Question

我正在尝试为字典编写日期插值算法：

每个词典条目都有N个键
条目的每个键都有相应的日期列表
密钥列表中不能复制任何日期
如果我要在处理后整理所有条目的列表，我最终会得到一个连续的每日日期列表
插值应仅在未出现在其他列表中的日期之间（参见下面的简化示例

实施例。 D = {＆＃34;条目＆＃34;：{＆＃34; list1＆＃34;：[1,2,9,12]，＆＃34; list2＆＃34;：[4,6]}}

Dfinal = {＆＃34; Entry＆＃34;：{＆＃34; list1＆＃34;：[1,2,3,7,8,9,10,11,12]，＆＃34; list2＆＃34;：[4,5,6]}}

在我的示例代码中，我将所有列表合并为一个，对该列表进行排序，然后获取每个日期的索引，然后我可以在外部索引相同的情况下进行插值。 [0] [1] [0] [2]当我点击另一个列表中存在的日期时，我遇到的问题是弄清楚如何在另一个列表上切换和插值。我几乎在那里，我只是不确定如何完成这个难题！

from datetime import date, timedelta as td
import datetime
import numpy as np
import calendar

#Interpolate Dates
def datesBetween(StartDate,EndDate):
    dates = []
    year1 = datetime.datetime.strptime(str(StartDate),'%d%b%Y').year
    year2 = datetime.datetime.strptime(str(EndDate),'%d%b%Y').year
    Month1 = datetime.datetime.strptime(str(StartDate),'%d%b%Y').month
    Month2 = datetime.datetime.strptime(str(EndDate),'%d%b%Y').month
    day1 = datetime.datetime.strptime(str(StartDate),'%d%b%Y').day
    day2 = datetime.datetime.strptime(str(EndDate),'%d%b%Y').day
    d1 = date(year1, Month1, day1)
    d2 = date(year2, Month2, day2)
    delta = d2 - d1
    dates.append(datetime.datetime.strptime(str(StartDate),'%d%b%Y').strftime('%d%b%Y'))
    d0 = datetime.datetime.strptime(str(StartDate),'%d%b%Y')
    for i in range(delta.days):
        d0 += datetime.timedelta(days=1)
        dates.append(d0.strftime('%d%b%Y'))
    return dates

#Sort Dates
def date_key(a):
    a = datetime.datetime.strptime(a, '%d%b%Y').date()
    return a


dates1 = ['28aug2017', '29aug2017', '30aug2017', '31aug2017', '01sep2017', '12sep2017','13sep2017', '14sep2017','20sep2017', '21sep2017', '25sep2017', '26sep2017','27sep2017', '28sep2017', '29sep2017', '02oct2017']

dates2 = ['05sep2017', '06sep2017', '07sep2017', '08sep2017', '11sep2017', '03oct2017']

dates3 = [ '15sep2017', '18sep2017', '19sep2017']


master = [dates1,dates2,dates3]

conc = []
for lst in master:
    conc = conc + lst

sortedDates = sorted(conc,key=date_key)

indexMain = []
indexSub = []

for d in sortedDates:
    indexMain.append(next((i for i, sublist in enumerate(master) if str(d) in sublist), -1))
    indexSub.append(next((sublist for sublist in master if str(d) in sublist)).index(str(d)))


#interpolation of the respective lists
i = 0
while i < len(indexMain)-1:

    d1 = master[indexMain[i]][indexSub[i]]
    d2 = master[indexMain[i+1]][indexSub[i+1]]

    if indexMain[i] == indexMain[i+1]:
        temp = datesBetween(str(d1),str(d2))
        for val in temp:
            if val not in master[indexMain[i]]:
                master[indexMain[i]].append(str(val))

    if indexMain[i] != indexMain[i+1]:
        temp = datesBetween(str(d1),str(d2))
        for val in temp[:len(temp)-1]:
            if val not in master[indexMain[i]]:
                master[indexMain[i]].append(str(val))

    i += 1    

print(master[0])
print(master[1])
print(master[2])

Answer 1

<强>解决

nextafterf

跨越N个python列表的日期插值

1 个答案: