快速计算在Numpy中设置的大数标准差的方法

def randomInterval(a,b):
    r = ((b-a)*float(random.random(1)) + a)
    return r 

N = 10e6
Sum = 0
x = []
for sample in range(0,int(N)):
    n = randomInterval(-5.,5.)
    while n == 5.0:
        n = randomInterval(-5.,5.) # since X is [-5,5)
    Sum += n
    x = np.append(x, n)

A = Sum/N

for sample in range(0,int(N)):
    summation = (x[sample] - A)**2.0

standard_deviation = np.sqrt((1./N)*summation)

import numpy as np
N = int(1e6)
a = np.random.uniform(-5,5,size=(N,))
standard_deviation = np.std(a)