如何在 Python 中实现不可变数据结构?

pythonserver side programmingprogrammingdata structure更新于 2026/2/14 21:00:17

问题

您需要在 Python 中实现不可变数据结构。

简介..

当您想要防止多个人同时在并行编程中修改一段数据时，不可变数据结构非常方便。可变数据结构(例如数组)可以随时更改，而不可变数据结构则不能。

如何做到这一点..

让我一步一步向您展示如何处理不可变和可变数据结构。

示例

# 步骤 01 - 创建可变数组。

# 定义一个数组
atp_players = ['Murray', 'Nadal', 'Djokovic']
print(f" *** Original Data in my array is - {atp_players}")

*** 我的数组中的原始数据是 -

['Murray', 'Nadal', 'Djokovic']

# 将球员姓名从 Murray 改为 Federer
atp_players[0] = 'Federer'
print(f" *** Modified Data in my array is - {atp_players}")

*** 修改后的数组数据为 -

['Federer', 'Nadal', 'Djokovic']

结论

我们已经能够像这样更改数组值，如果您是此数组的独家用户，这可能会很有用。然而，在实时生产中，多个程序可能会使用此数组进行更改，并可能导致意外数据。

另一方面，元组的行为略有不同，请看下面的示例。

# 步骤 02 - 尝试更改元组

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:

上面您看到的是，元组不能被修改，对吗?但是有一个例外，如果元组中有数组，则可以更改其值。

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。它是一个可以扩展以创建构造函数的类。让我们通过编程来理解。

# 以 Python 方式创建 Grandslam 标题的简单类。
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

您觉得这个类怎么样?它是不可变的吗?让我们通过将 Federer 更改为 Nadal 来检查一下。

stats.player = 'Nadal'
print(f" *** Stats has details as {stats.player} - {stats.titles}")

*** Stats has details as Nadal - 20

因此，毫不奇怪它是一个不可变的数据结构，因为我们能够将 Federer 更新为 Nadal。现在让我们创建一个带有 NamedTuple 的类，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

看来德约科维奇还得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 的详细信息为 Djokovic - 17

示例

最后，我将举一个例子，涵盖上面解释的所有内容。

对于这个例子，假设我们正在为蔬菜店编写软件。

from typing import Tuple

# 创建一个类来表示一次购买
class Prices(NamedTuple):
id: int
name: str
price: int # 价格(美元)

# 创建一个类来跟踪购买
class Purchase(NamedTuple):
purchase_id: int
items: Tuple[Prices]

# 创建蔬菜项目及其相应的价格
carrot = Prices(1, "carrot", 2)
tomato = Prices(2, "tomato", 3)
eggplant = Prices(3, "eggplant", 5)

# 现在假设我们的第一位顾客汤姆先生购买了胡萝卜和西红柿
tom_order = Purchase(1, (carrot, tomato))

# 知道我们需要向汤姆先生收取的总费用
total_cost = sum(item.price for item in tom_order.items)
print(f"*** Total Cost from Mr.Tom is - {total_cost}$")