CODEWITHSUNDEEP
python
David Tan
David Tan

Reputation: 311

Python: fast way to compute the average of several (same length) lists?

Is there a simple way to calculate the mean of several (same length) lists in Python? Say, I have [[1, 2, 3], [5, 6, 7]], and want to obtain [3,4,5]. This is to be doing 100000 times, so want it to be fast.

Upvotes: 17

Views: 34539

Answers (4)

Postan Grayhole
Postan Grayhole

Reputation: 11

Slightly modified version for smooth work with RGB pixels:

def average(*l):
  l=tuple(l)
  def divide(x): return x // len(l)
  return list(map(divide, map(sum, zip(*l))))
print(average([0,20,200],[100,40,100]))
>>> [50,30,150]

Upvotes: 0

arshajii
arshajii

Reputation: 129497

In case you're using numpy (which seems to be more appropriate here):

>>> import numpy as np
>>> data = np.array([[1, 2, 3], [5, 6, 7]])
>>> np.average(data, axis=0)
array([ 3.,  4.,  5.])

Upvotes: 32

l4mpi
l4mpi

Reputation: 5149

Extending NPEs answer, for a list containing n sublists which you want to average, use this (a numpy solution might be faster, but mine uses only built-ins):

def average(l):
    llen = len(l)
    def divide(x): return x / llen
    return map(divide, map(sum, zip(*l)))

This sums up all sublists and then divides the result by the number of sublists, producing the average. You could inline the len computation and turn divide into a lambda like lambda x: x / len(l), but using an explicit function and pre-computing the length should be a bit faster.

Upvotes: 2

NPE
NPE

Reputation: 500207

In [6]: l = [[1, 2, 3], [5, 6, 7]]

In [7]: [(x+y)/2 for x,y in zip(*l)]
Out[7]: [3, 4, 5]

(You'll need to decide whether you want integer or floating-point maths, and which kind of division to use.)

On my computer, the above takes 1.24us:

In [11]: %timeit [(x+y)/2 for x,y in zip(*l)]
1000000 loops, best of 3: 1.24 us per loop

Thus processing 100,000 inputs would take 0.124s.

Interestingly, NumPy arrays are slower on such small inputs:

In [27]: In [21]: a = np.array(l)

In [28]: %timeit (a[0] + a[1]) / 2
100000 loops, best of 3: 5.3 us per loop

In [29]: %timeit np.average(a, axis=0)
100000 loops, best of 3: 12.7 us per loop

If the inputs get bigger, the relative timings will no doubt change.

Upvotes: 6

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