揭秘,一些你可能不知道的 Python 小技巧
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463篇原創(chuàng)干貨,第一時(shí)間送達(dá)
來(lái)源:CSDN 參考鏈接:
https://levelup.gitconnected.com/python-tricks-i-can-not-live-without-87ae6aff3af8
01.集合
myword = "NanananaBatman"set(myword){'N', 'm', 'n', 'B', 'a', 't'}
# 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']
我們可以看到,“c”元素不再重復(fù)出現(xiàn)了。只有一個(gè)地方你需要注意,mylist 與 mynewlist 之間的元素順序可能會(huì)有所不同:
mylist = ["c", "c", "a","b"]mynewlist = list(set(mylist))# mynewlist is:['a', 'b', 'c']
下面,我們來(lái)進(jìn)一步深入。
假設(shè)某些實(shí)體之間有一對(duì)多的關(guān)系,舉個(gè)更加具體的例子:用戶與權(quán)限。通常,一個(gè)用戶可以擁有多個(gè)權(quán)限。現(xiàn)在假設(shè)某人想要修改多個(gè)權(quán)限,即同時(shí)添加和刪除某些權(quán)限,應(yīng)當(dāng)如何解決這個(gè)問(wèn)題?
# 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
總的來(lái)說(shuō),不要害怕使用集合,它們能幫助你解決很多問(wèn)題,更多詳情,請(qǐng)參考 Python 官方文檔。
02.日歷
當(dāng)開發(fā)與日期和時(shí)間有關(guān)的功能時(shí),有些信息可能非常重要,比如某一年的這個(gè)月有多少天。這個(gè)問(wèn)題看似簡(jiǎn)單,但是我相信日期和時(shí)間是一個(gè)非常有難度的話題,而且我覺得日歷的實(shí)現(xiàn)問(wèn)題非常多,簡(jiǎn)直就是噩夢(mèng),因?yàn)槟阈枰紤]大量的極端情況。
那么,究竟如何才能找出某個(gè)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;
某個(gè)月的第一天當(dāng)然非常簡(jiǎn)單,就是 1 號(hào)。但是,“某個(gè)月的第一天是周X”,如何使用這條信息呢?你可以很容易地查到某一天是周幾:
2)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):+ offset - 1) % 7])1 Thursday2 Friday3 Saturday4 Sunday...18 Sunday19 Monday20 Tuesday21 Wednesday22 Thursday23 Friday24 Saturday...28 Wednesday29 Thursday# which basically makes sense;
也許這段代碼不適合直接用于生產(chǎn),因?yàn)槟憧梢允褂?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'
# 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
03.枚舉有第二個(gè)參數(shù)
是的,枚舉有第二個(gè)參數(shù),可能很多有經(jīng)驗(yàn)的開發(fā)人員都不知道。下面我們來(lái)看一個(gè)例子:
mylist = ['a', 'b', 'd', 'c', 'g', 'e']for i, item in enumerate(mylist):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):item)# and now you will get:16 a17 b18 d19 c20 g21 e
第二個(gè)參數(shù)可以指定枚舉開始的地方,比如上述代碼中的 enumerate(mylist,16)。如果你需要處理偏移量,則可以考慮這個(gè)參數(shù)。
04.if-else 邏輯
你經(jīng)常需要根據(jù)不同的條件,處理不同的邏輯,經(jīng)驗(yàn)不足的開發(fā)人員可能會(huì)編寫出類似下面的代碼:
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
雖然這段代碼看上去也沒(méi)有那么糟,但是如果有 20 多個(gè)條件呢?
那么,究竟應(yīng)該怎樣處理呢?
from enum import IntEnumclass StatusE(IntEnum):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')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 中這種模式很常見,它可以讓代碼看起來(lái)更加整潔,尤其是當(dāng)方法非常龐大,而且需要處理大量條件時(shí)。
05.enum 模塊
enum 模塊提供了一系列處理枚舉的工具函數(shù),最有意思的是 Enum 和 IntEnum。我們來(lái)看個(gè)例子:
from enum import Enum, IntEnum, Flag, IntFlagclass MyEnum(Enum):FIRST ="first"SECOND ="second"THIRD ="third"class MyIntEnum(IntEnum):ONE = 1TWO = 2THREE = 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
MyEnum.FIRST == "first" # False# butMyIntEnum.ONE == 1 # True# to make first example to work:MyEnum.FIRST.value == "first" # True
from enum import Enumfrom django.utils.translation import gettext_lazy as _class MyEnum(Enum):FIRST ="first"SECOND ="second"THIRD ="third"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,它是一個(gè)交互式的命令行 shell,有點(diǎn)像 Python 解釋器。
首先,你需要安裝 iPython:
pip install 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 支持很多系統(tǒng)命令,比如 ls 或 cat,tab 鍵可以顯示提示,而且你還可以使用上下鍵查找前面用過(guò)的命令。更多具體信息,請(qǐng)參見官方文檔。
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