从嵌套字典创建列表

Question

I'd like to create a list that is composed of dictionary keys, but those dictionaries are nested within a list. For example, given the below 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"
            }
        }
    ]
}

I'd like to create list of all of the keys from each test_case's variables - Sample output would be:

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

I have code that accomplishes this, but it seems unnecessarily complicated...

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

You can use a list comprehension like so:

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

Note that the order of the keys for each inner dict is not guaranteed since dicts are not ordered in < Python 3.6

Answer 2

This list comprehension is more forgiving to malformed data (eg it would ignore data that did not contain variables ).

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']

Although the code above is efficient, this is more readable:

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