Question

I have the following string:

'{
    "key1": "val1",
    "key2": ["a","b", 3],
    "key3": {"foo": 27, "bar": [1, 2, 3]}
}'

I want to parse only one level so result should be a one level dictionary with key, and value should be just a string(don’t need to parse it)

For given string it should return following dictionary:

{
    "key1": "val1",
    "key2": "['a','b', 3]",
    "key3": "{'foo': 27, 'bar': [1, 2, 3]}"
}

Is there a fast way to do it? Without parsing whole string to json and convert all values back to strings.

Answer 1

I think you can solve this using regex, it is working for me:

import re
pattern = re.compile('"([a-zA-Z0-9]+)"\s*:\s*(".*"|\[.*\]|\{.*\})')    
dict(re.findall(pattern, json_string))

But I dont know if this is faster, you need try using your data.

[EDIT]

Yes, it is faster. I tried the scripts below and the regex version is 5 times faster.

using json module:

import json

val='''
{
    "key1": "val1",
    "key2": ["a","b", 3],
    "key3": {"foo": 27, "bar": [1, 2, 3]}
}
'''

for n in range(100000):
    dict((k,json.dumps(v)) for k,v in json.loads(val).items())

using regex:

import re

val='''{
    "key1": "val1",
    "key2": ["a","b", 3],
    "key3": {"foo": 27, "bar": [1, 2, 3]}
}'''

pattern = re.compile('"([a-zA-Z0-9]+)"\s*:\s*(".*"|\[.*\]|\{.*\})')    
for n in range(100000):
    dict(re.findall(pattern, val))