python sort list of json by value

Question

I have a file consists of JSON, each a line, and want to sort the file by update_time reversed.

sample JSON file:

{ "page": { "url": "url1", "update_time": "1415387875"}, "other_key": {} }
{ "page": { "url": "url2", "update_time": "1415381963"}, "other_key": {} }
{ "page": { "url": "url3", "update_time": "1415384938"}, "other_key": {} }

want output:

{ "page": { "url": "url1", "update_time": "1415387875"}, "other_key": {} }
{ "page": { "url": "url3", "update_time": "1415384938"}, "other_key": {} }
{ "page": { "url": "url2", "update_time": "1415381963"}, "other_key": {} }

my code:

#!/bin/env python
#coding: utf8

import sys
import os
import json
import operator

#load json from file
lines = []
while True:
    line = sys.stdin.readline()
    if not line: break
    line = line.strip()
    json_obj = json.loads(line)
    lines.append(json_obj)

#sort json
lines = sorted(lines, key=lambda k: k['page']['update_time'], reverse=True)

#output result
for line in lines:
    print line

The code works fine with sample JSON file, but if a JSON has no 'update_time', it will raise KeyError exception. Are there non-exception ways to do this?

Nathron · Accepted Answer · 2018-05-08 20:25:22Z

51

Write a function that uses try...except to handle the KeyError, then use this as the key argument instead of your lambda.

def extract_time(json):
    try:
        # Also convert to int since update_time will be string.  When comparing
        # strings, "10" is smaller than "2".
        return int(json['page']['update_time'])
    except KeyError:
        return 0

# lines.sort() is more efficient than lines = lines.sorted()
lines.sort(key=extract_time, reverse=True)

edited May 8, 2018 at 20:25

Nathron

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answered Nov 14, 2014 at 7:18

Ferdinand Beyer

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1 Comment

astrochun Over a year ago

This works but you should avoid using namespaces that are for built-in library.

alecxe · Accepted Answer · 2014-11-14 07:19:37Z

35

You can use dict.get() with a default value:

lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)

Example:

>>> lines = [
...     {"page": {"url": "url1", "update_time": "1415387875"}, "other_key": {}},
...     {"page": {"url": "url2", "update_time": "1415381963"}, "other_key": {}},
...     {"page": {"url": "url3", "update_time": "1415384938"}, "other_key": {}},
...     {"page": {"url": "url4"}, "other_key": {}},
...     {"page": {"url": "url5"}, "other_key": {}}
... ]
>>> lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)
>>> for line in lines:
...     print line
... 
{'other_key': {}, 'page': {'url': 'url1', 'update_time': '1415387875'}}
{'other_key': {}, 'page': {'url': 'url3', 'update_time': '1415384938'}}
{'other_key': {}, 'page': {'url': 'url2', 'update_time': '1415381963'}}
{'other_key': {}, 'page': {'url': 'url4'}}
{'other_key': {}, 'page': {'url': 'url5'}}

Though, I would still follow the EAFP principle that Ferdinand suggested - this way you would also handle cases when page key is also missing. Much easier to let it fail and handle it than checking all sorts of corner cases.

answered Nov 14, 2014 at 7:19

alecxe

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2 Comments

KARTHEEK GUMMALURI Over a year ago

how to assign json file to lines so that it must dynamically if I we 1 million lines then it will not load right so that's why

Hassan Raza Over a year ago

lines = sorted(json_string, key=lambda k: k['modified'], reverse=True) TypeError: string indices must be integers i am getting error.

gurel_kaynak · Accepted Answer · 2019-01-24 07:40:33Z

18

# sort json
lines = sorted(lines, key=lambda k: k['page'].get('update_time', 0), reverse=True)

edited Jan 24, 2019 at 7:40

answered Nov 14, 2014 at 7:21

gurel_kaynak

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Comments

Mrinal · Accepted Answer · 2020-01-30 20:43:07Z

3

def get_sortest_key(a: dict, o: dict):
    v = None
    k = None
    for key, value in a.items():
        if v is None:
            v = value
            k = key
            continue
        if v > value:
            v = value
            k = key
    o.update({k: v})
    a.pop(k)
    if a:
        get_sortest_key(a, o)
    else:
        return


def call(o):
    a = {'a': 9, 'b': 1, 'c': 3, 'k': 3, 'l': -1, 's': 100}
    z = get_sortest_key(a, o)
    print(o)


o={}    
call(o)

edited Jan 30, 2020 at 20:43

answered Jan 30, 2020 at 20:35

Mrinal

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1 Comment

Tom Johnson Over a year ago

Adding some explanation of how your code addresses the original problem would be useful.

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