(Integrand, 0, np.pi/2, args=(y))[0]



def fnIntegrate_x(ystart, yend, nsteps, ArrayInput = False):
    '''

    '''

    yarray = np.arange(ystart,yend, (yend - ystart)/float(nsteps))
    I = np.zeros(nsteps)
    if ArrayInput :
        I = Integrate_x(yarray)
    else :
        for i,y in enumerate(yarray) :

            I[i] = Integrate_x(y)

    return y, I




NCALLS_SET = 1000
NSETS = 10

SETS_t = np.zeros(NSETS)

for i in np.arange(NSETS) :

    XInputs = np.random.rand(NCALLS_SET, 2)

    t0 = time.time()
    for x in XInputs :
        Integrand(x[0], x[1])
    t1 = time.time()
    SETS_t[i] = (t1 - t0)/NCALLS_SET

print "Benchmarking Integrand - Single Values:"
print "NCALLS_SET: ", NCALLS_SET
print "NSETS: ", NSETS    
print "TimePerCall(s): ", np.mean(SETS_t) , np.std(SETS_t)/ np.sqrt(SETS_t.size)        
print "TotalTime: ",np.sum(SETS_t) * NCALLS_SET
'''
Benchmarking Integrand - Single Values:
NCALLS_SET:  1000
NSETS:  10
TimePerCall(s):  1.23999834061e-05 4.06987868647e-06
'''








NCALLS_SET = 1000
NSETS = 10

SETS_t = np.zeros(NSETS)

for i in np.arange(NSETS) :

    XInputs = np.random.rand(NCALLS_SET, 2)

    t0 = time.time()
    Integrand(XInputs[:,0], XInputs[:,1])
    t1 = time.time()
    SETS_t[i] = (t1 - t0)/NCALLS_SET

print "Benchmarking Integrand - Array Values:"
print "NCALLS_SET: ", NCALLS_SET
print "NSETS: ", NSETS    
print "TimePerCall(s): ", np.mean(SETS_t) , np.std(SETS_t)/ np.sqrt(SETS_t.size)        
print "TotalTime: ",np.sum(SETS_t) * NCALLS_SET
'''
Benchmarking Integrand - Array Values:
NCALLS_SET:  1000
NSETS:  10
TimePerCall(s):  2.00009346008e-07 1.26497018465e-07
'''












NCALLS_SET = 1000
NSETS = 100

SETS_t = np.zeros(NSETS)

for i in np.arange(NSETS) :


    t0 = time.time()
    fnIntegrate_x(0, 1, NCALLS_SET, False)
    t1 = time.time()
    SETS_t[i] = (t1 - t0)/NCALLS_SET

print "Benchmarking fnIntegrate_x - Single Values:"
print "NCALLS_SET: ", NCALLS_SET
print "NSETS: ", NSETS    
print "TimePerCall(s): ", np.mean(SETS_t) , np.std(SETS_t)/ np.sqrt(SETS_t.size)        
print "TotalTime: ",np.sum(SETS_t) * NCALLS_SET
'''
NCALLS_SET:  1000
NSETS:  100
TimePerCall(s):  0.000165750000477 8.61204306241e-07
TotalTime:  16.5750000477
'''








NCALLS_SET = 1000
NSETS = 100

SETS_t = np.zeros(NSETS)

for i in np.arange(NSETS) :


    t0 = time.time()
    fnIntegrate_x(0, 1, NCALLS_SET, True)
    t1 = time.time()
    SETS_t[i] = (t1 - t0)/NCALLS_SET

print "Benchmarking fnIntegrate_x - Array Values:"
print "NCALLS_SET: ", NCALLS_SET
print "NSETS: ", NSETS    
print "TimePerCall(s): ", np.mean(SETS_t) , np.std(SETS_t)/ np.sqrt(SETS_t.size)        

'''
****  Doesn't  work!!!! *****
Traceback (most recent call last):
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 159, in <module>
    fnIntegrate_x(0, 1, NCALLS_SET, True)
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 35, in fnIntegrate_x
    I = Integrate_x(yarray)
  File "C:\Users\JP\Documents\Python\TestingQuad\TestingQuad_v2.py", line 23, in Integrate_x
    return quad(Integrand, 0, np.pi/2, args=(y))[0]
  File "C:\Python27\lib\site-packages\scipy\integrate\quadpack.py", line 247, in quad
    retval = _quad(func,a,b,args,full_output,epsabs,epsrel,limit,points)
  File "C:\Python27\lib\site-packages\scipy\integrate\quadpack.py", line 312, in _quad
    return _quadpack._qagse(func,a,b,args,full_output,epsabs,epsrel,limit)
quadpack.error: Supplied function does not return a valid float.

'''

Answer 1

害怕我在这里以否定的方式回答我自己的问题。我不认为这是可能的。看起来像quad是用其他东西写的库的某种端口 - 因此内部的库定义了事情是如何完成的 - 所以如果不重新设计库本身，它可能不可能做我想要的。< / p>

对于有多D集成时序问题的其他人，我发现最好的方法是使用专用的集成库。我发现'cuba'似乎有一些非常高效的多D集成例程。

http://www.feynarts.de/cuba/

这些例程是用c语言编写的，所以我最终使用SWIG与他们交谈 - 最后也是为了提高效率在c中重写了我的被积函数 - 这加快了负载....

Answer 2

可以通过numpy.vectorize函数。我有这个问题很长一段时间，然后来到这个vectorize函数。

你可以像这样使用它：

vectorized_function = numpy.vectorize(your_function)

output = vectorized_function(your_array_input)

Answer 3

使用quadpy（属于我的项目）。它是完全矢量化的，因此可以处理任何形状的数组值函数，并且执行速度非常快。

Answer 4

我在从 -np.inf 到 np.inf 在所有维度上整合概率密度函数时遇到了这个问题。

我通过创建一个包含 *args 的包装函数、将 args 转换为一个 numpy 数组并集成包装函数来修复它。

我认为使用 numpy 的 vectorize 只会对所有值都相等的子空间进行积分。

这是一个例子：

from scipy.integrate import nquad
from scipy.stats import multivariate_normal

mean = [0., 0.]
cov = np.array([[1., 0.],
                [0., 1.]])

bivariate_normal = multivariate_normal(mean=mean, cov=cov)

def pdf(*args):
    x = np.array(args)
    return bivariate_normal.pdf(x)

integration_range = [[-18, 18], [-18, 18]]

nquad(pdf, integration_range)

Output: (1.000000000000001, 1.3429066352690133e-08)

如何使用数组输入调用python scipy quad

4 个答案: