In daily life, we can use many different formats to represent dates and times, for example, July 4, March 8, 2022, 22:00 or 2022 December 31, 23:59:59. They use a combination of integers and strings, or they can use floating point numbers to represent a day, a minute, etc. The various ways to represent time are really dazzling.
But fortunately, Python has the datetime module, which allows us to easily manipulate objects representing dates and times.
In today’s article, we will learn the following:
Let's do it!
As we have done before As you can see, representing date and time in programming is a very challenging thing. First, we must represent them in a standard, universally accepted format. Fortunately, the International Organization for Standardization (ISO) has developed a global standard, ISO 8601, which represents date and time-related objects as YYYY-MM-DD HH:MM:SS, with information ranging from the most important (year, YYYY) to least significant (seconds, SS). Each part of this format is represented as a four-digit or two-digit number.
The datetime module in Python has 5 main classes (part of the module):
Additionally, we will use the zoneinfo module, which gives us a more modern way of working with time zones, and the dateutil package, which contains many useful functions for working with dates and times.
Let’s import the datetime module and create our first date and time object:
# From the datetime module import date from datetime import date # Create a date object of 2000-02-03 date(2022, 2, 3)
Output:
datetime.date(2022, 2, 3)
In the above code , we imported the date class from the module, and then created a datetime.date object of February 3, 2022. Note that the sequence of numbers used to create this object is exactly the same as in ISO 8061 (but we omitted the 0 and only wrote the one-digit month and day).
The encoding of time is based on reality, so suppose we want to create an object of 2000-26-03:
# Create a date object of 2000-26-03 date(2000, 26, 3)
Output:
--------------------------------------------------------------------------- ValueErrorTraceback (most recent call last) Input In [2], in 1 # Create a date object of 2000-26-03 ----> 2 date(2000, 26, 3) ValueError: month must be in 1..12
us Getting ValueError: month must be in 1..12, no doubt there is no 26th month in the calendar, throwing an exception.
Let's see how to create a datetime.time object:
# From the datetime module import time from datetime import time # Create a time object of 05:35:02 time(5, 35, 2)
Output:
datetime.time(5, 35, 2)
Now if we want to What if it contains both date and time? We should use the datetime class:
# From the datetime module import datetime from datetime import datetime # Create a datetime object of 2000-02-03 05:35:02 datetime(2000, 2, 3, 5, 35, 2)
Output:
datetime.datetime(2000, 2, 3, 5, 35, 2)
As expected, we successfully created the datetime object. We can also pass keyword arguments to the datetime constructor more explicitly:
datetime(year=2000, month=2, day=3, hour=5, minute=35, second=2)
Output:
datetime.datetime(2000, 2, 3, 5, 35, 2)
If we only pass in three parameters (year, month and day), will an error be reported?
# Create a datetime object of 2000-02-03 datetime(2000, 2, 3)
Output:
datetime.datetime(2000, 2, 3, 0, 0)
We can see that there are two zeros (representing respectively) hours and minutes in the object. The seconds are also omitted.
In many situations we want to know the exact current time. You can use the now() method of the datetime class:
# Time at the moment now = datetime.now() now
Output:
datetime.datetime(2022, 8, 1, 0, 9, 39, 611254)
We get a datetime object, where the last number is microseconds.
