遍历列表中的元素并像这样更新总数:

def sum(a):
    total = 0
    index = 0
    while index < len(a):
        total = total + a[index]
        index = index + 1
    return total

对一串数字求和:

sum(list_of_nums)

使用列表推导式生成一个新列表，其中相邻元素在xs中求平均值:

[(x + y) / 2 for x, y in zip(xs, xs[1:])]

将所有相邻元素求和为一个值:

sum((x + y) / 2 for x, y in zip(xs, xs[1:]))

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]

一种简单的方法是使用iter_tools排列

# If you are given a list

numList = [1,2,3,4,5,6,7]

# and you are asked to find the number of three sums that add to a particular number

target = 10
# How you could come up with the answer?

from itertools import permutations

good_permutations = []

for p in permutations(numList, 3):
    if sum(p) == target:
        good_permutations.append(p)

print(good_permutations)

结果是:

[(1, 2, 7), (1, 3, 6), (1, 4, 5), (1, 5, 4), (1, 6, 3), (1, 7, 2), (2, 1, 7), (2, 3, 
5), (2, 5, 3), (2, 7, 1), (3, 1, 6), (3, 2, 5), (3, 5, 2), (3, 6, 1), (4, 1, 5), (4, 
5, 1), (5, 1, 4), (5, 2, 3), (5, 3, 2), (5, 4, 1), (6, 1, 3), (6, 3, 1), (7, 1, 2), 
(7, 2, 1)]

注意，顺序很重要——即1,2,7也表示为2,1,7和7,1,2。您可以通过使用集合来减少这种情况。

>>> a = range(10)
>>> sum(a)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: 'int' object is not callable
>>> del sum
>>> sum(a)
45

sum似乎已经在代码的某个地方定义了，并且覆盖了默认函数。所以我删除了它，问题解决了。

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