将函数应用于链表的 Pythonic 方式

Question

我想将函数应用于(链式)列表并尽可能以最优雅的方式保持结构。更准确地说，考虑

def fun(x, y):
    return x+y

list_1 = [[{'x': 3, 'y': 4}, {'x': 6, 'y': 5}], [{'x': 9, 'y': 4}, {'x': 1, 'y': 5}]]
list_2 = [{'x': 6, 'y': 4}, {'x': 5, 'y': 5}]

然后 list_1 的输出应该是 [[7, 11], [13, 6]] 而 list_2 [10, 10]。这可以通过使用

[[foo(**i) for i in this_sublist] for this_sublist in list_1]
[foo(**i) for i in list_2]

但是，我想避免区分不同的深度，而是有一个单一的声明。

Answer 1

递归方法:

def operate(v):
    if isinstance(v, list):
        return [operate(v) for v in v]  # or list(map(operate, v))
    elif isinstance(v, dict):
        # use sum or whatever function you need on v
        return sum(v.values())
    # implement whatever error handling logic you want in case v is neither 
    # a dict nor a list


list_1 = [[{'x': 3, 'y': 4}, {'x': 6, 'y': 5}], [{'x': 9, 'y': 4}, {'x': 1, 'y': 5}]]
list_2 = [{'x': 6, 'y': 4}, {'x': 5, 'y': 5}]
list_3 = [[[{'x': 3, 'y': 4}, {'x': 6, 'y': 5}], [{'x': 9, 'y': 4}, {'x': 1, 'y': 5}]]]
print([operate(v) for v in list_1])
print([operate(v) for v in list_2])
print([operate(v) for v in list_3])

输出

[[7, 11], [13, 6]]
[10, 10]
[[[7, 11], [13, 6]]]