If we only need today's date, we can use the today() method of the date class:
today = date.today() today
Output:
datetime.date(2022, 8, 1)
If we only need To need time, you must access the hour, minute, and second properties of the datetime.now() object and pass them to the time constructor:
time(now.hour, now.minute, now.second)
Output:
datetime.time(11, 33, 25)
We can also use the isocalendar() function to extract the week number and day number from a datetime object. It will return a three-item tuple containing the ISO year, week number, and working day number:
# isocalendar() returns a 3-item tuple with ISO year, week number, and weekday number now.isocalendar()
Output:
datetime.IsoCalendarDate(year=2022, week=7, weekday=1)
In ISO format, the week starts on Monday and ends on Ends Sunday. The days of the week are encoded by numbers from 1 (Monday) to 7 (Sunday). If we want to access one of these tuple elements, we need to use bracket notation:
# Access week number week_number = now.isocalendar()[1] week_number
Output:
7
在数据科学和一般编程中，我们主要使用以数十种不同格式存储为字符串的日期和时间，具体取决于地区、公司或我们需要的信息粒度。有时，我们需要日期和确切时间，但在其他情况下，我们只需要年份和月份。我们该如何从字符串中提取我们需要的数据，以便将其作为日期时间（日期、时间）对象来操作呢？
我们学习的第一个将日期字符串转换为日期对象的函数是 fromisoformat，我们这样称呼它是因为它使用 ISO 8601 格式（即 YYYY-MM-DD），让我们看一个例子：
# Convert a date string into a date object
date.fromisoformat("2022-12-31")Output:
datetime.date(2022, 12, 31)
ISO 格式也包含时间，但如果我们却不能将它传递给函数：
date.fromisoformat("2022-12-31 00:00:00")Output:
---------------------------------------------------------------------------
ValueErrorTraceback (most recent call last)
Input In [13], in
----> 1 date.fromisoformat("2022-12-31 00:00:00")
ValueError: Invalid isoformat string: '2022-12-31 00:00:00'当然，我们也可以进行逆向运算，将 datetime 对象转换为 ISO 格式的日期字符串，我们应该使用 isoformat()：
# Convert a datetime object into a string in the ISO format date(2022, 12, 31).isoformat()
Output:
'2022-12-31'
为了解决上述 ValueError 问题，我们可以使用 strptime() 函数，该函数可以将任意日期/时间字符串转换为日期时间对象。我们的字符串不一定需要遵循 ISO 格式，但我们应该指定字符串的哪一部分代表哪个日期或时间单位（年、小时等）。让我们看一个例子，首先，我们将使用严格的 ISO 格式将字符串转换为日期时间对象：
# Date as a string iso_date = "2022-12-31 23:59:58" # ISO format iso_format = "%Y-%m-%d %H:%M:%S" # Convert the string into a datetime object datetime.strptime(iso_date, iso_format)
Output:
datetime.datetime(2022, 12, 31, 23, 59, 58)
在第一行，我们创建一个日期/时间字符串。在第二行中，我们使用特殊代码指定字符串的格式，该代码包含一个百分号，后跟一个编码日期或时间单位的字符。最后，在第三行中，我们使用 strptime() 函数将字符串转换为日期时间对象。这个函数有两个参数：字符串和字符串的格式。
我们上面使用的代码还可以编码其他日期和时间单位，如工作日、月份名称、周数等。
代码 | 示例 | 说明 |
%A | Monday | 完整的工作日名称 |
%B | December | 全月名称 |
%W | 2 | The number of weeks (Monday is the first day) |
让我们再看几个使用其他格式的示例：
# European date as a string european_date = "31-12-2022" # European format european_format = "%d-%m-%Y" # Convert the string into a datetime object datetime.strptime(european_date, european_format)
Output:
datetime.datetime(2022, 12, 31, 0, 0)
如上所示，字符串已成功转换，但还有额外的零表示时间字段，让我们看一个使用其他代码的示例：
# Full month name date full_month_date = "12 September 2022" # Full month format full_month_format = "%d %B %Y" # Convert the string into a datetime object datetime.strptime(full_month_date, full_month_format)
Output:
datetime.datetime(2022, 9, 12, 0, 0)
还是可以正常转换，但是需要注意的是，我们定义的格式应该与日期字符串的格式相匹配。因此，如果我们有空格、冒号、连字符或其他字符来分隔时间单位，那么它们也应该在代码字符串中。否则，Python 将抛出 ValueError：
# Full month name date full_month_date = "12 September 2022" # Wrong format (missing space) full_month_format = "%d%B %Y" # Convert the string into a datetime object datetime.strptime(full_month_date, full_month_format)
Output:
---------------------------------------------------------------------------
ValueErrorTraceback (most recent call last)
Input In [18], in
5 full_month_format = "%d%B %Y"
7 # Convert the string into a datetime object
----> 8 datetime.strptime(full_month_date, full_month_format)
File ~/coding/dataquest/articles/using-the-datetime-package/env/lib/python3.10/_strptime.py:568, in _strptime_datetime(cls, data_string, format)
565 def _strptime_datetime(cls, data_string, format="%a %b %d %H:%M:%S %Y"):
566 """Return a class cls instance based on the input string and the
567 format string."""
--> 568 tt, fraction, gmtoff_fraction = _strptime(data_string, format)
569 tzname, gmtoff = tt[-2:]
570 args = tt[:6] + (fraction,)
File ~/coding/dataquest/articles/using-the-datetime-package/env/lib/python3.10/_strptime.py:349, in _strptime(data_string, format)
347 found = format_regex.match(data_string)
348 if not found:
--> 349 raise ValueError("time data %r does not match format %r" %
350(data_string, format))
351 if len(data_string) != found.end():
352 raise ValueError("unconverted data remains: %s" %
353 data_string[found.end():])
ValueError: time data '12 September 2022' does not match format '%d%B %Y'可以看到，即使缺少一个空格也可能导致错误！
在 Python 中，我们还可以使用 strftime() 函数将日期时间对象转换为字符串。它有两个参数：一个日期时间对象和输出字符串的格式。
# Create a datetime object
datetime_obj = datetime(2022, 12, 31)
# American date format
american_format = "%m-%d-%Y"
# European format
european_format = "%d-%m-%Y"
# American date string
print(f"American date string: {datetime.strftime(datetime_obj, american_format)}.")
