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, ...]
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
这个问题在这里得到了解答
a = [1,2,3,4]
sum(a)
sum(a)返回10
其他回答
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。
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]
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
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:]))
>>> 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。