如何根据另一个字典替换列表字典中键的值?

时间:2021-04-18 22:06:44

标签: python dictionary list-comprehension

以下是数据示例:

[{
  'name': 'Age', 
  'p_value': '<0.001', 
  'ks_score': '0.07', 
}, 
{
  'name': 'peer_LM_Mean_SelfAware', 
  'p_value': '<0.001', 
  'ks_score': '0.06', 
}]

我有超过 16,000 部字典,列表中有几个项目。我想要做的是从另一个由旧名称和新名称组成的字典中替换此字典中 name 的值。

col_rename_list = {'Male': 'Male, n (%)', 
                   'Age': 'Age, years', 
                   'Baby_Boomers__1946_1964_': 'Baby Boomers (1946-1964), n (%)',
                   'Generation_X__1965_1980_': 'Generation X (1965-1980), n (%)',
                   'Generation_Y___Millennials__1981_1996_': 'Generation Y/Millennials (1981-1996), n (%)', 
                   'Race_Asian': 'Asian, n (%)',
                   'peer_LM_Mean_SelfAware': 'Self-Awareness'
                  }

我该怎么做?我能想到的一种方法如下:

for d in dictionary_list:
    for k,v in d.items():
        if k == 'name':
            if col_rename_list.get(v) is not None:
                d[k] = col_rename_list.get(v)

这行得通,但有没有更好更有效的方法来做到这一点?

1 个答案:

答案 0 :(得分:1)

您可以将 dict.get 与默认参数一起使用:

lst = [
    {
        "name": "Age",
        "p_value": "<0.001",
        "ks_score": "0.07",
    },
    {
        "name": "peer_LM_Mean_SelfAware",
        "p_value": "<0.001",
        "ks_score": "0.06",
    },
]

col_rename_list = {
    "Male": "Male, n (%)",
    "Age": "Age, years",
    "Baby_Boomers__1946_1964_": "Baby Boomers (1946-1964), n (%)",
    "Generation_X__1965_1980_": "Generation X (1965-1980), n (%)",
    "Generation_Y___Millennials__1981_1996_": "Generation Y/Millennials (1981-1996), n (%)",
    "Race_Asian": "Asian, n (%)",
    "peer_LM_Mean_SelfAware": "Self-Awareness",
}

for d in lst:
    d["name"] = col_rename_list.get(d["name"], d["name"])

# pretty print:
from pprint import pprint
pprint(lst)

打印:

[{'ks_score': '0.07', 'name': 'Age, years', 'p_value': '<0.001'},
 {'ks_score': '0.06', 'name': 'Self-Awareness', 'p_value': '<0.001'}]