# European date string
print(f"European date string: {datetime.strftime(datetime_obj, european_format)}.")Output:
American date string: 12-31-2022. European date string: 31-12-2022.
我们采用相同的日期时间对象并将其转换为两种不同的格式。我们还可以指定其他格式，例如完整的月份名称后跟日期和年份。
full_month = "%B %d, %Y" datetime.strftime(datetime_obj, full_month)
Output:
'December 31, 2022'
另一种使用 strftime 的方法是将它放在 datetime 对象之后：
datetime_obj = datetime(2022, 12, 31, 23, 59, 59) full_datetime = "%B %d, %Y %H:%M:%S" datetime_obj.strftime(full_datetime)
Output:
'December 31, 2022 23:59:59'
在实际使用当中，如果我们想提取不同年份 12 月 31 日的工作日名称，strftime() 可能很方便：
# Extract the weekday name of December 31
weekday_format = "%A"
for year in range(2022, 2026):
print(f"Weekday of December 31, {year} is {date(year, 12, 31).strftime(weekday_format)}.")Output:
Weekday of December 31, 2022 is Saturday. Weekday of December 31, 2023 is Sunday. Weekday of December 31, 2024 is Tuesday. Weekday of December 31, 2025 is Wednesday.
在编程中，通常会看到以 Unix 时间戳格式存储的日期和时间，这种格式将任何日期表示为数字。一般情况时间戳是从 1970 年 1 月 1 日 00:00:00 UTC（协调世界时）开始的 Unix 纪元经过的秒数。我们可以使用 timestamp() 函数计算这个数字：
new_year_2023 = datetime(2022, 12, 31) datetime.timestamp(new_year_2023)
Output:
1672441200.0
1672441200 就是从 Unix 纪元开始到 2022 年 12 月 31 日之间的秒数。
我们可以使用 fromtimestamp() 函数执行逆运算：
datetime.fromtimestamp(1672441200)
Output:
datetime.datetime(2022, 12, 31, 0, 0)
有时我们可能想要计算两个日期之间的差异或对日期和时间执行其他算术运算。幸运的是，Python 的工具包中有许多工具可以执行此类计算。
我们可以执行的第一个操作是计算两个日期之间的差异。为此，我们使用减号：
# Instatiate two dates
first_date = date(2022, 1, 1)
second_date = date(2022, 12, 31)
# Difference between two dates
date_diff = second_date - first_date
# Function to convert datetime to string
def dt_string(date, date_format="%B %d, %Y"):
return date.strftime(date_format)
print(f"The number of days and hours between {dt_string(first_date)} and {dt_string(second_date)} is {date_diff}.")Output:
The number of days and hours between January 01, 2022 and December 31, 2022 is 364 days, 0:00:00
让我们看看 first_date - second_date 返回什么类型的对象：
type(date_diff)
Output:
datetime.timedelta
此对象的类型是 datetime.timedelta，它的名称中有 delta，指的是一个绿色字母 delta，在科学和工程中，描述了一种变化<实际上，这里它代表了时间上的变化（差异）。
如果我们只对两个日期之间的天数感兴趣怎么办？我们可以访问 timedelta 对象的不同属性，其中之一称为 .days。
print(f"The number of days between {dt_string(first_date)} and {dt_string(second_date)} is {(date_diff).days}.")Output:
The number of days between January 01, 2022 and December 31, 2022 is 364.
现在我们知道了 timedelta 对象，是时候介绍 timedelta() 函数了。它允许我们通过加减时间单位（如天、年、周、秒等）对时间对象执行许多算术运算。例如，我们可能想知道从现在起 30 天后是一周中的哪一天。为此，我们必须创建一个表示当前时间的对象和一个定义我们添加到其中的时间量的 timedelta 对象：
# Import timedelta
from datetime import timedelta
# Current time
now = datetime.now()
# timedelta of 30 days
one_month = timedelta(days=30)
# Day in one month/using dt_string function defined above
print(f"The day in 30 days is {dt_string(now + one_month)}.")Output:
The day in 30 days is March 16, 2022.
