AWS Sagemaker 输出如何读取分布在多行的多个 json 对象的文件

Question

我有一堆像这样的 json 文件

{"vector": [0.017906909808516502, 0.052080217748880386, -0.1460590809583664, ], "word": "blah blah blah"}
{"vector": [0.01027186680585146, 0.04181386157870293, -0.07363887131214142, ], "word": "blah blah blah"}
{"vector": [0.011699287220835686, 0.04741542786359787, -0.07899319380521774, ], "word": "blah blah blah"}

我可以阅读的内容

f = open(file_name)
data = []
for line in f:
   data.append(json.dumps(line))

但是我有另一个文件，输出是这样的

{
    "predictions": [[0.875780046, 0.124219939], [0.892282844, 0.107717164], [0.887681246, 0.112318777]
    ]
}
{
    "predictions": [[0.0, 1.0], [0.0, 1.0], [0.0, 1.0]
    ]
}
{
    "predictions": [[0.391415, 0.608585], [0.992118478, 0.00788147748], [0.0, 1.0]
    ]
}

即json 被格式化为多行，所以我不能简单地逐行读取 json。有没有一种简单的方法来解析这个？还是我必须编写一些东西将每个 json 对象逐行拼接在一起，并且 json.loads 是什么？

Answer 1

嗯，据我所知，很遗憾无法使用json.loads 加载JSONL 格式的数据。不过，一种选择是提供一个辅助函数，可以将其转换为有效的 JSON 字符串，如下所示：

import json

string = """
{
    "predictions": [[0.875780046, 0.124219939], [0.892282844, 0.107717164], [0.887681246, 0.112318777]
    ]
}
{
    "predictions": [[0.0, 1.0], [0.0, 1.0], [0.0, 1.0]
    ]
}
{
    "predictions": [[0.391415, 0.608585], [0.992118478, 0.00788147748], [0.0, 1.0]
    ]
}
"""


def json_lines_to_json(s: str) -> str:
    # replace the first occurrence of '{'
    s = s.replace('{', '[{', 1)

    # replace the last occurrence of '}
    s = s.rsplit('}', 1)[0] + '}]'

    # now go in and replace all occurrences of '}' immediately followed
    # by newline with a '},'
    s = s.replace('}\n', '},\n')

    return s


print(json.loads(json_lines_to_json(string)))

打印：

[{'predictions': [[0.875780046, 0.124219939], [0.892282844, 0.107717164], [0.887681246, 0.112318777]]}, {'predictions': [[0.0, 1.0], [0.0, 1.0], [0.0, 1.0]]}, {'predictions': [[0.391415, 0.608585], [0.992118478, 0.00788147748], [0.0, 1.0]]}]

注意：您的第一个示例实际上看起来不像是有效的 JSON（或者至少是我理解的 JSON 行）。特别是，由于最后一个数组元素后的尾随逗号，这部分似乎无效：

{"vector": [0.017906909808516502, 0.052080217748880386, -0.1460590809583664, ], ...}

为确保调用辅助函数后有效，您还需要删除结尾的逗号，因此每一行的格式如下：

{"vector": [0.017906909808516502, 0.052080217748880386, -0.1460590809583664 ], ...},

---

似乎还有一个similar question，他们建议在换行符上拆分并在每一行上调用json.loads；实际上，在每个对象上多次调用 json.loads 应该（稍微）性能较低，而不是在列表中调用一次，如下所示。

from timeit import timeit
import json


string = """\
{"vector": [0.017906909808516502, 0.052080217748880386, -0.1460590809583664 ], "word": "blah blah blah"}
{"vector": [0.01027186680585146, 0.04181386157870293, -0.07363887131214142 ], "word": "blah blah blah"}
{"vector": [0.011699287220835686, 0.04741542786359787, -0.07899319380521774 ], "word": "blah blah blah"}\
"""


def json_lines_to_json(s: str) -> str:

    # Strip newlines from end, then replace all occurrences of '}' followed
    # by a newline, by a '},' followed by a newline.
    s = s.rstrip('\n').replace('}\n', '},\n')

    # return string value wrapped in brackets (list)
    return f'[{s}]'


n = 10_000

print('string replace:        ', timeit(r'json.loads(json_lines_to_json(string))', number=n, globals=globals()))
print('json.loads each line:  ', timeit(r'[json.loads(line) for line in string.split("\n")]', number=n, globals=globals()))

结果：

string replace:         0.07599360000000001
json.loads each line:   0.1078384