Question

我需要打印a到b范围内的c的所有项目

01/08/2017, 02/08/2017, 03/08/2017, 04/08/2017, 05/08/2017, 06/08/2017, 07/08/2017, 08/08/2017, 09/08/2017, 10/08/2017, 11/08/2017, 12/08/2017, 13/08/2017, 14/08/2017, 15/08/2017, 16/08/2017, 17/08/2017, 18/08/2017, 19/08/2017, 20/08/2017, 21/08/2017, 22/08/2017, 23/08/2017, 24/08/2017, 25/08/2017, 26/08/2017, 27/08/2017, 28/08/2017

a = '01/08/2017'
b = '28/08/2017'
c = ['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', '24/08/2017', '25/08/2020', '26/08/2020', '27/08/2020', '28/08/2020']

我的答案应该是：

['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', '24/08/2017']

Answer 1

您可以使用datetime模块：

import datetime
def to_datetime(d):
 day, month, year = map(int, d.split('/'))
 return datetime.datetime(year, month, day, 0, 0, 0)

a = '01/08/2017'
b = '28/08/2017'
_a = to_datetime(a)
_b = to_datetime(b)
c = ['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', '24/08/2017', '25/08/2017', '26/08/2017', '27/08/2017', '28/08/2017']
for i in c:
  if _a <= to_datetime(i) <= _b:
    print(i)

输出：

Answer 2

如果您愿意使用第三方库，则可以使用熊猫：

import pandas as pd

s = pd.to_datetime(pd.Series(c))
res = s[s.between(a, b)].tolist()

print(res)

[Timestamp('2017-01-08 00:00:00'),
 Timestamp('2017-08-20 00:00:00'),
 Timestamp('2017-08-21 00:00:00'),
 Timestamp('2017-08-22 00:00:00'),
 Timestamp('2017-08-23 00:00:00'),
 Timestamp('2017-08-24 00:00:00')]

Answer 3

将日期转换为datetime个对象并进行直接比较：

from datetime import datetime as dt
FMT = "%d/%m/%Y"

[date for date in c 
 if  dt.strptime(a, FMT) 
  <= dt.strptime(date, FMT) 
  <= dt.strptime(b, FMT)]

#['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', 
# '24/08/2017', '25/08/2017', '26/08/2017', '27/08/2017', '28/08/2017']

Answer 4

您可以使用条件列表理解：

import datetime as dt

f = '%d/%m/%Y'  # Date format.

>>> [date for date in c 
     if dt.datetime.strptime(a, f) 
     <= dt.datetime.strptime(date, f) 
     <= dt.datetime.strptime(b, f)]
['01/08/2017',
 '20/08/2017',
 '21/08/2017',
 '22/08/2017',
 '23/08/2017',
 '24/08/2017',
 '25/08/2017',
 '26/08/2017',
 '27/08/2017',
 '28/08/2017']

Answer 5

如果您要的是某种解决方案，并且没有依赖项，那么该解决方案会更容易理解。

a = '01/08/2017'
b = '28/08/2017'
c = ['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', '24/08/2017', '25/08/2020', '26/08/2020', '27/08/2020', '28/08/2020']

a = a.split("/") #break requirements into components
b = b.split("/")
answerList = [] # form the final list structure

for i in range(0, len(c)): #iterate through all of c
    cHolder = c[i].split("/") #break c into components
    componentValidator = 0 # simple all-or-nothing counter

    for j in range(3): # go through every component
        if (int(a[j]) <= int(cHolder[j]) <= int(b[j])): # check ranges
            componentValidator = componentValidator + 1 # for each component

    if componentValidator == 3: # if all correct
        answerList.append(c[i]) # add to final answers
print(answerList) # print final answers

输出：

['01/08/2017', '20/08/2017', '21/08/2017', '22/08/2017', '23/08/2017', '24/08/2017']

使用Python从日期范围中检查项目列表

5 个答案: