这篇文章主要讲解了“有哪些Python小技巧”，文中的讲解内容简单清晰，易于学习与理解，下面请大家跟着小编的思路慢慢深入，一起来研究和学习“有哪些Python小技巧”吧！
专注于为中小企业提供网站建设、成都做网站服务,电脑端+手机端+微信端的三站合一,更高效的管理,为中小企业甘南免费做网站提供优质的服务。我们立足成都，凝聚了一批互联网行业人才，有力地推动了千余家企业的稳健成长，帮助中小企业通过网站建设实现规模扩充和转变。
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。
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
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。
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。
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：
