基于日期存在性的python列表dict排序

我的清单如下：

lst = [{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}, {'dt': u'NA'}, {'dt': u'December 1st'}, {'dt': u'Feb 1, 2018'}, {'dt': u'January 31'}, {'dt': u'Oct 31, 2017'}, {'dt': u'July 11'}, {'dt': u'Jan 22, 2018'},{'dt': u'Thursday'}]

lst.sort(key=lambda item:item['dt'], reverse=True)
print lst

[{'dt': u'Thursday'}, {'dt': u'Oct 31, 2017'}, {'dt': u'NA'}, {'dt': u'July 11'}, {'dt': u'January 31'}, {'dt': u'Jan 22, 2018'}, {'dt': u'Feb 1, 2018'}, {'dt': u'December 1st'}, {'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}]

但我的输出应该如下所示

[{'dt': u'Feb 1, 2018'},{'dt': u'Jan 22, 2018'},{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'},{'dt': u'Oct 31, 2017'},{'dt': u'December 1st'},{'dt': u'July 11'}, {'dt': u'January 31'},{'dt': u'Thursday'},{'dt': u'NA'}]

元素应按以下顺序排序：

• 带有月、日和年的第一个字符串（从现在到过去）
• 然后是那些只有月份和日（12月至1月）的
• 然后使用

  'NA'

---

如果您更明确地知道要刮取的日期是如何格式化的，那么答案可能会更清楚一些：

from datetime import datetime
import time
import re


def key(v):
    s = v['dt']
    s = re.sub('(?<=1)st|(?<=2)nd|(?<=3)rd|(?<=[04-9])th', '', s)

    try:
        dt = datetime.strptime(s, '%b %d, %Y')
        return time.mktime(dt.timetuple())+1, 0, 0
    except ValueError:
        pass

    try:
        dt = datetime.strptime(s, '%B %d')
        return 0, dt.timetuple().tm_yday+1, 0
    except ValueError:
        pass

    try:
        i = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'].index(s)
        return 0, 0, i+1
    except ValueError:
        pass

    return 0, 0, 0


lst = [{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}, {'dt': u'NA'}, {'dt': u'December 1st'}, {'dt': u'Feb 1, 2018'}, {'dt': u'January 31'}, {'dt': u'Oct 31, 2017'}, {'dt': u'July 11'}, {'dt': u'Jan 22, 2018'},{'dt': u'Thursday'}]
lst.sort(key=key, reverse=True)

从日期时间导入日期时间
导入时间
进口稀土
def键（v）：
s=v['dt']
s=re.sub（“（？如果您更明确地知道要刮取的日期是如何格式化的，那么答案可能会更清楚一些：

from datetime import datetime
import time
import re


def key(v):
    s = v['dt']
    s = re.sub('(?<=1)st|(?<=2)nd|(?<=3)rd|(?<=[04-9])th', '', s)

    try:
        dt = datetime.strptime(s, '%b %d, %Y')
        return time.mktime(dt.timetuple())+1, 0, 0
    except ValueError:
        pass

    try:
        dt = datetime.strptime(s, '%B %d')
        return 0, dt.timetuple().tm_yday+1, 0
    except ValueError:
        pass

    try:
        i = ['Sunday', 'Monday', 'Tuesday', 'Wednesday', 'Thursday', 'Friday', 'Saturday'].index(s)
        return 0, 0, i+1
    except ValueError:
        pass

    return 0, 0, 0


lst = [{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}, {'dt': u'NA'}, {'dt': u'December 1st'}, {'dt': u'Feb 1, 2018'}, {'dt': u'January 31'}, {'dt': u'Oct 31, 2017'}, {'dt': u'July 11'}, {'dt': u'Jan 22, 2018'},{'dt': u'Thursday'}]
lst.sort(key=key, reverse=True)

从日期时间导入日期时间
导入时间
进口稀土
def键（v）：
s=v['dt']
s=re.sub（“（？以下是针对您的问题的另一个硬编码解决方案：

from datetime import datetime
import calendar
import re

def datesort(item):
    item = re.sub('st|nd|rd|th', '', item['dt'])
    try:
        return datetime.strptime(item, '%b %d, %Y')
    except ValueError:
        try:
            return datetime.strptime(item + ', 1903', '%b %d, %Y')
        except ValueError:
            try:
                return datetime.strptime(item + ', 1903', '%B %d, %Y')
            except ValueError:
                try:
                    return datetime(1902, 1, list(calendar.day_name).index(item) + 1)
                except ValueError:
                    return datetime(1901,1,1)

lst = [{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}, {'dt': u'NA'}, {'dt': u'December 1st'}, {'dt': u'Feb 1, 2018'}, {'dt': u'January 31'}, {'dt': u'Oct 31, 2017'}, {'dt': u'July 11'}, {'dt': u'Jan 22, 2018'},{'dt': u'Thursday'}]

sorted(lst, key=datesort, reverse=True)
#[{'dt': u'Feb 1, 2018'},{'dt': u'Jan 22, 2018'},{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'},{'dt': u'Oct 31, 2017'},{'dt': u'December 1st'},{'dt': u'July 11'}, {'dt': u'January 31'},{'dt': u'Thursday'},{'dt': u'NA'}]

以下是针对您的问题的另一个硬编码解决方案：

from datetime import datetime
import calendar
import re

def datesort(item):
    item = re.sub('st|nd|rd|th', '', item['dt'])
    try:
        return datetime.strptime(item, '%b %d, %Y')
    except ValueError:
        try:
            return datetime.strptime(item + ', 1903', '%b %d, %Y')
        except ValueError:
            try:
                return datetime.strptime(item + ', 1903', '%B %d, %Y')
            except ValueError:
                try:
                    return datetime(1902, 1, list(calendar.day_name).index(item) + 1)
                except ValueError:
                    return datetime(1901,1,1)

lst = [{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'}, {'dt': u'NA'}, {'dt': u'December 1st'}, {'dt': u'Feb 1, 2018'}, {'dt': u'January 31'}, {'dt': u'Oct 31, 2017'}, {'dt': u'July 11'}, {'dt': u'Jan 22, 2018'},{'dt': u'Thursday'}]

sorted(lst, key=datesort, reverse=True)
#[{'dt': u'Feb 1, 2018'},{'dt': u'Jan 22, 2018'},{'dt': u'Dec 22, 2017'}, {'dt': u'Dec 14, 2017'}, {'dt': u'Dec 12, 2017'},{'dt': u'Oct 31, 2017'},{'dt': u'December 1st'},{'dt': u'July 11'}, {'dt': u'January 31'},{'dt': u'Thursday'},{'dt': u'NA'}]

为什么要在2018年2月20日（即今天）之前对2018年2月1日进行排序？您正在按字母顺序对字符串进行排序，请将字符串转换为某种日期格式，以便能够对其进行排序。当您有
“12月1日”
时，现在是哪一年？当前？听起来您需要先将值标准化，将字符串更改为日期格式，然后按日期排序。为什么
“2018年2月1日”
在“2月20日”之前排序（即今天）？要按字母顺序对字符串进行排序，请将字符串转换为某种日期格式，以便能够对其进行排序。当您有
“12月1日”
，现在是哪一年？听起来您需要先将值标准化，将字符串更改为日期格式，然后按日期排序。