揭露,一些你可能不知 道的 Python 小技巧
发布时间:2022-11-30 12:40:47 所属栏目:语言 来源:
导读: 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。 01.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
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。 01.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 around mylist = ["a", "b", "c","c"] # let's make a set out of it myset = 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。 02.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 calendar calendar.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 2021 calendar.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 February calendar.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 monthrange for day in range(1, 29): print(day,weekdays[(day + offset - 1) % 7]) 1 Thursday 2 Friday 3 Saturday 4 Sunday ... 18 Sunday 19 Monday 20 Tuesday 21 Wednesday 22 Thursday 23 Friday 24 Saturday ... 28 Wednesday 29 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 datetime mydate = 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) #False calendar.isleap(2024) #True # or checking how many days will be leap days for given yearspan: calendar.leapdays(2021, 2026) # 1 calendar.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 03.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 a 1 b 2 d 3 c 4 g 5 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 a 17 b 18 d 19 c 20 g 21 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。 04.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 = 1 IN_PROGRESS = 2 CLOSED = 3 def 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 status handle_status_change(2) #Handling in progress status handle_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 IntEnum class StatusE(IntEnum): OPEN = 1 IN_PROGRESS = 2 CLOSED = 3 def 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 status handle_status_change(StatusE.IN_PROGRESS.value) # Handling in progress status handle_status_change(StatusE.CLOSED.value) # Handling closed status handle_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。 05.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, IntFlag class 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 # # 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') #, get enum by value MyEnum['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 # but MyIntEnum.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 Enum from 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()) 06.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 information IPython 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。 (编辑:拼字网 - 核心网) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |
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