如果我们查看 timedelta 函数的帮助页面 (help(timedelta))，我们会看到它有以下参数：days=0, seconds=0, microseconds=0, milliseconds=0, minutes=0, hours= 0，周=0。因此我们还可以练习在日期中添加或减去其他时间单位。例如，我们可以计算 2030 年新年前 12 小时的时间：
# New year 2030
new_year_2030 = datetime(2030, 1, 1, 0, 0)
# timedelta of 12 hours
twelve_hours = timedelta(hours=12)
# Time string of 12 hours before New Year 2023
twelve_hours_before = (new_year_2030 - twelve_hours).strftime("%B %d, %Y, %H:%M:%S")
# Print the time 12 hours before New Year 2023
print(f"The time twelve hours before New Year 2030 will be {twelve_hours_before}.")Output:
The time twelve hours before New Year 2030 will be December 31, 2029, 12:00:00.
我们还可以组合 timedelta() 函数的多个参数来计算出更具体的时间。例如，从现在起 27 天 3 小时 45 分钟后的时间是多少？
# Current time
now = datetime.now()
# Timedelta of 27 days, 3 hours, and 45 minutes
specific_timedelta = timedelta(days=27, hours=3, minutes=45)
# Time in 27 days, 3 hours, and 45 minutes
twenty_seven_days = (now + specific_timedelta).strftime("%B %d, %Y, %H:%M:%S")
print(f"The time in 27 days, 3 hours, and 45 minutes will be {twenty_seven_days}.")Output:
The time in 27 days, 3 hours, and 45 minutes will be March 13, 2022, 15:18:39.
我们可以从帮助页面中看到该功能不允许我们使用几个月或几年。为了克服这个限制，我们可以使用 dateutil 包中的 relativedelta 函数。此函数与 timedelta() 非常相似，但它扩展了更多的功能。
例如，我们想从当前时间中减去 2 年 3 个月 4 天 5 小时：
# Import relativedelta
from dateutil.relativedelta import relativedelta
# Current time
now = datetime.now()
# relativedelta object
relative_delta = relativedelta(years=2, months=3, days=4, hours=5)
two_years = (now - relative_delta).strftime("%B %d, %Y, %H:%M:%S")
print(f"The time 2 years, 3 months, 4 days, and 5 hours ago was {two_years}.")Output:
The time 2 years, 3 months, 4 days, and 5 hours ago was November 10, 2019, 06:33:40.
我们还可以使用 relativedelta() 来计算两个日期时间对象之间的差异：
relativedelta(datetime(2030, 12, 31), now)
Output:
relativedelta(years=+8, months=+10, days=+16, hours=+12, minutes=+26, seconds=+19, microseconds=+728345)
这些算术运算可能看起来非常抽象和不切实际，但实际上，它们在许多应用程序中都很有用。
比如说，我们脚本中的某个操作应该只在特定日期前 30 天执行。我们可以定义一个保存当前时间的变量，并为其添加一个 30 天的 timedelta 对象，如果今天是这一天，就会触发相关操作！
还有，假设我们正在使用 pandas 处理数据集，其中一列包含一些日期。想象一下，我们有一个数据集，其中保存着我们公司一年中的每一天的利润。我们想要创建另一个数据集，该数据集将保存距当前日期正好一年的日期，并预测每一天的利润，此时我们一定会在日期上使用算术计算！
下面我们来看一看时区，它们可以有不同的形式。我们还应该知道，一些地区实施夏令时 (DST)，而另一些地区则没有。
Python 区分两种类型的日期和时间对象：naive 和 aware。一个 naive 对象不保存任何有关时区的信息，而 aware 对象则保存了它。
首先，让我们看一个 naive 时间对象：
# Import tzinfo from datetime import tzinfo # Naive datetime naive_datetime = datetime.now() # Naive datetime doesn't hold any timezone information type(naive_datetime.tzinfo)
Output:
NoneType
从 Python 3.9 开始，使用 Internet Assigned Numbers Authority 数据库实现了时区的具体实现，实现此功能的模块称为 zoneinfo。
让我们使用 zoneinfo，特别是 ZoneInfo 类创建一个感知日期时间对象，它是 datetime.tzinfo 抽象类的一个实现：
# Import ZoneInfo
from zoneinfo import ZoneInfo
utc_tz = ZoneInfo("UTC")
# Aware datetime object with UTC timezone
dt_utc = datetime.now(tz=utc_tz)
# The type of an aware object implemented with ZoneInfo is zoneinfo.ZoneInfo
type(dt_utc.tzinfo)Output:
zoneinfo.ZoneInfo
对于拥有 aware 的 datetime 对象具有有关时区的信息（实现为 zoneinfo.ZoneInfo 对象）。
让我们看几个例子，我们想确定中欧和加利福尼亚的当前时间。
首先，我们可以在 `zoneinfo 中列出所有可用的时区：
import zoneinfo # Will return a long list of timezones (opens many files!) zoneinfo.available_timezones()
Output:
{'Africa/Abidjan',
'Africa/Accra',
'Africa/Addis_Ababa',
'Africa/Algiers',
'Africa/Asmara',
'Africa/Asmera',
'Africa/Bamako',
'Africa/Bangui',
'Africa/Banjul',
'Africa/Bissau',
'Africa/Blantyre',
'Africa/Brazzaville',
'Africa/Bujumbura',
'Africa/Cairo',
'Africa/Casablanca',
'Africa/Ceuta',
'Africa/Conakry',
'Africa/Dakar',
'Africa/Dar_es_Salaam',
'Africa/Djibouti',
'Africa/Douala',
...
