Python：嵌套If循环

这让我抓狂： 我试图根据月份从列表中求和，我尝试了一些事情，但急需指导

我正在努力： 第1-12个月

从列表中迭代读取PlanWeek4值，然后求和

然后根据求和值计算平滑平均值

这是我的密码：

G_counter = 1
j = i
m = 1
Plantotal = 0
PlanMonth = 0
DFD = []
EC_PlanData = [500,500.... etc] # 52 values

PlanWeek = range(j,j+3)
Month = range(m,13,1)

## Define Variables
ym, xh, xm, N1, Z1, N2, Z2 = 0,0,0,0,0,0,0

for month in Month:      # for each month 1 - 13
    for i,e in enumerate(list_1):      # read through list
        PlanMonth = PlanMonth + i+3    # sum 4 weekly values
        DFD.append(PlanMonth)          # append sum value to DFD list
        if i == 1:                     # if G_counter = 1, which it always is
            IFX.append(PlanMonth)      # also append to IFX list

    Plantotal= Plantotal+PlanMonth     # calculations here on are
    for i,e in enumerate(DFD):         # evaluated after appending above
        y = e

    ym = Plantotal / m                 # These are calculating a smoothing average
    xh = xh + m
    xm = xh / m      
    N1 = (m-xm) * (y-ym)
    Z1 = (m-xm) * (m-xm)
    N2 = N2 + N1
    Z2 = Z2 + Z1

    if Z2 == 0:                        # Decision on FC value
        FC = 0                         # This or
    else:
        FC = ym -(N2/Z2)*xm + (N2/Z2)*(m+1) # This

    J +=4                              # Advances on 4 time periods
    M +=1                              # Advances on 1 Month
    PlanMonth = 0                      # Resets PlanMonth variable

---

你必须意识到12不能除以52，而且每个月也没有4周的时间。因此，为了给出一个例子，你可以微调以得到你想要的，我定义了一个星期属于它的星期四属于的同一个月。这与ISO 8601对一年中第一周的定义非常吻合。如果还有一周，我就把这一周加到十二月

import datetime
from itertools import groupby

def get_week(date):
    return date.isocalendar()[1]

def group_by_month(weeks, year):
    """
    Group a list containing one item per week, starting with week 1, by month.

    If there are too few items to fill a year, stop after last item.
    If there are more items than weeks in the year, stop before new year.
    """
    day = datetime.timedelta(days=1)
    week = datetime.timedelta(days=7)

    # Find first Thursday (it's in week 1 by ISO 8601)
    date = datetime.date(year, 1, 1)
    while date.weekday() != 3:
        date += day

    # Create list of one day from each week
    thursdays = []
    while date.year == year:
        thursdays.append(date)
        date += week

    # Check if the last day is in the last week and if not, 
    # add the week of the last day
    last = tursdays[-1]
    if get_week(last.replace(day=31)) != get_week(last):
        # this will not be a Thursday, but what the hey
        thursdays.append(last.replace(day=31))

    # The thursdays are already sorted by month, so 
    # it's OK to use groupby without sorting first
    for k, g in groupby(zip(weeks, thursdays), key=lambda x: x[1].month):
        yield [x[0] for x in g]

list_1 = [500] * 52

print map(sum, group_by_month(list_1, 2012))

结果:

[2000, 2000, 2500, 2000, 2500, 2000, 2000, 2500, 2000, 2000, 2500, 2000]

---

您还应意识到一个事实，即年份可能包含，如果是，您必须提供53项清单，而不是52项清单。如果不这样做，第53周将被忽略。

PlanMonth最初来自哪里？我希望你不是真的在那个列表中写了500 52次，列表2和列表3在哪里？到底是什么问题？“计划周”的目的是什么？不清楚你想要实现什么。建议用伪代码描述问题。您能指定任务吗？您是否只需要为给定月份的第3个列表元素求和？第1个月的PlanMonth是指从1到4的值的总和。第2个月的PlanMonth是指从5到8等值的总和，因此列表2和列表3变为13个总和值。我希望你不是真的在那张单子上写了500次52次。。不，这是自动生成的，但为了验证起见，我将数字固定为500。@Manang Student请通过单击链接更新您的答案，并重新编排以使其更清晰，并在评论中回答问题。还要确保示例代码可以作为独立脚本运行，而不会引发异常。哦，哇，这太棒了，从初学者的角度来看，我有很多东西要学习。但这确实有效。在我看到这个之前，我已经转载了我原来的帖子。至于53周的情况，我知道，在Excel工作了很多年；，但你一举解决了这个问题。我会看看我是否可以继续你提供的平滑平均部分的工作。很好的回答。顺便说一句，@manengstudent可以简化与其他python代码编写者的沟通，我建议您看看。阅读符合几乎所有人都知道并遵循的风格的代码更容易，对于python来说，就是这样。