Calculating arithmetic mean (one type of average) in Python [duplicate]

This question already has answers here: Finding the average of a list (24 answers) Closed last year.

Is there a built-in or standa开发者_运维知识库rd library method in Python to calculate the arithmetic mean (one type of average) of a list of numbers?

---

I am not aware of anything in the standard library. However, you could use something like:

def mean(numbers):
    return float(sum(numbers)) / max(len(numbers), 1)

>>> mean([1,2,3,4])
2.5
>>> mean([])
0.0

In numpy, there's numpy.mean().

---

NumPy has a numpy.mean which is an arithmetic mean. Usage is as simple as this:

>>> import numpy
>>> a = [1, 2, 4]
>>> numpy.mean(a)
2.3333333333333335

---

Use statistics.mean:

import statistics
print(statistics.mean([1,2,4])) # 2.3333333333333335

It's available since Python 3.4. For 3.1-3.3 users, an old version of the module is available on PyPI under the name stats. Just change statistics to stats.

---

You don't even need numpy or scipy...

>>> a = [1, 2, 3, 4, 5, 6]
>>> print(sum(a) / len(a))
3

---

Use scipy:

import scipy;
a=[1,2,4];
print(scipy.mean(a));

---

Instead of casting to float you can do following

def mean(nums):
    return sum(nums, 0.0) / len(nums)

or using lambda

mean = lambda nums: sum(nums, 0.0) / len(nums)

UPDATES: 2019-12-15

Python 3.8 added function fmean to statistics module. Which is faster and always returns float.

Convert data to floats and compute the arithmetic mean.

This runs faster than the mean() function and it always returns a float. The data may be a sequence or iterable. If the input dataset is empty, raises a StatisticsError.

fmean([3.5, 4.0, 5.25])

4.25

New in version 3.8.

---

from statistics import mean
avarage=mean(your_list)

for example

from statistics import mean

my_list=[5,2,3,2]
avarage=mean(my_list)
print(avarage)

and result is

3.0

---

If you're using python >= 3.8, you can use the fmean function introduced in the statistics module which is part of the standard library:

>>> from statistics import fmean
>>> fmean([0, 1, 2, 3])
1.5

It's faster than the statistics.mean function, but it converts its data points to float beforehand, so it can be less accurate in some specific cases.

You can see its implementation here

---

def avg(l):
    """uses floating-point division."""
    return sum(l) / float(len(l))

Examples:

l1 = [3,5,14,2,5,36,4,3]
l2 = [0,0,0]

print(avg(l1)) # 9.0
print(avg(l2)) # 0.0

---

def list_mean(nums):
    sumof = 0
    num_of = len(nums)
    mean = 0
    for i in nums:
        sumof += i
    mean = sumof / num_of
    return float(mean)

---

The proper answer to your question is to use statistics.mean. But for fun, here is a version of mean that does not use the len() function, so it (like statistics.mean) can be used on generators, which do not support len():

from functools import reduce
from operator import truediv
def ave(seq):
    return truediv(*reduce(lambda a, b: (a[0] + b[1], b[0]), 
                           enumerate(seq, start=1), 
                           (0, 0)))

---

I always supposed avg is omitted from the builtins/stdlib because it is as simple as

sum(L)/len(L) # L is some list

and any caveats would be addressed in caller code for local usage already.

Notable caveats:

1. non-float result: in python2, 9/4 is 2. to resolve, use float(sum(L))/len(L) or from __future__ import division

2. division by zero: the list may be empty. to resolve:

   if not L:
       raise WhateverYouWantError("foo")
   avg = float(sum(L))/len(L)
   

---

Others already posted very good answers, but some people might still be looking for a classic way to find Mean(avg), so here I post this (code tested in Python 3.6):

def meanmanual(listt):

mean = 0
lsum = 0
lenoflist = len(listt)

for i in listt:
    lsum += i

mean = lsum / lenoflist
return float(mean)

a = [1, 2, 3, 4, 5, 6]
meanmanual(a)

Answer: 3.5