一些你可能不知道的 Python 小技巧!

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文章转自:CSDN  参考链接:

https://levelup.gitconnected.com/python-tricks-i-can-not-live-without-87ae6aff3af8

一些你可能不知道的 Python 小技巧!
在本文中,我们来看一看日常工作中经常使用的一些 Python 小技巧。

集合

开发人员常常忘记 Python 也有集合数据类型,大家都喜欢使用列表处理一切。
集合(set)是什么?简单来说就是:集合是一组无序事物的汇集,不包含重复元素。
如果你熟练掌握集合及其逻辑,那么很多问题都可以迎刃而解。举个例子,如何获取一个单词中出现的字母?
myword = "NanananaBatman"set(myword){'N', 'm', 'n', 'B', 'a', 't'}
就这么简单,问题解决了,这个例子就来自 Python 的官方文档,大可不必过于惊讶。
再举一个例子,如何获取一个列表的各个元素,且不重复?
# first you can easily change set to list and other way aroundmylist = ["a", "b", "c","c"]# let's make a set out of itmyset = set(mylist)# myset will be:{'a', 'b', 'c'}# and, it's already iterable so you can do:for element in myset:    print(element)# but you can also convert it to list again:mynewlist = list(myset)# and mynewlist will be:['a', 'b', 'c']
我们可以看到,“c”元素不再重复出现了。只有一个地方你需要注意,mylist 与 mynewlist 之间的元素顺序可能会有所不同:
mylist = ["c", "c", "a","b"]mynewlist = list(set(mylist))# mynewlist is:['a', 'b', 'c']
可以看出,两个列表的元素顺序不同。
下面,我们来进一步深入。
假设某些实体之间有一对多的关系,举个更加具体的例子:用户与权限。通常,一个用户可以拥有多个权限。现在假设某人想要修改多个权限,即同时添加和删除某些权限,应当如何解决这个问题?
# this is the set of permissions before change;original_permission_set = {"is_admin","can_post_entry", "can_edit_entry", "can_view_settings"}# this is new set of permissions;new_permission_set = {"can_edit_settings","is_member", "can_view_entry", "can_edit_entry"}# now permissions to add will be:new_permission_set.difference(original_permission_set)# which will result:{'can_edit_settings', 'can_view_entry', 'is_member'}# As you can see can_edit_entry is in both sets; so we do notneed# to worry about handling it# now permissions to remove will be:original_permission_set.difference(new_permission_set)# which will result:{'is_admin', 'can_view_settings', 'can_post_entry'}# and basically it's also true; we switched admin to member, andadd# more permission on settings; and removed the post_entrypermission
总的来说,不要害怕使用集合,它们能帮助你解决很多问题,更多详情,请参考 Python 官方文档。

日历

当开发与日期和时间有关的功能时,有些信息可能非常重要,比如某一年的这个月有多少天。这个问题看似简单,但是我相信日期和时间是一个非常有难度的话题,而且我觉得日历的实现问题非常多,简直就是噩梦,因为你需要考虑大量的极端情况。
那么,究竟如何才能找出某个月有多少天呢?
import calendarcalendar.monthrange(2020, 12)# will result:(1, 31)# BUT! you need to be careful here, why? Let's read thedocumentation:help(calendar.monthrange)# Help on function monthrange in module calendar:# monthrange(year, month)#     Return weekday (0-6~ Mon-Sun) and number of days (28-31) for#    year, month.# As you can see the first value returned in tuple is a weekday,# not the number of the first day for a given month; let's try# to get the same for 2021calendar.monthrange(2021, 12)(2, 31)# So this basically means that the first day of December 2021 isWed# and the last day of December 2021 is 31 (which is obvious,cause# December always has 31 days)# let's play with Februarycalendar.monthrange(2021, 2)(0, 28)calendar.monthrange(2022, 2)(1, 28)calendar.monthrange(2023, 2)(2, 28)calendar.monthrange(2024, 2)(3, 29)calendar.monthrange(2025, 2)(5, 28)# as you can see it handled nicely the leap year;
某个月的第一天当然非常简单,就是 1 号。但是,“某个月的第一天是周X”,如何使用这条信息呢?你可以很容易地查到某一天是周几:
calendar.monthrange(2024, 2)(3, 29)# means that February 2024 starts on Thursday# let's define simple helper:weekdays = ["Monday", "Tuesday","Wednesday", "Thursday", "Friday","Saturday", "Sunday"]# now we can do something like:weekdays[3]# will result in:'Thursday'# now simple math to tell what day is 15th of February 2020:offset = 3  # it's thefirst value from monthrangefor day in range(1, 29):    print(day,weekdays[(day + offset - 1) % 7])1 Thursday2 Friday3 Saturday4 Sunday...18 Sunday19 Monday20 Tuesday21 Wednesday22 Thursday23 Friday24 Saturday...28 Wednesday29 Thursday# which basically makes sense;
也许这段代码不适合直接用于生产,因为你可以使用 datetime 更容易地查找星期:
from datetime import datetimemydate = datetime(2024, 2, 15)datetime.weekday(mydate)# will result:3# or:datetime.strftime(mydate, "%A")'Thursday'
总的来说,日历模块有很多有意思的地方,值得慢慢学习:
# checking if year is leap:calendar.isleap(2021)  #Falsecalendar.isleap(2024)  #True# or checking how many days will be leap days for given yearspan:calendar.leapdays(2021, 2026) # 1calendar.leapdays(2020, 2026) # 2# read the help here, as range is: [y1, y2), meaning that second# year is not included;calendar.leapdays(2020, 2024) # 1

枚举有第二个参数

是的,枚举有第二个参数,可能很多有经验的开发人员都不知道。下面我们来看一个例子:
mylist = ['a', 'b', 'd', 'c', 'g', 'e']for i, item in enumerate(mylist):    print(i, item)# Will give:0 a1 b2 d3 c4 g5 e# but, you can add a start for enumeration:for i, item in enumerate(mylist, 16):    print(i, item)# and now you will get:16 a17 b18 d19 c20 g21 e
第二个参数可以指定枚举开始的地方,比如上述代码中的 enumerate(mylist,16)。如果你需要处理偏移量,则可以考虑这个参数。

if-else 逻辑

你经常需要根据不同的条件,处理不同的逻辑,经验不足的开发人员可能会编写出类似下面的代码:
OPEN = 1IN_PROGRESS = 2CLOSED = 3def handle_open_status():    print('Handling openstatus')def handle_in_progress_status():    print('Handling inprogress status')def handle_closed_status():    print('Handling closedstatus')def handle_status_change(status):    if status == OPEN:       handle_open_status()    elif status ==IN_PROGRESS:        handle_in_progress_status()    elif status == CLOSED:       handle_closed_status()handle_status_change(1)  #Handling open statushandle_status_change(2)  #Handling in progress statushandle_status_change(3)  #Handling closed status
虽然这段代码看上去也没有那么糟,但是如果有 20 多个条件呢?
那么,究竟应该怎样处理呢?
from enum import IntEnumclass StatusE(IntEnum):    OPEN = 1    IN_PROGRESS = 2    CLOSED = 3def handle_open_status():    print('Handling openstatus')def handle_in_progress_status():    print('Handling inprogress status')def handle_closed_status():    print('Handling closedstatus')handlers = {    StatusE.OPEN.value:handle_open_status,   StatusE.IN_PROGRESS.value: handle_in_progress_status,    StatusE.CLOSED.value:handle_closed_status}def handle_status_change(status):    if status not inhandlers:         raiseException(f'No handler found for status: {status}')    handler =handlers[status]    handler()handle_status_change(StatusE.OPEN.value)  # Handling open statushandle_status_change(StatusE.IN_PROGRESS.value)  # Handling in progress statushandle_status_change(StatusE.CLOSED.value)  # Handling closed statushandle_status_change(4)  #Will raise the exception
在 Python 中这种模式很常见,它可以让代码看起来更加整洁,尤其是当方法非常庞大,而且需要处理大量条件时。

enum 模块

enum 模块提供了一系列处理枚举的工具函数,最有意思的是 Enum 和 IntEnum。我们来看个例子:
from enum import Enum, IntEnum, Flag, IntFlagclass MyEnum(Enum):    FIRST ="first"    SECOND ="second"    THIRD ="third" class MyIntEnum(IntEnum):    ONE = 1    TWO = 2    THREE = 3# Now we can do things like:MyEnum.FIRST  #<MyEnum.FIRST: 'first'># it has value and name attributes, which are handy:MyEnum.FIRST.value  #'first'MyEnum.FIRST.name  #'FIRST'# additionally we can do things like:MyEnum('first')  #<MyEnum.FIRST: 'first'>, get enum by valueMyEnum['FIRST']  #<MyEnum.FIRST: 'first'>, get enum by name
使用 IntEnum 编写的代码也差不多,但是有几个不同之处:
MyEnum.FIRST == "first"  # False# butMyIntEnum.ONE == 1  # True# to make first example to work:MyEnum.FIRST.value == "first"  # True
在中等规模的代码库中,enum 模块在管理常量方面可以提供很大的帮助。
enum 的本地化可能有点棘手,但也可以实现,我用django快速演示一下:
from enum import Enumfrom django.utils.translation import gettext_lazy as _class MyEnum(Enum):    FIRST ="first"    SECOND ="second"    THIRD ="third"    @classmethod    def choices(cls):        return [            (cls.FIRST.value, _('first')),            (cls.SECOND.value, _('second')),            (cls.THIRD.value, _('third'))         ]# And later in eg. model definiton:some_field = models.CharField(max_length=10,choices=MyEnum.choices())

iPython

iPython 就是交互式 Python,它是一个交互式的命令行 shell,有点像 Python 解释器。
首先,你需要安装 iPython:
pip install ipython
接下来,你只需要在输入命令的时候,将 Python 换成 ipython:
# you should see something like this after you start:Python 3.8.5 (default, Jul 28 2020, 12:59:40)Type 'copyright', 'credits' or 'license' for more informationIPython 7.18.1 -- An enhanced Interactive Python. Type '?' forhelp.In [1]:
ipython 支持很多系统命令,比如 ls 或 cat,tab 键可以显示提示,而且你还可以使用上下键查找前面用过的命令。更多具体信息,请参见官方文档。

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一些你可能不知道的 Python 小技巧!
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