If I have a list like this:
results=[-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439,
0.53459687, -1.34069996, -1.61042692, -4.032
Numpy is indeed the most elegant and fast way to do it.
I think the actual question was about how to access the individual elements of a list to do such a calculation yourself, so below an example:
results=[-14.82381293, -0.29423447, -13.56067979, -1.6288903, -0.31632439,
0.53459687, -1.34069996, -1.61042692, -4.03220519, -0.24332097]
import numpy as np
print 'numpy variance: ', np.var(results)
# without numpy by hand
# there are two ways of calculating the variance
# - 1. direct as central 2nd order moment (https://en.wikipedia.org/wiki/Moment_(mathematics))divided by the length of the vector
# - 2. "mean of square minus square of mean" (see https://en.wikipedia.org/wiki/Variance)
# calculate mean
n= len(results)
sum=0
for i in range(n):
sum = sum+ results[i]
mean=sum/n
print 'mean: ', mean
# calculate the central moment
sum2=0
for i in range(n):
sum2=sum2+ (results[i]-mean)**2
myvar1=sum2/n
print "my variance1: ", myvar1
# calculate the mean of square minus square of mean
sum3=0
for i in range(n):
sum3=sum3+ results[i]**2
myvar2 = sum3/n - mean**2
print "my variance2: ", myvar2
gives you:
numpy variance: 28.8223642606
mean: -3.731599805
my variance1: 28.8223642606
my variance2: 28.8223642606