Numpy mean AND variance from single function?

I don't think NumPy provides a function that returns both the mean and the variance.

However, SciPy provides the function scipy.stats.norm.fit() which returns the mean and standard deviation of a sample. The function is named after its more specific purpose of fitting a normal distribution to a sample.

Example:

>>> import scipy.stats
>>> scipy.stats.norm.fit([1,2,3])
(2.0, 0.81649658092772603)

Note that fit() does not apply Bessel's correction to the standard deviation, so if you want that correction, you have to multiply by the appropriate factor.

You can also avoid the subtraction by making use of the relation between mean, variance and power of a signal:

In [7]: import numpy as np

In [8]: a = np.random.rand(1000)

In [9]: %%timeit
   ...: a.mean()
   ...: a.var()
   ...: 
10000 loops, best of 3: 24.7 us per loop

In [10]: %%timeit
    ...: m = a.mean()
    ...: np.mean((a-m)**2)
    ...: 
100000 loops, best of 3: 18.5 us per loop

In [11]: %%timeit
    ...: m = a.mean()
    ...: power = np.mean(a ** 2)
    ...: power - m ** 2
    ...: 
100000 loops, best of 3: 17.3 us per loop

In [12]: %%timeit
    ...: m = a.mean()
    ...: power = np.dot(a, a) / a.size
    ...: power - m ** 2
    ...: 
100000 loops, best of 3: 9.16 us per loop

You can't pass a known mean to np.std or np.var, you'll have to wait for the new standard library statistics module, but in the meantime you can save a little time by using the formula:

In [329]: a = np.random.rand(1000)

In [330]: %%timeit
   .....: a.mean()
   .....: a.var()
   .....: 
10000 loops, best of 3: 80.6 µs per loop

In [331]: %%timeit
   .....: m = a.mean()
   .....: np.mean((a-m)**2)
   .....: 
10000 loops, best of 3: 60.9 µs per loop

In [332]: m = a.mean()

In [333]: a.var()
Out[333]: 0.078365856465916137

In [334]: np.mean((a-m)**2)
Out[334]: 0.078365856465916137

If you really are trying to speed things up, try np.dot to do the squaring and summing (since that's what a dot-product is):

In [335]: np.dot(a-m,a-m)/a.size
Out[335]: 0.078365856465916137

In [336]: %%timeit
   .....: m = a.mean()
   .....: c = a-m
   .....: np.dot(c,c)/a.size
   .....: 
10000 loops, best of 3: 38.2 µs per loop