给定一个数字列表,例如:

[1, 2, 3, 4, 5, ...]

我如何计算它们的总和:

1 + 2 + 3 + 4 + 5 + ...

我如何计算他们的两两平均值:

[(1+2)/2, (2+3)/2, (3+4)/2, (4+5)/2, ...]

当前回答

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)

其他回答

尝试使用列表理解。喜欢的东西:

new_list = [(old_list[i] + old_list[i+1])/2 for i in range(len(old_list-1))]

生成器是一种简单的编写方法:

from __future__ import division
# ^- so that 3/2 is 1.5 not 1

def averages( lst ):
    it = iter(lst) # Get a iterator over the list
    first = next(it)
    for item in it:
        yield (first+item)/2
        first = item

print list(averages(range(1,11)))
# [1.5, 2.5, 3.5, 4.5, 5.5, 6.5, 7.5, 8.5, 9.5]

使用简单的列表-理解和求和:

>> sum(i for i in range(x))/2. #if x = 10 the result will be 22.5

解决这个问题最简单的方法是:

l =[1,2,3,4,5]
sum=0
for element in l:
    sum+=element
print sum
>>> 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。