Question

我有一个由五列组成的数据框：唯一ID，YEAR-WEEK_nb类型的事件列表，列表的长度，min_date和max_date。

我想做的是计算min_date和max_date之间的距离（以周为单位）。例如，在第一行中，我们将有0000-8，持续8周。

我创建了两个函数：

def min_date(date):
    return min(date)

def max_date(date):
    return 

week["date_min"] = week["week_year"].apply(min_date)
week["date_max"] = week["week_year"].apply(max_date)

我认为我应该做的是

def max_minus_min_date(date):
    return max(date) - min(date)

但是我不知道如何将字符串转换为我的时间戳类型。

您可以在数据集下面找到它作为字典：

dico= {'ID': {0: '3014-4298-ae43-14-4298-4298-a',
  1: '002445cc-a38d-5c25bb06e4c47406c',
  2: '4d1e-a5559-000b0601f33b1d',
  3: '002445cc-00305e69-7c76a-77b32aba5dec'},
 'year_week': {0: ['2018-15', '2018-16', '2018-23'],
  1: ['2019-39', '2019-40', '2019-41', '2019-42', '2019-43'],
  2: ['2018-1',
   '2018-12',
   '2018-2',
   '2018-23',
   '2018-24',
   '2018-25',
   '2018-26',
   '2018-27',
   '2018-3',
   '2018-36',
   '2018-38',
   '2018-39',
   '2018-4',
   '2018-40',
   '2018-41',
   '2018-42',
   '2018-45',
   '2018-47',
   '2018-48',
   '2018-49',
   '2018-50',
   '2018-51',
   '2018-6',
   '2018-7',
   '2018-8',
   '2019-12',
   '2019-13',
   '2019-15',
   '2019-16',
   '2019-17',
   '2019-18',
   '2019-20',
   '2019-21',
   '2019-22',
   '2019-23',
   '2019-24',
   '2019-25',
   '2019-26',
   '2019-27',
   '2019-28',
   '2019-29',
   '2019-3',
   '2019-30',
   '2019-31',
   '2019-32',
   '2019-33',
   '2019-34',
   '2019-35',
   '2019-36',
   '2019-37',
   '2019-38',
   '2019-4',
   '2019-5',
   '2019-6',
   '2019-7',
   '2019-8'],
  3: ['2018-36',
   '2018-38',
   '2018-39',
   '2018-40',
   '2018-41',
   '2018-42',
   '2018-43',
   '2018-44',
   '2018-45',
   '2018-46',
   '2018-47',
   '2018-48',
   '2018-49',
   '2018-52',
   '2019-10',
   '2019-11',
   '2019-2',
   '2019-3',
   '2019-4',
   '2019-5',
   '2019-6',
   '2019-7',
   '2019-8',
   '2019-9']},
 'length_list': {0: 3, 1: 5, 2: 56, 3: 24},
 'date_min': {0: '2018-15', 1: '2019-39', 2: '2018-1', 3: '2018-36'},
 'date_max': {0: '2018-23', 1: '2019-43', 2: '2019-8', 3: '2019-9'}}

Answer 1

IIUC，您需要将星期数和年份转换为日期时间

首先让我们这样做，然后是周差异。

date_format = '%Y-%W-%w'

s1 =  pd.to_datetime((df['date_min'] + '-0'),format=date_format)
s2 = pd.to_datetime((df['date_max'] + '-0'),format=date_format)

week_diff = (s2 - s1) / np.timedelta64(1,'W')

df['week_diff'] = week_diff

print(df[['date_min','date_max','week_diff']])

  date_min date_max  week_diff
0  2018-15  2018-23        8.0
1  2019-39  2019-43        4.0
2   2018-1   2019-8       60.0
3  2018-36   2019-9       26.0

Answer 2

从字符串中提取星期的最快方法如下：

import re

re_week = re.compile("(?P<year>\d{4})\-(?P<week>\d{1,2})")  
# captures both 2019-1 and 2019-42 (one and two digit week number)

def extract_year_week(datestr):
    mtch = re_week.match(datestr)
    return int(mtch.group('year')), int(mtch.group('week'))

一旦有了它，就可以使用Datetime from year and week number的转换来修改最小/最大函数：

def max_minus_min_date(date):
    dt_conv = [extract_year_week(dt) for dt in date]
    sdt = dt_conv.sort(key=lambda yw: yw[0] + yw[1])
    maxd = datetime.strptime('{0} {1} 0'.format(*sdt[-1]), "%Y %W %w")
    mind = datetime.strptime('{0} {1} 0'.format(*sdt[0]), "%Y %W %w")
    return (maxd - mind).days / 7

Answer 3

尝试一下：

首先格式化日期

import datetime
mx = '2019-5'
mn = '2019-3'
mxDate = datetime.datetime.strptime(mx + '-1', "%Y-%W-%w")
mnDate = datetime.datetime.strptime(mn + '-1', "%Y-%W-%w")

然后计算周数差异

weeks = (mxDate - mnDate ).days / 7
print(weeks)

计算熊猫数据帧中的持续时间

3 个答案: