揭露，一些你可能不知道的 Python 小技巧
发布时间：2021-11-04 18:51:12 所属栏目：语言 来源：互联网
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：集合是一组无序事物的汇集，不包含重复元素。 如果你熟练掌握集合及
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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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