'build/etc/localtime'}现在我们可以使用 ZoneInfo 来确定不同区域的当前时间：
# Function to convert datetime into ISO formatted time
def iso(time, time_format="%Y-%m-%d %H:%M:%S"):
return time.strftime(time_format)
# CET time zone
cet_tz = ZoneInfo("Europe/Paris")
# PST time zone
pst_tz = ZoneInfo("America/Los_Angeles")
# Current time in Central Europe
dt_cet = datetime.now(tz=cet_tz)
# Current time in California
dt_pst = datetime.now(tz=pst_tz)
print(f"Current time in Central Europe is {iso(dt_cet)}.")
print(f"Current time in California is {iso(dt_pst)}.")Output:
Current time in Central Europe is 2022-02-14 11:33:42. Current time in California is 2022-02-14 02:33:42.
让我们打印 datetime.now(tz=cet_tz)：
print(datetime.now(tz=cet_tz))
Output:
2022-02-14 11:33:43.048967+01:00
我们看到有 +01:00，它表示 UTC 偏移量。事实上，CET 时区比 UTC 早一小时。
此外，ZoneInfo 类处理夏令时。例如，我们可以将一天（24 小时）添加到 DST 更改发生的一天。
# Define timedelta time_delta = timedelta(days=1) # November 5, 2022, 3 PM november_5_2022 = datetime(2022, 11, 5, 15, tzinfo=pst_tz) # Note that we should use tzinfo in the datetime construct print(november_5_2022) # Add 1 day to November 5, 2022 print(november_5_2022 + time_delta)
Output:
2022-11-05 15:00:00-07:00 2022-11-06 15:00:00-08:00
正如我们所见，偏移量从 -07:00 变为 -08:00，但时间保持不变（15:00）。
New Your City 的时代广场在新年前夜吸引了成千上万的人。让我们应用到目前为止所学的一切来为时代广场除夕创建一个倒数计时器！
在这里，我们将使用 dateutil 包中的 tz，它允许我们设置本地时区来演示 dateutil 包的实用程序。但是我们也可以使用 zoneinfo 中的 build/etc/localtime 时区来做同样的事情。
from zoneinfo import ZoneInfo
from dateutil import tz
from datetime import datetime
def main():
"""Main function."""
# Set current time in our local time zone
now = datetime.now(tz=tz.tzlocal())
# New York City time zone
nyc_tz = ZoneInfo("America/New_York")
# New Year 2023 in NYC
new_year_2023 = datetime(2023, 1, 1, 0, 0, tzinfo=nyc_tz)
# Compute the time left to New Year in NYC
countdown = relativedelta(new_year_2023, now)
# Print time left to New Year 2023
print(f"New Year in New Your City will come on: {new_year_2023.strftime('%B %-d, %Y %H:%M:%S')}.")
print(f"Time left to New Year 2023 in NYC is: {countdown.months} months, {countdown.days} days, {countdown.hours} hours, {countdown.minutes} minutes, {countdown.seconds} seconds.")
if __name__ == "__main__":
main()Output:
New Year in New Your City will come on: January 1, 2023 00:00:00.
Time left to New Year 2023 in NYC is: 10 months, 17 days, 18 hours, 26 minutes, 16 seconds.
我们将代码包装在 main() 函数中，现在我们可以在 .py 文件中使用它。在这个脚本中，我们处理了时区，创建了一个 datetime 对象，使用 strftime() 将其转换为字符串，甚至访问了 relativedelta 对象的时间属性！
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