迭代Python中的一系列日期

Question

我有以下代码来执行此操作,但我怎样才能做得更好？现在我认为它比嵌套循环更好,但是当你在列表理解中有一个生成器时,它开始得到Perl-one-liner.

day_count = (end_date - start_date).days + 1
for single_date in [d for d in (start_date + timedelta(n) for n in range(day_count)) if d <= end_date]:
    print strftime("%Y-%m-%d", single_date.timetuple())

笔记

• 我实际上并没有用它来打印.这只是为了演示目的.
• 在start_date和end_date变量是datetime.date因为我不需要时间戳对象.(它们将用于生成报告).

样本输出

对于开始日期2009-05-30和结束日期2009-06-09:

2009-05-30
2009-05-31
2009-06-01
2009-06-02
2009-06-03
2009-06-04
2009-06-05
2009-06-06
2009-06-07
2009-06-08
2009-06-09

Answer 1

为什么有两个嵌套迭代？对我来说,它只产生一次迭代产生相同的数据列表:

for single_date in (start_date + timedelta(n) for n in range(day_count)):
    print ...

并且不存储任何列表,只迭代一个生成器.此外,发电机中的"if"似乎是不必要的.

毕竟,线性序列应该只需要一个迭代器而不是两个迭代器.

与John Machin讨论后更新:

也许最优雅的解决方案是使用生成器函数来完全隐藏/抽象日期范围内的迭代:

from datetime import timedelta, date

def daterange(start_date, end_date):
    for n in range(int ((end_date - start_date).days)):
        yield start_date + timedelta(n)

start_date = date(2013, 1, 1)
end_date = date(2015, 6, 2)
for single_date in daterange(start_date, end_date):
    print(single_date.strftime("%Y-%m-%d"))

注意:为了与内置range()函数保持一致,此迭代在到达之前停止end_date.因此,对于包容性迭代,请使用第二天,就像您一样range().

Answer 2

这可能更清楚:

from datetime import date, timedelta

start_date = date(2019, 1, 1)
end_date = date(2020, 1, 1)
delta = timedelta(days=1)
while start_date <= end_date:
    print (start_date.strftime("%Y-%m-%d"))
    start_date += delta

Answer 3

使用dateutil图书馆:

from datetime import date
from dateutil.rrule import rrule, DAILY

a = date(2009, 5, 30)
b = date(2009, 6, 9)

for dt in rrule(DAILY, dtstart=a, until=b):
    print dt.strftime("%Y-%m-%d")

这个python库有许多更高级的功能,一些非常有用,比如relative deltas-并且被实现为一个容易包含在项目中的单个文件(模块).

Answer 4

熊猫一般非常适合时间序列,并且直接支持日期范围.

import pandas as pd
daterange = pd.date_range(start_date, end_date)

然后,您可以遍历日期范围以打印日期:

for single_date in daterange:
    print (single_date.strftime("%Y-%m-%d"))

它还有很多选项可以让生活更轻松.例如,如果您只想要工作日,则只需交换bdate_range.见http://pandas.pydata.org/pandas-docs/stable/timeseries.html#generating-ranges-of-timestamps

Pandas的强大功能实际上是它的数据帧,它支持矢量化操作(非常像numpy),可以非常快速,轻松地操作大量数据.

编辑:您也可以完全跳过for循环并直接打印,这样更容易,更有效:

print(daterange)

Answer 5

import datetime

def daterange(start, stop, step=datetime.timedelta(days=1), inclusive=False):
  # inclusive=False to behave like range by default
  if step.days > 0:
    while start < stop:
      yield start
      start = start + step
      # not +=! don't modify object passed in if it's mutable
      # since this function is not restricted to
      # only types from datetime module
  elif step.days < 0:
    while start > stop:
      yield start
      start = start + step
  if inclusive and start == stop:
    yield start

# ...

for date in daterange(start_date, end_date, inclusive=True):
  print strftime("%Y-%m-%d", date.timetuple())

这个功能比你严格要求的更多,支持负步骤等.只要你分解你的范围逻辑,那么你不需要单独的day_count,最重要的是当你从多个函数调用函数时,代码变得更容易阅读地方.

Answer 6

为什么不尝试:

import datetime as dt

start_date = dt.datetime(2012, 12,1)
end_date = dt.datetime(2012, 12,5)

total_days = (end_date - start_date).days + 1 #inclusive 5 days

for day_number in range(total_days):
    current_date = (start_date + dt.timedelta(days = day_number)).date()
    print current_date

Answer 7

这是我能想到的最易读的解决方案.

import datetime

def daterange(start, end, step=datetime.timedelta(1)):
    curr = start
    while curr < end:
        yield curr
        curr += step

Answer 8

为了完整起见，Pandas 也有一个period_range用于处理越界时间戳的函数：

import pandas as pd

pd.period_range(start='1/1/1626', end='1/08/1627', freq='D')

Answer 9

显示从今天开始的最后n天:

import datetime
for i in range(0, 100):
    print((datetime.date.today() + datetime.timedelta(i)).isoformat())

输出:

2016-06-29
2016-06-30
2016-07-01
2016-07-02
2016-07-03
2016-07-04

Answer 10

import datetime

def daterange(start, stop, step_days=1):
    current = start
    step = datetime.timedelta(step_days)
    if step_days > 0:
        while current < stop:
            yield current
            current += step
    elif step_days < 0:
        while current > stop:
            yield current
            current += step
    else:
        raise ValueError("daterange() step_days argument must not be zero")

if __name__ == "__main__":
    from pprint import pprint as pp
    lo = datetime.date(2008, 12, 27)
    hi = datetime.date(2009, 1, 5)
    pp(list(daterange(lo, hi)))
    pp(list(daterange(hi, lo, -1)))
    pp(list(daterange(lo, hi, 7)))
    pp(list(daterange(hi, lo, -7))) 
    assert not list(daterange(lo, hi, -1))
    assert not list(daterange(hi, lo))
    assert not list(daterange(lo, hi, -7))
    assert not list(daterange(hi, lo, 7)) 

Answer 11

for i in range(16):
    print datetime.date.today() + datetime.timedelta(days=i)

Answer 12

Numpy的arange功能可以应用于日期:

import numpy as np
from datetime import datetime, timedelta
d0 = datetime(2009, 1,1)
d1 = datetime(2010, 1,1)
dt = timedelta(days = 1)
dates = np.arange(d0, d1, dt).astype(datetime)

使用astypeis来转换numpy.datetime64为datetime.datetime对象数组.

Answer 13

您可以使用 pandas 库简单可靠地生成两个日期之间的一系列日期

import pandas as pd

print pd.date_range(start='1/1/2010', end='1/08/2018', freq='M')

您可以通过将 freq 设置为 D、M、Q、Y（每日、每月、每季度、每年）来更改生成日期的频率