简短而简单:

def ave(x,y):
  return (x + y) / 2.0

map(ave, a[:-1], a[1:])

这是它的样子:

>>> a = range(10)
>>> map(ave, a[:-1], a[1:])
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

由于Python在处理两个列表上的映射时有些愚蠢，你必须截断列表a[:-1]。如果你使用itertools.imap，它会像你期望的那样工作:

>>> import itertools
>>> itertools.imap(ave, a, a[1:])
<itertools.imap object at 0x1005c3990>
>>> list(_)
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

简短而简单:

def ave(x,y):
  return (x + y) / 2.0

map(ave, a[:-1], a[1:])

这是它的样子:

>>> a = range(10)
>>> map(ave, a[:-1], a[1:])
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

由于Python在处理两个列表上的映射时有些愚蠢，你必须截断列表a[:-1]。如果你使用itertools.imap，它会像你期望的那样工作:

>>> import itertools
>>> itertools.imap(ave, a, a[1:])
<itertools.imap object at 0x1005c3990>
>>> list(_)
[0.5, 1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5]

使用成对的itertools食谱:

import itertools
def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = itertools.tee(iterable)
    next(b, None)
    return itertools.izip(a, b)

def pair_averages(seq):
    return ( (a+b)/2 for a, b in pairwise(seq) )

我只使用带有map()的

a = [1,2,3,4,5,6,7,8,9,10]
b = map(lambda x, y: (x+y)/2.0, fib[:-1], fib[1:])
print b

本着itertools的精神。灵感来自双人食谱。

from itertools import tee, izip

def average(iterable):
    "s -> (s0,s1)/2.0, (s1,s2)/2.0, ..."
    a, b = tee(iterable)
    next(b, None)
    return ((x+y)/2.0 for x, y in izip(a, b))

例子:

>>>list(average([1,2,3,4,5]))
[1.5, 2.5, 3.5, 4.5]
>>>list(average([1,20,31,45,56,0,0]))
[10.5, 25.5, 38.0, 50.5, 28.0, 0.0]
>>>list(average(average([1,2,3,4,5])))
[2.0, 3.0, 4.0]

试试以下方法:

mylist = [1, 2, 3, 4]   

def add(mylist):
    total = 0
    for i in mylist:
        total += i
    return total

result = add(mylist)
print("sum = ", result)