從嵌套字典創建列表

Question

我想創建一個由字典鍵組成的列表，但是那些字典嵌套在列表中。 例如，給定以下JSON：

{
    "test_cases": [{
            "name": "first request",
            "request": {
                "url": "{{env_base_url}}/v2/597649d3110000be08b1be84/{{env_userName}}"
            },
            "variables": {
                "var1": "some stuff",
                "var2": "some other stuff"
            }
        },
        {
            "name": "second request",
            "request": {
                "url": "{{env_base_url}}/v2/597649d3110000be08b1be84/{{env_userName}}"
            },
            "variables": {
                "var3": "some new stuff",
                "var4": "some other new stuff"
            }
        }
    ]
}

我想從每個test_case的變量創建所有鍵的列表-示例輸出為：

list = ['var1', 'var2', 'var3', 'var4]

我有完成此任務的代碼，但似乎不必要地復雜...

test_var_names = [list(test['variables'].keys()) for test in data['test_cases']]
i =0
while i < len(test_var_names):
    test_var_name = test_var_names[i]
    if isinstance(test_var_name, list):
        for item in test_var_name:
            test_var_names.append(item)
        test_var_names.remove(test_var_name)
        i = i-1
    i += 1
print (test_var_names)

Answer 1

您可以像這樣使用列表理解 ：

lst = [k for d in dct['test_cases'] for k in d['variables'].keys()]
# ['var1', 'var2', 'var4', 'var3']

請注意，不能保證每個內部dict的鍵順序，因為dict在<Python 3.6中沒有排序

Answer 2

這種列表理解更能容忍格式錯誤的數據（例如，它將忽略不包含variables數據）。

d = {
    "test_cases": [{
            "name": "first request",
            "request": {
                "url": "{{env_base_url}}/v2/597649d3110000be08b1be84/{{env_userName}}"
            },
            "variables": {
                "var1": "some stuff",
                "var2": "some other stuff"
            }
        },
        {
            "name": "second request",
        },
        {
            "name": "third request",
            "request": {
                "url": "{{env_base_url}}/v2/597649d3110000be08b1be84/{{env_userName}}"
            },
            "variables": {
                "var3": "some new stuff",
                "var4": "some other new stuff"
            }
        }
    ]
}

>>> [var 
     for sublist in [case.get('variables', {}).keys() for case in d.get('test_cases', [])]
     for var in sublist]
['var1', 'var2', 'var4', 'var3']

盡管上面的代碼很有效，但是可讀性更高：

variables = []
items = d.get('test_cases')
for item in items:
    variables.extend(item.get('variables', {}).keys())