当您只有最后一层的键、值列表时,如何填充嵌套dict?

我希望我正确理解了你的问题。可以使用递归替换键的值:

dct = {
    "layer1": {
        "layer2": {"layer3": {"a": 0, "b": 0, "c": 0}},
        "layer4": {"d": 0, "e": 0},
    }
}

dct2 = {"a": 1, "b": 2, "c": 3, "d": 4, "e": 5}


def replace(d):
    if isinstance(d, dict):
        for k in d & dct2.keys():
            d[k] = dct2[k]

        for k in d - dct2.keys():
            replace(d[k])

    elif isinstance(d, list):
        for i in d:
            replace(i)


replace(dct)
print(dct)

打印:

{
    "layer1": {
        "layer2": {"layer3": {"a": 1, "b": 2, "c": 3}},
        "layer4": {"d": 4, "e": 5},
    }
}

另一种使用递归的方法,实际上只需要一个写循环。我发现这更容易理解。它也不会改变原来的dict:

data = {"layer1": {"layer2": {"layer3": {"a": None, "b": None, "c": None}, "layer4": {"d": None, "e": None}}}}
values = {"a": 1, "b": 2, "c": 3, "d": 4, "e": 5}

def fill_data(data):
    out = {}

    for key, value in data.items():
        if isinstance(value, dict):
            out[key] = fill_data(value)
        else:
            out[key] = values.get(key, value)

    return out

print(fill_data(data))

{'layer1': {'layer2': {'layer3': {'a': 1, 'b': 2, 'c': 3}, 'layer4': {'d': 4, 'e': 5}}}}