convert strings into python dict

Question

I have a string which is as follows

my_string = '"sender" : "md-dgenie", "text" : "your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!", "time" : "1439155575925", "name" : "John"'

I want to construct a dict from the above string. I tried something as suggested here

split_text = my_string.split(",")
for i in split_text :
    print i

then I got output as shown below:

"sender" : "md-dgenie"
 "text" : "your dudegenie code is 6632. welcome to the world of dudegenie! your wish
 my command!"     ### finds "," here and splits it here too.
 "time" : "1439155803426"
 "name" : "p"

I want output as key pair values of a dictionary as given below :

my_dict = { "sender" : "md-dgenie",
     "text" : "your dudegenie code is 6632. welcome to the world of dudegenie! your wish, my command!",
     "time" : "1439155803426",
     "name" : "p" }

Basically I want to skip that "," from the sentence and construct a dict. Any suggestions would be great! Thanks in advance!

Are you sure that is how you want to format my_string? I can't see how that can be a valid string, but then again, python is a weird language so there might be some obscure string formatting I don't know about. — tobloef
– tobloef, Commented Aug 14, 2015 at 19:59
Please correct the code you have posted. Your my_string is not a string, or anything else. It is syntactically invalid. — Two-Bit Alchemist
– Two-Bit Alchemist, Commented Aug 14, 2015 at 19:59
@TobLoef : Sorry! my bad! I forgot to single quotes to my_string. I have updated it. — d-coder
– d-coder, Commented Aug 14, 2015 at 20:00

tzaman · Accepted Answer · 2015-08-14 20:12:48Z

8

Your string is almost already a python dict, so you could just enclose it in braces and then evaluate it as such:

import ast
my_dict = ast.literal_eval('{{{0}}}'.format(my_string))

edited Aug 14, 2015 at 20:12

answered Aug 14, 2015 at 20:01

tzaman

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7 Comments

d-coder Over a year ago

Oooh! How is this different from eval ?

tzaman Over a year ago

@d-coder ast.literal_eval just parses the string and returns it if it evaluates to a literal. It doesn't actually execute it, so it's not vulnerable to injection attacks.

d-coder Over a year ago

Cool! It worked like a charm! I would give you badge for now because I think I can't up vote it. :P

tzaman Over a year ago

@d-coder you can accept the answer by clicking on the check mark (you might have to wait a bit first.)

d-coder Over a year ago

Finally one upvote from my side and accepted your answer! :)

|

vks · Accepted Answer · 2015-08-14 20:00:29Z

2

my_string =' "sender" : "md-dgenie", "text" : "your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!", "time" : "1439155575925", "name" : "John"'
import re
print dict(re.findall(r'"([^"]*)"\s*:\s*"([^"]*)"',my_string))

You can do it via finding tuples using re.findall and passing it to dict

answered Aug 14, 2015 at 20:00

vks

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4 Comments

d-coder Over a year ago

Thanks! but I think I will go with @tzaman answer. I bad at regex!

d-coder Over a year ago

There you go :) regex freaks me out!

d-coder Over a year ago

I have one more question actually. Should I update this one or post another ? I want to know how store time in milliseconds to be stored in django models field ?

tzaman Over a year ago

@d-coder if your new question is unrelated to parsing dictionary-like strings, post it separately.

Padraic Cunningham · Accepted Answer · 2015-08-14 20:22:52Z

1

You could have also split on ", and stripped the whitespace and ":

my_string = '"sender" : "md-dgenie", "text" : "your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!", "time" : "1439155575925", "name" : "John"'
print(dict(map(lambda x:x.strip('" ') ,s.split(":")) for s in my_string.split('",')))

{'name': 'John', 'time': '1439155575925', 'sender': 'md-dgenie', 'text': 'your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!'}

answered Aug 14, 2015 at 20:22

Padraic Cunningham

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Comments

jstevenco · Accepted Answer · 2015-08-15 01:29:41Z

0

A different take using dict comprehension, simpler regex and zip(*[iter()]*n):

import re
my_string = '"sender" : "md-dgenie", "text" : "your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!", "time" : "1439155575925", "name" : "John"'
{k:v for k,v in zip(*[iter(re.findall(r'"(.+?)"',my_string))]*2)}

{'text': 'your dudegenie code is 6326. welcome to the world of dudegenie! your wish, my command!', 'sender': 'md-dgenie', 'name': 'John', 'time': '1439155575925'}

edited Aug 15, 2015 at 1:29

answered Aug 15, 2015 at 1:14

jstevenco

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