[英]How to get an item from a dictionary which contains list of dictionaries
我有包含字典列表的字典:
dict = {"key0": "some_value", "key1": 1, "value": [{"key2": "some_value", "key3": "some_value", "key4": {"key5": 0}, "key6": {"key7": {"key8": "some_value", "id": "some_id"}, "key9": {"key10": "some_value", "id": "some_id_2"}}}]}
我正在努力尋找獲取“id”字段的方法,以防我發現“key7”或“key9”存在。 就像是:
for key in dict.items():
if 'value' in dict.keys():
if 'key7' in dict.values():
print("I want to print key7 id value here")
預期 output: some_id
如果有人會遇到這種情況,請找到一種訪問項目的方法:
my_dict = {"key0": "some_value", "key1": 1, "value": [{"key2": "some_value", "key3": "some_value", "key4": {"key5": 0}, "key6": {"key7": {"key8": "some_value", "id": "some_id"}, "key9": {"key10": "some_value", "id": "some_id_2"}}}]}
然后是這樣的:
for k, v in my_dict.items():
if isinstance(v, list):
value_list = v
if not value_list:
pass
elif value_list is not None:
for vl in value_list:
print(vl['key6']['key7']['id'])
print(vl['key6']['key9']['id'])
它遍歷字典,直到字典的值是列表,然后如果值是列表,使用列表理解,您可以訪問元素,它將 output:
some_id
some_id_2
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