我正在尝试创建一个通用函数来替换嵌套字典的键中的点。我有一个非泛型函数,可深入3个层次,但是必须有一种方法可以实现这种泛型。任何帮助表示赞赏!到目前为止,我的代码:

output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}

def print_dict(d):
    new = {}
    for key,value in d.items():
        new[key.replace(".", "-")] = {}
        if isinstance(value, dict):
            for key2, value2 in value.items():
                new[key][key2] = {}
                if isinstance(value2, dict):
                    for key3, value3 in value2.items():
                        new[key][key2][key3.replace(".", "-")] = value3
                else:
                    new[key][key2.replace(".", "-")] = value2
        else:
            new[key] = value
    return new

print print_dict(output)

更新:要回答我自己的问题,我使用json提出了一个解决方案object_hooks:
import json

def remove_dots(obj):
    for key in obj.keys():
        new_key = key.replace(".","-")
        if new_key != key:
            obj[new_key] = obj[key]
            del obj[key]
    return obj

output = {'key1': {'key2': 'value2', 'key3': {'key4 with a .': 'value4', 'key5 with a .': 'value5'}}}
new_json = json.loads(json.dumps(output), object_hook=remove_dots)

print new_json

最佳答案

是的,存在更好的方法:

def print_dict(d):
    new = {}
    for k, v in d.iteritems():
        if isinstance(v, dict):
            v = print_dict(v)
        new[k.replace('.', '-')] = v
    return new

(编辑:这是递归的,更多关于Wikipedia。)

关于python - Python递归替换嵌套字典键中的字符?,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/11700705/

10-12 20:07