Python regex query to parse very simple dictionary

Question

I am new to regex module and learning a simple case to extract key and values from a simple dictionary.

the dictionary can not contain nested dicts and any lists, but may have simple tuples

MWE

import re

# note: the dictionary are simple and does NOT contains list, nested dicts, just these two example suffices for the regex matching.
d = "{'a':10,'b':True,'c':(5,'a')}" # ['a', 10, 'b', True, 'c', (5,'a') ]
d = "{'c':(5,'a'), 'd': 'TX'}" # ['c', (5,'a'),  'd', 'TX']

regexp = r"(.*):(.*)" # I am not sure how to repeat this pattern separated by ,

out = re.match(regexp,d).groups()
out

Don't use regex for this. Use ast.literal_eval.

trincot
– trincot

2022-11-11 15:30:56 +00:00
Commented Nov 11, 2022 at 15:30 — trincot
– trincot, Commented Nov 11, 2022 at 15:30

trincot · Accepted Answer · 2022-11-11 16:19:15Z

2

You should not use regex for this job. When the input string is valid Python syntax, you can use ast.literal_eval.

Like this:

import ast
# ...
out = ast.literal_eval(d)

Now you have a dictionary object in Python. You can for instance get the key/value pairs in a (dict_items) list:

print(out.items())

Regex

Regex is not the right tool. There will always be cases where some boundary case will be wrongly parsed. But to get the repeated matches, you can better use findall. Here is a simple example regex:

regexp = r"([^{\s][^:]*):([^:}]*)(?:[,}])"
out = re.findall(regexp, d)

This will give a list of pairs.

edited Nov 11, 2022 at 16:19

answered Nov 11, 2022 at 15:35

trincot

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7 Comments

Swifty Over a year ago

To OP: note that your d object is not a dictionary, but a string representation of a dict; if you made it with d = str(d), better work on the original dictionary!

dallascow Over a year ago

I have first tried that way eval(d), but sometimes I may forget to give quotes around variables, then the literal eval will either fail or pick up values assigned to that variable. So, I am going the harder way of regex. get list and in the end make dictionary.

dallascow Over a year ago

bad example: "a:10" then a can pick up already defined value, instead of 'a'

dallascow Over a year ago

That's the reason i am looking for regex and if the dictionary fails later, I will catch that error with Exceptions.

trincot Over a year ago

"I may forget quotes": but then you could introduce ambiguity. For instance, 1 and '1' are not the same value, nor are True and 'True', ...Etc. You just should not "forget". Syntax rules are there for good reason, not to annoy people.

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Swifty · Accepted Answer · 2022-11-11 16:49:10Z

1

Regex would be hard (perhaps impossible, but I'm not versed enough to say confidently) to use because of the ',' nested in your tuples. Just for the sake of it, I wrote (regex-less) code to parse your string for separators, ignoring parts inside parentheses:

d = "{'c':(5,'a',1), 'd': 'TX', 1:(1,2,3)}" 

d=d.replace("{","").replace("}","")
indices = []
inside = False
for i,l in enumerate(d):
    if inside:
        if l == ")":
            inside = False
            continue
        continue
    if l == "(":
        inside = True
        continue
    if l in {":",","}:
        indices.append(i)
indices.append(len(d))
parts = []
start = 0
for i in indices:
    parts.append(d[start:i].strip())
    start = i+1

parts
#  ["'c'", "(5,'a',1)", "'d'", "'TX'", '1', '(1,2,3)']

answered Nov 11, 2022 at 16:49

Swifty

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1 Comment

Swifty Over a year ago

After reading @trincot 's addition to their answer, obviously not impossible :) Nicely done!

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