假设我有一个清单：
y = ['1', '2', '3', '4','5','6','7','8','9','10']
我想创建一个计算移动n天平均值的函数。
因此，如果n为5，我希望我的代码计算前1-5，添加它并找到平均值，这将是3.0，然后继续到2-6，计算平均值，这将是4.0，然后3-7,4-8,5-9,6-10。
我不想计算前n-1天，因此从第n天开始，它将计算前几天。
def moving_average(x:'list of prices', n):
for num in range(len(x)+1):
print(x[num-n:num])
这似乎打印出我想要的东西：
[]
[]
[]
[]
[]
['1', '2', '3', '4', '5']
['2', '3', '4', '5', '6']
['3', '4', '5', '6', '7']
['4', '5', '6', '7', '8']
['5', '6', '7', '8', '9']
['6', '7', '8', '9', '10']
但是，我不知道如何计算这些列表中的数字。有什么想法吗？
答案 0 :(得分：21)
旧版本的Python文档中有一个很棒的滑动窗口生成器itertools examples：
from itertools import islice
def window(seq, n=2):
"Returns a sliding window (of width n) over data from the iterable"
" s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ... "
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
使用你的移动平均线是微不足道的：
from __future__ import division # For Python 2
def moving_averages(values, size):
for selection in window(values, size):
yield sum(selection) / size
针对您的输入运行此命令（将字符串映射到整数）会给出：
>>> y= ['1', '2', '3', '4','5','6','7','8','9','10']
>>> for avg in moving_averages(map(int, y), 5):
... print(avg)
...
3.0
4.0
5.0
6.0
7.0
8.0
要为“不完整”设置返回None第一次n - 1次迭代，只需展开moving_averages函数：
def moving_averages(values, size):
for _ in range(size - 1):
yield None
for selection in window(values, size):
yield sum(selection) / size
答案 1 :(得分：6)
虽然我喜欢Martijn's answer，就像乔治一样，我想知道通过使用运行求和而不是一次又一次地应用sum()这是不是更快相同的数字。
此外，在加速阶段将None值设为默认值的想法很有意思。实际上，可能存在许多可以设想移动平均线的不同场景。让我们将平均值的计算分为三个阶段：
average := sum(x[iteration_counter-window_size:iteration_counter])/window_size window_size - 1＆＃34;平均值＆＃34;号。这是一个接受
的功能None）或提供部分平均值以下是代码：
from collections import deque
def moving_averages(data, size, rampUp=True, rampDown=True):
"""Slide a window of <size> elements over <data> to calc an average
First and last <size-1> iterations when window is not yet completely
filled with data, or the window empties due to exhausted <data>, the
average is computed with just the available data (but still divided
by <size>).
Set rampUp/rampDown to False in order to not provide any values during
those start and end <size-1> iterations.
Set rampUp/rampDown to functions to provide arbitrary partial average
numbers during those phases. The callback will get the currently
available input data in a deque. Do not modify that data.
