Error: index 2 is out of bounds for axis 0 with size 2

Question

This is my first time here. Apologies, I'm quite new to python. I'm trying to use a monte carlo method to compute an integral in 8D as part of my uni coursework. However, when I run the program, I get the error described in the title on line when i define x0[i]. It seems to only allow i to run up to 2, but I want to investigate the error as N is varied so I need to be able to use N>2. Any help would be appreciated

Jake

a=0          #Defining Limits of Integration 
b=np.pi/8
N=4    #Number of random numbers generated in each dimension

for i in range(N):
     x0rand[i]=((np.pi)/8)*np.random.uniform(0,1) #Generates N random numbers 8D between 0 and pi/8
     x1rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x2rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x3rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x4rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x5rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x6rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x7rand[i]=((np.pi/8))*np.random.uniform(0,1)

def func(x0,x1,x2,x3,x4,x5,x6,x7):  #Defining the integrand
    return (10**6)*np.sin(x0+x1+x2+x3+x4+x5+x6+x7)

integral = 0.0
for i in range(N):
integral += func(x0rand[i],x1rand[i],x2rand[i],x3rand[i],x4rand[i],x5rand[i],x6rand[i],x7rand[i])

answer=(((b-a)**8)/N)*integral
print ('The Integral is:', answer)

IndexError                                Traceback (most recent call last)
 in ()
  4 
  5 for i in range(N):
  ----> 6   x0rand[i]=((np.pi)/8)*np.random.uniform(0,1) #Generates N random numbers 8D between 0 and pi/8
  7   x1rand[i]=((np.pi/8))*np.random.uniform(0,1)
  8   x2rand[i]=((np.pi/8))*np.random.uniform(0,1)

 IndexError: index 2 is out of bounds for axis 0 with size 2

user_2413201 · Answer

The posted program isn't the complete code which has run, otherwise you'd have got the message NameError: name 'x0rand' is not defined. You must have defined and dimensioned x0rand before, and that with size 2 rather than N.

To (re-)define x0rand etc. appropriately, you can put

x0rand, x1rand, x2rand, x3rand, x4rand, x5rand, x6rand, x7rand = np.zeros((8, N))

before the loop.

user_3194618 · Answer

Unfortunately this code doesn't work, anyway i suggest to use the power of numpy:

Some HIT

for i in range(N):
     x0rand[i]=((np.pi)/8)*np.random.uniform(0,1) #Generates N random numbers 8D between 0 and pi/8
     x1rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x2rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x3rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x4rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x5rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x6rand[i]=((np.pi/8))*np.random.uniform(0,1)
     x7rand[i]=((np.pi/8))*np.random.uniform(0,1)

can be initialized in vectorial mode, you can see size parameter.

xrand = np.random.uniform(0,1,size=(N,))

The function, itself can become more efficient. Calling it will be more readable.

def func(arr):  #Defining the integrand
    return (10**6)*np.sin(arr.sum())

func(xrand)

Error: index 2 is out of bounds for axis 0 with size 2

2 Answers