Question

我目前正在使用图表遍历。 find_path路径递增例如path = path + [start]返回[30, 3, 5, 8]，而find_path2返回路径像path +=[start]一样递增，返回每个值find_path2都有的列表已遍历，[30, 3, 4, 0, 1, 2, 5, 8]。

我猜+=的行为与追加方法的行为一样path = path + [start] 重新分配？

有人可以解释path +=[start]之间的区别和path = path + [start]

def find_path2(graph, start, end, path=[]):
    path += [start]
    if start == end:
        return path
    if not start in graph:
        return None
    for node in graph[start]:
        if node not in path:
            newpath = find_path2(graph, node, end, path)
            if newpath: return newpath
    return None

def find_path(graph, start, end, path=[]):
    path = path + [start]
    if start == end:
        return path
    if not start in graph:
        return None
    for node in graph[start]:
        if node not in path:
            newpath = find_path(graph, node, end, path)
            if newpath: return newpath
    return None

graph = {
    0: [],
    1: [],
    2: [],
    3: [4,2,5],
    4: [0,1],
    5:[8],
    10:[5,6],
    6:[],
    30:[3]
}
print find_path(graph,30,8) #[30, 3, 5, 8]
print find_path2(graph,30,8) #[30, 3, 4, 0, 1, 2, 5, 8]

Answer 1

一个创建一个新列表（+），另一个修改原始列表（+ =）：

In [28]: path = [30, 3, 4, 0, 1, 2, 5, 8]

In [29]: id(path)
Out[29]: 140580805174120

In [30]:  path = path + [7] # assignment = new list

In [31]: id(path)
Out[31]: 140580805174840 

In [32]: path = [30, 3, 4, 0, 1, 2, 5, 8]

In [33]: id(path)
Out[33]: 140580805175416

In [34]:  path += [7] # same as list.extend, modifies list

In [35]: id(path)
Out[35]: 140580805175416

如果您调用该函数两次，您的mutable default arg会导致您遇到麻烦，您应该使用None作为arg值，并在第一次调用该函数时将其设置为空列表，而不是递归调用：

def find_path2(graph, start, end, path=None):
    if path is None:
        path = []

这实际上与+=和list = list + [value]之间的差异相同，如果每次调用函数时都没有创建新对象，则相同的对象/列表用于重复调用

Answer 2

path = path + [start]创建一个新列表。不更新对列表的现有引用，而+=就地修改列表。
Python有__iadd__所以+=不仅仅是语法糖。

Python中列表中+ =的含义

2 个答案: