Question

我有一堆没有年份的日期，格式如下：

Thu Apr 10
Mon Mar 28

在python中有一个简单的方法来确定这些日期的来源吗？

Answer 1

当然会出现一年以上有效的情况，但假设你想在晚些时候服用，如果有这种情况，那么这个算法应该可以正常工作。

For each string
    Take the Month(Apr) and the Date(10) 
    For each year from this year down to whenever you'd like to stop
        Make a datetime object with the Month, Date, and Year
        If the datetime object's .weekday is the same as the Day you have (Thu)
            You've found the right year

或者在python中，它可能看起来像这样：

import datetime

with open("yourFile") as f:
for line in f:  

    day = #get the day from the line  (0-30)
    month = #get the number representing the month from the line (0-11)

    year = 2016
    while True:
        testDate = datetime.date(year, month, day)
        weekday = #turn the string "Thu" or "Mon" into the number value 
                  #it represents (0-6)
        if testDate.weekday == weekday:
            #You have found a matching year!!!
        else:
            year = year + 1

Answer 2

以下代码应该有效，您必须选择一年开始，然后选择step（-1为时间，1为前进及时）它会给你第一年它遇到的条件为真：

import datetime
weekdays = {'Mon': 0, 'Tue': 1, 'Wed': 2, 'Thu': 3, 'Fri': 4, 'Sat': 5, 'Sun': 6}
months = {'Jan': 1, 'Feb': 2, 'Mar': 3, 'Apr': 4, 'May': 5, 'Jun': 6,
          'Jul': 7, 'Aug': 8, 'Sep': 9, 'Oct': 10, 'Nov': 11, 'Dec': 12}
dates = ['Thu Apr 10', 'Mon Mar 28']
startYear = 2016
step = -1
years = []

for date in dates:
  [weekDay, month, day] = date.split(' ')
  day = int(day)
  year = startYear
  while True:
    if datetime.date(year, months[month], day).weekday() == weekdays[weekDay]:
      years.append(year)
      break;
    else:
      year = year + step

计算工作日，月和日的年份

2 个答案: