Calculating arithmetic mean (one type of average) in Python

问题

Is there a built-in or standard library method in Python to calculate the arithmetic mean (one type of average) of a list of numbers?

回答1:

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().

回答2:

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

回答3:

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.

回答4:

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

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

回答5:

Use scipy:

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

回答6:

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)

回答7:

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

回答8:

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

回答9:

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)

回答10:

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:

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

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

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

回答11:

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)))

回答12:

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

来源：https://stackoverflow.com/questions/7716331/calculating-arithmetic-mean-one-type-of-average-in-python

标签

python

math

statistics

average

mean