使用jq的答案确实很有用，但如果你仍然想用Python来实现（正如问题所显示的那样），你可以用内置的json模块来实现。

import json
from io import StringIO
in_json = StringIO("""[{
    "key01": "value01",
    "key02": "value02",

    "keyN": "valueN"
},
{
    "key01": "value01",
    "key02": "value02",

    "keyN": "valueN"
},
{
    "key01": "value01",
    "key02": "value02",

    "keyN": "valueN"
}
]""")

result = [json.dumps(record) for record in json.load(in_json)]  # the only significant line to convert the JSON to the desired format

print('\n'.join(result))

{"key01": "value01", "key02": "value02", "keyN": "valueN"}
{"key01": "value01", "key02": "value02", "keyN": "valueN"}
{"key01": "value01", "key02": "value02", "keyN": "valueN"}

• 我在这里使用StringIO和print只是为了使示例更容易在本地测试。

作为替代方案，您可以使用Python jq binding将其与the other answer组合。

如果你想摆脱Python，可以使用jq：

$ cat a.json 
[{
    "key01": "value01",
    "key02": "value02",
    "keyN": "valueN"
},
{
    "key01": "value01",
    "key02": "value02",
    "keyN": "valueN"
},
{
    "key01": "value01",
    "key02": "value02",
    "keyN": "valueN"
}
]

$ cat a.json | jq -c '.[]'
{"key01":"value01","key02":"value02","keyN":"valueN"}
{"key01":"value01","key02":"value02","keyN":"valueN"}
{"key01":"value01","key02":"value02","keyN":"valueN"}

我使用的迭代器是'.[]'来遍历数组，-c将每个JSON对象放在一行中。
资源：

• https://stedolan.github.io/jq/manual/
• https://github.com/stedolan/jq

获取JSON文件，转换成ND-JSON文件。

import json

with open("results-20190312-113458.json", "r") as read_file:
    data = json.load(read_file)
result = [json.dumps(record) for record in data]
with open('nd-proceesed.json', 'w') as obj:
    for i in result:
        obj.write(i+'\n')

希望这对某人有帮助。

with open('out.json', 'w') as f:
  for obj in objs:
    json.dump(obj, f)
    f.write('\n')

如果文件很大，你可能需要使用更快的模块，比如msgspec或orjson，因为他们使用的是c（或者c++，我不完全知道，但有更快）
示例：

• 读

with open('test.json', 'rb') as f:
    j = msgspec.json.decode(f.read())
    
j = b'\n'.join(map(msgspec.json.encode, j))

• 写

with open('test.json', 'wb') as f:
    f.write(j)