How do I extract a dictionary in a Python string?

Question

example_string = "Bla bla {"value": "1"} bla bla"

Is there a way I can search through a string with the goal of extracting only the dictionary(s) without using regular expressions?

I am trying to extract a dictionary in a script tag of a website.

EDIT : I added the exact string below. Didn't want to use regex simply because I am not familiar with it and was looking for an easier way (performance is not an issue). I tried the suggested solution with regex with no succes.

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BTW"},"extra_attributes":{"additional_cost":{"1212907":0,"1280407":0,"1280899":0,"1280900":0,"1289700":0,"1289750":0},"additional_cost_description":{"1212907":null,"1280407":null,"1280899":null,"1280900":null,"1289700":null,"1289750":null},"inriver_item_art_number":{"1212907":"036553024","1280407":"036553014","1280899":"036353002","1280900":"036503006","1289700":"036453006","1289750":"A50704007"},"inriver_item_package_qty":{"1212907":"0","1280407":"0","1280899":"0","1280900":"0","1289700":"0","1289750":"0"},"inriver_item_sales_unit":{"1212907":"STUK","1280407":"STUK","1280899":"STUK","1280900":"STUK","1289700":"STUK","1289750":"STUK"},"inriver_item_supplier_art_nr":{"1212907":"6-30710-PU-0-1","1280407":"6-32158-05-0-1","1280899":"6-30710-AQ-0-1","1280900":"6-30710-PT-0-1","1289700":"6-32159-01-0-1","1289750":"6-32234-03-0-1"},"level_price":{"1212907":{"id":"54257","product_id":"1212907","a_price":"30.4700","b_price":"29.3400","c_price":"25.7900","d_price":"24.4800","e_price":"46.6900","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 21:32:55","_base_price_group":"B","_base_price":"29.3400","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"29.3400","C":"25.7900","D":"24.4800"}},"_final_price_group":"B","_final_price":"29.3400","promo_price":null},"1280407":{"id":"179868","product_id":"1280407","a_price":"34.5300","b_price":"32.2800","c_price":"29.1500","d_price":"26.9000","e_price":"56.9700","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 22:18:16","_base_price_group":"B","_base_price":"32.2800","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"32.2800","C":"29.1500","D":"26.9000"}},"_final_price_group":"B","_final_price":"32.2800","promo_price":null},"1280899":{"id":"179601","product_id":"1280899","a_price":"28.3000","b_price":"27.2600","c_price":"23.9600","d_price":"22.9600","e_price":"52.5100","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 22:18:16","_base_price_group":"B","_base_price":"27.2600","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"27.2600","C":"23.9600","D":"22.9600"}},"_final_price_group":"B","_final_price":"27.2600","promo_price":null},"1280900":{"id":"179602","product_id":"1280900","a_price":"31.7500","b_price":"30.5800","c_price":"26.8800","d_price":"25.5100","e_price":"48.6700","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 22:18:16","_base_price_group":"B","_base_price":"30.5800","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"30.5800","C":"26.8800","D":"25.5100"}},"_final_price_group":"B","_final_price":"30.5800","promo_price":null},"1289700":{"id":"194219","product_id":"1289700","a_price":"32.0700","b_price":"29.8600","c_price":"26.3300","d_price":"24.2200","e_price":"45.7600","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 22:25:57","_base_price_group":"B","_base_price":"29.8600","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"29.8600","C":"26.3300","D":"24.2200"}},"_final_price_group":"B","_final_price":"29.8600","promo_price":null},"1289750":{"id":"194253","product_id":"1289750","a_price":"44.9200","b_price":"42.0000","c_price":"37.9200","d_price":"35.0000","e_price":"74.1100","preh":null,"staffle_a":"1.0000","staffle_b":"10.0000","staffle_c":"50.0000","staffle_d":"100.0000","updated_at":"2018-08-14 22:25:57","_base_price_group":"B","_base_price":"42.0000","_tier_price_options":{"group":{"B":"10.0000","C":"50.0000","D":"100.0000"},"price":{"B":"42.0000","C":"37.9200","D":"35.0000"}},"_final_price_group":"B","_final_price":"42.0000","promo_price":null}},"name":{"1212907":"6-30710-PU-0-1 SECURY EUROPA R4 55\/72\/8\/16 1050  ( KRUK )  4 ROLTAPPEN","1280407":"6-32158-05-0-1 EUROPA R4 55-92-8 F16 (1050) (KRUK BED.) 4 ROLNOKKEN","1280899":"6-30710-AQ-0-1 SECURY EUROPA 35-92-8-F16 R4 1050 (KRUK) 4 ROL (EX 036 35 30 01)","1280900":"6-30710-PT-0-1 SECURY EUROPA R4 50\/72\/8\/16 1050 (KRUK) 4 ROL. \/TAND","1289700":"6-32159-01-0-1 GU SECURY EUROPA S 45-92-8 F16 R4 - 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"""

Answer 1

Assuming there's never more than one dictionary in the string, and there are no other braces, and the dictionary is in a JSON-compatible format (all of which are true for your example, but I have no idea if they're true for all of your real data), you can just use simple string operations to extract it:

prefix, openbrace, rest = s.partition('{')
j, closebrace, suffix = rest.rpartition('}')
if openbrace and closebrace:
    j = openbrace + j + closebrace
    d = json.loads(j)

If those assumptions aren't true, it gets a lot harder.

First, you need to extract matching brace pairs. Something like this:

braces = 0
start = None
for i, c in enumerate(s):
    if c == '{':
        if not braces: start = I
        braces += 1
    elif c == '}':
        braces -= 1
        if not braces:
            yield s[start:I+1]

Except that if there can be braces inside strings, you need to skip them—and you need to handle it in a way that works for strings in JavaScript or whatever your source language actually is, including the appropriate backslash-escaped quote rules and so on. And if your string can have HTML entity escapes for braces (and quotes) you need to deal win that, too.

And then you have to parse each dict. If they're, say, strings of arbitrary JavaScript code, you can't just parse them as JSON, or as Python dicts, because neither of those will handle perfectly valid JS objects like, say, {abc: 1} (notice no quotes around the key), so you'll need to write your own parser.

And if the dicts aren't even necessarily literals—eg, {abc: spam} , where spam is the name of a variable, or a nontrivial expression, or something like that—then you'll need to use a JS interpreter, feed it the appropriate environment, then interpret the code.

Answer 2

Without using a regex, you have to iterate over the string while you see the first { and then save everything after that to the closing } (or use str.partition for that).

If there is a dict in the dict itself, you have also to count opening parantheses and only stop saving if a corresponding closing parenthesis was found.

If there are unmatched parantheses it is a problem anyway, not simple to solve. Better stick to a solution with a regex: https://stackoverflow.com/a/51861961/2648551

Answer 3

you can use a stack to find matching braces without using a regex

import ast
def find_dicts(s):
    stack = []
    buffer=""
    for ch in s:
        if ch == "{":
            buffer += "{"
            stack.append(ch)
        elif ch == "}":
            stack.pop(-1)
            buffer += "}"
            if not stack:
              yield ast.literal_eval(buffer)
              buffer = ""   
        elif stack:
            buffer += ch

print(list(find_dicts("""Bla bla {"value": "1"} bla bla""")            ))

I think this should also handle nested dicts https://repl.it/repls/JuicyWoodenTranslation

as well as more than one dict in the string