What is the fastest way to find the average for a list of tuples in Python, each tuple containing a pair of namedtuples?

Question

import numpy as numpy
from collections import namedtuple
from random import random

Smoker    = namedtuple("Smoker", ["Female","Male"])
Nonsmoker = namedtuple("Nonsmoker", ["Female","Male"])

LST = [(Smoker(random(),random()),Nonsmoker(random(),random())) for i in range(100)]

So I have a long list whose elements are tuples. Each tuple contains a pair of namedtuples. What is the fastest way to find the average of this list? Ideally the result is still of the same structure, that is, (Smoker(Female=w,Male=x),Nonsmoker(Female=y,Male=z))..

grizzly = Smoker(np.mean([a.Female for a,b in LST]),np.mean([a.Male for a,b in LST]))
panda = Nonmoker(np.mean([b.Female for a,b in LST]),np.mean([b.Male for a,b in LST]))
result = (grizzly, panda)

Answer 1

np.mean has to convert the list to an array, which takes time. Python sum saves time:

In [6]: %%timeit
   ...: grizzly = Smoker(np.mean([a.Female for a,b in LST]),np.mean([a.Male for
   ...: a,b in LST]))
   ...: panda = Nonsmoker(np.mean([b.Female for a,b in LST]),np.mean([b.Male for
   ...:  a,b in LST]))
   ...: result = (grizzly, panda)
   ...: 
   ...: 
158 µs ± 597 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

In [9]: %%timeit
   ...: n=len(LST)
   ...: grizzly = Smoker(sum([a.Female for a,b in LST])/n,sum([a.Male for a,b in
   ...:  LST])/n)
   ...: panda = Nonsmoker(sum([b.Female for a,b in LST])/n,sum([b.Male for a,b i
   ...: n LST])/n)
   ...: result = (grizzly, panda)
   ...: 
   ...: 
46.2 µs ± 37.4 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Both produce the same result (to within a small epsilon):

In [8]: result
Out[8]: 
(Smoker(Female=0.5383695316982974, Male=0.5493854404111675),
 Nonsmoker(Female=0.4913454565011218, Male=0.47143788469638825))

If you could collect the values in one array, possibly (n,4) shape, then the mean will be fast. For one time calculation it probably isn't worth it -

In [11]: M = np.array([(a.Female, a.Male, b.Female, b.Male) for a,b in LST])
In [12]: np.mean(M, axis=0)
Out[12]: array([0.53836953, 0.54938544, 0.49134546, 0.47143788])

In [13]: timeit M = np.array([(a.Female, a.Male, b.Female, b.Male) for a,b in LST])
128 µs ± 1.22 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
In [14]: timeit np.mean(M, axis=0)
21.9 µs ± 371 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Since named tuples can be accessed like regular tuples, we can make an array directly from LST:

In [16]: np.array(LST).shape
Out[16]: (100, 2, 2)
In [17]: np.array(LST).mean(axis=0)
Out[17]: 
array([[0.53836953, 0.54938544],
       [0.49134546, 0.47143788]])

But timing isn't encouraging:

In [18]: timeit np.array(LST).mean(axis=0)
1.26 ms ± 7.92 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)

I can also make a structured array from your list - with nested dtypes:

In [26]: dt = np.dtype([('Smoker', [('Female','f'),('Male','f')]),('Nonsmoker',[
    ...: ('Female','f'),('Male','f')])])
In [27]: M=np.array(LST,dt)
In [28]: M['Smoker']['Female'].mean()
Out[28]: 0.53836954

Curiously timing is relatively good:

In [29]: timeit M=np.array(LST,dt)
40.6 µs ± 243 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

But I have to take each mean separately, or else convert it to an unstructured array first.

I could make a (n,4) float array from the structured one with a view or a recfunctions utility:

In [53]: M1 = M.view([('f0','f',(4,))])['f0']
In [54]: M1.shape
Out[54]: (100, 4)
In [55]: M2=rf.structured_to_unstructured(M)