如何从更大的字典中获取仅包含键和 GID 的字典？

Question

This returns a list of all the gids. I would like it so that all gids are reachable by their key. eg
{0: "23172371321","123421412343",1" "12312412214","123124124"} .
The GID are accessible by the code task_gid_dict[j][i]["gid"]

i = 0
j = 0
task_gid_dict_2 = {}
for i in range(len(task_gid_dict)):
    while True:
        try:
            task_gid.append(task_gid_dict[j][i]["gid"])
            i = i+1
        except:
            j = j+1
            break
    task_gid_dict_2[j] = task_gid
        
task_gid_dict_2
        

At then moment it looks like this task_gid_dict = {0: [{'gid': '1199729685867432', 'name': 'SAMPLE', 'resource_type': 'task', 'resource_subtype': 'default_task'}, ...

EXTRA:

task_detail = {}

for k,i in task_gid_dict.items():
    task_detail[i] = client.tasks.get_task(task_gid_2[k][i], {'param': 'value', 'param': 'value'}, opt_pretty=True)
    
task_detail

Answer 1

Your output will look a little different from what you have up top. A dict stores a single value for a single key, but the value can be a list object, so instead of 0: 123456789, 987654321, you'll have 0: [123456789, 987654321].

task_gid_dict_2 = {}
# iterate through key value pairs of a dictionary
for k, v in task_gid_dict.items():
  # initialize the key in the new dict with an empty list
  task_gid_dict_2[k] = []
  # iterate through the sub-dictionaries in each value
  for sub_dict in v:
    # Add the gid from the sub-dict to your new list
    task_gid_dict_2.append(sub_dict['gid'])

That gets you what you want and I think makes the most sense, but if you want to take it a step further, you could do it with a combined dict and list comprehension as well:

task_gid_2 = {k: [sub_d['gid'] for sub_d in v] for k, v in task_gid_dict.items()}