我以前的Python值如何在此字典中被覆盖？

Question

我正在尝试更新字典，但是以前的值一直被覆盖。这是我正在使用的代码的简化版本。

selected_session = ""
encounters_to_add = []
session_encounters = {}

selected_session = "Session 1: Paint it Black"
encounters_to_add.append("Meet the Town (Narrative)")
session_encounters[selected_session] = encounters_to_add

encounters_to_add.append("Barn Bandits (Structured)")
session_encounters[selected_session] = encounters_to_add

encounters_to_add.clear()
selected_session = "Session 2: Myster Man"
encounters_to_add.append("A Crime Most Foul (Social)")
session_encounters[selected_session] = encounters_to_add

print(session_encounters)

我想回来

{'Session 1: Paint it Black': ['Meet the Town (Narrative)', 'Barn Bandits (Structured)'], 'Session 2: Mystery Man': ['A Crime Most Foul (Social)']}

相反，我回来了

{'Session 1: Paint it Black': ['A Crime Most Foul (Social)'], 'Session 2: Myster Man': ['A Crime Most Foul (Social)']}

我不明白以前的值是如何被覆盖的。由于我使用的是新密钥，因此应该在我脑海中将值添加到该密钥中。

Answer 1

那是因为您只是存储对在第二行中创建的列表实例的 引用 ：

# Creates a new instance of a list where
# 'encounters_to_add' is the reference to this list
encounters_to_add = []  
# Stores only a copy of the 'reference' in your dictionary
session_encounters[selected_session] = encounters_to_add

这不是列表的副本 -似乎就是您所假设的-而是对该列表的引用的副本。

您对encounters_to_add所做的任何事情都会影响session_encounters[selected_session]，因为

session_encounters[selected_session] = encounters_to_add
session_encounters[selected_session] is encounters_to_add # is True

或更清楚一点：

session_encounters['A'] = encounters_to_add
session_encounters['B'] = encounters_to_add

session_encounters['A'] is encounters_to_add # is True
session_encounters['B'] is encounters_to_add # is also True

---

您需要两个实例，例如：

encounters_to_add_1 = []  # An instance of a list
encounters_to_add_2 = []  # Another list instance

session_encounters = {}

selected_session = "Session 1: Paint it Black"
encounters_to_add_1.append("Meet the Town (Narrative)")
encounters_to_add_1.append("Barn Bandits (Structured)")
session_encounters[selected_session] = encounters_to_add_1   

selected_session = "Session 2: Myster Man"
encounters_to_add_2.append("A Crime Most Foul (Social)")
session_encounters[selected_session] = encounters_to_add_2

print(session_encounters)

输出：

{'Session 1: Paint it Black': ['Meet the Town (Narrative)', 'Barn Bandits (Structured)'], 'Session 2: Myster Man': ['A Crime Most Foul (Social)']}

---

这是另一个示例-尝试找出此处发生的情况：

my_list = list()
my_dict = {i: my_list for i in range(3)}
my_dict[0].append('Hello World!')
print(my_dict)

my_dict = {i: list() for i in range(3)}
my_dict[0].append('Hello World!')
print(my_dict)

你为什么得到：

{0: ['Hello World!'], 1: ['Hello World!'], 2: ['Hello World!']}
{0: ['Hello World!'], 1: [], 2: []}

Answer 2

您只有3次插入字典的一个列表，第二次插入将覆盖第一个，因此实际上同一列表最后仅出现两次。在添加之间也要清除它，这就是为什么丢失两个原始条目的原因。只需在print(session_encounters)调用之后添加一个clear()。

逐步引导您完成示例：

selected_session = ""
encounters_to_add = []
session_encounters = {}

selected_session = "Session 1: Paint it Black"
encounters_to_add.append("Meet the Town (Narrative)")
session_encounters[selected_session] = encounters_to_add

encounters_to_add.append("Barn Bandits (Structured)")
session_encounters[selected_session] = encounters_to_add

encounters_to_add.clear()

selected_session = "Session 2: Myster Man"
encounters_to_add.append("A Crime Most Foul (Social)")
session_encounters[selected_session] = encounters_to_add

解决方案很简单：始终创建新列表：

session_encounters = {}

selected_session = "Session 1: Paint it Black"
encounters_to_add = ["Meet the Town (Narrative)", "Barn Bandits (Structured)"]
session_encounters[selected_session] = encounters_to_add

selected_session = "Session 2: Myster Man"
encounters_to_add  ["A Crime Most Foul (Social)"]
session_encounters[selected_session] = encounters_to_add

print(session_encounters)

请注意，您也可以使用collections.defaultdict使其更加简单明了：

from collections import defaultdict

session_encounters = defaultdict(list)
session_encounters["Session 1: Paint it Black"].append("Meet the Town (Narrative)")
session_encounters["Session 1: Paint it Black"].append("Barn Bandits (Structured)")
session_encounters["Session 2: Myster Man"].append("A Crime Most Foul (Social)")

print(session_encounters)

Answer 3

在python中，有两种传递/分配值的方式，按引用或按值。

1. 传递引用：该对象收到对该对象的隐式引用，指向该值的原始位置。默认情况下，可变对象将通过引用传递。 （例如列表）
2. 传递值：将创建该值的本地副本并将其链接到该对象。默认情况下，不可变对象将按值传递（例如，整数，字符串等）

     

   更新（贷记为 MSeifert ）：请注意，默认情况下，Python始终会按引用分配值。上述传递值实际上是由于某些对象类型的不可变性而发生的行为。阅读评论以获取更多详细信息/示例。

您面临的问题是，您将encounters_to_add列表（通过引用）传递到了字典sessions_encounters中，然后使用encounters_to_add.clear()方法重置了对象。这影响了所有对象关联（包括与"Session 1: Paint it Black"字典内的键session_encounters关联的值）。

要解决此问题，您需要按照以下代码按值传递encounters_to_add

  

请注意，唯一的区别在于第session_encounters[selected_session] = encounters_to_add[:]行的赋值，其中[:]将指示python创建列表的本地副本（按值传递）而不是通过引用传递它。

selected_session = ""
encounters_to_add = []
session_encounters = {}

selected_session = "Session 1: Paint it Black"
encounters_to_add.append("Meet the Town (Narrative)")

encounters_to_add.append("Barn Bandits (Structured)")
session_encounters[selected_session] = encounters_to_add[:]

encounters_to_add.clear()
selected_session = "Session 2: Myster Man"
encounters_to_add.append("A Crime Most Foul (Social)")
session_encounters[selected_session] = encounters_to_add[:]

print(session_encounters)

这将输出

{'Session 1: Paint it Black': ['Meet the Town (Narrative)', 'Barn Bandits (Structured)'], 'Session 2: Myster Man': ['A Crime Most Foul (Social)']}