"""
d = deque()
running_sum = 0.0
data = iter(data)
# rampUp
for count in range(1, size):
try:
val = next(data)
except StopIteration:
break
running_sum += val
d.append(val)
#print("up: running sum:" + str(running_sum) + " count: " + str(count) + " deque: " + str(d))
if rampUp:
if callable(rampUp):
yield rampUp(d)
else:
yield running_sum / size
# steady
exhausted_early = True
for val in data:
exhausted_early = False
running_sum += val
#print("st: running sum:" + str(running_sum) + " deque: " + str(d))
yield running_sum / size
d.append(val)
running_sum -= d.popleft()
# rampDown
if rampDown:
if exhausted_early:
running_sum -= d.popleft()
for (count) in range(min(len(d), size-1), 0, -1):
#print("dn: running sum:" + str(running_sum) + " deque: " + str(d))
if callable(rampDown):
yield rampDown(d)
else:
yield running_sum / size
running_sum -= d.popleft()
它似乎比Martijn的版本快一点 - 但它更优雅。这是测试代码：
print("")
print("Timeit")
print("-" * 80)
from itertools import islice
def window(seq, n=2):
"Returns a sliding window (of width n) over data from the iterable"
" s -> (s0,s1,...s[n-1]), (s1,s2,...,sn), ... "
it = iter(seq)
result = tuple(islice(it, n))
if len(result) == n:
yield result
for elem in it:
result = result[1:] + (elem,)
yield result
# Martijn's version:
def moving_averages_SO(values, size):
for selection in window(values, size):
yield sum(selection) / size
import timeit
problems = [int(i) for i in (10, 100, 1000, 10000, 1e5, 1e6, 1e7)]
for problem_size in problems:
print("{:12s}".format(str(problem_size)), end="")
so = timeit.repeat("list(moving_averages_SO(range("+str(problem_size)+"), 5))", number=1*max(problems)//problem_size,
setup="from __main__ import moving_averages_SO")
print("{:12.3f} ".format(min(so)), end="")
my = timeit.repeat("list(moving_averages(range("+str(problem_size)+"), 5, False, False))", number=1*max(problems)//problem_size,
setup="from __main__ import moving_averages")
print("{:12.3f} ".format(min(my)), end="")
print("")
输出：
Timeit
--------------------------------------------------------------------------------
10 7.242 7.656
100 5.816 5.500
1000 5.787 5.244
10000 5.782 5.180
100000 5.746 5.137
1000000 5.745 5.198
10000000 5.764 5.186
现在可以通过此函数调用解决原始问题：
print(list(moving_averages(range(1,11), 5,
rampUp=lambda _: None,
rampDown=False)))
输出：
[None, None, None, None, 3.0, 4.0, 5.0, 6.0, 7.0, 8.0]
答案 2 :(得分：1)
使用sum和map功能。
print(sum(map(int, x[num-n:num])))
Python 3中的map函数基本上是 lazy 版本：
[int(i) for i in x[num-n:num]]
我确信你可以猜出sum函数的作用。
答案 3 :(得分：1)
避免重新计算中间数的方法..
list=range(0,12)
def runs(v):
global runningsum
runningsum+=v
return(runningsum)
runningsum=0
runsumlist=[ runs(v) for v in list ]
result = [ (runsumlist[k] - runsumlist[k-5])/5 for k in range(0,len(list)+1)]
打印结果
[2,3,4,5,6,7,8,9]
make that runs（int（v））.. then .. repr（runsumlist [k] - runsumlist [k-5]）/ 5） 如果你蚂蚁携带数字字符串..
没有全局的Alt：
list = [float[x] for x in range(0,12)]
nave = 5
movingave = sum(list[:nave]/nave)
for i in range(len(list)-nave):movingave.append(movingave[-1]+(list[i+nave]-list[i])/nave)
print movingave
即使您输入的值是整数
，也一定要做浮动数学运算[2.0,3.0,4.0,5.0,6.0,7.0,8.0,9,0]
答案 4 :(得分：0)
另一个解决方案是itertools食谱pairwise()。您可以将其扩展为nwise()，它为您提供了滑动窗口（如果iterable是生成器，则可以工作）：
def nwise(iterable, n):
ts = it.tee(iterable, n)
for c, t in enumerate(ts):
next(it.islice(t, c, c), None)
return zip(*ts)
def moving_averages_nw(iterable, n):
yield from (sum(x)/n for x in nwise(iterable, n))
>>> list(moving_averages_nw(range(1, 11), 5))
[3.0, 4.0, 5.0, 6.0, 7.0, 8.0]
虽然短iterable的设置成本相对较高，但此成本会降低影响，但数据集的时间越长。这使用sum()，但代码相当优雅：
Timeit MP cfi *****
--------------------------------------------------------------------------------
10 4.658 4.959 7.351
100 5.144 4.070 4.234
1000 5.312 4.020 3.977
10000 5.317 4.031 3.966
100000 5.508 4.115 4.087
1000000 5.526 4.263 4.202
10000000 5.632 4.326 4.242