如何在Python中实现不可变的数据结构？
问题
您需要在Python中实现不可变的数据结构。
介绍..
当您要防止多个人同时在并行编程中修改一条数据时，不可变数据结构非常方便。可变数据结构（例如Array）可以随时更改，而可变数据结构则不能更改。
怎么做..
让我逐步向您展示如何处理不可变和可变的数据结构。
示例
# STEP 01 - Create a Mutable array.
# Define an array
atp_players = ['Murray', 'Nadal', 'Djokovic']
print(f" *** Original Data in my array is - {atp_players}")***我数组中的原始数据是-
['Murray', 'Nadal', 'Djokovic']
# Changing the player name from Murray to Federer
atp_players[0] = 'Federer'
print(f" *** Modified Data in my array is - {atp_players}")***我数组中的修改数据是-
['Federer', 'Nadal', 'Djokovic']
结论
我们已经能够像更改数组值一样，如果您是该数组的独占用户，这可能会很有用。但是，在实时生产中，多个程序可能正在使用此数组进行更改，并可能导致意外的数据。
另一方面，元组的行为有所不同，请看下面的例子。
# STEP 02 - Try changing a Tuple
try:
atp_players_tuple = ('Murray', 'Nadal', 'Djokovic')
print(f" *** Original Data in my tuple is - {atp_players_tuple}")
atp_players_tuple[0] = 'Federer'
except Exception as error:
print(f" *** Tried modifying data but ended up with - {error}")*** Original Data in my tuple is - ('Murray', 'Nadal', 'Djokovic')
*** Tried modifying data but ended up with - 'tuple' object does not support item assignment结论：
您在上面看到的是，元组不能被修改对吗？但是，有一个例外，如果元组中包含数组，则可以更改值。
atp_players_array_in_tuple = (['Murray'], ['Nadal'], ['Djokovic'])
print(f" *** Original Data in my tuple with arrays is - {atp_players_array_in_tuple}")
atp_players_array_in_tuple[0][0] = 'Federer'
print(f" *** Modified Data in my tuple with arrays is - {atp_players_array_in_tuple}")*** Original Data in my tuple with arrays is - (['Murray'], ['Nadal'], ['Djokovic']) *** Modified Data in my tuple with arrays is - (['Federer'], ['Nadal'], ['Djokovic'])
那么如何保护数据呢？嗯，只需将数组转换为元组即可。
try:
atp_players_tuple_in_tuple = (('Murray'), ('Nadal'), ('Djokovic'))
print(f" *** Original Data in my tuple is - {atp_players_tuple_in_tuple}")
atp_players_tuple_in_tuple[0] = 'Federer'
except Exception as error:
print(f" *** Tried modifying data in my tuple but ended up with - {error}")*** Original Data in my tuple is - ('Murray', 'Nadal', 'Djokovic')
*** Tried modifying data in my tuple but ended up with - 'tuple' object does not support item assignment还有更多..Python有一个很棒的内置工具，名为NamedTuple。您可以扩展该类以创建构造函数。让我们以编程方式理解。
# Create a simple class on Grandslam titles in pythons way.
class GrandSlamsPythonWay:
def __init__(self, player, titles):
self.player = player
self.titles = titles
stats = GrandSlamsPythonWay("Federer", 20)
print(f" *** Stats has details as {stats.player} - {stats.titles}")*** Stats has details as Federer - 20
您如何看待这个课程，这是一成不变的吗？让我们将Federer更改为Nadal进行检查。
stats.player = 'Nadal'
print(f" *** Stats has details as {stats.player} - {stats.titles}")*** Stats has details as Nadal - 20
因此，这毫无疑问是一个不变的数据结构，因为我们能够将Federer更新为Nadal。现在，让我们使用NamedTuple创建一个类，看看它的默认行为是什么。
from typing import NamedTuple
class GrandSlamsWithNamedTuple(NamedTuple):
player: str
titles: int
stats = GrandSlamsWithNamedTuple("Federer", 20)
print(f" *** Stats has details as {stats.player} - {stats.titles}")
stats.player = 'Djokovic'
print(f" *** Stats has details as {stats.player} - {stats.titles}")*** Stats has details as Federer - 20
---------------------------------------------------------------------------
AttributeError Traceback (most recent call last)
in
10 print(f" *** Stats has details as {stats.player} - {stats.titles}")
11
---> 12 stats.player = 'Djokovic'
13 print(f" *** Stats has details as {stats.player} - {stats.titles}")
AttributeError: can't set attribute德约科维奇似乎不得不再等一段时间才能获得20个大满贯冠军。
但是，我们可以使用_replace方法进行复制并在_replace期间更新值。
djokovic_stats = stats._replace(player="Djokovic", titles=17)
print(f" *** djokovic_stats has details as {djokovic_stats.player} - {djokovic_stats.titles}")*** djokovic_stats has details as Djokovic - 17
示例
最后，我将举一个涵盖以上所有内容的示例。
对于此示例，假设我们正在为蔬菜店编写软件。
from typing import Tuple
# Create a Class to represent one purchase
class Prices(NamedTuple):
id: int
name: str
price: int # Price in dollars
# Create a Class to track the purchase
class Purchase(NamedTuple):
purchase_id: int
items: Tuple[Prices]
# Create vegetable items and their corresponding prices
carrot = Prices(1, "carrot", 2)
tomato = Prices(2, "tomato", 3)
eggplant = Prices(3, "eggplant", 5)
# Now let say our first cusotmer Mr.Tom had purchased carrot and tomato
tom_order = Purchase(1, (carrot, tomato))
# Know the total cost we need to charge Mr.Tom
total_cost = sum(item.price for item in tom_order.items)
print(f"*** Total Cost from Mr.Tom is - {total_cost}$")输出结果
*** Total Cost from Mr.Tom is - 5$