将嵌套键名存储在变量中

编程问答 2022-09-30

问题描述

我有一个字典foo，其中可能包含0..x个键，其中一些键是嵌套的。我有一个键列表，如果它们存在于foo中，我想使用不同的键名将它们的值添加到不同的字典bar中。由于我不知道foo中是否存在每个键，因此需要检查字典中是否存在它们，如果不存在，则将bar中的值分配为None存在。例如：

foo = {"ex1": "1","ex2": "2","ex3": {"ex3nest": "3"},"ex4": "4","ex5": "5"}

field_mappings = {"bar1": "ex1","bar2": "ex2","bar4": "ex4"}

bar = {}
for key,value in field_mappings.items():
    bar[key] = foo[value] if value in foo else None

bar的值将正确为：

{"bar1": "1","bar2": "2","bar4": "4"}

不过，我想向bar添加一个包含嵌套ex3nest键值的键，例如：

field_mappings = {"bar1": "ex1","bar3": key for ex3nest here,"bar4": "ex4"}

我看不到可以分配给bar3键的任何值，该值可以插入到循环中的value中，以插入ex3nest的值。

在我尝试执行操作时，是否有办法在字符串中存储嵌套键名，或者我的方法一开始是完全错误的？

解决方法

嘿，如果我对问题的理解正确，可以使用此代码，假设您在嵌套字典中有一个键：

foo = {"ex1": "1","ex2": "2","ex3": {"ex3nest": "3"},"ex4": "4","ex5": "5"}

field_mappings = {"bar1": "ex1","bar2": "ex2","bar3": "ex3","bar4": "ex4"}

bar = {}
for key,value in field_mappings.items():
    if value in foo:
        if type(foo[value])==type({}):
            bar[key] = foo[value].keys()[0]
        else:
            bar[key] = foo[value]
   else:
       bar[key] = None

enter image description here

或如果嵌套字典中有多个键，则使用此键：

bar = {}
for key,value in field_mappings.items():
    if value in foo:
        if type(foo[value])==type({}):
            bar[key] = ",".join(foo[value].keys())
        else:
            bar[key] = foo[value]
     else:
        bar[key] = None

enter image description here

,

我相信有更好的方法，但是我将使用ex3.ex3nest样式作为键（从js对象中借用的样式）。

foo = {
    "ex1": "1","ex3": {
        "ex3nest": "3"
    },"ex5": "5","ex6": {
        "ex6nest1": {
            "ex6nest2": "6"
        }
    },"ex8": {
        "ex8nest1": {
            "ex8nest2": "8"
        }
    },}

field_mappings = {
    "bar1": "ex1","bar3": "ex3.ex3nest","bar4": "ex4","bar5": "ex5","bar6": "ex6.ex6nest1.ex6nest2",# 2 levels deep
    "bar7": "ex7",# doesn't exists
    "bar8": "ex8.INVALID_KEY",# oops!
}

bar = {}
for key,value in field_mappings.items():
    nested_keys = value.split(".")
    
    bar_value = foo
    for curr_key in nested_keys:
        bar_value = bar_value.get(curr_key,None)
        
        if bar_value is None:
            break

    bar[key] = bar_value

print(bar)

输出

{'bar1': '1','bar2': '2','bar3': '3','bar4': '4','bar5': '5','bar6': '6','bar7': None,'bar8': None}
dictionarypythonpython-3.x